6
th
Edition
GMAT Algebra Guide ®
Joern Meissner
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Algebra Guide
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Algebra Guide
About the Company Manhattan Review’s origin can be traced directly back to an Ivy League MBA classroom in 1999. While teaching advanced quantitative subjects to MBAs at Columbia Business School in New York City, Professor Dr. Joern Meissner developed a reputation for explaining complicated concepts in an understandable way. Remembering their own less-than-optimal experiences preparing for the GMAT, Prof. Meissner’s students challenged him to assist their friends, who were frustrated with conventional GMAT preparation options. In response, Prof. Meissner created original lectures that focused on presenting GMAT content in a simplified and intelligible manner, a method vastly different from the voluminous memorization and so-called tricks commonly offered by others. The new approach immediately proved highly popular with GMAT students, inspiring the birth of Manhattan Review. Since its founding, Manhattan Review has grown into a multi-national educational services firm, focusing on GMAT preparation, MBA admissions consulting, and application advisory services, with thousands of highly satisfied students all over the world. The original lectures have been continuously expanded and updated by the Manhattan Review team, an enthusiastic group of master GMAT professionals and senior academics. Our team ensures that Manhattan Review offers the most timeefficient and cost-effective preparation available for the GMAT. Please visit www.ManhattanReview.com for further details.
About the Founder Professor Dr. Joern Meissner has more than 25 years of teaching experience at the graduate and undergraduate levels. He is the founder of Manhattan Review, a worldwide leader in test prep services, and he created the original lectures for its first GMAT preparation class. Prof. Meissner is a graduate of Columbia Business School in New York City, where he received a PhD in Management Science. He has since served on the faculties of prestigious business schools in the United Kingdom and Germany. He is a recognized authority in the areas of supply chain management, logistics, and pricing strategy. Prof. Meissner thoroughly enjoys his research, but he believes that grasping an idea is only half of the fun. Conveying knowledge to others is even more fulfilling. This philosophy was crucial to the establishment of Manhattan Review, and remains its most cherished principle.
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Contents 1 Welcome 2 Concepts of Algebra 2.1 Identities . . . . . . . . 2.2 Linear Equations . . . 2.3 Quadratic Equations . 2.4 Inequalities . . . . . . 2.5 Functions and graphs
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3 4 6 14 19 28
3 Practice Questions 3.1 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Data Sufficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 Answer-key 97 4.1 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 4.2 Data Sufficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 5 Solutions 101 5.1 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 5.2 Data Sufficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 6 Talk to Us
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Chapter 1
Welcome Dear Students, Here at Manhattan Review, we constantly strive to provide you the best educational content for standardized test preparation. We make a tremendous effort to keep making things better and better for you. This is especially important with respect to an examination such as the GMAT. A typical GMAT aspirant is confused with so many test-prep options available. Your challenge is to choose a book or a tutor that prepares you for attaining your goal. We cannot say that we are one of the best, it is you who has to be the judge. There are umpteen numbers of books on Algebra for GMAT preparation. What is so different about this book? The answer lies in its approach to deal with the questions. Solution of each question is dealt with in detail. There are many questions that have been solved through alternate approaches. The objective is to understand questions from multiple aspects. Few seemingly scary questions have been solved through Logical Deduction or through Intuitive approach. The book has a great collection of 150 GMAT-like questions: 100 PS and 50 DS. Apart from books on ‘Number Properties’, ‘Word Problem’, ‘Algebra’, ‘Arithmetic’, ‘Geometry’, ‘Permutation and Combination’, and ‘Sets and Statistics’ which are solely dedicated to GMAT-QA-PS & DS, the book on ‘GMAT-Math Essentials’ is solely dedicated to develop your math fundamentals. Another publication ‘GMAT Quantitative Ability Question Bank’ boasts of a collection of 500 GMAT like questions. The Manhattan Review’s ‘GMAT-Algebra’ book is holistic and comprehensive in all respects. Should you have any comments or questions, please feel free to write to us at
[email protected]. Happy Learning! Professor Dr. Joern Meissner & The Manhattan Review Team
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Algebra Guide – Concepts
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Chapter 2
Concepts of Algebra
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Algebra Guide – Concepts
2.1
Identities
Let us first look at a few basic terms. Polynomial: A polynomial is an expression of variables that can only have rational coefficients. For example, x 3 + 2xy 2 + 6z is a polynomial. Order and Degree: The Order of a polynomial refers to the number of variables involved in the polynomial. For example, x 2 + 2xy is a polynomial of order ‘2’ since there are two variables, x and y. The Degree of a polynomial refers to the highest value of the exponent of any variable in the polynomial. For example, x 2 + 2xy has degree ‘2’ since the highest exponent is 2. Identities: An identity is a statement equating two expressions which are equal for all values of the variables involved. For example, 5x = 2x + 3x is an identity, as 5x and 2x + 3x are equal for all values for x. Some important identities are as follows: • (a ± b)2 = a2 ± 2ab + b2 1 1 2 = x2 + 2 ± 2 • x± x x • a2 − b2 = (a + b) (a − b) • (a ± b)3 = a3 ± 3a2 b + 3ab2 ± b3 = a3 ± b3 ± 3ab(a ± b) Let us take an example: 1 1 = 2, what is the value of x 4 + 4 ? x x 1 2 1 We have: x + = 22 => x 2 + 2 + 2 = 4 x x 1 => x 2 + 2 = 4 − 2 = 2 x 1 1 2 = 22 => x 4 + 4 + 2 = 4 => x 2 + 2 x x 1 => x 4 + 4 = 4 − 2 = 2. x If x +
Let us take another example: (0.21 + 0.17)2 + (0.21 − 0.17)2 What is the value of ? 0.212 + 0.172 Let a = 0.21 and b = 0.17
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Thus, we have: (0.21 + 0.17)2 + (0.21 − 0.17)2 0.212 + 0.172 =
2 2 (a + b) + (a − b) 2 2 (a + b )
a2 + 2ab + b2 + a2 − 2ab + b2 a2 + b 2 2 a2 + b2 = a2 + b 2
=
=2
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2.2
Algebra Guide – Concepts
Linear Equations
Linear equations are equations in which the highest exponent/index of the variables is 1, i.e. the degree is 1. For example, 2x + 5 = 11 is a linear equation with a single variable, x, whose index is one. Similarly, 3x + 5y = 9 is a linear equation with two variables, x and y, and the indices of both are 1. Solving linear equations with a single variable: While solving any linear equation with a single term, we need to group all terms containing the variable on the left and all other constant terms on the right. While shifting terms from left to right as addition or subtraction, the sign of the terms are negated. Similarly, while shifting terms from left to right as a multiplication or division, the processes are interchanged. Let us take an example: 2x + 1 3 −4= −x 3 2 We group all terms with x to the left and the other constant terms to the right. Thus, the term −x, when taken to the left becomes +x, similarly, the term −4, when taken to the right, becomes +4: 2x + 1 3 +x = +4 3 2 Simplifying the left and right sides: 3+2×4 (2x + 1) + 3x = 3 2 11 5x + 1 = => 3 2 The ‘3’ on the left and the ‘2’ on the right are both ‘divisions’, and hence, on changing sides, they become ‘multiplication’: => 2 (5x + 1) = 11 × 3 => 10x + 2 = 33 We take the constant ‘2’ to the right: => 10x = 33 − 2 = 31 The ‘10’ as a product on the left is taken to the right as a division: => x =
31 10
Solving Linear Equations with two variables: There are two possible ways of solving such equations:
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• Method of substitution: From the first equation, we express one variable in terms of the other and substitute in the second equation. • Solving simultaneously: Using both equations simultaneously we eliminate one of the two variables. Let us take an example: 2x + 3y = 15 . . . (i) 5x − 2y = 28 . . . (ii) • Solving by the method of substitution: 15 − 3y From (i): x = 2 Substituting in (ii): 5 15 − 3y − 2y = 28 2 The above can be solved as discussed above, for a single variable: => 75 − 15y − 4y = 56 => −19y = −19 => y = 1 Substituting y = 1 in the expression for x, we have: x=
15 − 3 × 1 =6 2
• Solving simultaneously: We need to eliminate one of the variables by multiplying the equations with a suitable constant and then adding/subtracting them: Equation (i) × 2 + equation (ii) × 3: 4x + 6y = 30 15x − 6y = 84 => 19x = 114 => x = 6 Substituting x = 6 in (i): 2 × 6 + 3y = 15 => y =
15 − 12 =1 3
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Algebra Guide – Concepts
Graphically, a pair of linear equations with two variables represents two straight lines, the solution of which represents their point of intersection. Let us take two general equations: ax + by = c . . . (i) Ax + By = C . . . (ii) Here, a, b, c, A, B, C are the constants and x, y are the variables. There are three important results possible:
•
a b c = = : This implies that the equations are identical equations, i.e. they are overlapping A B C lines and hence, they have infinite solutions.
•
b c a = 6= : This implies that the equations are equations of parallel lines and since parallel A B C lines never intersect, there exists no possible solution.
•
a b 6= : This refers to the case of intersecting lines, i.e. a unique solution can be obtained from A B the two equations.
◦
b c a 6= = : Since the ratio of coefficients of y is the same as the ratio of the constant A B C terms, the value of y in the equation is zero.
◦
b a c 6= = : Since the ratio of coefficients of x is the same as the ratio of the constant B A C terms, the value of x in the equation is zero.
Solving for x and y from a single equation given they are integers: Let us take an example: Find all the possible values of x and y if 6x + 32y = 102, where x, y are positive integers: • First we need to reduce the coefficients to their lowest terms. Thus, we have: 3x + 16y = 51 • We find any possible integer solution of the above equation (it is not important to keep both x and y positive at this stage). We can see that x = 17 and y = 0 obviously satisfies. • It is clear that we need to increase the value of y to make it positive. Thus, we add the coefficient of x to the value of y and simultaneously, we subtract the coefficient of y from the value of x to generate the next possible integer solution. Thus, we get:
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Algebra Guide – Concepts x 17 17 − 16 = 1 1 − 16 = −15
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y 0 0+3=3 3+3=6
We see that on repeating the same process a second time, the value of x becomes negative and hence, inadmissible. Thus, there is only one possible solution: x = 1, y = 3. Determining the value of an expression involving the variables without solving: Let us take an example: 2x + 5y = 19 . . . (i) x + 2y = 8 . . . (ii) We need to determine the values of: (1) 3x + 7y (2) x + 3y (3) 4x + 9y For (1): Simply adding (i) and (ii): 3x + 7y = 27 For (2): Simply subtracting (ii) from (i): x + 3y = 11 For (3): (i) + 2 × (ii): 4x + 9y = 35 Additional solved problems: (1) In the GCAT test, for every correct answer, a student is awarded ‘+3’ marks and for every wrong answer, the student is awarded ‘−1’ marks. There are 40 questions and the student must answer all questions. I. If the student gets 40 marks, how many questions did he get correct? II. How many different scores can the student get, if he randomly marks the answers? Explanation: Let the student got x correct and y wrong answers. I. Thus, we have: x + y = 40 (There are 40 questions and he answered all questions) 3x − y = 40 (Each correct answer is of 3 marks and each wrong is of −1 mark) Adding the above two equations: 4x = 80 => x = 20
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Algebra Guide – Concepts II. If the student answers all questions randomly, x changes from 0 to 40 and y, correspondingly changes from 40 to 0. Thus, there are 41 possible values of the score: 120, 116, . . . . . . . . . − 36, −40.
(2) A and B have a number of marbles. The total number of marbles is 24. If A gives B 6 marbles, the ratio of the number of marbles with A and B is reversed. How many marbles does A have? Explanation: Let the number of marbles with A be x. Thus, the number of marbles with B = (24 − x). After A gives 6 marbles to B, we have: Number of marbles with A = (x − 6). Number of marbles with B = (24 − x + 6) = (30 − x). Since the ratio of the number of marbles with A and B is reversed, we have: x−6 24 − x = 30 − x x => x 2 − 6x = 720 − 54x + x 2 => 48x = 720 => x = 15 Alternate approach: As we can see, the above approach is a bit complicated. However, we can use a much simpler logic. After A gives B 6 marbles, the total number of marbles with A and B remain unchanged. Since the ratio of the number of marbles has reversed, we can simply say that the number of marbles with A and B have been interchanged. Thus, we have: x − 6 = 24 − x => x = 15
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(3) Abe has 20 coins with him. A few of the coins are dimes (a dime is equal to 10 cents) and the rest are quarters (a quarter is equal to 25 cents). If the total amount with him is $4.25, how many dimes and quarters does he have separately? (Note: 100 cents make a dollar) Explanation: Let us solve this problem in three different ways. Approach 1: Using two variables Let the number of dimes be x and the number of quarters be y. Thus, we have: x + y = 20 . . . (i) Total worth of dimes is 10x cents and the total worth of quarters is 25y cents. Since the total amount with him is $4.25, i.e. 425 cents, we have: 10x + 25y = 425 => 2x + 5y = 85 . . . (ii) (ii) – (i) × 2: 3y = 45 => y = 15 => x = 5 Approach 2: Using one variable Let the number of dimes be x. Since there are 20 coins, the number of quarters = (20 − x) Since the total amount is 425 cents, we have: 10x + 25 (20 − x) = 425 => 10x − 25x = 425 − 500 => −15x = −75
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Algebra Guide – Concepts => x = 5 => Number of quarters = 20 − 5 = 15 Approach 3: Using No variables Let us assume that all his coins are dimes. Thus, he has 20 dimes, which account for 20 × 10 = 200 cents. However, he actually has 425 cents. Thus, there is a shortage of 425 − 200 = 225 cents. Thus, we can conclude that some of his dimes are actually quarters. In that case, each such dime would increase in value from 10 cents to 25 cents, i.e. an increase of 25 − 10 = 15 cents. The number of times such a conversion from dime to quarter required to fulfill the shortage of 225 cents =
225 = 15 times 15
Thus, of the 20 dimes, 15 need to be converted to quarters and the rest should remain as dimes. Thus, number of quarters is 15 and the number of dimes is 5. (4) Joe purchases 7 identical gift-packs from Shop A and 4 identical gift-packs from Shop B. Each gift pack has a number of pens. The total number of pens in the 11 gift-packs is 26. If he wants additional 18 pens, what is the maximum number of gift-packs, similar to the ones he purchased initially, he needs to buy? Explanation: Let the number of pens present in gift-packs from Shop A = x Let the number of pens present in gift-packs from Shop B = y Thus, total number of pens = 7x + 4y
Thus, we have: 7x + 4y = 26 Since x and y must be positive integers, 4y is even. Thus, 7x must also be even since the sum of 7x and 4y is 26, an even number.
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Since 7x is even, x must be even. Thus, the only value of x = 2 (any value of x greater than 3 is not possible since 7x would exceed 26 and y would become negative). Thus, we have: y=
26 − 7 × 2 =3 4
Thus, the gift-packs from Shop A have 2 pens each and those from Shop B have 3 pens each. Thus, to maximize the number of gift-packs, he should purchase the ones from Shop A. Thus, to buy 18 pens, the maximum number of gift-packs required =
18 =9 2
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2.3
Algebra Guide – Concepts
Quadratic Equations
An equation is said to be quadratic when the variable contains the highest exponent of ‘2’, thus a x 2 + b x + c is a quadratic expression in one variable i.e. of x . Solving a Quadratic Equation by Factorization: A quadratic equation can be solved quickly if it can be written in the form of a product of two linear equations. Let us take an example: 2x 2 − 7x + 6 = 0 The given equation can be written as: 2x 2 − 4x − 3x + 6 = 0 While dividing the middle term in two parts, it should be seen that the product of the two terms equals the product of the first and last terms of the quadratic. Thus, here we have: (−4x) × (−3x) = 2x 2 × (6) = 12x 2 => 2x (x − 2) − 3 (x − 2) = 0 => (2x − 3) (x − 2) = 0 This results in two separate linear equations: 2x − 3 = 0 OR x−2=0 => x =
3 OR x = 2 2
Thus, each quadratic equation can be written in the form x − p roots of the quadratic equation.
x − q = 0, where p and q are the
Roots of a Quadratic Equation: After suitable reduction, every quadratic equation can be written in the form: ax 2 + b x + c = 0 The solution or the roots of the above equation is given by: q q 2 2 ac ac − b + b − 4a − b − b − 4a p= and q = , where p and q are the roots of the equation. a a 2a 2a Let us see how:
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ax 2 + bx + c = 0 Dividing throughout by a: x2 +
c b x+ =0 a a
Let us combine the first two terms of the above equation to form a perfect square: b b 2 b 2 c + − + =0 2a 2a 2a a 2 2 2 b b − 4ac b c => x + = − = 2a 2a a 4a2
=> x 2 + 2x
Taking square roots on both sides: s √ b b2 − 4ac b2 − 4ac => x + =± =± 2 2a 4a 2a √ 2 b b − 4ac => x = − ± 2a 2a √ 2 −b ± b − 4ac => x = 2a Let us find the roots of the previous equation 2x 2 − 7x + 6 = 0 using the above formula: We have: a = 2, b = −7, c = 6 Thus, the roots are: −b ±
√
q − (−7) ± (−7)2 −4 (2) (6) 7−1 7+1 3 7±1 b2 − 4ac = = OR = OR 2 = 2a 2 (2) 4 4 4 2
Discriminant: 2 ac , i.e. the quantity under the square root, is called the discriminant The value given by D ( 4))= b − 4a and depending upon its value we can determine the nature of the roots of the quadratic equations:
• If D ≥ 0: The roots are real • If D = 0: The roots are real and equal • If D > 0: The roots are real and unequal • If D < 0: The roots are imaginary • If D is a perfect square: The roots are real and rational • If D is not a perfect square: The roots are real and irrational (one root is conjugate of the other, √ i.e. the roots are of the form a ± b) Let us take some example • In the equation x 2 + 2x + 1 = 0, D = b2 − 4ac = 22 − 4 (1) (1) = 0 => The roots are real and equal. © 1999–2016 Manhattan Review
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Algebra Guide – Concepts • In the equation x 2 +2x +2 = 0, D = b2 −4ac = 22 −4 (1) (2) = −4 < 0 => The roots are complex.
Sum and Product of the roots: If the roots of the quadratic equation ax 2 + bx + c = 0 are p and q, we have: ax 2 + bx + c ≡ a x − p
x−q
=> ax 2 + bx + c ≡ a x 2 − x p + q + pq => ax 2 + bx + c ≡ ax 2 − a p + q x + apq Comparing the coefficient of x and the constant term, we have: b = −a p + q and c = apq => p+q=−
c b and pq= a a
Expressing a quadratic equation in terms of its roots: Suppose we have to form the equation whose roots are p and q. Thus, we have: x − p = 0 and x − q = 0 => x − p
x−q =0
=> x 2 − x p + q + pq = 0 => x 2 − (Sum of roots) x + (Product of roots) = 0 Let us take an example: The equation having roots −2 and 5 is: x 2 − (−2 + 5) x + (−2) (5) = 0 => x 2 − 3x − 10 = 0 Additional solved problems: (1) A man can swim in still water (without any current) at a rate of 4 miles per hour. He undertakes to swim from point A to point B and back, in a river which has a current of its own. If the distance between the points A and B is 15 miles, and the total time he takes for the trip is 8 hours, what is the rate of flow of the river current in miles per hour? Note: While swimming against the river, the man’s normal swimming rate would be reduced by the amount of the rate of the current, similarly, while going along with the flow of the river, his speed would be increased by the rate of the current. Explanation:
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Let the rate of the river current be r miles per hour. Since the man makes a round trip, one way he would be swimming against the river, while the other way, he would be swimming with the river. Thus, while swimming against the river, the rate of the man = (4 − r ) miles per hour Thus, time taken to cover 15 miles =
15 4−r
hours.
While swimming with the river, the rate of the man = (4 + r ) miles per hour Thus, time taken to cover 15 miles =
15 4+r
hours.
Since the total time is 8 hours, we have: 15 15 + =8 4−r 4+r => 15 (4 + r + 4 − r ) = 8 42 − r 2
=> 8r 2 = 8 × 42 − 15 × 8 = 128 − 120 => 8r 2 = 8 => r 2 = 1 => r = 1
(2) A man buys a number of pieces of chocolates for $24. If the price of a piece of chocolate increases by $2, he can buy 1 piece of chocolate less for the same amount. What is the price of 1 piece of chocolate? Explanation: Let the price of a piece of chocolate be $x. Thus, the number of chocolate pieces the man can buy for $24 =
24 x
New price of a piece of chocolate = $ (x + 2). Thus, the number of chocolate pieces the man can now buy for $24 =
24 x+2
Since the number of pieces of chocolate is 1 less than previously, we have: 24 24 = −1 x+2 x =>
24 24 − =1 x x+2
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Algebra Guide – Concepts
=>
24 (x + 2 − x) =1 x (x + 2)
=> x (x + 2) = 48 Since 48 = 6 × 8, we have: x=6 Alternately, we can solve the quadratic: x (x + 2) = 48 => x 2 + 2x − 48 = 0 => (x + 8) (x − 6) = 0 => x = 6
(3) The total cost, in dollars, of manufacturing n items of a product is given by 2n2 + 30 . The selling price of each item is fixed at $36. What is the number of items that must be sold so as to have maximum profit? Explanation: Total cost of manufacturing n items = $ 2n2 + 30 . Selling price of each item = $36. Thus, total selling price of n items = $ (36n). Thus, profit earned = Selling price – Cost price = 36n − 2n2 + 30
= −2 n2 − 18n + 15
= −2 n2 − 18n + 81 − 66
= −2(n − 9)2 + 132 The profit would be maximized if the negative square term, i.e. −2(n − 9)2 becomes ‘0’, which happens when n = 9. The corresponding maximum value of the profit is $132.
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2.4
19
Inequalities
Inequalities: It deals with cases where variables or numbers are less than or more than other variables or numbers. The following symbols are used: >: More than y Adding z to both sides: x + z > y + z Subtracting z from both sides: x − z > y − z (2) In any given inequality, we can multiply or divide any other ‘positive’ quantity on both sides keeping the inequality the same. For example: If we have: x > y Multiplying a positive quantity z on both sides: x × z > y × z y x > Dividing by a positive quantity z on both sides: z z (3) In any given inequality, we can multiply or divide any other ‘negative’ quantity on both sides, thus, reversing the inequality. For example: If we have: x > y Multiplying a negative quantity (−z) on both sides: −x × z < −y × z x y Dividing by a negative quantity (−z) on both sides: − < − z z (4) Any two inequalities having the same inequality can be added. For example: If we have: x > y and p > q Adding the inequalities, we have: x + p > y + q (5) In order to subtract one inequality from another, the former has to be negated, thus reversing the inequality and then the inequalities can be added. For example: If we have x > y and p > q Negating the former inequality, we have: −x < −y => −y > −x Thus, on adding the inequalities p > q and −y > −x, we have: p−y >q−x (6) Any term with an even exponent is always non-negative, its minimum value being ‘0’. Thus, we have: x 2 ≥ 0, y 4 ≥ 0, z6 ≥ 0, etc. Thus, if we have: x 5 y 2 z > 0, we have: => x 4 y 2 xz > 0
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Algebra Guide – Concepts Since x 4 y 2 is always non-negative, we have: => xz > 0 => x > 0 and y > 0 OR x < 0 and y < 0
(7) Inequality of ratio of two quantities and the inequality of the product of those quantities are similar. x >0 For example: If we have: y x => × y 2 > 0 (since y 2 > 0) y => xy > 0 Note: However, there is a slightly different result for “greater than or equal to” or “less than or equal to” type of inequalities, as shown: If we have:
x ≥0 y
x × y2 ≥ 0 y => xy ≥ 0
=>
On careful observation, it can be seen that y = 0 is a possible solution for xy ≥ 0, however, not x so in the case of ≥ 0. y (1) Number line based inequalities: There are four important regions on a number line, as shown below:
IV
III
II
I
−∞ − 1 0 1 ∞
Some properties of the above four regions: ◦ Region I: A number x in the region satisfies 1 < x < ∞ Higher the exponent of x, higher is the value of the term and vice versa. Thus, we have: x < x 2 < x 3 . . . For example: 2 < 22 < 23 √ √ √ √ 3 x > x > 3 x . . . For example: 2 > 2 > 2 ◦ Region II: A number x in the region satisfies 0 < x < 1 Higher the exponent of x, smaller is the value of the term and vice-versa. www.manhattanreview.com
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Thus, we have: 2 3 1 1 1 > > 2 2 s s2 √ √ 1 1 3 1 x < x < 3 x . . . For example: < < 2 2 2 x > x 2 > x 3 . . . For example:
◦ Region III: A number x in the region satisfies −1 < x < 0 • For odd exponents (the values are always negative): Higher the exponent of x, higher is the value of the term and vice versa. Thus, we have: 1 3 1 5 1 < − x < x 3 < x 5 . . . For example: − < − 2 s2 s 2 √ √ 1 1 1 3 5 x > 3 x > 5 x . . . For example: − > − > − 2 2 2 • For even exponents (the values are always positive): Higher the exponent of x, smaller is the value of the term and vice versa. Thus, we have: 1 2 1 4 1 6 x 2 > x 4 > x 6 . . . For example: − > − > − 2 2 2 (Note: Square roots and fourth roots, etc. are not possible for negative numbers) An even exponent results in a positive value, which will always be greater than the value resulting from an odd exponent, which is always negative. ◦ Region IV: A number x in the region satisfies −∞ < x < −1 • For odd exponents (the values are always negative): Higher the exponent of x, smaller is the value of the term and vice versa. Thus, we have: x > x 3 > x 5 . . . For example: −2 > (−2)3 > (−2)5 x<
√ 3
x<
√ 5
x . . . For example: −2 <
√ 3
−2 <
√ 5
−2
• For even exponents (the values are always positive): Higher the exponent of x, higher is the value of the term and vice versa. Thus, we have: x 2 < x 4 < x 6 . . . For example: (−2)2 < (−2)4 < (−2)6 (Note: Square roots and fourth roots, etc. are not possible for negative numbers) An even exponent results in a positive value, which will always be greater than the value resulting from an odd exponent, which is always negative. (2) Quadratic inequalities: We have the following two rules: ◦ x 2 < k2 => −k < x < k © 1999–2016 Manhattan Review
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Algebra Guide – Concepts ◦ x 2 > k2 => x < −k OR x > k Any quadratic can be converted to one of the above two forms. Let us take an example: x 2 − 8x + 12 < 0 => x 2 − 2 × x × 4 + 42 − 42 + 12 < 0 => (x − 4)2 < 4 => −2 < x − 4 < 2 => 2 < x < 6
Alternately, we can use the rules below: If (x − k) and (x − m) are the factors of a quadratic, and k > m, we have: ◦ (x − k) (x − m) > 0 => x > k OR x < m Thus, x is greater than the greatest root OR smaller than the smallest root. ◦ (x − k) (x − m) < 0 => m < x < k Thus, x lies between the two roots. Let us take an example: x 2 − 8x + 12 < 0 => (x − 2) (x − 6) < 0 Since the roots are 2 and 6, with 6 being the greater root, we have: => 2 < x < 6 Modulus or Absolute value: Modulus of a number is a function that returns the magnitude of the number ignoring the sign. Thus, we have: |2| = |−2| = 2 Hence, we have: |x| = 2 = > x=±2 Another way of interpreting the modulus of a number is distance of the number from a point on the number line. For example, |x − 3| denotes the distance of x from the point ‘3’ on the number line.
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Thus, |x − 3| = 4 implies the distance of the point x from the point ‘3’ on the number line is 4 units. Since the distance can be measured on either side of ‘3’, we attain the points 3 + 4 = 7 and 3 − 4 = −1 on the number line, as shown below.
4 −1
4 3 7
Thus, using the above concept, we have: |x − a| = b => x − a = ±b => x = a ± b (1) Modulus based inequality: We have the following two rules: ◦ |x − a| > b => x − a > b OR x − a < −b => x > a + b OR x < a − b Alternately, the given inequality implies that the distance of x from a is greater than b units. If the distance were equal to b units, we would have got the points (a + b) and (a − b). Since the distance should be greater than b units, x should overshoot the above two points, i.e. x is greater than the greatest point or x is less than the smallest point. Thus, we have: x > a + b OR x < a − b ◦ |x − a| < b => −b < x − a < b => a − b < x < a + b Alternately, the given inequality implies that the distance of x from a is lesser than b units. If the distance were equal to b units, we would have got the points (a + b) and (a − b). Since the distance should be lesser than b units, x should remain within the above two points. Thus, we have: a − b < x < a + b • For any two positive numbers a and b, we always have: (a+b) √ ≥ ab 2 • For a given sum of two or more quantities, the product of the quantities is maximized if the quantities are made equal. Similarly, for a given product of two or more quantities, the sum of the quantities is minimized if the quantities are made equal. Let us take a few examples:
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Algebra Guide – Concepts ◦ What is the maximum value of a × b if 2a + 3b = 20? Since we need to maximize a × b, we need to maximize a and b. Since 2a + 3b = 20, the product 2a × 3b would be maximized if 2a = 3b => 2a + 2a = 20 => a = 5 => 2a = 3b = 10 Thus, the maximum value of 2a × 3b = 10 × 10 = 100 => The maximum value of a × b =
100 50 (2a) × (3b) = = 6 6 3
◦ What is the minimum perimeter of a rectangle having an area of 100? Let the length and width of the rectangle be l and w, respectively. Thus, the area of a rectangle = l × w = 100 We need to minimize the perimeter, i.e. 2 (l + w). Thus, we need to minimize the value of (l + w). The value of (l + w) will be minimized if l = w => l × l = 100 => l = 10 => l = w = 10 Thus, minimum perimeter = 2 (10 + 10) = 40. Additional solved problems: (1) What are the integer values of x satisfying
2x − 3 x+1
≤ 0?
Explanation: 2x − 3 x+1 Thus, we have:
We have:
≤0
Case I: 2x − 3 ≤ 0 AND x + 1 > 0 3 => x ≤ AND x > −1 2 3 => −1 < x ≤ 2 => The integer values of x are: 0 and 1. Case II: 2x − 3 ≥ 0 AND x + 1 < 0 3 => x ≥ AND x < −1, which is not a possible scenario 2 www.manhattanreview.com
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Thus, the only possible integer values of x are: 0 and 1.
(2) What are the integer values of x satisfying x 2 − 6 < 3? Explanation: We have: x 2 − 6 < 3 => −3 < x 2 − 6 < 3 => 3 < x 2 < 9 Case I: x 2 < 9 => −3 < x < 3 => The possible integer values of x are: −2, −1, 0, 1, 2 Case II: x 2 > 3 √ √ => x > 3 OR x < − 3 => The possible integer values of x are: · · · − 4, −3, −2, 2, 3, 4, . . . Thus, the common set of values of x are: −2 and 2.
(3) Which of the following is true about x if x 2 > x > x 3 ? (A) x > 1 (B) 0 < x < 1 (C) −1 < x < 0 (D) x < −1 Explanation: We know that there are four major regions on the number line: I. x > 1 II. 0 < x < 1 III. −1 < x < 0 IV. x < −1 Let us pick 1 number from each region and check whether it satisfies the given inequality: • x = 2 : 22 > 2 ≯ 23 2 3 1 1 1 1 • x= : ≯ > 2 2 2 2 © 1999–2016 Manhattan Review
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Algebra Guide – Concepts
• x=−
1 1 2 1 3 1 : − >− ≯ − 2 2 2 2
• x = −2 : (−2)2 > −2 > (−2)3 – Satisfies Alternate approach: Since x 2 is greater than both x, a term with a lower exponent and x 3 , a term with a higher exponent, the value of x must be negative. Since x, a term with a smaller odd exponent, is greater than x 3 , a term with a higher odd exponent, the value of x must be less than −1. Hence, correct answer is option D.
(4) If −5 < x < 3, −9 < y < −5, −13 < p < 8, −4 < q < 6 and 1 ≤ r ≤ 8, where x, y, p, q and r are integers, what are the range of values of I. x + 2y II. 2x − y III. p × q q IV. r Explanation: I. We have: −9 < y < −5 => −18 < 2y < −10 => −5 − 18 < x + 2y < 3 − 10 => −23 < x + 2y < −7 II. We have: −5 < x < 3 => −10 < 2x < 6 => −10 − 9 < 2x + y < −5 + 6 => −19 < 2x + y < 1 III. We have: −13 < p < 8 and −4 < q < 6 => −12 ≤ p ≤ 7 and −3 ≤ q ≤ 5 Thus, the maximum value of p × q = The greater among {(−12) × (−3)} and {(7) × (5)} = 36 Also, the minimum value of p × q = The smaller among {(−12) × (5)} and {(7) × (−3)} = −60 Thus, we have: −60 ≤ pq ≤ 36
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IV. We have: −4 < q < 6 => −3 ≤ q ≤ 5 and 1 ≤ r ≤ 8 q To determine the maximum value of , we must keep the ratio positive. Since r is entirely r positive, we must take the positive value of q = 5 and the least positive value of r = 1. q 5 Thus, the maximum value of = = 5 r 1 q Again, to determine the minimum value of , we must keep the ratio negative. Since r is r entirely positive, we must take the negative value of q = −3 and the least positive value of r = 1. (Note: the minimum value of
q is the negative number with the largest magnitude) r
Thus, the minimum value of
q −3 = = −3 r 1
Thus, we have: −3 ≤
q ≤5 r
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Algebra Guide – Concepts
2.5
Functions and graphs of functions
If y is expressed in terms of x, such that, for any x, a unique value of y is obtained, the expression of y in terms of x is called a function of x. Functions of x are usually denoted by symbols of the form f (x) , g (x) , h(x), etc. Some examples of functions: • f (x) = 2x + 5 • f (x) = x 2 − 2x + 1 1 x √ • f (x) = x + 1
• f (x) =
• g (x) =
x−2 x2 + 3
• g (x) = 1 The idea of a function can be represented by a simple diagram as shown below:
Input 𝑥
𝑓 𝑥
Output 𝑦
Method of substitution in functions: For any given f (x), the value of f (a) is determined by substituting x = a on the right side of the equality. Let us take an example: f (x) = 2x + 1 Thus, we have: • f (1) = 2 (1) + 1, i.e. we replace x with 1 on the right side of the equality => f (1) = 3 • f x 2 = 2 x 2 + 1 = 2x 2 + 1 • f (2x + 1) = 2 (2x + 1) + 1 = 4x + 3 Domain and Range: The Domain refers to the set of values of x that can be used in the function. The Range refers to the set of values of f (x) obtained using the above values of x.
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Let us take an example: f (x) = x 2 − 1 Let the set of values of x to be used be {−1, 0, 1}. Thus, we have: • f (−1) = (−1)2 − 1 = 0 • f (0) = 02 − 1 = −1 • f (1) = 12 − 1 = 0 Here, Domain = {−1, 0, 1} and Range = {−1, 0} To determine the Domain for a function, two things need to considered: • For any term under a square-root or fourth-root, etc., the term should be non-negative. √ For example: f (x) = x − 1 We have: x − 1 ≥ 0 => x ≥ 1 Thus, the Domain is: 1 ≤ x < ∞ • For any term in the denominator, the term must be non-zero. 2x For example: f (x) = x−3 We have: x − 3 6= 0 => x 6= 3 Thus, the Domain is: x is any real number except 3 => −∞ < x < 3 OR 3 < x < ∞ Let us take an example: f (x) = √
x+3 2x−4−4
Thus, we have: • 2x − 4 ≥ 0 => x ≥ 2 . . . (i) √ √ • 2x − 4 − 4 6= 0 => 2x − 4 6= 4 => 2x − 4 6= 16 => x 6= 10 . . . (ii) Thus, the Domain: 2 ≤ x < 10 OR 10 < x < ∞ Composite functions: For any two functions f (x) and g (x), the functions defined as f (f (x)), f g (x) , g g (x) and g (f (x)) are composite functions. Let us take an example:
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Algebra Guide – Concepts
f (x) = 2x + 1 and g (x) = x 2 − 1 Thus, we have: • f (f (x)) = 2 (f (x)) + 1 = 2 (2x + 1) + 1 = 4x + 3 • f g (x) = 2 g (x) + 1 = 2 x 2 − 1 + 1 = 2x 2 − 1 2 2 • g g (x) = g (x) + 1 = x 2 + 1 + 1 = x 4 + 2x 2 + 2 2
• g (f (x)) = (f (x)) + 1 = (2x + 1)2 + 1 = 4x 2 + 4x + 2 Special case in a composite function: If f (x) and g (x) are two functions and it is observed that f g (x) = g (f (x)) = x, we have:
Input 𝑎
𝑓 𝑥
Output 𝑏
Input 𝑏
𝑔 𝑥
Output 𝑎
Thus, we have: If f (a) = b => g (b) = a Note: Such functions f (x) and g (x) are inverse functions of one another. Let us take an example: If f (x) = (x + 1)3 − 1 and g (x) =
√ 3 x + 1 − k, such that f g (x) = g (f (x)), what is the value of k?
The normal way of solving, by evaluating the composite functions f g (x) and g (f (x)), is complicated. Instead, we use the above method: f (1) = (1 + 1)3 − 1 = 7 => g (7) = 1 =>
√ 3
7 + 1 − k = 1 => 2 − k = 1
=> k = 1 Periodic function: A function f (x) is periodic if there exists a number n so that f (x + n) = f (x) for all x. Here, n is the period of the function. Let us take an example: If f (x + 3) = f (x + 2) − f (x + 1), what is the value of n if f (1) = −f (1 + n)? We have: f (x + 3) = f (x + 2) − f (x + 1) Substituting different values of x:
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• x = 0 : f (3) = f (2) − f (1) . . . (i) • x = 1 : f (4) = f (3) − f (2) . . . (ii) Adding (i) and (ii): f (3) + f (4) = f (3) − f (1) => f (1) = −f (4) => f (1) = −f (1 + 3) => n = 3 Piece-wise functions: Functions which have different expressions over different values of x are piece-wise functions. Some examples are shown below: • Modulus function: f (x) = |x|: ◦ f (x) = x, if x ≥ 0, ◦ f (x) = −x, if x < 0 The graph of f (x) = |x| is shown:
Y
O
X
• Greatest Integer Function: f (x) = [x]: It is a function that returns the greatest integer less than or equal to x. Thus, we have: ◦ [1.23]: The greatest integer less than or equal to 1.23, i.e. the greatest integer among 1, 0, −1, −2, · · · = 1 ◦ [1]: The greatest integer less than or equal to 1, i.e. the greatest integer among 1, 0, −1, −2, · · · = 1 ◦ [−1.23]: The greatest integer less than or equal to −1.23, i.e. the greatest integer among −2, −3, −4, · · · = −2
◦ [−1]: The greatest integer less than or equal to −1, i.e. the greatest integer among −1, −2, −3, −4, · · · = −1 • Least Integer Function: f (x) = {x}: It is a function that returns the least integer greater than or equal to x. Thus, we have:
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Algebra Guide – Concepts ◦ {1.23}: The least integer greater than or equal to 1.23, i.e. the least integer among 2, 3, 4, · · · = 2 ◦ {1}: The least integer greater than or equal to 1, i.e. the least integer among 1, 2, 3, 4, · · · = 1 ◦ {−1.23}: The least integer greater than or equal to −1.23, i.e. the least integer among −1, 0, 1, 2, · · · = −1 ◦ {−1}: The least integer greater than or equal to −1, i.e. the least integer among −1, 0, 1, 2, · · · = −1
• Max-Min function:
◦ f (x) = max(a, b) implies that f (x) = a if a > b OR f (x) = b if b > a ◦ f (x) = min(a, b) implies that f (x) = a if a < b OR f (x) = b if b < a
Let us take an example: If f (x) = min(6x − 8, x 2 ), for what integer values of x is f (x) = x 2 ? Since f (x) = min(6x − 8, x 2 ) = x 2 , we have: x 2 < 6x − 8 => x 2 − 6x + 8 < 0 => (x − 2) (x − 4) < 0 => 2 < x < 4 Thus, the only integer value of x = 3
Properties of graphs of functions:
• The graph of f x+p is obtained by shifting the graph of f (x) by p units left • The graph of f x−p is obtained by shifting the graph of f (x) by p units right • The graph of f (x) +p is obtained by shifting the graph of f (x) by p units up • The graph of f (x) −p is obtained by shifting the graph of f (x) by p units down • The graph of f (−x) is obtained by reflecting the graph of f (x) about the Y-axis • The graph of −f (x) is obtained by reflecting the graph of f (x) about the X-axis
Graphs of some quadratic functions: • f (x) = x 2 : www.manhattanreview.com
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33
Y
X
• f (x) = x 2 + 1:
Y
0, 1
X
• f (x) = (x − 1)2 :
Y
1, 0
X
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Algebra Guide – Concepts
Additional solved problems: √ √ √ f (4) 1 1 1 f (3) f (2) 4 3 √ ? (1) If f (x) = x + x + x + 1 + √ + √ + , what is the value of + + 3 4 1 1 x x x f 1 f f 4 3 2 Explanation: We have: √ 4
√ 3
√
1 1 1 x+1+ √ + √ + √ 3 4 x x x s s s 1 1 1 1 1 1 1 => f = 4 + 3 + +1+ s + s + s x x x x 1 3 1 4 1 x x x √ √ √ 1 1 1 1 = √ + √ + √ +1+ x+ 3x+ 4x => f 4 3 x x x x 1 f (x) => f = f (x) => = 1 1 x f x
f (x) =
=>
x+
x+
f (3) f (4) f (2) = = =1 1 1 1 f f f 2 3 4
Thus, we have: f (4) f (3) f (2) + + 1 1 1 f f f 4 3 2 =1+1+1=3
(2) If f (x) =
(A)
4x , what is the value of f (a) + f (1 − a)? +2
4x
1 2
(B) 1 (C) 2 Explanation: We have: f (x) =
4x 4x + 2
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Algebra Guide – Concepts
=> f (a) =
35
4a 4a + 2
Also, we have:
f (1 − a) =
=
4 4a 4 +2 4a
4(1−a) 4(1−a) + 2
=
4 1 a a 4+2×4 4 4a
=
4a 4 × a 4 2 (2 + 4a )
=
2 2 + 4a
Thus, we have: f (a) + f (1 − a) =
4a 2 4a + 2 + = 4a + 2 2 + 4a 4a + 2
=1 Alternate approach: Let a = 1 => 1 − a = 0:
f (1) = f (0) =
41 4 2 = = +2 6 3
41
1 4o = +2 3
4o
=> f (1) + f (0) =
2 1 + =1 3 3
Since the answer options are constant values, the answer must be Option B.
(3) Are the following functions the same? (A) f (x) = x x (B) g (x) = x × x × x × x × · · · × x (x times) Explanation:
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36
Algebra Guide – Concepts (A) We have f (x) = x x For example: If x = 3 => f (x) = 33 = 27 If x = −1 => f (−1) = (−1)−1 = −1 1 If x = => f 2
12 1 1 1 1 = = √ ≈= = 0.71 2 2 1.4 2
Thus, f (x) = x x is valid for all real values of x. (B) We have: g (x) = x × x × x × x × · · · × x (x times) If x is multiplied for x times, the result obtained is x x . Thus, apparently, f (x) and g (x) appear to be identical. However, we are multiplying x for x times, which makes sense only when x is a positive 1 1 integer. (Multiplying for times or multiplying −1 for −1 times makes no sense) 2 2 For example: g (3) = 3 × 3 × 3 (3 is multiplied for 3 times) = 27 Thus, g (x) is valid only for positive integer values of x. Thus, f (x) is not the same as g (x). (4) If (2, 1) are the coordinates of a point on the graph of f (x), what would be the coordinates of that point for the function −f (x) + 1? Explanation: To modify f (x) to −f (x) + 1, we follow the following steps: • f (x) → −f (x): The graph is reflected about the X-axis. Thus, the Y-coordinate of the point would be negated. Thus, the coordinates of −f (x) = (2, −1) • −f (x) → −f (x) + 1: The graph is shifted ‘up’ by 1 unit. Thus, the Y-coordinate of th point would increase by ‘1’. Thus, the final coordinates of −f (x) + 1 = (2, −1 + 1) = (2, 0)
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Algebra Guide – Concepts
37
In the GMAT, only two kinds of questions asked: Problem Solving and Data Sufficiency.
Problem Solving Problem solving (PS) questions may not be new to you. You must have seen these types of questions in your school or college days. The format is as follows: There is a question stem and is followed by options, out of which, only one option is correct or is the best option that answers the question correctly. PS questions measure your skill to solve numerical problems, interpret graphical data, and assess information. These questions present to you five options and no option is phrased as “None of these“. Mostly the numeric options, unlike algebraic expressions, are presented in an ascending order from option A through E, occasionally in a descending order until there is a specific purpose not to do so.
Data Sufficiency For most of you, Data Sufficiency (DS) may be a new format. The DS format is very unique to the GMAT exam. The format is as follows: There is a question stem followed by two statements, labeled statement (1) and statement (2). These statements contain additional information. Your task is to use the additional information from each statement alone to answer the question. If none of the statements alone helps you answer the question, you must use the information from both the statements together. There may be questions which cannot be answered even after combining the additional information given in both the statements. Based on this, the question always follows standard five options which are always in a fixed order. (A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the question asked. (B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the question asked. (C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked. (D) EACH statement ALONE is sufficient to answer the question asked. (E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed. In the next chapters, you will find 150 GMAT-like quants questions. Best of luck!
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Algebra Guide – Concepts
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Chapter 3
Practice Questions
39
40
Algebra Guide – Questions
3.1 1.
Problem Solving
The price of 19 chocolates and 21 pens is $29, while the price of 21 chocolates and 19 pens is $31. What is the price of 1 chocolate? (A)
$0.50
(B)
$1.00
(C)
$1.25
(D)
$1.50
(E)
$2.00
Solve yourself:
2.
Abe’s age is equal to the sum of the ages of his son and a 12-year old daughter. If Abe’s son is elder to Abe’s daughter, and the average age of Abe and his two children ten years ago was 20 years, what is Abe’ present age? (A)
30 years
(B)
33 years
(C)
39 years
(D)
45 years
(E)
51 years
Solve yourself:
3.
3 apples, 3 guavas and 4 bananas, together cost $10. Also, 3 apples, 2 guavas and 4 bananas together cost $9. What is the total cost of 9 apples, 8 guavas and 12 bananas? (A)
26
(B)
29
(C)
30
(D)
32
(E)
37
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Algebra Guide – Questions
41
Solve yourself:
4.
A person has a few cents and a few dollars such that the total amount is a dollars and b cents, where b < 100. After spending $3.50, he was left with 2b dollars and 64 cents. What is the value of (a + b)? (A)
14
(B)
28
(C)
32
(D)
46
(E)
64
Solve yourself:
5.
1 . If 8 1 however, 2 is subtracted from both numerator and denominator, the fraction decreases by . 4 What is the value of the original fraction? In a fraction, if 4 is added to both numerator and denominator, the fraction increases by
(A) (B) (C) (D) (E)
7 8 3 4 1 2 1 4 3 16
Solve yourself:
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42 6.
Algebra Guide – Questions If 7x − 2y = 12, 4x + y = 9 and 2x + 5y = K, what is the value of K? (A)
9
(B)
10
(C)
11
(D)
12
(E)
13
Solve yourself:
7.
x y 1 = = , where A, B and C are positive integers and the A B C greatest common divisor of A, B and C is 1, what is the value of (A + B + C)?
If 2x + 3y = 7, 5x + 3y = 13 and (A)
1
(B)
2
(C)
3
(D)
4
(E)
8
Solve yourself:
8.
The sum of the digits of a two-digit number is 5. The ratio of 20 less than the number and 12 3 more than the number is . What is the product of the digits of the number? 11 (A) 0 (B)
1
(C)
4
(D)
5
(E)
6
Solve yourself:
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Algebra Guide – Questions 9.
43
x, y and z satisfy the following set of equations: 3x + 7y − 11z = 0 6x − y − 7z = 0 3x + y − kz = 0 What is the value of k? (A) (B) (C) (D) (E)
1 4 3 2 7 3 5
Solve yourself:
10.
A group of children have a number of pens, such that each child has at least one pen. If one of the children, Ann, takes 1 pen from each of the other, the number of pens with her would be thrice the number of children in the group. If the total number of pens among the children is 42, which of the following could be the number of children in the group, so that it can be ensured that Ann has the greatest number of pens? I.
5
II.
9
III.
15
(A)
Only I
(B)
Only II
(C)
Only III
(D)
Both I and II
(E)
Both II and III
Solve yourself:
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44 11.
Algebra Guide – Questions A ball is thrown from the top of a building. The distance, in feet, covered by the ball in t seconds after it was dropped is given by 15t 2 . If the distance covered by the ball in the t th second after it was dropped was 225 feet. What is the value of t? (A)
3
(B)
4
(C)
7
(D)
8
(E)
10
Solve yourself:
12.
A ball is thrown up from a height of 3 feet above the ground. The distance, in feet, of the ball from the ground is given by h = 3 + 24t − 4t 2 , where t = time in seconds. What is the maximum height above the ground reached by the ball? (A)
32 feet
(B)
33 feet
(C)
35 feet
(D)
36 feet
(E)
39 feet
Solve yourself:
13.
The number of units sold, N, of a new product is expected to follow the relation: N = 120 − P , where P is the selling price per unit. If the cost of manufacturing any number of units of the new product is constant, and equal to $2000, what should be the maximum selling price of each unit so that there is neither profit nor loss? (A)
$20
(B)
$60
(C)
$80
(D)
$100
(E)
$110
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Algebra Guide – Questions
45
Solve yourself:
14.
What is the minimum positive integer value of p so that x 2 − px + 8p = 0 has real and unequal roots? (A)
12
(B)
24
(C)
27
(D)
32
(E)
33
Solve yourself:
15.
If one of the solutions of x 2 − px + 12 = 0 is x = (A) (B) (C) (D) (E)
16.
3 , what is the value of p? 2
3 2 8 19 2 57 4 15
If y = x 2 + kx + l intersects the X-axis at (4, 0) and the Y-axis at (0, 64), what is the value of k? (A)
−80
(B)
−40
(C)
−20
(D)
20
(E)
80
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46
17.
Algebra Guide – Questions
The inside dimensions of a rectangular steel frame, as shown in the diagram below, having uniform width of x inches, are 12 inches by 8 inches. If the area of the frame is 44 square inches, what is the total perimeter of the frame? 12 𝑥
8 𝑥
(A)
98 inches
(B)
49 inches
(C)
48 inches
(D)
24 inches
(E)
8 inches
Solve yourself:
18.
A roller-coaster track is designed in the form of a parabolic arch, whose height, in feet, above the ground is given as h = −kd (d − 20), where k is a positive number and d represents the distance along the length of the track measured from the left-most point where the arch starts. If the arch reaches to a maximum height of 30 feet, what is the value of k? (A) (B) (C) (D) (E)
3 10 2 5 3 4 4 5 3 2
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Algebra Guide – Questions
19.
47
For f (x) = x 2 + bx + c, f (1) = 0. If f (3) = 2f (5), what is the value of k such that f (k) = 0, k 6= 1? (A) (B) (C) (D) (E)
10 3 37 7 17 3 7 19 2
Solve yourself:
20.
If the roots of the fourth degree equation x 4 − 2x 3 + x 2 + x + 3 = 0 are p, q, r and s, what is the value of 2 + p 2 + q (2 + r ) (2 + s)? (A)
−37
(B)
−9
(C)
9
(D)
37
(E)
40
Solve yourself:
21.
Which of the following is the correct solution of the inequality x + I. II.
2 ≤ 3? x
x>2 1≤x≤2
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Algebra Guide – Questions III.
x ? − 9x + 18 2
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Algebra Guide – Questions
49
Solve yourself:
24.
Which of the following is the correct solution of the inequality x 3 − 9x 5 > 0? I. II. III.
0 b
If a = 6x − 8 and b = x 2 , for how many integer values of x will f (a, b) = b? (A)
None
(B)
One
(C)
Two
(D)
Three
(E)
Four
Solve yourself:
60.
If f (x) = |x − 1|, g (x) = x + a and f g (−3) = 2, what is the sum of the possible values of a? (A)
2
(B)
6
(C)
8
(D)
9
(E)
10
Solve yourself:
61.
For what value of n the equation below have no possible solution? 4 (2x − 1) + 3 (x − 2) = n (x + 2) − 3 (x + 1) (A)
(D)
−4 7 − 2 7 2 7
(E)
14
(B) (C)
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Algebra Guide – Questions
65
Solve yourself:
62.
If the equation x 2 − px + 12 = 0 has exactly one root common with the equation x 2 − 6x + 9 = 0, what is the value of p? (A)
−10
(B)
−7
(C)
4
(D)
7
(E)
10
Solve yourself:
63.
(0.23 + 0.52)2 − 0.232 + 0.522 What is the value of ? 0.752 − 0.292 (A)
0.3
(B)
0.4
(C)
0.5
(D)
2.0
(E)
4.0
Solve yourself:
64.
If p and q, p > q are the roots of x 2 − x − 12 = 0, the roots of the equation x 2 − 50x + 49 = 0 are (A)
2p and 3q
(B)
p 2 and q2
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66
Algebra Guide – Questions (C) (D) (E)
p + q and p − q 2 2 p + q and p − q p 2 + q2 and p 2 − q2
Solve yourself:
65.
If a and b are non-zero integers such that a2 + 2b2 + 2a + b = 0 and a2 + b2 = 2ab, what is the value of (a + b)? (A)
−2
(B)
−1
(C)
0
(D)
1
(E)
2
Solve yourself:
66.
If x + y = 2 and z = x 2 + y 2 , what is the minimum value of z?
(B)
1 2 1
(C)
2
(D)
4
(E)
6
(A)
Solve yourself:
67.
If q and 2q are the roots of the equation k x 2 − x + x + 1 = 0, where q 6= 0, which of the following is a correct relation between k and q?
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Algebra Guide – Questions
(A) (B) (C) (D) (E)
k=
67
q 2
1 1 + 3q k = 2q − 1 1 k= 2q2 k = q2 + 1 k=
Solve yourself:
68.
f (x) is a quadratic polynomial such that f (1) = 1 and f (2) = 2. If f (3) = 5, what is the value of f (0)? (A)
−6
(B)
−2
(C)
0
(D)
1
(E)
2
Solve yourself:
69.
For all positive integers x > 2, let f (x) be defined as f (x) = (−1)x × f (x − 1) × f (x − 2). If f (1) = −f (2) = 1, what is the value of f (23)? (A)
−1
(B)
0
(C)
1
(D)
3
(E)
7
Solve yourself:
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68
70.
Algebra Guide – Questions
For all values of x, let f (x) = (A)
−
3 2
(C)
1 3 1
(D)
p
(E)
pq
(B)
2x + 1 . If f p = q, what is the value of f q ? x−2
Solve yourself:
71.
What is the area bounded by f (x) = |x − 1| − x, the X and Y axes?
(C)
1 4 1 2 1
(D)
2
(E)
3 2
(A) (B)
Solve yourself:
72.
If f (x) = ax 2 + bx such that f (1) = f (−1) + 2, what is the value of f (3) − f (−3)? (A)
1
(B)
2
(C)
4
(D)
6
(E)
8
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Algebra Guide – Questions
73.
If x = 1 + (A) (B)
−2 √ − 3
(C)
0
(D)
1 √
(E)
√
69
3, what is the value of x 2 − 2x − 2 ?
3
Solve yourself:
√
74.
If x is a positive integer such that 24 (A)
0
(B)
1
(C)
2
(D)
4
(E)
6
x
√
+ 10
x
√
= 26
x
, what is the value of x?
Solve yourself:
75.
The function f (x) = 2x 2 − 7x + 6 is positive for all values of x except when x lies in a particular range. How many integer values of x lie within that range? (A)
None
(B)
One
(C)
Two
(D)
Three
(E)
Four
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Algebra Guide – Questions
Solve yourself:
76.
12 − x Which of the following is the correct range of values of x such that 3 ≤ 2? (A)
2≤x≤6
(B)
4 ≤ x ≤ 10
(C)
8 ≤ x ≤ 16
(D)
6 ≤ x ≤ 18
(E)
4 ≤ x ≤ 24
Solve yourself:
77.
How many integer values of x satisfy x 2 − 4 |x| + 3 ≤ 0? (A)
Two
(B)
Three
(C)
Four
(D)
Six
(E)
Eight
Solve yourself:
78.
What is the maximum value of (A)
1
(B)
2
(C)
3
(D)
4
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12 ? x 2 − 12x + 40
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Algebra Guide – Questions (E)
71
6
Solve yourself:
79.
If a and b are distinct roots of x 2 − px − q = 0, which of the following denotes the value of (a + b)? (A)
p
(B)
q
(C)
p+q
(D)
p−q
(E)
pq
Solve yourself:
80.
Let [x] denote the greatest integer value less than or equal to x for all positive integers x, such x+1 x+2 that 1 ≤ x ≤ 4. Which of the following expressions denotes the value of + + 2 4 x+4 ? 8 (A)
2x − 2
(B)
2x − 1
(C)
x
(D)
x+1
(E)
x+2
Solve yourself:
81.
If f (x) = x 3 − kx 2 + 2x, and f (−x) = −f (x), what is the value of f (1 − k)?
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72
Algebra Guide – Questions (A)
−3
(B)
0
(C)
1
(D)
3
(E)
4
Solve yourself:
82.
The function f is defined as f (x) = ax 2 + bx + c. If f (2) = f (3) = 0, and f (4) = 2, what is the value of (a + b + c)? (A)
1
(B)
2
(C)
8
(D)
11
(E)
12
Solve yourself:
83.
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets 5 than machine Y. At these rates, if the two machines together produce w widgets in 3 days, 4 how many days would it take machine X alone to produce 2w widgets? (A)
4
(B)
6
(C)
8
(D)
10
(E)
12
Solve yourself:
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Algebra Guide – Questions 84.
73
If f (x − 1) = 2x 2 − 3x + 3, what is the expression for f (x)? (A)
2x 2 + x + 2
(B)
2x 2 − x + 1
(C)
2x 2 + 2x − 3
(D)
2x 2 − 2x + 4
(E)
2x 2 − 4x + 2
Solve yourself:
85.
If f (x) = a (A)
−1
(B)
0
(C)
1
(D)
b−a
(E)
a+b
x−b a−b
+b
x−a f (a) + f (b) , what is the value of ? b−a f (a + b)
Solve yourself:
86.
If h (x) = 2px+1 , what is the value of (A)
2−1
(B)
1
(C)
2
(D)
2p
(E)
2a+b
h (a) × h (b) ? h (a + b)
Solve yourself:
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74 87.
Algebra Guide – Questions If f (x) = ax 2 + bx + c, and f (x + 1) = f (x) + x + 1, what is the value of (a + b)?
(D)
1 2 1 3 2 3
(E)
4
(A) (B) (C)
Solve yourself:
88.
If f (x) = x 2 +3 and g (x) = 3f (x), what is the greatest possible value of f (x − 1) if g (x) = 84? (A)
11
(B)
19
(C)
29
(D)
39
(E)
41
Solve yourself:
89.
If the function f is defined by f (x) = x 2 1 − x 2 (A)
(1 − x)2
(B)
(1 − x)3
(C)
(1 + x) (1 − x)2
(D)
(1 − x) (1 + x)2 2 1 − x2
(E)
2
for all values of x, then
f
f (x) √ = 1−x
Solve yourself:
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Algebra Guide – Questions 90.
75
If the cost of a dozen eggs were reduced by x cents, a buyer would pay one cent less for (x + 1) eggs than he would pay if the cost of a dozen eggs were increased by x cents. What is the value of x? (A)
1
(B)
2
(C)
3
(D)
4
(E)
6
Solve yourself:
91.
If a and b are real numbers such that a percent of (a − 2b) when added to b percent of b, the value obtained is 0, then which of the following statements is true? I.
a=b
II.
a+b =0
III.
a−b =1
(A)
Only I
(B)
Only II
(C)
Only III
(D)
Only I and III
(E)
Only II and III
Solve yourself:
92.
If xy 6= 0 and x 2 y 2 − xy = 6, which of the following could be y in terms of x? I.
6 x
II.
−
III.
3 x
(A)
Only I
2 x
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76
Algebra Guide – Questions (B)
Only II
(C)
Only I and II
(D)
Only I and III
(E)
Only II and III
Solve yourself:
93.
1
y=
1
2+
2+ 2+
1 1 ...∞
The expression of y above extends to infinity. Which of the following is the correct equivalent of y?
(A) (B) (C) (D) (E)
1 3 √
2−1 √ 2− 2 1 √ 2 1
Solve yourself:
94.
If x 6= 0, what quantity can be added to result?
(D)
x x+1 x+1 x x−1 − (x + 1)
(E)
−
(A) (B) (C)
x+1 x
or multiplied with
x−1 x
to obtain the same
x2 x−1
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Algebra Guide – Questions
77
Solve yourself:
95.
The sum of the squares of two positive numbers x and y is 20. If the sum of their reciprocals is 2, what is the product of the numbers?
(D)
1 2 2 5 2 3
(E)
4
(A) (B) (C)
Solve yourself:
96.
If (A) (B) (C) (D) (E)
√ 3 m+2 x √ = √ , what is x 3 , if expressed in terms of m? 3 3 x+1 m+2+ m−2 m+2 m−2 m−2 m+2 m+2 m m m−2 m3 + 2 m3 − 2
Solve yourself:
97.
3
How many solutions of x are possible, if x −3 + x − 2 = 2?
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78
Algebra Guide – Questions (A)
One
(B)
Two
(C)
Four
(D)
Five
(E)
Six
Solve yourself:
98.
If the function f is defined by f (x) = (A)
f (x) 3f (x) − 1
(B)
2f (x) f (x) + 1
(C)
f (x) f (x) + 1
(D)
3f (x) 2f (x) + 1
(E)
2f (x) f (x) − 1
x , what is f (2x) in terms of f (x)? x−2
Solve yourself:
99.
If g (x) = x + 4 for all x and f (x) = f (x) = g (x)? (A)
−5
(B)
−3
(C)
−2
(D)
−1
(E)
−
2x 2 − 5 for all x < 0, then for what value of x is x
1 2
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Algebra Guide – Questions
79
Solve yourself:
100.
The function f is defined such that f (n) = f (n − 1) × f (n − 2), for all n > 2. If f (1) = 1 and f (2) = −1, what is the value of f (33) + f (34)? (A)
−2
(B)
−1
(C)
0
(D)
1
(E)
2
Solve yourself:
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80
Algebra Guide – Questions
3.2
Data Sufficiency
Data sufficiency questions have five standard options. They are listed below and will not be repeated for each question. (A)
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the question asked.
(B)
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the question asked.
(C)
both the statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
(D)
EACH statement ALONE is sufficient to answer the question asked.
(E)
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
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Algebra Guide – Questions 101.
81
A man has three sons, among whom he distributed his wealth. Is the share of the eldest son greater than $12000? (1)
The eldest son received the greatest share.
(2)
The total wealth was $30000.
Solve yourself:
102.
Is x > y? (1)
2x + 3y − z = 0
(2)
7x − 3y − 2z = 0
Solve yourself:
103.
If z is positive, is x > z? (1)
2x + 3y − z = 0
(2)
7x − 3y − 2z = 0
Solve yourself:
104.
What is Harry’s present age? (1)
When Harry joined school with Ron, five years ago, sum of their ages was 13 years.
(2)
When Harry joined school with Ron, five years ago, Harry was one year younger to Ron.
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105.
Algebra Guide – Questions
What is Harry’s present age? (1)
When Harry joined school, five years ago, he was half as old as his elder brother.
(2)
The age of Harry’s elder brother is five years less than twice Harry’s age.
Solve yourself:
106.
A man purchased few oranges and few apples. What is the total price of one orange and one apple? (1)
The man purchased 30 fruits, consisting of only apples and oranges and the total price came to $40.
(2)
Had the man purchased as many apples as oranges and as many oranges as apples, he would have paid $80.
Solve yourself:
107.
What is the sum of digits of a four-digit number abcd, where a is the first digit, b is the second digit, c is the third digit and d is the fourth digit and the sum of the first three digits equals the fourth digit? (1)
The sum of its third digit and twice its second digit equals 10 times its first digit.
(2)
The sum of the first and last digits equals 5 times the second digit.
Solve yourself:
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Algebra Guide – Questions 108.
83
The individual ages of Ann and Bob are whole numbers such that each of them is a two-digit number. What is Ann’s age? (1)
Ann’s age is such that if the digits of her age are reversed, Bob’s age is obtained.
(2)
Bob’s age exceeds Ann’s age by
1 of the combined ages of Ann and Bob. 11
Solve yourself:
109.
The price of an apple is $1 and the price of an orange is $2. A man purchased few apples and few oranges. What is the number of oranges and apples purchased by the man? (1)
The man spent $41 to purchase the fruits.
(2)
Had the man purchased as many apples as oranges and as many oranges as apples, he would have saved half the cost of an orange.
Solve yourself:
110.
If a = b = 1, what is the value of c? p 1 1 q 1 1 (1) = a+ and = b+ q 2 b r 2 c r 1 1 (2) = c+ p 2 a Solve yourself:
111.
What is the value of q? (1) p + 2q p + 3q = 35 (2)
p and q are positive integers.
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Algebra Guide – Questions
Solve yourself:
112.
What is the largest possible integer value of k? (1)
The roots of x 2 − kx + 35 = 0 and the roots of x 2 − 7x + k = 0 are real numbers.
(2)
k>0
Solve yourself:
113.
Is
1 2a 1 + = ? x+1 y +1 a+b
(1)
ax 2 = by
(2)
ay 2 = bx
Solve yourself:
114.
P, Q, R and S are four cakes, is the total cost of P and S less than that of Q and R? (1)
The cost of R is half of the cost of S.
(2)
The cost of Q is twice that of R, which in turn, is costlier than P.
Solve yourself:
115.
If m and n are integers, what is the largest possible value of m?
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Algebra Guide – Questions
(1) (2)
85
2 1 1 + = m n 10 m and n are positive integers.
Solve yourself:
116.
A, B and C have some marbles with them. Does C have at least 50 percent of all the marbles with them together? (1)
The sum of the number of marbles with A and B together is 40 percent less than that with B and C together.
(2)
The sum of the number of marbles with B and C together is 20 percent less than that with A and C together.
Solve yourself:
117.
A group of friends wanted to purchase a camera by sharing the cost equally. What is the price of the camera? (1)
The price of the camera is between $210 and $230.
(2)
If two friends back out, the remaining friends have to contribute $1 extra to purchase the camera.
Solve yourself:
118.
A man has a certain amount of money in $1 and $10 bills. Can the total number of bills with him be nine? (1)
The number of $1 bills multiplied by the number of $10 bills is equal to the total money (in dollars) with him.
(2)
The number of $1 bills is greater than eight.
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Algebra Guide – Questions
Solve yourself:
119.
The letters A, B, C, D and E represent distinct numbers among 2, 4, 5, 6 and 10, not necessarily in the same order. What is the value of B? (1)
A+C=E
(2)
A+E=B
Solve yourself:
120.
If x, y and z are positive numbers and 5x + 7y + 8z = k, can the value of k equal 20? (1)
2x + 3y + 4z = 10
(2)
3x + 4y + 5z = 12
Solve yourself:
121.
Can the value of k equal 20? (1)
5x + 7y + 9z = k
(2)
x, y and z are positive integers
Solve yourself:
122.
If f (x) = x 3 − 4x + p, is 0 < p < 3?
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Algebra Guide – Questions (1)
f (0) × f (1) < 0
(2)
f (0) > 0
87
Solve yourself:
123.
If f (x) = (a − x n ) (1)
n=m
(2)
a=1
1 m
and f (1) = p, what is the value of f p ?
Solve yourself:
a, b, x, y, . . .
124.
1 of the previous term. What is the value of Each term of the above sequence is 9 more than 3 x−y ? (1)
a = 54
(2)
y=
5 x 6
Solve yourself:
125.
If f(n+1) (x) = f(n) (x) + 1, for all positive integer values of n = 1, 2, 3, . . . , what is the value of f(4) (x)? (1)
f1 (x) = 0
(2)
f5 (x) = 4
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Algebra Guide – Questions
Solve yourself:
126.
If f (n + 3) = f (n + 2) × f (n), for all positive integer values of n = 1, 2, 3, . . . , what is the value of f (5) − f (3)? (1)
f (1) = f (2) = 1
(2)
f (3) = −1
Solve yourself:
127.
Aluminum costs $2 per kilogram, and copper costs $4 per kilogram. If 10 kilograms of alloy K consists of x kilograms of aluminum and y kilograms of copper, is x > y? (1)
y >3
(2)
The cost of 10 kilograms of alloy K is less than $30.
Solve yourself:
128.
What is the age of P, if the average (arithmetic mean) age of P, Q, and R is 24 years? (1)
The difference between the ages of P and Q is 6 years and R is 6 years elder to Q.
(2)
Among the three, P is the youngest and R is the oldest.
Solve yourself:
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Algebra Guide – Questions 129.
89
What is the age of P, if the average (arithmetic mean) age of P, Q, and R is 24 years and no two among the three have the same age? (1)
The difference between the ages of P and Q is 6 years and R is 6 years elder to Q.
(2)
Among the three, P is the youngest and R is the oldest.
Solve yourself:
130.
A, B and C have a total of $60 with them. Does A have the highest amount? (1)
A and B together have 40 percent more than what C has.
(2)
B has $6 more than what A and C together have.
Solve yourself:
131.
Is the present age of B more than double the age of A? (1)
B’s age was double the age of A, four years back.
(2)
B is 20 years older than A.
Solve yourself:
132.
Jill bought only apples worth $0.30 each and oranges worth $0.58 each. How many apples did she buy? (1)
She bought $8.80 worth of apples and oranges.
(2)
She bought an equal number of apples and oranges.
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90
133.
Algebra Guide – Questions
Juan bought some paperback books that cost $8 each and some hardcover books that cost $25 each. If Juan bought more than 10 paperback books, how many hardcover books did he buy? (1)
The total cost of the hardcover books that Juan bought was at least $150.
(2)
The total cost of all the books that Juan bought was less than $260.
Solve yourself:
134.
Some computers at a certain company are Brand X and the rest are Brand Y. If there are a total of 880 computers at the company, how many of the computers are Brand Y? (1)
Ratio of the number of Brand Y computers to the number of Brand X computers at the company is 5 to 6.
(2)
The number of Brand X computers is greater than the number of Brand Y computers at the company by 80.
Solve yourself:
135. + d e f
x p q
y
z m
n
The figure above represents an addition table where four entries, p, q, m and n are shown; for example, d + x = p. What is the value of (m + n)? (1)
d + y = −3
(2)
e + z = 12
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Algebra Guide – Questions
91
Solve yourself:
136.
The total charge to rent a car for one day from Company J consists of a fixed charge of $15.00 and a charge of $0.20 per mile driven. The total charge to rent a car for one day from Company K consists of a fixed charge of $20.00 and a charge of $0.10 per mile driven. Is the total charge to rent a car from Company J for one day and drive it for x miles less than $25.00? (1)
The total charge to rent a car from Company K for one day and drive it for x miles is less than $25.00
(2)
x < 50
Solve yourself:
137.
What was the cost of a certain telephone call? (1)
The call lasted 8 minutes.
(2)
The cost of the first minute of the call was $0.32, which was twice the cost of each minute of the call after the first.
Solve yourself:
138.
Yesterday, Nan parked her car at a certain parking garage that charges more for the first hour than for each additional hour. If Nan’s total parking charge at the garage yesterday was $3.75, for how many hours of parking was she charged? (1)
Parking charges at the garage are $0.75 for the first hour and $0.50 for each additional hour or fraction of an hour.
(2)
If the charge for the first hour had been $1.00, Nan’s total parking charge would have been $4.00.
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Algebra Guide – Questions
Solve yourself:
139.
A piece of chalk, 8 cm long is broken into three pieces whose lengths, in cm, are distinct integers. What is the product of the length of the three pieces? (1)
The length of the longest piece is equal to the sum of the lengths of the other two pieces.
(2)
The length of the shortest piece is 1 cm.
Solve yourself:
140.
Barbara sells two products, A and B. She earns commissions of $12 per unit on product A sold and $5 per unit on product B sold. If her total commission was $300, how many units of product A did she sell? (1)
Her commission from the sale of product A was at least $120.
(2)
She sold 8 more units of product A than units of product B.
Solve yourself:
141.
What is the total cost of 1 apple, 1 orange and 1 lemon? (1)
The total cost of 5 apples, 4 oranges and 3 lemons is Rs. 130
(2)
The total cost of 3 apples, 4 oranges and 5 lemon is Rs. 110
Solve yourself:
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Algebra Guide – Questions 142.
93
If the sum of the ages of A and B is 70 years, what is the age of A? (1)
A, at present, is twice as old as B was x years ago.
(2)
B, at present, has the same age as A had x years ago.
Solve yourself:
143.
If x and y are positive integers, what is the value of xy? (1)
2x+2y = 8
(2)
23x+2y = 32
Solve yourself:
144.
Joe purchased a number of cakes priced at $13 each and a number of biscuit packets priced at $7 each. What is the total number of cakes and biscuit packets purchased by Joe? (1)
The total price of items purchased is $33.
(2)
If the count of cakes and the count of biscuit packets purchased had been interchanged, the total price of the items would have been $27.
Solve yourself:
145.
Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle? (1)
A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2)
C paid a total of $16 to A and B for his share.
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Algebra Guide – Questions
Solve yourself:
146.
In an experiment with n bacteria, it was found that each bacteria weighed 10−12 grams. Each of the n bacteria gave birth to n new bacteria, each of which also weighed 10−12 grams. What was the value of n? (1) (2)
1 of the total weight of all bacteria. 16 The total weight of all bacteria was 24 10−11 grams.
The first n bacteria weighed
Solve yourself:
147.
If p is an integer, what is the value of p? (1)
pq = 4
(2)
q − 2p = 7
Solve yourself:
148.
If a2 − b2 = 20, what is value of a? (1)
a and b are positive integers.
(2)
a + b = 10
Solve yourself:
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Algebra Guide – Questions
149.
What is the value of
95 a ? b
(1)
a2 + b = 2a b
(2)
(a − 2)2 + |b − 2| = 0
Solve yourself:
150.
What is the value of (1) (2)
1 1 + ? a b
1−a 1−b + =2 a b ab 1 = a+b 4
Solve yourself:
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Algebra Guide – Answer Key
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Chapter 4
Answer-key
97
98
4.1
Algebra Guide – Answer Key
Problem Solving
(1) C
(23) C
(45) C
(2) D
(24) E
(46) E
(3) B
(25) E
(47) C
(4) D
(26) D
(48) C
(5) B
(27) E
(6) A
(28) C
(7) D
(29) B
(8) B
(30) C
(9) E
(31) C
(10) B
(32) C
(11) D
(33) B
(12) E
(34) C
(49) B (50) E (51) D (52) D (53) E (54) B (55) B (56) D (57) C
(13) D
(35) D (58) D
(14) E
(36) C (59) B
(15) C
(37) D
(16) C
(38) B
(17) A
(39) C
(18) A
(40) A
(19) C
(41) B
(64) D
(20) D
(42) A
(65) A
(21) E
(43) C
(66) C
(22) C
(44) D
(67) D
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(60) C (61) E (62) D (63) C
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Algebra Guide – Answer Key
99
(68) E
(79) A
(90) B
(69) C
(80) C
(91) A
(70) D
(81) D
(92) E
(71) A
(82) B
(93) B
(72) D
(83) E
(94) D
(73) C
(84) A
(95) C
(74) D
(85) C
(96) A
(75) B
(86) C
(97) A
(76) D
(87) B
(98) B
(77) D
(88) D
(99) D
(78) C
(89) D
(100) C
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100
4.2
Algebra Guide – Answer Key
Data Sufficiency
(101) E
(118) D
(135) C
(102) E
(119) C
(136) D
(103) C
(120) A
(137) C
(104) C
(121) C
(138) A
(105) E
(122) A
(106) C
(123) D
(107) C
(124) D
(108) C
(125) D
(109) C
(126) A
(110) C
(127) B
(111) C
(128) C
(112) A
(129) A
(113) C
(130) B
(114) C
(131) A
(147) C
(115) A
(132) A
(148) D
(116) C
(133) C
(149) D
(117) C
(134) D
(150) D
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(139) A (140) B (141) C (142) C (143) D (144) D (145) C (146) D
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Chapter 5
Solutions
101
102
Algebra Guide – Solutions
5.1 1.
Problem Solving
Let the price of 1 chocolate and 1 pen be $x and $y, respectively. Thus, we have: 19x + 21y = 29 . . . (i) 21x + 19y = 31 . . . (ii) Adding (i) and (ii): 40x + 40y = 60 => x + y =
3 . . . (iii) 2
Subtracting (i) from (ii): 2x − 2y = 2 => x − y = 1 . . . (iv) Adding (iii) and (iv): 3 5 = 2 2 5 => x = = 1.25 4
2x = 1 +
Thus, the price of a chocolate is $1.25 The correct answer is option C.
2.
We know that the average age of the Abe and his two children ten years ago was 20 years. Thus, at present, the age of each must have increased by 10 years; hence their average age must have been 10 + 20 = 30 years. Thus, the present total age of Abe and his two children = 30 × 3 = 90 years. Since Abe’s age is equal to the sum of the ages of his two children, we have: Abe’s age + Sum of ages of his two children = 90 years => 2 × (Abe’s age) = 90 years => Abe’s age =
90 = 45 years. 2
Alternate approach: Let the present age of Abe’s son = x years. Present age of Abe’s daughter = 12 years. www.manhattanreview.com
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Algebra Guide – Solutions
103
Thus, Abe’s present age = (x + 12) years. Thus, total present age of Abe and his two children = (2x + 24) years. Ten years back, each must have been 10 years younger, thus the total age would have been 10 × 3 = 30 years less. Thus, total age of Abe and his two children = (2x + 24 − 30) = (2x − 6) years. Since their average age was 20 years, their total age = 20 × 3 = 60 years. Thus, we have: 2x − 6 = 60 => x = 33 Thus, Abe’s present age = (x + 12) = 45 years. The correct answer is option D.
3.
Let the cost of one apple, one guava and one banana be $x, $y and $z, respectively. Thus, we have: 3x + 3y + 4z = 10 . . . (i) 3x + 2y + 4z = 9 . . . (ii) We need to determine the value of 9x + 8y + 12z . Adding (i) and (ii): 6x + 5y + 8z = 19 . . . (iii) Adding (i) and (iii): 9x + 8y + 12z = 29 Alternate approach: We can see that if (i) is multiplied by 3: 9x + 9y + 12z = 30 => 9x + 8y + 12z = 30 − y, i.e. the required value is less than 30. Similarly, if (ii) is multiplied by 3: 9x + 6y + 12z = 27 => 9x + 8y + 12z = 27 + 2y, i.e. the required value is greater than 27. Thus, the required answer lies between 27 and 30, and hence, must be 29 (the only possible option).
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Algebra Guide – Solutions The correct answer is option B.
4.
Initial amount = a dollars, b cents. Amount spent = $3.50 = 3 dollars, 50 cents. Final amount = 2b dollars, 64 cents. There are two possibilities: •
b > 50: b − 50 = 64 => b = 114 However, we know that b < 100 Hence, this scenario is not possible.
•
b < 50: Since b is less than 50, in order to properly subtract 50 cents from b, we need to borrow 100 cents from a dollars, so that the initial sum of money becomes (a − 1) dollars and (100 + b) cents. Thus, we have: 100 + b − 50 = 64 => b = 14
Considering the subtraction of the dollars: (a − 1) − 3 = 2b => a − 4 = 2 × 14 = 28 => a = 32 Thus, we have: Initial amount = $32.14, amount spent = $3.50, final amount = $28.64 Thus, we have: a + b = 32 + 14 = 46. The correct answer is option D.
5.
Let the fraction be
x . y
Thus, we have: x+4 x 1 = + y +4 y 8
=>
x+4 x 1 − = y +4 y 8
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Algebra Guide – Solutions
=>
4 y −x 1 = y y +4 8
=>
1 y −x = . . . (i) y y +4 32
105
x−2 x 1 = − y −2 y 4
=>
x−2 1 x − = y y −2 4
=>
2 y −x 1 = y y −2 4
=>
y −x 1 = . . . (ii) y y −2 8
Dividing (i) by (ii): y −2 1 = => 4y − 8 = y + 4 y +4 4 => y = 4 Substituting y = 4 in (i): 4−x 1 = 4 (4 + 4) 32 => 4 − x = 1 => x = 1 Thus, the required fraction =
x 3 = . y 4
Alternate approach: We can use the options and find the value of the fraction that satisfies the given condition. For example, working with Option B: Assuming the original fraction to be • •
3 : 4
3+4 3 7 3 1 − = − = – Satisfies 4+4 4 8 4 8 3 1 1 3 3−2 − = − = – Satisfies 4 4−2 4 2 4
The correct answer is option B. © 1999–2016 Manhattan Review
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106 6.
Algebra Guide – Solutions We solve the first two equations and substitute the values of x and y in the third equation. Thus, we have: 7x − 2y = 12 . . . (i) 4x + y = 9 . . . (ii) 2x + 5y = K . . . (iii) Multiplying (ii) by 2 and adding with (i): 7x + 8x = 12 + 18 => x = 2 Substituting x = 2 in (ii): 8 + y = 9 => y = 1 Substituting x = 2 and y = 1 in (iii): K = 2 × 2 + 5 × 1 = 9. The correct answer is option A.
7.
We have: 2x + 3y = 7 . . . (i) 5x + 3y = 13 . . . (ii) y 1 x = = . . . (iii) A B C Subtracting (i) from (ii): 3x = 6 => x = 2 Substituting x = 2 in (i): 3y = 7 − 4 => y = 1 Substituting x = 2 and y = 1 in (iii): 2 1 1 = = A B C Since A, B and C are positive integers and the greatest common divisor (GCD) of A, B and C is 1, we can see that the only possible sets of values are: A = 2, B = C = 1 (Note: Other values: A = 4, B = 2, C = 2; etc. (not possible since the GCD should be 1); A = 1 1 1 , B = , C = ; etc. (not possible since A, B, C are positive integers). 2 4 4 => A + B + C = 2 + 1 + 1 = 4. The correct answer is option D.
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Algebra Guide – Solutions 8.
107
Let the two-digit number be N = 10x + y, where x and y are the tens and unit digit of N, respectively. Thus, we have: x + y = 5 . . . (i) The fraction formed: 10x + y − 20 3 = 10x + y + 12 11 => 110x + 11y − 220 = 30x + 3y + 36 => 80x + 8y = 256 => 8 10x + y = 256 => 10x + y =
256 = 32 8
Thus, we have: N = 10x + y = 32 Note: One should NOT solve the equations 10x +y = 32 and x +y = 5 and determine the values of x and y, since, finally, the value of 10x + y has to be determined. Observe that: 10x + y = 32 = 10 × 3 + 2 => x = 3 and y = 2 satisfies x + y = 5. Thus, the product of the digits of the number = 3 × 2 = 6. Alternate approach: Since the sum of digits of the number is 5, the possible numbers are: 50, 41, 32, 23 and 14 We know that the ratio of 20 less than the number and 12 more than the number is
3 . 11
Working with the above numbers: • • •
30 3 50 − 20 = 6= – Does not satisfy 50 + 12 62 11 41 − 20 21 3 N = 41 => = 6 = – Does not satisfy 40 + 12 52 11 32 − 20 12 3 N = 32 => = = – Satisfies 32 + 12 44 11 N = 50 =>
Thus the product of the digits = 3 × 2 = 6 The correct answer is option B.
9.
We have: 3x + 7y = 11z . . . (i)
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108
Algebra Guide – Solutions 6x − y = 7z . . . (ii) 3x + y = kz . . . (iii) Multiplying (i) with 2 and subtracting (ii) from it: 14y − −y = 22z − 7z => y = z Substituting y = z in (ii): 6x = 7z + y = 7z + z => x =
4 z 3
4 Substituting x = z and y = z in (iii): 3 4 3 z + z = kz 3 => 5z = kz => k = 5. The correct answer is option E.
10.
Let the number of children in the group be n. Let the number of pens initially with Ann be x. If Ann takes 1 pen from each of the other children, she would have taken (n − 1) pens from the others, and hence, the total number of pens with her = (x + n − 1). Since the number of pens with her would be thrice the number of children in the group, we have: x + n − 1 = 3n => x = 2n + 1 . . . (i) Since the total number of pens with n children is 42, and Ann alone has x = (2n + 1) pens, the number of pens with the other (n − 1) children = 42 − (2n + 1) = (41 − 2n). Since Ann has the greatest number of pens, none of the other children can have (2n + 1) pens. Working with the options: Statement I: n = 5 Number of pens with Ann = 2n+1 = 11 Number of pens with the other 4 children = 41−2n = 31 Assuming that each of the 3 children has a minimum of 1 pen, then the 4th child would have 31 − 3 = 28 pens, which is greater than the number of pens (11) Ann has. – Does not satisfy
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Algebra Guide – Solutions
109
Statement II: n = 9 Number of pens with Ann = 2n+1 = 19 Number of pens with the other 8 children = 41−2n = 23 If any one other child has 19 pens, then the number of pens with the remaining 7 children would be 23 − 19 = 4, which is not possible since each child should have at least one pen. Thus, Ann definitely has the greatest number of pens. – Satisfies Statement III: n = 15 Number of pens with Ann = 2n+1 = 31 Number of pens with the other 14 children = 41−2n = 11 Since each child must have at least one pen, we cannot have a scenario where 14 children have a total of 11 pens with them. – Does not satisfy Note: Though it appears while working with the first two options, that the value of n should be on the higher side, it is not prudent to assume that the largest value of n in the options would also satisfy the conditions. The correct answer is option B.
11.
The distance covered by the ball in the t th second = (Distance covered in t seconds) – (Distance covered in (t − 1) seconds) = 15t 2 − 15(t − 1)2 o n = 15 t 2 − (t − 1)2 = 15 (t − (t − 1)) (t + (t − 1)) = 15 (2t − 1) Thus, we have: 15 (2t − 1) = 225 => 2t − 1 = 15 => t = 8 The correct answer is option D.
12.
Height of the ball above the ground = 3 + 24t − 4t 2 = −4 t 2 − 6t + 3 = −4 t 2 − 2 × t × 3 + 3 = −4 t 2 − 2 × t × 3 + 32 − 32 + 3
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Algebra Guide – Solutions n o = −4 (t − 3)2 − 9 + 3 = −4(t − 3)2 + 39 Since (t − 3)2 is a perfect square term, it is always non-negative. Thus, when it is multiplied with (−4), it becomes non-positive. n o Thus, the height of the ball from the ground is 39, reduced by the amount −4(t − 3)2 . Thus, the height of the ball above the ground will be the maximum when the negative term, i.e. n o −4(t − 3)2 becomes zero, which happens when: −4(t − 3)2 = 0 => t = 3 The corresponding value of the height (maximum value) is thus, 39 feet. The correct answer is option E.
13.
For N units sold at price P , total selling price (in dollars) = P × N = P (120 − P ) Total cost incurred = $2000 Since there is neither profit nor loss, we have: Cost price = Selling price => 2000 = P (120 − P ) => P 2 − 120P + 2000 = 0 => P 2 − 100P − 20P + 2000 = 0 => (P − 100) (P − 20) = 0 => P = 20 or 100 Thus, the maximum selling price = $100. The correct answer is option D.
14.
In a quadratic ax 2 +bx +c = 0, the roots are real and unequal, if the discriminant, i.e. b2 −4ac > 0. Thus, for the quadratic x 2 − px + 8p = 0, we have: −p
2
− 4 (1) 8p > 0
=> p 2 − 32p > 0 => p 2 − 2 × p × 16 + 162 − 162 > 0 www.manhattanreview.com
© 1999–2016 Manhattan Review
Algebra Guide – Solutions => p − 16
2
111
> 162
Thus, we have: p − 16 > 16 => p > 32 OR p − 16 < −16 => p < 0 Thus, the minimum positive integer value of p = 33. The correct answer is option E.
15.
We know that x = Substituting x =
3 satisfies the equation x 2 − px + 12 = 0 2
3 in the equation, we have: 2
2 3 3 −p + 12 = 0 2 2
=>
3 9 57 p = + 12 = 2 4 4
=> p =
2 57 19 × = 3 4 2
Alternate approach: Let the other solution be x = k. Thus, we have: 3 x − px + 12 ≡ x − (x − k) 2 2
3 3 => x − px + 12 ≡ x − x k + + k 2 2 2
2
Comparing coefficients: 12 =
2 3 k => k = 12 × = 8 2 3
p =k+
3 3 19 =8+ = 2 2 2
The correct answer is option C. © 1999–2016 Manhattan Review
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Algebra Guide – Solutions
16.
To determine the point where y = x 2 + kx + l intersects the Y-axis, we need to substitute x = 0. Since the point of intersection on the Y-axis is (0, 64), we have: 0 + 0 + l = 64 => l = 64 . . . (i) We also know that y = x 2 + kx + l intersects the X-axis at (4, 0). Thus, substituting x = 4 and y = 0 in the equation, we have: 42 + 4k + l = 0 Substituting l = 64: 16 + 4k + 64 = 0 => k = −20 The correct answer is option C.
17. 12 𝑥
8 𝑥
Area of the frame = (Area of outer rectangle) – (Area of inner rectangle) = (12 + 2x) (8 + 2x) − 12 × 8 = 24x + 16x + 4x 2 = 40x + 4x 2 Since the area is 44 square inches, we have: 40x + 4x 2 = 44 => x 2 + 10x − 11 = 0 => (x + 11) (x − 1) = 0 => x = −11 or 1 => x = 1 (Negative value for the width is not possible) Thus, the required perimeter = (Perimeter of outer rectangle) + (Perimeter of inner rectangle) www.manhattanreview.com
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113
= 2 {(12 + 2x) + (8 + 2x)} + 2 (12 + 8) = 2 (14 + 10) + 40 = 98 The correct answer is option A.
18.
We know that the height of the arch above the ground is zero at the two ends of the arch, as shown in the diagram below: Parbolic arch 30 feet
𝑑 =0
𝑑 = 20
Thus, we have: h = 0 => −kd (d − 20) = 0 => d = 0 or 20 The arch reaches its maximum height of 30 feet at the middle, i.e. for d =
0 + 20 = 10 2
Thus, we have: d = 10, h = 30 => h = −kd (d − 20) => 30 = −k × 10 (10 − 20) => k =
3 10
Alternate approach: h = −kd (d − 20) = −k d2 − 20d o n = −k (d − 10)2 − 100 = −k(d − 10)2 + 100k In order to maximize the height, the negative term, i.e.
n o −k(d − 10)2 should be ‘0’, i.e. d = 10
and the corresponding maximum value is 100k. © 1999–2016 Manhattan Review
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114
Algebra Guide – Solutions Thus, we have: 100k = 30 => k =
3 10
The correct answer is option A.
19.
We know that f (1) = 0, i.e. x = 1 is a solution of x 2 + bx + c = 0. We need to determine k, where f (k) = 0, k 6= 1, i.e. we need to determine the other solution of x 2 + bx + c = 0. Since 1 and k are the solutions of x 2 + bx + c = 0, we have: x 2 + bx + c ≡ (x − 1) (x − k) Since f (3) = 2f (5), we have: (3 − 1) (3 − k) = 2 (5 − 1) (5 − k) => 2 (3 − k) = 8 (5 − k) => 3 − k = 20 − 4k => k =
17 3
The correct answer is option C.
20.
Since p, q, r and s are the roots of x 4 − 2x 3 + x 2 + x + 3 = 0, we have: x 4 − 2x 3 + x 2 + x + 3 ≡ x − p
x − q (x − r ) (x − s)
Substituting x = −2 above, we have: (−2)4 − 2(−2)3 + (−2)2 + (−2) + 3 = −2 − p => 16 + 16 + 4 + 1 = 2 + p => 2 + p
−2 − q (−2 − r ) (−2 − s)
2 + q (2 + r ) (2 + s)
2 + q (2 + r ) (2 + s) = 37
The correct answer is option D.
21.
We have: x+
2 ≤3 x
=> x +
2 −3≤0 x
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Algebra Guide – Solutions
=>
x 2 + 2 − 3x ≤0 x
=>
(x − 2) (x − 1) ≤0 x
115
Thus, the possible scenarios are: •
(x − 2) (x − 1) ≥ 0 AND x < 0 (a)
x − 2 ≥ 0 AND x − 1 ≥ 0 AND x < 0 => x ≥ 2 AND x ≥ 1 AND x < 0 – Not possible
(b)
x − 2 ≤ 0 AND x − 1 ≤ 0 AND x < 0
=> x ≤ 2 AND x ≤ 1 AND x < 0 => x < 0 . . . (i) •
(x − 2) (x − 1) ≤ 0 AND x > 0 (a)
x − 2 ≥ 0 AND x − 1 ≤ 0 AND x > 0 => x ≥ 2 AND x ≤ 1 AND x > 0 – Not possible
(b)
x − 2 ≤ 0 AND x − 1 ≥ 0 AND x > 0
=> x ≤ 2 AND x ≥ 1 AND x > 0 => 1 ≤ x ≤ 2 . . . (ii) Thus, the solution, from (i) and (ii), is: x0 x+7
=>
x+3+x+7 >0 x+7
=>
2 (x + 5) >0 x+7
=>
x+5 >0 x+7
Thus, we have the following scenarios: (a)
x + 5 > 0 AND x + 7 > 0 => x > −5 AND x > −7 => x > −5
(b)
x + 5 < 0 AND x + 7 < 0 => x < −5 AND x < −7 => x < −7
Thus, we have: x > −5 OR x < −7 . . . (i) •
x+3
x+3 −1
117
x+3−x−7 −4x + 7 < 0
=>
1 >0 x+7
=> x + 7 > 0
=> x > −7 . . . (ii) Thus, from (i) and (ii), we finally have: x > −5 OR x < −7 – Shown by the black arrows AND x > −7 – Shown by the gray arrow Placing them on a number line, we have:
−7
−5
Thus, the overlapping region is: x > −5 The correct answer is option C.
23.
We have: x−3 1 > x 2 − 9x + 18 2
=>
1 x−3 − >0 (x − 3) (x − 6) 2
Cancelling (x − 3) from numerator and denominator, we have: 1 1 − >0 x−6 2
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Algebra Guide – Solutions
=>
2 − (x − 6) >0 2 (x − 6)
=>
−x + 8 >0 x−6
=>
x−8 0 => x < 8 AND x > 6 Thus, the only integer value of x = 7
•
x − 8 > 0 AND x − 6 < 0 => x > 8 AND x < 6 – Not possible
Alternate approach: x−3 1 > 2 (x − 3) (x − 6)
=>
1 1 > 2 (x − 6)
Plugging in the options: 1 1 ≯ – Does not satisfy 5−6 2
•
Option A: x = 5:
•
Option B: x = 6: Division by ‘0’ is not possible – Does not satisfy
• • •
1 1 > – Satisfies 7−6 2 1 1 Option A: x = 8: ≯ – Does not satisfy 8−6 2 1 1 Option A: x = 9: ≯ – Does not satisfy 9−6 2 Option A: x = 7:
The correct answer is option C.
24.
We have: x 3 − 9x 5 > 0 => x 3 1 − 9x 2 > 0
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119
=> x 2 x 1 − 9x 2 > 0 Cancelling x 2 , since it is always positive: x {(1 − 3x) (1 + 3x)} > 0 Thus, we have the following scenarios: •
•
x > 0 AND (1 − 3x) (1 + 3x) > 0 (a)
x > 0 AND 1 − 3x > 0 AND 1 + 3x > 0 1 1 => x > 0 AND x < AND x > − 3 3 1 => 0 < x < . . . (i) 3
(b)
x > 0 AND 1 − 3x < 0 AND 1 + 3x < 0 1 1 => x > 0 AND x > AND x < − – Not possible 3 3
x < 0 AND (1 − 3x) (1 + 3x) < 0 (a)
x < 0 AND 1 − 3x > 0 AND 1 + 3x < 0 1 1 => x < 0 AND x < AND x < − 3 3 1 => x < − . . . (ii) 3
(b)
x < 0 AND 1 − 3x < 0 AND 1 + 3x > 0 1 1 => x < 0 AND x > AND x > − – Not possible 3 3
Thus, from (i) and (ii), we have: 0 3x + 2y = 125 . . . (i) The least value of x + y occurs when x is the largest, i.e. 41 (since 125 when divided by 3 gives 41 as quotient and 2 as remainder). Thus, 2y = 2, i.e. the value of y is 1.
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121
Alternatively, since the coefficient of x is ‘3’, which is greater than the coefficient of y which is ‘2’, least value of x + y would occur if we maximize the value of x or minimize the value of y. => x + y = 41 + 1 = 42 (least) The largest value of x + y occurs when y is the largest, i.e. 62 (since 125 when divided by 2 gives 62 as quotient and 1 as remainder). However, in that case, x becomes a fraction. However, if we take y = 61, then we have 3x = 3, i.e. the value of x is 1. => x + y = 61 + 1 = 62 (largest) Since the dealer does not wish to purchase greater than 42 items, we have: x + y ≤ 42 Thus, the only possible solution is: x + y = 42. Thus, the minimum number of total items that the dealer can purchase is 42. The correct answer is option E.
28.
Let the number of chicken burgers and the number of vegetable burgers sold be x and y, respectively. Thus, total price of the burgers = $ 8x + 5y . Since potato fires were ordered with burgers 50 times, each time the discount on a burger being $1, total discount on the burgers = $50. Thus, total revenue generated = $ 8x + 5y − 50 . Thus, we have: 8x + 5y − 50 ≤ 1110 => 8x + 5y ≤ 1160 . . . (i) Since the total number of burgers sold was at least 160, we have: x + y ≥ 160 => 5x + 5y ≥ 800 => − 5x + 5y ≤ −800 . . . (ii) Thus, adding (i) and (ii): 8x + 5y − 5x + 5y < 1160 − 800 => 3x ≤ 360
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Algebra Guide – Solutions => x ≤ 120 Thus, the maximum number of chicken burgers sold was 120. The correct answer is option C.
29.
Let the number of sedans and the number of minivans used be s and m, respectively. Since the number of people is at least 48, we have: 4s + 8m ≥ 48 => s + 2m ≥ 12 . . . (i) Since the total amount available is $600, we have: 60s + 80m ≤ 600 => 3s + 4m ≤ 30 => −3s − 4m ≥ −30 . . . (ii) From (i) × 2 + (ii): −s ≥ −6 . . . (iii) Adding (i) and (iii): 2m ≥ 6 => m ≥ 3 Thus, the minimum number of minivans required is 3. The correct answer is option B.
30.
Let the number of pencils and the number of erasers in a gift box be x and y, respectively. Thus, we have: x + y ≤ 12 . . . (i) x ≥ y + 3 . . . (ii) From (ii): Assuming x = y + 3 and substituting in (i), we have: y + 3 + y ≤ 12 => 2y ≤ 9 => y ≤ 4.5 If x is greater than y + 3 , the value of y will reduce even further to accommodate the higher value of x.
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123
Thus, the maximum value of y is 4 (Since y is an integer). The correct answer is option C.
31.
Let the radius and the height of the cylinder be r and h, respectively. Thus, we have: Maximum and maximum height = 10 + 1 = 11 inches and 10 − 1 = 9 inches, respectively. Maximum and minimum diameter = 3 + 1 = 4 inches and 3 – 1 = 2 inches, respectively. Thus, maximum and minimum radius =
4 2 = 2 inches and = 1 inches, respectively. 2 2
Thus, maximum volume = π (2)2 (11) = 44π cubic inches. Also, minimum volume = π (1)2 (9) = 9π cubic inches. Thus, required difference = 44π − 9π = 35π cubic inches. The correct answer is option C.
32.
|x − a| refers to the distance of x from a. Thus, y is the sum of distances of x from the points −3, 4 and 7. Thus, the sum of the distances will be minimized if one of the distances becomes ‘0’, i.e. x takes a value −3, 4 or 7: •
x= −3: y = |−3 + 3| + |−3 − 4| + |−3 − 7| = 0 + 7 + 10 = 17
•
x = 4: y = |4 + 3| + |4 − 4| + |4 − 7| = 7 + 0 + 3 = 10
•
x = 7: y = |7 + 3| + |7 − 4| + |7 − 7| = 10 + 3 + 0 = 13
Thus, the minimum value of y = 10. The correct answer is option C.
33.
We have: 2 x − 15 ≤ 6 => −6 ≤ x 2 − 15 ≤ 6 => 9 ≤ x 2 ≤ 21 Thus, possible integer values of x are: −4, −3, 3, 4. Thus, there are four possible integer values of x.
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Algebra Guide – Solutions The correct answer is option B.
34.
|x − a| refers to the distance of x from a. Thus, |x − 3| + |x − 9| is the sum of distances of x from 3 and 9.
6
3
9
Thus, we have: •
If x is anywhere between 3 and 9, inclusive, the sum of the distances of x from 3 and 9 would always be 6, i.e. less than 8. Thus, possible integer values of x are: 3, 4, 5, 6, 7, 8 and 9.
•
If x is d distance to the left of 3: Distance of x from 3 = d Distance of x from 9 = d + 6 Thus, we have: d + (d + 6) ≤ 8 => d ≤ 1 Thus, only possible integer value of x = 3 − 1 = 2.
•
If x is D distance to the right of 9: Distance of x from 9 = D Distance of x from 3 = D + 6 Thus, we have: (D + 6) + D ≤ 8 => D ≤ 1 Thus, only possible integer value of x = 9 + 1 = 10.
Thus, the integer values of x are: 2, 3, 4, 5, 6, 7, 8, 9 and 10. Thus, there are nine possible integer values of x. The correct answer is option C.
35.
We have: |x + 15| = 3 |x − 15|
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125
=> x + 15 = ±3 (x − 15) Thus, we have: •
x + 15 = 3 (x − 15) => x + 15 = 3x − 45 => 2x = 60 => x = 30
•
x + 15 = −3 (x − 15) => x + 15 = −3x + 45 => 4x = 30 => x = 7.5
Thus, sum of the possible values of x = 30 + 7.5 = 37.5 The correct answer is option D.
36.
Since x and y are integers, x + y and x − y are also integers. Also, since x and y are non-negative integers, we must have: x + y ≥ x − y . . . (i) x + y ≥ 0 . . . (ii) Once the values of x + y
and x − y
are determined, the two equations can be solved to
obtain the values of x and y. Thus, the possible scenarios are:
|x+y|
|x−y|
x+y
x−y
x
y
6
0
6
0
3
3
5
1
5
1
3
2
5
1
5
−1
2
3
4
2
4
2
3
1
4
2
4
−2
1
3
3
3
3
3
3
0
3
3
3
−3
0
3
Thus, there are seven possible pairs of solutions of x, y . © 1999–2016 Manhattan Review
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126
Algebra Guide – Solutions The correct answer is option C.
37.
We have: x + 5 x + 3 ≥ x − 3 x − 5
=>
|x + 5| |x + 3| ≥ |x − 3| |x − 5|
Since the denominators are absolute values, we may cross multiply keeping the inequality unchanged: => |x + 5| × |x − 5| ≥ |x + 3| × |x − 3| => |(x + 5) (x − 5)| ≥ |(x + 3) (x − 3)| => x 2 − 25 ≥ x 2 − 9 Thus, we have: •
x 2 − 25 ≥ x 2 − 9 => −25 > −9 – Not possible
•
x 2 − 25 ≤ − x 2 − 9 => x 2 − 25 ≤ −x 2 + 9 => 2x 2 ≤ 34 => x 2 ≤ 17 √ √ => − 17 ≤ x ≤ 17 => −4._ ≤ x ≤ 4._
Thus, the possible integer values of x are: −4, −3, −2, −1, 0, 1, 2, 3, 4. However, since (x − 3) is in the denominator of the original inequality, x cannot take the value 3. Thus, there are eight possible integer values of x. The correct answer is option D.
38.
We have: x x − 1 < 1 Since the denominator is an absolute value, we may cross multiply keeping the inequality unchanged:
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127
=> |x| < |x − 1| => − (x − 1) < x < x − 1 => −x + 1 < x < x − 1 => −x + 1 < x OR x < x − 1 (which is not possible) => −x + 1 < x => 1 < 2x => x <
1 2
Thus, the greatest possible integer value of x = 0. Alternately, one could simply substitute the options and among those options which satisfy the inequality, the greatest value of x which satisfies the equation would be the answer. The correct answer is option B.
39.
We know that: |a| + |b| ≥ |a + b| •
The equality holds if both a and b are of the same sign
•
The inequality holds if a and b are of opposite signs
Thus, we have: |x + 2| + |2x − 1| > |3x + 1| => |x + 2| + |2x − 1| > |(x + 2) + (2x − 1)| Thus, we have: •
x + 2 > 0 AND 2x − 1 < 0 => x > −2 AND x <
1 1 => −2 < x < . . . (i) 2 2
OR •
x + 2 < 0 AND 2x − 1 > 0 => x < −2 AND x >
1 – Not possible 2
Thus, from (i), the greatest possible integer value of x = 0. Alternately, one could simply substitute the options and among those options which satisfy the inequality, the greatest value would be the answer. The correct answer is option C. © 1999–2016 Manhattan Review
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128 40.
Algebra Guide – Solutions We have: |x − 1| > |x| + 3 Thus: •
If x ≥ 0: |x| = x Also, x > x − 1 Thus, the LHS is definitely smaller than the RHS. Thus, there is no possible positive value of x.
•
If x < 0: |x| = −x Also, x − 1 < 0 => |x − 1| = − (x − 1) => − (x − 1) > −x + 3 => 1 > 3, which is not true Thus, there is no possible negative value of x.
Thus, there is no possible value of x. The correct answer is option A.
41.
f (x) = 2x 4 − 3x 2 + 1 The roots of f (x) = 0 are p, q, r and s. Thus, we have: 2x 4 − 3x 2 + 1 = 0 Let k = x 2 2k2 − 3k + 1 = 0 => 2k2 − 2k − k + 1 = 0 => 2k (k − 1) − 1 (k − 1) = 0 => (2k − 1) (k − 1) = 0 1 or 1 2 1 => x 2 = or 1 2
=> k =
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129
1 => x = ± √ or ±1 2 1 1 Thus, the roots p, q, r and s are: − √ , √ , −1, and 1 2 2 1 1 => p + q + r + s = − √ + √ + (−1) + 1 = 0 2 2
The correct answer is option B.
42.
To find the points of intersection, we need to equate the two functions and determine the value of x (that satisfies both the equations). The value of x gives the X-coordinate value of the point of intersection. On substituting the value of x in either of the two functions, we get the Y-coordinate of the point of intersection. Thus, we have: f (x) = g (x) => x 2 + 4 = => x 2 −
x2 +3 2
x2 = −1 2
=> x 2 = −2 However, the square of any number must always be non-negative. Since x 2 comes to be negative, the values of x are imaginary, i.e. there are no real points of intersection. The correct answer is option A.
43.
We are given that: •
f p, q = 0, if p < q
•
f p, q = f p − q, q + 1 if p ≥ q
The above functions determine the quotient when p is divided by q. For example: •
If p = 10 and q = 12: since 10 cannot be divided by 12 (since 10 < 12), the quotient obtained is ‘0’.
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Algebra Guide – Solutions •
If p = 10 and q = 3, the quotient obtained is ‘3’. Let us see how the function gives the result as 3: f (10, 3) = f (10 − 3, 3) + 1 = f (7, 3) + 1 f (7, 3) + 1 = f (7 − 3, 3) + 1 + 1 = f (4, 3) + 2 f (4, 3) + 2 = f (4 − 3, 3) + 1 + 2 = f (1, 3) + 3 f (1, 3) + 3 = 0 + 3 (since 1 < 3) = 3
Since we have: f p, q = 3, we have: p when divided by q gives a quotient 3. Working with the options: •
Statement I: f (7, 2): 7, when divided by 2, gives a quotient 3 (and remainder 1) – Satisfies Alternately: f (7, 2) = f (7 − 2, 2) + 1 = f (5, 2) + 1 f (5, 2) + 1 = f (5 − 2, 2) + 1 + 1 = f (3, 2) + 2 f (3, 2) + 2 = f (3 − 2, 2) + 1 + 2 = f (1, 2) + 3 f (1, 2) + 3 = 0 + 3 = 3
•
Statement II: f (10, 3): 10, when divided by 3, gives a quotient 3 (and remainder 1) – Satisfies
•
Statement III: f (17, 6): 17, when divided by 6, gives a quotient 2 (and remainder 5) – Does not satisfy
The correct answer is option C.
44.
We are given that: •
f p, q = f p, q − 1 + p if q > 1
•
f p, q = 0 if q = 0
The above functions determine the product when p is multiplied with q. For example: •
If p = 10 and q = 3, the product obtained is 30. Let us see how the function gives the result as 30: f (10, 3) = f (10, 3 − 1) + 10 = f (10, 2) + 10 f (10, 2) + 10 = f (10, 2 − 1) + 10 + 10 = f (10, 1) + 20 f (10, 1) + 20 = f (10, 1 − 1) + 10 + 20 = f (10, 0) + 30
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131
f (10, 0) + 30 = 0 + 30 = 30 Since we have: f p, q = 12, we have: p when multiplied with q gives 12. Working with the statements: •
Statement I: f (6, 2): 6, when multiplied with 2 gives 12 – Satisfies Alternately: f (6, 2) = f (6, 2 − 1) + 6 = f (6, 1) + 6 f (6, 1) + 6 = f (6, 1 − 1) + 6 + 6 = f (6, 0) + 12 f (6, 0) + 12 = 0 + 12 = 12
•
Statement II: f (4, 3): 4, when multiplied with 3 gives 12 – Satisfies
•
Statement III: f (6, 3): 6, when multiplied with 3 gives 18 – Does not satisfy
The correct answer is option D.
45.
We are given that: •
f p, q = p if p < q
•
f p, q = f p − q, q if p > q
•
f p, q = 0 if p = q
The above function determines the remainder when p is divided by q. For example: •
If p = 10 and q = 12: since 10 cannot be divided by 12 (since 10 < 12), the remainder obtained is ‘10’.
•
If p = 10 and q = 3, the remainder obtained is ‘1’. Let us see how the function gives the result as 1: f (10, 3) = f (10 − 3, 3) = f (7, 3) f (7, 3) = f (7 − 3, 3) = f (4, 3) f (4, 3) = f (4 − 3, 3) = f (1, 3) f (1, 3) = 1 (since 1 < 3)
•
If p = 12 and q = 3, the remainder obtained is ‘0’. Let us see how the function gives the result as 0: f (12, 3) = f (12 − 3, 3) = f (9, 3) f (9, 3) = f (9 − 3, 3) = f (6, 3) f (6, 3) = f (6 − 3, 3) = f (3, 3) f (3, 3) = 0 (since 3 = 3)
Since we have: f p, q = 7, we have: p when divided by q leaves a remainder 7. Working with the options: © 1999–2016 Manhattan Review
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Algebra Guide – Solutions •
Statement I: f (12, 5): 12, when divided by 5, leaves a remainder 2 (and quotient 2) – Does not satisfy
Alternately: f (12, 5) = f (12 − 5, 5) = f (7, 5) f (7, 5) = f (7 − 5, 5) = f (2, 5) f (2, 5) = 2 •
Statement II: f (25, 8): 25, when divided by 8, leaves a remainder 1 (and quotient 3) – Does not satisfy
•
Statement III: f (27, 20): 27, when divided by 20, leaves a remainder 7 (and quotient 1) – Satisfies
The correct answer is option C.
46.
Working with the options: •
f p, q = Option A/B/C: g p, q
p p+q
! ×
1 p+q q
! =
p p+q
! ×
q p+q
! =
pq p+q
2
Thus, options A, B and C are incorrect. •
Option D/E: f p, q × g p, q =
p p+q
! ×
p+q q
! =
p q
Thus, option D is incorrect, and option E is correct. The correct answer is option E.
47.
We know that [x] denotes the greatest integer less than or equal to x. Thus, for example, [2.3] equals the greatest integer less than or equal to 2.3: The integers less than 2.3 are: 2, 1, 0, −1, etc. Among these, the greatest is 2. Thus, [2.3] = 2. We know that x is a positive integer. Thus, we have: x + 1 ≤ x + x (the equality holds only if x = 1) => x + 1 ≤ 2x =>
2x ≥ 1 . . . (i) x+1
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133
Also, we have: x+1>x x < 2 . . . (ii) x+1 =>
Thus, from (i) and (ii): 1≤
2x = [1] = 1 x+1 x+1 2x 2x is a number between 1 and 2 => =1 For all other x: x+1 x+1
For x = 1:
Thus, each of the 100 terms, i.e. f (1) , f (2) . . . f (100), has the value ‘1’. Hence, the sum is 1 + 1 + 1 . . . 1 (100 times) = 100. The correct answer is option C. Alternate Approach: We know that for positive numbers, Thus,
x x 2 + 4x ≤ 0 => x (x + 4) ≤ 0 => −4 ≤ x ≤ 0 Thus, the only non-negative integer value of x is x = 0 Thus, there is only ‘one’ possible value of x. The correct answer is option B.
50.
We know that [x] denotes the greatest integer less than or equal to x. Thus, for example: •
[2.3] equals the greatest integer less than or equal to 2.3: The integers less than 2.3 are: 2, 1, 0, −1, etc. Among these, the greatest is 2. Thus, [2.3] = 2.
•
[−2.3] equals the greatest integer less than or equal to −2.3: The integers less than −2.3 are: −3, −4, −5, etc. Among these, the greatest is −3. Thus, [−2.3] = −3.
We have: x 2 + 4x − 12 = 0 => x 2 + 6x − 2x − 12 = 0 => (x + 6) (x − 2) = 0 => x = −6 or 2 Thus, the values of p and q are −6 and 2 Since we need to maximize f p − f q , we assign p = 2 and q = −6: f (2) = 6 f (−6) = − = [−1.5] = −2 4 © 1999–2016 Manhattan Review
2 = [0.5] = 0 4
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136
Algebra Guide – Solutions => f p − f q
= f (2) − f (−6)
= 0 − (−2) = 2
The correct answer is option E.
51. Y
O
X 4, 0
f (1) refers to the value of y at x = 1, which is ‘3’ units to the left of x = 4.
Since the graph is symmetric about the line x = 4, the value of y for a value of x ‘3’ units to the left of x = 4 is the same as the value of y for a value of x ‘3’ units to the right of x = 4, i.e. x =4+3=7
Thus, we have:
f (7) = f (1) = 9
The correct answer is option D.
52.
From the graph, it is clear that the shaded region depicts g (x) ≥ f (x), since the line for g (x) is above the line for f (x).
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137
Y 𝑓 𝑥 𝑥=2
O
𝑥=5
𝑔 𝑥
X 2, 0
4, 0
The other intersecting point would be given by equating g (x) with f (x) . => x 2 − 6x + 8 = x − 2 => x 2 − 7x − 10 = 0 => x 2 − 5x − 2x − 10 = 0 => x = 2 or 5 The shaded region is bounded by the lines x = 2 and x = 5. Thus, the integer values of x are: 2, 3, 4 and 5. Thus, there are ‘four’ possible values of x. Alternate approach: We have: f (x) ≤ g (x) => x 2 − 6x + 8 ≤ x − 2 => x 2 − 7x + 10 ≤ 0 => (x − 2) (x − 5) ≤ 0 => 2 ≤ x ≤ 5 Integer values for x are, 2, 3, 4 and 5. The correct answer is option D. © 1999–2016 Manhattan Review
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138
Algebra Guide – Solutions
53. Y 15 𝑓 𝑥
𝑔 𝑥
10
5
O
1 2 3
X
From the graphs, we observe that:
•
f (x) satisfies the coordinates: (0, 2) , (1, 5) , (2, 10)
•
g (x) satisfies the coordinates: (0, 1) , (1, 2) , (2, 5) , (3, 10)
•
f (x) − g (x) satisfies the coordinates: (0, 1) , (1, 3) , (2, 5)
Working with the options:
•
Option A: Does not satisfy the coordinate (0, 2)
•
Option B: Does not satisfy the coordinates (0, 2) or (2, 10)
•
Option C: Does not satisfy the coordinates (0, 1) or (1, 2) or (3, 10)
•
Option D: Does not satisfy the coordinates (0, 1) or (1, 2) or (3, 10)
•
Option E: Satisfies the coordinates (0, 1) , (1, 3) and (2, 5)
The correct answer is option E.
54.
From the graph, we have:
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139
Y 𝑔 𝑥
𝑓 𝑥 X
O
1 2 1
−1
−2 1 −3
Area of triangle ABC
=
1 × Base × Height 2
=
1 1 ×1×1= 2 2
Alternate approach:
g (0) = 2 × 0 − 3 = −3 => Coordinates of C are (0, −3)
f (0) = 0 − 2 = −2 => Coordinates of B are (0, −2)
Thus BC = −2 − (−3) = 1
Equating g (x) and f (x):
2x − 3 = x − 2 => x = 1
Thus height of the triangle ABC = 1
Thus, area of triangle ABC =
1 1 ×1×1= 2 2
The correct answer is option B.
55. © 1999–2016 Manhattan Review
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140
Algebra Guide – Solutions Y
4
O
2 4
X
The above quadratic has roots 2 and 4. Thus, we have: f (x) = ax 2 + bx + c = a (x − 2) (x − 4) Since the graph intersects the Y-axis at (0, 4), we have: f (0) = 4 => a (0 − 2) (0 − 4) = 4 => 8a = 4 => a =
1 2
1 (x − 2) (x − 4) 2 1 1 => f (x) = x 2 − 6x + 8 = x 2 − 3x + 4 2 2 => f (x) =
Thus, we have: ax 2 + bx + c =
1 2 x − 3x + 4 2
1 , b = −3, c = 4 2 3 => a + b + c = 2 => a =
The correct answer is option B. Alternate approach: We know that c is the y-intercept. As the graph intersects the y axis at (0, 4), we have: c = 4 The graph intersects X-axis at (2, 0) and (4, 0) so we have: 0 = a × 22 + b × 2 + 4 www.manhattanreview.com
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141
=> 4a + 2b = −4 => 2a + b = −2 . . . (i) Also, we have: 0 = a × 42 + b × 4 + 4 => 16a + 4b = −4 => 4a + b = −1... (ii) Subtracting (i) from (ii) we get, 2a = 1 => a =
1 2
Plugging this value in (i) we get, b = −3 Thus, a + b + c =
56.
3 1 + (−3) + 4 = 2 2
We have: 2f (x) + f (−x) = 2x + 1 Substituting x = 1: 2f (1) + f (−1) = 2 × 1 + 1 => 2f (1) + f (−1) = 3 . . . (i) Substituting x = −1: 2f (−1) + f (− (−1)) = 2 (−1) + 1 => 2f (−1) + f (1) = −1 . . . (ii) From (i) × 2 – (ii): 4f (1) − f (1) = 3 × 2 − (−1) => 3f (1) = 7 => f (1) =
7 3
The correct answer is option D.
57.
We have: f (2x − 1) = x 2 + 3x . . . (i) Since we need the value of f (5), we need to use a suitable value of x so that we have: 2x − 1 = 5 => x = 3
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142
Algebra Guide – Solutions Thus, substituting x = 3 in (i): f (2 × 3 − 1) = 32 + 3 × 3 => f (5) = 18 The correct answer is option C.
58.
We have: f (x) = ax 2 + bx + c . . . (i) Also, we have: f (x + 1) = f (x) + x + 1 . . . (ii) Substituting x = 0 in (ii): f (0 + 1) = f (0) + 0 + 1 => f (1) = f (0) + 1 => a 12 + b (1) + c = a 02 + b (0) + c + 1 => a + b = 1 . . . (iii) Substituting x = −1 in (ii): f (−1 + 1) = f (−1) + (−1) + 1 => f (0) = f (−1) => a 02 + b (0) + c = a (−1)2 + b (−1) + c => a − b = 0 => a = b . . . (iv) Substituting a = b in (iii): a+a=1 => a =
1 2
The correct answer is option D.
59.
We have: f (a, b) = b => a > b Since a = 6x − 8 and b = x 2 , we have: => 6x − 8 > x 2
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143
=> x 2 − 6x + 8 < 0 => (x − 2) (x − 4) < 0 => 2 < x < 4 Thus, the only integer value of x = 3. The correct answer is option B.
60.
We have: f (x) = |x − 1| and g (x) = x + a => g (−3) = −3 + a => f g (−3) = f (−3 + a) = |(−3 + a) − 1| = |a − 4| Since f g (−3) = 2, we have: |a − 4| = 2 => a − 4 = ±2 => a = 4 ± 2 = 6 or 2 Thus, sum of the possible values of a = 6 + 2 = 8 The correct answer is option C.
61.
The above is a linear equation. On simplifying the above, we would finally get a form: kx + l = nx + m, where k, l, n and m are constants In the above, if k = n, the terms with x will cancel out. Thus, if l = m, i.e. the constant terms are equal, it implies that the above equation is true for any value of x. However, if l 6= m, the above equation would not be true for any value of x, i.e. there will be no solution. We have: 4 (2x − 1) + 3 (x − 2) = n (x + 2) − 3 (x + 1) => n (x + 2) = 4 (2x − 1) + 3 (x − 2) + 3 (x + 1) => nx + 2n = 14x − 7 Thus, we should have: nx = 14x AND 2n 6= −7
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Algebra Guide – Solutions => n = 14 (which satisfies 2n 6= −7) The correct answer is option E.
62.
We have: x 2 − 6x + 9 = 0 => (x − 3)2 = 0 => x = 3 Thus, the above equation has only one root. Thus, x 2 − px + 12 = 0 must have one root as x = 3 Thus, substituting x = 3 in x 2 − px + 12 results in ‘0’ => 32 − 3p + 12 = 0 => 3p = 21 => p = 7 The correct answer is option D.
63.
We have: (0.23 + 0.52)2 − 0.232 + 0.522 0.752 − 0.292
Considering the numerator: (0.23 + 0.52)2 − 0.232 + 0.522 = 0.232 + 2 × 0.23 × 0.52 + 0.522 − 0.232 + 0.522 = 2 × (0.23 × 0.52) Considering the denominator: 0.752 − 0.292 = (0.75 − 0.29) (0.75 + 0.29) = (0.46) (1.04) = (2 × 0.23) (2 × 0.52) = 4 × (0.23 × 0.52) Thus, we have: www.manhattanreview.com
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145
(0.23 + 0.52)2 − 0.232 + 0.522 0.752 − 0.292
=
2 × (0.23 × 0.52) 4 × (0.23 × 0.52)
=
2 = 0.5 4
The correct answer is option C.
64.
We have: x 2 − x − 12 = 0 => (x − 4) (x + 3) = 0 => x = 4 or −3 Since p and q are the roots, with p > q, we have: p = 4, q = −3 Again, we have: x 2 − 50x + 49 = 0 => (x − 1) (x − 49) = 0 => x = 1 or 49 Thus, the roots are 1 and 49, which have to be expressed in terms of p and q. Working with the options, we find that in Option D: p+q
2
= (4 + (−3))2 = 1
2
= (4 − (−3))2 = 49
AND p−q
The correct answer is option D.
65.
We have: a2 + 2b2 + 2a + b = 0 . . . (i) a2 + b2 = 2ab . . . (ii) => a2 − 2ab + b2 = 0
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146
Algebra Guide – Solutions => (a − b)2 = 0 => a − b = 0 => a = b Substituting a = b in (i): a2 + 2a2 + 2a + a = 0 => 3 a2 + a = 0 => a (a + 1) = 0 => a = 0 or −1 Since a is non-zero, we have: a = b = −1 => a + b = −2 The correct answer is option A.
66.
We have: x+y =2 => y = 2 − x Thus, we have: z = x2 + y 2 = x 2 + (2 − x)2 = 2x 2 − 4x + 4 = 2 x 2 − 2x + 2 = 2 x 2 − 2x + 1 + 2 = 2(x − 1)2 + 2 The minimum value of z occurs when the square term, i.e. (x − 1)2 = 0 (which occurs at x = 1). Thus, the minimum value of z = 0 + 2 = 2. Alternate approach: If sum of two variables is a constant, the sum of their squares would be the minimum if they are equal.
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147
Let us see why: We have: x + y = 2 => x = 2 − y Thus, we have: x 2 + y 2 = 2−y
2
+ y2
= 4 − 4y + 2y 2 = 2 y 2 − 2y + 4 =2
n
o 2 y −1 −1 +4
=2 y −1
2
+2
2 The above would have the minimum value if the square term, i.e. y − 1 = 0 (which occurs if y = 1 => x = 2 − y = 1. The corresponding minimum value is 0 + 2 = 2). Thus, we have: x = y => x = 1 => z = 12 + 12 = 2 The correct answer is option C.
67.
Since q and 2q are the roots, the quadratic equation can be represented as: x − q
x − 2q = 0
=> x 2 − 3qx + 2q2 = 0 . . . (i) The given quadratic equation is: k x2 − x + x + 1 = 0 => kx 2 − x (k − 1) + 1 = 0 1 k−1 x + = 0 . . . (ii) => x 2 − k k Comparing (i) and (ii): k−1 = 3q k 1 => 1 − = 3q k 1 => = 1 − 3q k 1 => k = . . . (iii) 1 − 3q 1 = 2q2 k => k =
1 . . . (iv) 2q2
The correct answer is option D. © 1999–2016 Manhattan Review
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148 68.
Algebra Guide – Solutions Let f (x) = ax 2 + bx + c Thus, we have: f (1) = 1 => a + b + c = 1 . . . (i) f (2) = 2 => 4a + 2b + c = 2 . . . (ii) f (3) = 5 => 9a + 3b + c = 5 . . . (iii) From (ii) – (i): 3a + b = 1 . . . (iv) From (iii) – (ii): 5a + b = 3 . . . (v) From (v) – (iv): 2a = 2 => a = 1 From (iv): b = 1 − 3 = −2 From (i): c = 1 − 1 − (−2) = 2 Thus, we have: f (x) = x 2 − 2x + 2 => f (0) = 2 Alternate approach: We have: f (1) = 1 => f (1) − 1 = 0 f (2) = 2 => f (2) − 2 = 0 Let the function g (x) = f (x) − x Thus, we see that: g (1) = f (1) − 1 = 0 AND g (2) = f (2) − 2 = 0 Thus, 1 and 2 are the roots of g (x) => g (x) = k (x − 1) (x − 2) => f (x) − x = k (x − 1) (x − 2) => f (x) = k (x − 1) (x − 2) + x Since f (3) = 5, we have: k (3 − 1) (3 − 2) + 3 = 5 => 2k = 2 => k = 1 => f (x) = (x − 1) (x − 2) + x => f (0) = (0 − 1) (0 − 2) + 0 = 2
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149
The correct answer is option E.
69.
We have: f (1) = −f (2) = 1 => f (1) = 1 and f (2) = −1 . . . (i) Also, we have: f (x) = (−1)x × f (x − 1) × f (x − 2) . . . (ii) Substituting x = 3, 4 . . . etc. in (ii), we have: •
x = 3: f (3) = (−1)3 × f (2) × f (1) = 1
•
x = 4: f (4) = (−1)4 × f (3) × f (2) = −1
•
x = 5: f (5) = (−1)5 × f (4) × f (3) = 1
Thus, we observe that the values of f (x) alternate between 1 and −1, with the even values of x resulting in −1 while the odd values of x resulting in 1. Thus, we have: f (23) = 1 (since 23 is odd) The correct answer is option C.
70.
We have: 2p + 1 =q f p = p−2 2q + 1 => f q = q−2 ! 2p + 1 2 +1 p−2 = 2p + 1 −2 p−2 =
4p + 2 + p − 2 2p + 1 − 2p + 4
=
5p =p 5
The correct answer is option D.
71.
The intercept on the X-axis is obtained from f (x) = 0 => |x − 1| − x = 0 => |x − 1| = x => x − 1 = x or −x
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150
Algebra Guide – Solutions Since x − 1 6= x, we have: x − 1 = −x => x =
1 2
Thus, f (x) intersects the X-axis at
1 , 0 . 2
The intercept on the Y-axis is obtained by substituting x = 0 in f (x) => f (0) = |0 − 1| − 0 = 1 Thus, f (x) intersects the Y-axis at (0, 1). Thus, the required area is the area of the triangle formed by the points
1 , 0 , (0, 1) and 2
(0, 0), as shown in the diagram below: Y 0,1
0,0
$ ,0 %
X
Thus, required area =
1 1 1 ×1× = 2 2 4
The correct answer is option A.
72.
We have: f (x) = ax 2 + bx Since f (1) = f (−1) + 2, we have: a 12 + b (1) = a(−1)2 + b (−1) + 2 => a + b = a − b + 2 => b = 1 => f (x) = ax 2 + x o n => f (3) − f (−3) = a 32 + 3 − a(−3)2 + (−3) = (9a + 3) − (9a − 3) = 6 The correct answer is option D.
73.
We have:
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Algebra Guide – Solutions x =1+
√
151
3
=> x − 1 =
√
3
Squaring both sides: (x − 1)2 =
√ 2 3
=> x 2 − 2x + 1 = 3 => x 2 − 2x − 2 = 0 Alternate approach: Substituting x = 1 +
√
3 in the expression, we have:
x 2 − 2x − 2 = 1+
√ 2 √ 3 −2 1+ 3 −2
√ √ =1+2 3+3−2−2 3−2 =0 The correct answer is option C.
74.
Since the numbers have
√ √ x as their exponent, x must be an integer, i.e. x must be a perfect
square (if x is not a perfect square, then each term will be irrational and the equation would never hold). The only possible options are A, B and D. Working with the options: •
Option A: x = 0 =>
√
x=0
Thus, each term becomes 1 and we get: 1 + 1 = 1, which is not true. – Does not satisfy •
Option B: x = 1 =>
√
x=1
Thus, the equation becomes: 24 + 10 = 26, which is not true. – Does not satisfy •
Option D: x = 4 =>
√
x=2
Thus, the equation becomes: 242 + 102 = 262 , i.e. 576 + 100 = 676, which is true. – Satisfies. The correct answer is option D. Alternate approach: We have: © 1999–2016 Manhattan Review
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152
Algebra Guide – Solutions √
24
x
+ 10
√ x
= 26
√ x
The numbers 24, 10 and 26 form a Pythagorean triplet, since the numbers are in the ratio 12, 5 and 13, which form a Pythagorean triplet, i.e. 122 + 52 = 132 Thus, we have: 242 + 102 = 262 =>
√
x=2
=> x = 4
75.
We need to determine the range of values of x for whichf (x) ≤ 0 => 2x 2 − 7x + 6 ≤ 0 => 2x 2 − 3x − 4x + 6 ≤ 0 => x (2x − 3) − 2 (2x − 3) ≤ 0 => (x − 2) (2x − 3) ≤ 0 3 => 2 (x − 2) x − ≤0 2 3 => (x − 2) x − ≤0 2 3 => ≤ x ≤ 2 2 Thus, the only one integer value of x = 2 The correct answer is option B.
76.
We have: 12 − x 3 ≤2
=>
|12 − x| ≤2 |3|
=>
|12 − x| ≤2 3
=> |12 − x| ≤ 6 => |− (x − 12)| ≤ 6 => |x − 12| ≤ 6 => −6 ≤ x − 12 ≤ 6 www.manhattanreview.com
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153
Adding 12 throughout: => −6 + 12 ≤ x − 12 + 12 ≤ 6 + 12 => 6 ≤ x ≤ 18 Alternately, from |x − 12| ≤ 6, we can say that the distance of x from 12 is 6 units, resulting in the extremities 12 + 6 = 18 and 12 – 6 = 6. The correct answer is option D.
77.
We know that: •
|x| = x if x ≥ 0
•
|x| = −x if x < 0
We have: x 2 − 4 |x| + 3 ≤ 0 •
Let x < 0, i.e. |x| = −x: x 2 − 4 |x| + 3 ≤ 0 => x 2 − 4 (−x) + 3 ≤ 0 => x 2 + 4x + 3 ≤ 0 => (x + 1) (x + 3) ≤ 0 => −3 ≤ x ≤ −1 Thus, the integer values of x are: −3, −2, −1, i.e. there are ‘three’ values . . . (i)
•
Let x ≥ 0, i.e. |x| = x: x 2 − 4 |x| + 3 ≤ 0 => x 2 − 4x + 3 ≤ 0 => (x − 1) (x − 3) ≤ 0 => 1 ≤ x ≤ 3 Thus, the integer values of x are: 1, 2, 3, i.e. there are ‘three’ values . . . (ii)
Thus, there are ‘six’ possible values of x. The correct answer is option D. 78.
We have to maximize the value of
12 . x 2 − 12x + 40
The value will be maximized if the value of the denominator is minimized. © 1999–2016 Manhattan Review
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154
Algebra Guide – Solutions Thus, we have: x 2 − 12x + 40 = x 2 − 2 × x × 6 + 62 + 4 = (x − 6)2 + 4 The above value is minimized if the square term, i.e. (x − 6)2 = 0 (which occurs at x = 6). Thus, the minimum value of the denominator is ‘4’. Thus, the maximum value of the expression =
12 =3 4
The correct answer is option C.
79.
We have: Since a is a root of x 2 − px − q = 0, we have: a2 − ap − q = 0 . . . (i) Since b is a root of x 2 − px − q = 0, we have: b2 − bp − q = 0 . . . (ii) Subtracting (i) from (ii): b2 − a2 − bp + ap = 0 => (b − a) (b + a) − p (b − a) = 0 => (b − a) b + a − p = 0 => b − a = 0 OR b + a − p = 0 => b = a OR b + a = p Since a and b are distinct roots, a 6= b => a + b = p The correct answer is option A.
80.
We know that [x] denotes the greatest integer less than or equal to x. Thus, for example, [2.3] equals the greatest integer less than or equal to 2.3: The integers less than 2.3 are: 2, 1, 0, −1, etc.
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155
Among these, the greatest is 2. Thus, [2.3] = 2. Since x is a positive integer and 1 ≤ x ≤ 4, we have: x+1 x+2 x+4 1+1 1+2 1+4 3 5 + + = + + = [1] + + = 2 4 8 2 4 8 4 8 1+0+0=1 x+1 x+2 x+4 2+1 2+2 2+4 3 6 x = 2: + + = + + = + [1] + = 2 4 8 2 4 8 2 8 1+1+0=2 x+1 x+2 x+4 3+1 3+2 3+4 5 7 x = 3: + + = + + = [2] + + = 2 4 8 2 4 8 4 8 2+1+0=3 x+1 x+2 x+4 4+1 4+2 4+4 5 6 8 x = 4: + + = + + = + + = 2 4 8 2 4 8 2 4 8 2+1+1=4
•
•
•
•
x = 1:
Thus, we see that the value of the expression is the same as the value of x assumed. x+1 x+2 x+4 Thus, the value of + + =x 2 4 8 The correct answer is option C. Alternate approach: x+2 x+4 x+1 + + =1 Check for = 1 : 2 4 8 Plug in x = 1 in all answer options and we get the answer as 1 in only B and C, rest all options gets eliminated. x+1 x+2 x+4 Check for = 2 : + + =2 2 4 8 Plug in x = 2 in B and C and we get the answer as 2 in only C.
81.
f (x) = x 3 − kx 2 + 2x
=> f (−x) = (−x)3 − k(−x)2 + 2 (−x)
= −x 3 − kx 2 − 2x
Since f (−x) = −f (x), we have:
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156
Algebra Guide – Solutions −x 3 − kx 2 − 2x = −(x 3 − kx 2 + 2x)
=> 2kx 2 = 0
=> k = 0
Thus, we have:
f (x) = x 3 + 2x
=> f (1 − k) = f (1) = 1 + 2 = 3
The correct answer is option D.
82.
Since f (2) = f (3) = 0, the roots of f (x) = ax 2 + bx + c must be x = 2 or 3.
Thus, we have:
f (x) = ax 2 + bx + c
= a (x − 2) (x − 3)
Since f (4) = 2, we have:
f (4) = a (4 − 2) (4 − 3) = 2a
=> 2a = 2
=> a = 1
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157
Thus, we have:
f (x) = 1 × (x − 2) (x − 3) = x 2 − 5x + 6
Comparing coefficients:
ax 2 + bx + c ≡ x 2 − 5x + 6
=> a = 1, b = −5, c = 6
=> a + b + c = 2
The correct answer is option B.
83.
Let machine Y takes a days to produce w widgets.
Thus, machine X takes (a + 2) days to produce w widgets.
Thus, total widgets produced by machines X and Y in 1 day =
w w + a a+2
=w
=
1 1 + a a+2
w (2a + 2) a (a + 2)
Thus, total widgets produced by machines X and Y in 3 days =
3w (2a + 2) a (a + 2)
Thus, we have:
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158
Algebra Guide – Solutions 3w (2a + 2) 5 = w a (a + 2) 4
=>
3 (2a + 2) 5 = a (a + 2) 4
=> 24a + 24 = 5a2 + 10a
=> 5a2 − 14a − 24 = 0
=> 5a2 − 20a + 6a − 24 = 0
=> (a − 4) (5a + 6) = 0
Since a is positive, we have:
a=4
Thus, time taken by machine X to produce w widgets = (a + 2) = 6 days
Thus, time taken by machine X to produce 2w widgets = 2 × 6 = 12 days
The correct answer is option E.
84.
f (x − 1) = 2x 2 − 3x + 3
We need to express the RHS in terms of (x − 1).
Since f (x − 1) is a quadratic with coefficient of x 2 as 2, the expression for f (x − 1) must have a 2(x − 1)2 term present.
2(x − 1)2 = 2 x 2 − 2x + 1 = 2x 2 − 4x + 2
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159
Thus, we can express:
2x 2 − 3x + 3 = 2x 2 − 4x + 2 + (x + 1)
Thus, we have:
f (x − 1) = 2(x − 1)2 + (x + 1)
=> f (x − 1) = 2(x − 1)2 + (x − 1) + 2
Thus, we have successfully expressed the RHS in terms of (x − 1)
=> f (x) = 2x 2 + x + 2
Alternate Approach:
f (x − 1) = 2x 2 − 3x + 3
f (x − 1) = 2((x − 1) + 1)2 − 3 ((x − 1) + 1) + 3
Thus, we have expressed the RHS in terms of (x − 1).
Similarly, f (x) = 2(x + 1)2 − 3(x + 1) + 3
=> f (x) = 2 x 2 + 2x + 1 − 3x − 3 + 3
=> f (x) = 2x 2 + 4x + 2 − 3x
=> f (x) = 2x 2 + x + 2
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160
Algebra Guide – Solutions The correct answer is option A.
85.
We have:
f (x) = a
x−b a−b
=> f (a) = a
f (a + b) = a
=
+b
a−b a−b
b−b f (b) = a a−b
x−a b−a
+b
a−a b−a
b−a +b b−a
a+b−b a−b
= a + 0 = a . . . (i)
= 0 + b = b . . . (ii)
+b
a+b−a b−a
a2 b2 a2 b2 a2 − b 2 + = − = = a + b . . . (iii) a−b b−a a−b a−b a−b
Thus, we have:
f (a) + f (b) a+b = =1 f (a + b) a+b
The correct answer is option C.
Alternate Approach:
A smarter approach could be to assume convenient, smart values of a & b.
Say, a = 1 & b = 2
Thus,
1−2 1−1 f (a = 1) = 1 × +2× =1 1−2 2−1
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Algebra Guide – Solutions f (b = 2) = 1 ×
161
2−2 2−1 +2× =2 1−2 2−1
3−2 3−1 f (a + b = 3) = 1 × +2× =3 1−2 2−1
Thus, we have:
f (a) + f (b) 1+2 =1 = f (a + b) 3
86.
We have:
h (x) = 2px+1
=> h (a) = 2pa+1
h (b) = 2pb+1
h (a + b) = 2p(a+b)+1 = 2pa+pb+1
Thus, we have:
h (a) × h (b) h (a + b)
=
2pa+1 × 2pb+1 2pa+pb+1
=
2pa+pb+2 2pa+pb+1
= 2(pa+pb+2)−(pa+pb+1)
=2
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162
Algebra Guide – Solutions The correct answer is option C.
Alternate Approach:
A smarter approach could be to assume convenient, smart values of a, b, & p.
Say, a = 0, b = −1, p = −1
Thus,
h (a = 0) = 2−1×0+1 = 2
h (b = −1) = 2−1×−1+1 = 4
h (a + b = 0 − 1 = −1) = 4
2×4 h (a) × h (b) = =2 h (a + b) 4 87.
We have:
f (x) = ax 2 + bx + c . . . (i)
f (x + 1) = f (x) + x + 1 . . . (ii)
Substituting x = 0 in (i):
f (0) = c . . . (iii)
Substituting x = 1 in (i):
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163
f (1) = a + b + c . . . (iv)
Substituting x = 0 in (ii):
f (0 + 1) = f (0) + 1
=> f (1) = f (0) + 1
=> a + b + c = c + 1 . . . using (iii) and (iv)
=> a + b = 1
The correct answer is option B.
88.
We have:
f (x) = x 2 + 3
g (x) = 3f (x) = 3 x 2 + 3
Since g (x) = 84, we have:
3 x 2 + 3 = 84
=> x 2 = 25
=> x = ±5
Thus, we have:
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164
Algebra Guide – Solutions f (x − 1) = (x − 1)2 + 3
If x = 5: f (x − 1) = (5 − 1)2 + 3 = 19
If x = −5: f (x − 1) = (−5 − 1)2 + 3 = 39
Thus, the maximum possible value of f (x − 1) = 39
The correct answer is option D.
89.
f (x) = x 2 1 − x 2
=> f
√
2
√ √ 2 2 2 1−x = 1−x 1− 1−x
= (1 − x) (1 − (1 − x))2
= (1 − x) x 2
2 f (x) x2 1 − x2 √ = => (1 − x) x 2 f 1−x
=
2 1 − x2 1 − x 2 (1 − x) (1 + x) = (1 − x) (1 − x)
= 1 − x 2 (1 + x)
= (1 − x) (1 + x) (1 + x)
= (1 − x) (1 + x)2
The correct answer is option D. www.manhattanreview.com
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Algebra Guide – Solutions 90.
165
Let the cost of a dozen eggs be d cents.
Cost of a dozen eggs if the price were reduced by x cents = (d − x) cents.
Thus, cost of 1 egg =
d−x cents. 12
Thus, cost of (x + 1) eggs =
(d − x) (x + 1) cents . . . (i) 12
Cost of a dozen eggs if the price were increased by x cents = (d + x) cents.
Thus, cost of 1 egg =
d+x cents. 12
Thus, cost of (x + 1) eggs =
(d + x) (x + 1) cents . . . (ii) 12
Thus, from (i) and (ii), we have:
(d + x) (x + 1) (d − x) (x + 1) − =1 12 12
=> dx + d + x 2 + x − dx + d − x 2 − x = 12
=> 2x 2 + 2x − 12 = 0
=> x 2 + x − 6 = 0
=> (x + 3) (x − 2) = 0
Since x must be positive, we have:
x=2
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166
Algebra Guide – Solutions The correct answer is option B.
91.
We know that if a percent of (a − 2b) when added to b percent of b, the value obtained is 0
=>
b a × (a − 2b) + ×b =0 100 100
=>
a2 − 2ab b2 + =0 100 100
=>
a2 − 2ab + b2 =0 100
=> a2 − 2ab + b2 = 0
=> (a − b)2 = 0
=> a − b = 0
=> a = b
Thus, only statement I is correct.
The correct answer is option A.
92.
x 2 y 2 − xy = 6
=> x 2 y 2 − xy − 6 = 0
=> x 2 y 2 − 3xy + 2xy − 6 = 0
=> xy xy − 3 + 2 xy − 3 = 0
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Algebra Guide – Solutions => xy + 2
167
xy − 3 = 0
=> xy = −2 OR 3
=> y = −
2 3 OR x x
The correct answer is option E.
Alternate Approach:
We can assume a convenient value of x and cross-check with the options for the value of y.
Say, x = 1, then
Option I: y =
6 =6 1
Since xy = 1 × 6 = 6 6= 0, it satisfies the first condition, but x 2 y 2 − xy = 12 × 62 − 1 × 6 = 30 6= 6, hence we discard this option.
Option II: y =
−2 = −2 1
Since xy = 1×−2 = −2 6= 0, it satisfies the first condition, and x 2 y 2 −xy = 12 ×(−2)2 −1×−2 = 4 + 2 = 6, hence we accept this option.
Option III: y =
3 =3 1
Since xy = 1 × 3 = 3 6= 0, it satisfies the first condition, and x 2 y 2 − xy = 12 × 32 − 1 × 3 = 9 − 3 = 6, hence we accept this option. © 1999–2016 Manhattan Review
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168
93.
Algebra Guide – Solutions 1
y=
1
2+
2+ 2+
1 1 ...∞
Taking reciprocal on both sides:
=>
=>
=>
1 =2+ y
1 −2= y
1 2+ 2+
1 1 ...∞
1 2+ 2+
=y
1 1 ...∞
1 −2=y y
=> 1 − 2y = y 2
=> y 2 + 2y = 1
=> y 2 + 2y + 1 = 1 + 1
=> y + 1
2
=2
√ => y + 1 = ± 2
√ => y = ± 2 − 1
Since y consists of all positive terms, y > 0
=> y =
√
2−1
The correct answer is option B. www.manhattanreview.com
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Algebra Guide – Solutions 94.
169
Let the required quantity be k.
Thus, according to the problem, we have:
x−1 x+1 =k k+ x x
x−1 x+1 −k= x x
x−1 x+1 −1 = x x
=> k
=> k
1 x+1 => k − = x x => k = −x
x+1 x
=> k = − (x + 1)
The correct answer is option D.
95.
Since the sum of the squares of x and y is 20, we have:
x 2 + y 2 = 20 . . . (i)
Since the sum of the reciprocals of x and y is 2, we have:
1 1 + =2 x y
=>
x+y =2 xy
=> x + y = 2xy
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170
Algebra Guide – Solutions Squaring both sides:
x+y
2
= 2xy
2
=> x 2 + y 2 + 2xy = 4x 2 y 2
Substituting x 2 + y 2 = 20 from (i):
20 + 2xy = 4x 2 y 2
=> 2x 2 y 2 − xy − 10 = 0
=> 2x 2 y 2 − 5xy + 4xy − 10 = 0
=> 2xy − 5 xy + 2 = 0
=> xy =
5 OR − 2 2
Since x and y are positive numbers, their product must also be positive
=> xy =
5 2
The correct answer is option C.
96.
We have: √ 3 x m+2 √ = √ 3 3 x+1 m+2+ m−2
Taking reciprocal on both sides:
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Algebra Guide – Solutions x+1 = x
√ 3
171
√ 3 m+2+ m−2 √ 3 m+2
Subtracting 1 from both sides:
x+1 −1= x
=>
√ 3
1 = x
√ 3
√ 3 m+2+ m−2 √ −1 3 m+2
m+2+
√ √ 3 3 m−2 − m+2 √ 3 m+2
√ 3 m−2 1 => = √ 3 x m+2 √ 3 m+2 => x = √ 3 m−2
Cubing both sides:
x3 =
m+2 m−2
The correct answer is option A.
97.
We have:
3
x −3 + x − 2 = 2
3
Let x − 2 = a
3 2 => x −3 = x − 2 = a2
=> a2 + a = 2
=> a2 + a − 2 = 0
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172
Algebra Guide – Solutions => (a + 2) (a − 1) = 0
=> a = −2 OR 1
3
=> x − 2 = −2 OR 1
=>
1
= −2 OR 1
3
x2
Taking reciprocal on both sides:
3
=> x 2 = − 12 OR 1
=> x 3
21
Since x 3
= − 12 OR 1
21
represents the square root of a number x 3 , it must be positive (since the square
root, by definition only takes the positive root)
Thus, we have:
x3
12
=1
Squaring both sides:
=> x 3 = 1
Taking cube-root on both sides:
=> x = 1
Thus, there is only one solution for x.
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Algebra Guide – Solutions
173
The correct answer is option A.
98.
f (x) =
x . . . (i) x−2
Let f (x) = y
=> y =
x x−2
=> xy − 2y = x
=> xy − x = 2y
=> x y − 1 = 2y
=> x =
2y . . . (ii) y −1
From (i), we have:
f (2x) =
=
2x 2x − 2
x x−1
Substituting the value of x from (ii), we have:
f (2x) =
=
2y y −1
!
2y −1 y −1 !
!
2y y −1
2y − y + 1 y −1
!
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174
Algebra Guide – Solutions
=
2y y +1
Substituting y = f (x), we have:
f (2x) =
2f (x) f (x) + 1
The correct answer is option B.
99.
f (x) =
2x 2 − 5 x
g (x) = x + 4
Since f (x) = g (x), we have:
2x 2 − 5 =x+4 x
=> 2x 2 − 5 = x 2 + 4x
=> x 2 − 4x − 5 = 0 => (x + 1) (x − 5) = 0 => x = −1 OR 5
However, we know that f (x) is only defined for x < 0
Thus, the only possible value of x = −1
The correct answer is option D.
100.
We have:
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Algebra Guide – Solutions
175
f (n) = f (n − 1) × f (n − 2), for all n > 2
We also have:
•
f (1) = 1
•
f (2) = −1
Thus, using the above, we have:
•
For n = 3 : f (3) = f (2) × f (1) = (−1) × 1 = −1
•
For n = 4 : f (4) = f (3) × f (2) = (−1) × (−1) = 1
•
For n = 5 : f (5) = f (4) × f (3) = 1 × (−1) = −1
•
For n = 6 : f (6) = f (5) × f (4) = (−1) × 1 = −1
Thus, we observe the different values of f (n) follow the sequence: (1, −1, −1)
Thus, when k is an integer, we have: •
If n = 3k + 1 => f (n) = 1
•
If n = 3k + 2 => f (n) = −1
•
If n = 3k => f (n) = −1
=> f (33) + f (34)
≡ f (3k) + f (3k + 1)
= −1 + 1
=0
The correct answer is option C. © 1999–2016 Manhattan Review
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176
Algebra Guide – Solutions
5.2
Data Sufficiency
Data sufficiency questions have five standard options. They are listed below and will not be repeated for each question. (A)
Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the question asked.
(B)
Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the question asked.
(C)
both the statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
(D)
EACH statement ALONE is sufficient to answer the question asked.
(E)
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
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Algebra Guide – Solutions 101.
177
Let the shares of the three sons be $x, $y and $z, with the eldest son’s share being $x. We need to determine whether x > 12000. From statement 1: The total wealth is not known. – Insufficient From statement 2: Though the total wealth is known, the manner it was divided among the three sons is not known. – Insufficient Thus, from both the statements together: x + y + z = 30000 x > y and x > z If we have: x = y = z, then value of each =
30000 = 10000 3
Since x is greater than y or z, we can have: •
x = 10002, y = z = 9999
•
x = 20000, y = z = 5000
Thus, the value of x may or may not exceed 12000. Thus, there is no unique answer. – Insufficient The correct answer is option E.
102.
From statement 1: We only have one equation with three unknowns. – Insufficient From statement 2: We only have one equation with three unknowns. – Insufficient Thus, from both the statements together: Since we need to compare x and y, we eliminate z from the two equations: 2x + 3y = z => 4x + 6y = 2z . . . (i) 7x − 3y = 2z . . . (ii) From (i) and (ii):
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178
Algebra Guide – Solutions 4x + 6y = 7x − 3y => 3x = 9y => x = 3y Thus, it apparently seems that x > y. However, x and y are variables, whose exact values cannot be determined from the equations. Thus, x and y may be positive or negative. If x and y are positive, x = 3y => x > y However, if x and y are negative, x = 3y => x < y Thus, there is no unique answer. – Insufficient The correct answer is option E.
103.
From statement 1: We only have one equation with three unknowns. – Insufficient From statement 2: We only have one equation with three unknowns. – Insufficient Thus, from both the statements together: Since we need to compare x and z, we eliminate y from the two equations: 2x + 3y − z = 0 . . . (i) 7x − 3y − 2z = 0 . . . (ii) By adding (i) & (ii): 9x = 3z => z = 3x Thus, it apparently seems that x < z. Again, we know that x and z are variables, whose exact values cannot be determined from the equations. It is given that z is positive, but x may be positive or negative. If x is positive, z = 3x => x < z We know that z is positive, so by the equality z = 3x, we can deduce that x is also positive, thus x ≯ z – Sufficient The correct answer is option C.
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Algebra Guide – Solutions 104.
179
Let Harry’s present age be x years. From statement 1: Harry’s age 5 years ago = (x − 5) years. Sum of ages of Harry and Ron = 13 years. Thus, Ron’s age 5 years ago = 13 − (x − 5) = (18 − x) years . . . (i) There is no information using which the value of x can be determined. – Insufficient From statement 2: Harry’s age 5 years ago = (x − 5) years. Harry is 1 year younger to Ron. Thus, Ron’s age 5 years back = (x − 5) + 1 = (x − 4) years . . . (ii) There is no information using which the value of x can be determined. – Insufficient Thus, from both the statements together: Equating (i) and (ii): 18 − x = x − 4 => x = 11 – Sufficient The correct answer is option C.
105.
Let Harry’s present age be x years. From statement 1: Let the present age of Harry’s brother be y years. Harry’s age 5 years ago = (x − 5) years. Age of Harry’s brother 5 years ago = y − 5 years. We know that five years ago, Harry was half as old as his elder brother. Thus, we have: x−5=
1 y −5 2
=> 2x = y + 5 . . . (i) Since we only have one equation with two unknowns, the value of x cannot be determined. – Insufficient © 1999–2016 Manhattan Review
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180
Algebra Guide – Solutions From statement 2: Let the present age of Harry’s brother be y years. We know that the age of Harry’s elder brother is five years less than twice Harry’s age. Thus, we have: y = 2x − 5 => 2x = y + 5 . . . (ii) Since we only have one equation with two unknowns, the value of x cannot be determined. – Insufficient Thus, from both the statements together: From equations (i) and (ii), we find that they are identical equations. Hence, the value of x cannot be determined. – Insufficient The correct answer is option E.
106.
Let the number of apples and the number of oranges be a and b, respectively. Let the price of one apple and the price of one orange be $x and $y, respectively. We need to determine the price of one apple and one orange, i.e. the value of x + y . From statement 1: Since 30 fruits were purchased, we have: a + b = 30 . . . (i) Since the total price was $40, we have: ax + by = 40 . . . (ii) Since we have two equations with four unknowns, the value of x + y cannot be determined. – Insufficient From statement 2: We know that on interchanging the number of apples and oranges, the total price is $80. Thus, we have: bx + ay = 80 . . . (iii) Since we have one equation with four unknowns, the value of x + y cannot be determined. – Insufficient
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Algebra Guide – Solutions
181
Thus, from both the statements together: Adding (ii) and (iii): x (a + b) + y (a + b) = 120 => x + y (a + b) = 120 . . . (iv) Substituting a + b = 30 from (i) in (iv), we have: => x + y × 30 = 120 => x + y = 4 – Sufficient The correct answer is option C.
107.
We know that the four-digit number is abcd, where a, b, c and d are the digits of the number. Since the sum of the first three digits equals the fourth digit, we have: a + b + c = d . . . (i) We need to determine the value of (a + b + c + d). Since a + b + c = d, we have: a + b + c + d = 2d. Thus, the required value we need to determine is 2d. From statement 1: We know that the sum of its third digit and twice its second digit equals 10 times its first digit => c + 2b = 10a . . . (ii) Only from (i) and (ii), the value of d cannot be determined. – Insufficient From statement 2: We know that the sum of the first and last digits equals 5 times the second digit => a + d = 5b . . . (iii) Only from (i) and (iii), the value of d cannot be determined. – Insufficient Thus, from both the statements together: Substituting d = a + b + c from (i) in (iii): a + (a + b + c) = 5b => 2a + c = 4b . . . (iv) From (ii): 2b = 10a − c
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182
Algebra Guide – Solutions => 4b = 20a − 2c . . . (v) Substituting the value of 4b from (v) in (iv): 2a + c = 20a − 2c => 3c = 18a => c = 6a Since a, b, c and d are single digit numbers, we have: The only possible value of a = 1 (a 6= 0 as this will make the number a three-digit number) => c = 6 . . . (vi) From (ii) and (vi): 6 + 2b = 10 => b = 2 . . . (vii) From (i), (vi) and (vii): d = a + b + c = 1 + 2 + 6 = 9 . . . (viii) Thus, the required answer = 2d = 18. – Sufficient The correct answer is option C.
108.
Let Ann’s age be 10x + y years, where x and y are the digits of the two-digit value of Ann’s age. From statement 1: We know that Ann’s age is such that if the digits of her age are reversed, Bob’s age is obtained. Thus, Bob’s age = 10y + x years . . . (i) However, we cannot solve for x and y. – Insufficient From statement 2: We know that Bob’s age exceeds Ann’s age by
1 of the combined ages of Ann and Bob. 11
Assuming Bob’s age to be n years, we have: 1 n − 10x + y = n + 10x + y 11 => 11n − 11 10x + y = n + 10x + y => 10n = 12 10x + y => n =
12 10x + y . . . (ii) 10
However, we cannot solve for x and y. – Insufficient Thus, from both the statements together: www.manhattanreview.com
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Algebra Guide – Solutions
183
From (i) and (ii), we have: 10y + x =
12 10x + y 10
=> 50y + 5x = 60x + 6y => 44y = 55x =>
x 4 = y 5
Since x and y are single digit numbers, the only possible values of x and y are: x = 4, y = 5 Thus, Ann’s age = 10x + 5 = 45 years. The correct answer is option C.
109.
Let the number of apples and the number of oranges be a and b, respectively. The price of one apple and the price of one orange are $1 and $2, respectively. We need to determine the number of apples and oranges, i.e. the value of (a + b). From statement 1: Since the man spent $41 to purchase the fruits, we have: a + 2b = 41 . . . (i) Multiple integer solutions are possible from the above equation, for example: •
a = 21, b = 10
•
a = 11, b = 15, etc.
Thus, the value of (a + b) cannot be uniquely determined. – Insufficient From statement 2: We know that had the man purchased as many apples as oranges and as many oranges as apples, 2 he would have saved half the cost of an orange, i.e. $ = $1. 2 Thus, we have: b + 2a = (a + 2b) − 1 => a = b − 1 . . . (ii) Multiple integer solutions are possible from the above equation. Thus, the value of (a + b) cannot be uniquely determined. – Insufficient © 1999–2016 Manhattan Review
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184
Algebra Guide – Solutions Thus, from both the statements together: Substituting the value of a from (ii) in (i): (b − 1) + 2b = 41 => b = 14 Thus, from (ii): a = 14 − 1 = 13 => a + b = 13 + 14 = 27 – Sufficient The correct answer is option C.
110.
We know that a = b = 1 From statement 1: We have: p 1 1 = a+ . . . (i) q 2 b 1 1 q = b+ . . . (ii) r 2 c To determine the value of c, we either need to eliminate p, q and r or we need to have those values. Since p, q and r cannot be eliminated from (i) and (ii) and neither are those values known, the answer cannot be determined. – Insufficient From statement 2: We have: r 1 1 = c+ . . . (iii) p 2 a Since there are three unknowns (p, r and c), the value of c cannot be determined using a single equation. – Insufficient Thus, from both the statements together: Multiplying (i), (ii) and (iii): 3 q r 1 1 1 1 p × × = a+ b+ c+ q r p 2 b c a 1 1 1 1 => 1 = × 1 + 1+ c+ 8 1 c 1
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Algebra Guide – Solutions
=>
185
1+c 1 ×2× × (1 + c) = 1 8 c
=> (1 + c)2 = 4c => 1 + 2c + c 2 = 4c => 1 − 2c + c 2 = 0 => (1 − c)2 = 0 => c = 1 – Sufficient The correct answer is option C.
111.
From statement 1: p + 2q
p + 3q = 35 . . . (i)
Since there are two unknowns, the value of q cannot be determined. – Insufficient From statement 2: We only know that p and q are positive integers. – Insufficient Thus, from both the statements together: From (i), we have: p + 2q
p + 3q = 35
Since p and q are positive integers, p + 2q and p + 3q are also positive integers and hence, are factors of 35. Also, we have: p + 2q < p + 3q; ; since p and q are positive integers Thus, we have: •
p + 3q = 7 AND p + 2q = 5: Solving, we have: q = 2, p = 1
•
p + 3q = 35 AND p + 2q = 1: Solving, we have: q = 34, p = −67 (not possible since q should be a positive integer)
Thus, we have: q = 2 – Sufficient The correct answer is option C.
112.
From statement 1: Since the roots of x 2 − kx + 35 = 0 are real, we have: (−k)2 − 4 (1) (35) ≥ 0
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186
Algebra Guide – Solutions => k2 ≥ 140 => k ≥
√
√ 140 OR k ≤ − 140
=> k ≥ 11. _ OR k ≤ −11. _ . . . (i) Again, since the roots of x 2 − 7x + k = 0 are real, we have: (−7)2 − 4 (1) (k) ≥ 0 => 4k ≤ 49 => k ≤
49 4
=> k ≤ 12.25 . . . (ii) Thus, from (i) and (ii), we have: The greatest possible integer value of k = 12 – Sufficient From statement 2: There is not enough information to determine the exact value of k. – Insufficient The correct answer is option A.
113.
From statement 1: ax 2 = by =>
b x2 = . . . (i) y a
However, the required value of
1 1 + x+1 y +1
!
1 1 + x+1 y +1
!
cannot be determined. – Insufficient
From statement 2: ay 2 = bx =>
b y2 = . . . (ii) x a
However, the required value of
cannot be determined. – Insufficient
Thus, from both the statements together: Equating (i) and (ii): x2 y2 = y x => x 3 = y 3 => x = y Thus, from (i): www.manhattanreview.com
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187
x2 b = x a => x = y =
b a
Thus, we have: 1 1 2 + = x+1 y +1 x+1 2 2a = b a+b +1 a Thus, the answer is ‘Yes’. – Sufficient
=
The correct answer is option C.
114.
Let the cost of P, Q, R and S be p, q, r and s, respectively. We need to determine whether: p + s < q + r From statement 1: We have: r =
s 2
=> s = 2r . . . (i) There is no information about the prices of the other cakes. – Insufficient From statement 2: q = 2r . . . (ii) p < r . . . (iii) There is no information about the price of S. – Insufficient Thus, from both the statements together: From (i) and (iii), we have: p + s = p + 2r < r + 2r => p + s < 3r . . . (iv) From (ii): q + r = 2r + r => q + r = 3r . . . (v) Thus, from (iv) and (v): p+s 10 (n + 2m) = mn => 10n = mn − 20m => m (n − 20) = 10n => m =
10n n − 20
n => m = 10 n − 20
(n − 20) + 20 = 10 n − 20
= 10 1 +
20 n − 20
= 10 +
200 n − 20
Since m is an integer, 200 should be divisible by (n − 20). 200 Since we need the largest possible value of m, the value of should be maximized, n − 20 i.e. the value of (n − 20) should be minimized (but kept positive, as we expect the largest
possibility). Thus, we have: n − 20 = 1 => n = 21 => m = 10 +
200 = 210 – Sufficient 21 − 20
From statement 2: There is not enough information to determine the exact value of m. – Insufficient The correct answer is option A.
116.
Let the number of marbles with A, B and C be a, b and c, respectively. c We need to determine whether: × 100 ≥ 50 a+b+c From statement 1: We know that the sum of the number of marbles with A and B together is 40 percent less than that with B and C together 40 => a + b = 1 − (b + c) 100
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189
3 (b + c) 5 a+b 3 => = . . . (i) b+c 5 => a + b =
We cannot determine the number of marbles with C as a percent with the total marbles. – Insufficient From statement 2: We know that the sum of the number of marbles with B and C together is 20 percent less than that with A and C together 20 => b + c = 1 − (a + c) 100 4 => b + c = (a + c) 5 5 a+c = . . . (ii) => b+c 4 We cannot determine the number of marbles with C as a percent with the total marbles. – Insufficient Thus, from both the statements together: We can see that (b + c) is in the denominator of both (i) and (ii). In (i), it corresponds to 5 and in (ii), it corresponds to 4. Let (b + c) = LCM of 5 and 4 = 20
=>
3 12 a+c 5 25 a+b = = AND = = b+c 5 20 b+c 4 20
Thus, we have: a + b = 12, b + c = 20, a + c = 25 57 2 57 33 Subtracting the value of (a + b) from the above: c = − 12 = 2 2 Adding the above: 2 (a + b + c) = 57 => a + b + c =
Thus, the required value 33 2 c = × 100 = 57 × 100 a+b+c 2 =
33 × 100 57
>
30 × 100 = 50% – Sufficient 60
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190
Algebra Guide – Solutions The correct answer is option C.
117.
Let the price of the camera = $x. Let the number of friends be n. From statement 1: We know that: 210 < x < 230 . . . (i) The exact value of x cannot be determined. – Insufficient From statement 2: Price each friend had to pay initially = $
x n
After 2 friends back out, number of friends left = (n − 2) x Thus, price each friend has to pay now = $ n−2 Since each of the remaining friends has to pay $1 extra, we have: x x − =1 n−2 n => x
=> x
1 1 − n−2 n
n − (n − 2) n (n − 2)
=1
=1
=> 2x = n (n − 2) . . . (ii) The value of x cannot be determined. – Insufficient Thus, from both the statements together: Since 210 < x < 230, we have: 420 < 2x < 460 => 420 < n (n − 2) < 460 The value of n (n − 2) is close to n2 , and it lies between 420 and 460. We know that 202 = 400, 20 × 21 = 420, 20 × 22 = 440 and 20 × 23 = 460 Thus, we have: n = 22 www.manhattanreview.com
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191
Thus, from (ii):
=> x =
n (n − 2) 22 × 20 = = 220 – Sufficient 2 2
The correct answer is option C.
118.
Let the number of $1 bills and $10 bills be x and y, respectively. We need to determine whether: x + y can equal nine. From statement 1: Total amount of money with him = $ x + 10y
Since the number of $1 bills multiplied by the number of $10 bills is equal to the total money (in dollars) with him, we have: xy = x + 10y => x y − 1 = 10y => x =
10y y −1
=> x + y =
10y + y 2 − y 9y + y 2 10y +y = = y −1 y −1 y −1
Let x + y = 9
=>
9y + y 2 =9 y −1
=> 9y + y 2 = 9y − 9 => y 2 = −9, which is not possible Thus, x + y cannot be 9. – Sufficient From statement 2: We know that the number of $1 bills is greater than 8, i.e. the minimum number is 9. Thus, the total number of bills with him cannot be 9 (it must be greater than 9). – Sufficient The correct answer is option D. © 1999–2016 Manhattan Review
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192 119.
Algebra Guide – Solutions From statement 1: We have: A+C=E Since the numbers are 2, 4, 5, 6 and 10, and we have 2 + 4 = 6 and 4 + 6 = 10, the possible cases are: A
C
E
2
4
6
. . . (i)
4
2
6
. . . (ii)
4
6
10
. . . (iii)
6
4
10
. . . (iv)
However, nothing is mentioned about B. – Insufficient From statement 2: We have: A+E=B Since the numbers are 2, 4, 5, 6 and 10, and we have 2 + 4 = 6 and 4 + 6 = 10, the possible cases are: A
E
B
2
4
6
. . . (v)
4
2
6
. . . (vi)
4
6
10
. . . (vii)
6
4
10
. . . (viii)
Thus, the value of B cannot be uniquely determined. – Insufficient Thus, from both the statements together: From the possibilities tabulated above, we have: •
If A = 2, from (i) and (v), both C and E have to be 4, which is not possible since each letter represents a distinct digit. Thus, A 6= 2
•
If A = 6, from (iv) and (viii), both C and E have to be 4, which is not possible since each letter represents a distinct digit. Thus, A 6= 6
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Algebra Guide – Solutions •
193
If A = 4, then we have: (a)
From (iii): C = 6 and E = 10. But from (v) to (viii), it is clear that E 6= 10. Thus, C 6= 6
(b)
From (ii) and (vii): C = 2, E = 6 and B = 10
Thus, the only possible scenario: A = 4, B = 10, C = 2, E = 6 and hence, D = 5. Thus, we have B = 10. – Sufficient The correct answer is option C.
120.
From statement 1: We have: 2x + 3y + 4z = 10 Multiplying throughout by 2: 4x + 6y + 8z = 20 Thus, we have: k = 5x + 7y + 8z = 4x + 6y + 8z + x + y = 20 + x + y Since x and y are positive, we have: k > 20 Thus, the answer to the question is ‘No’. – Sufficient From statement 2: We have: 3x + 4y + 5z = 12 Multiplying throughout by 2: 6x + 8y + 10z = 24 Thus, we have: k = 5x + 7y + 8z = 6x + 8y + 10z − x + y + 2z
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194
Algebra Guide – Solutions = 24 − x + y + 2z
Since x, y and z are positive, we have: k < 24 Thus, the value of k can be greater than 20, equal to 20 or less than 20. Thus, the answer cannot be uniquely determined. – Insufficient The correct answer is option A.
121.
From statement 1: We have no information about the values of x, y, & z. k may nor may not be 20. Thus, the answer cannot be uniquely determined. – Insufficient From statement 2: We have no information about the values of x, y, z, & k. k may nor may not be 20. Thus, the answer cannot be uniquely determined. – Insufficient Thus, from both the statements together: From statement 2, the minimum values of x, y and z are 1 each, thus the minimum value of k = 5x + 7y + 9z = 5 × 1 + 7 × 1 + 9 × 1 = 21. Thus, k CANNOT be equal to 20. – Sufficient The correct answer is option C.
122.
We have: f (x) = x 3 − 4x + p From statement 1: We have: f (0) = 0 − 0 + p = p Also, f (1) = 1 − 4 + p = p − 3 Since f (0) × f (1) < 0, we have: p p−3 0 < p < 3 The answer to the question is ‘Yes’. – Sufficient From statement 2:
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195
We have: f (0) = p Since f (0) > 0, we have: p>0 Thus, p may be greater than 3 or equal to 3 or less than 3. Thus, the answer cannot be uniquely determined. – Insufficient The correct answer is option A. 1 m f (x) = (a − x n )
123.
1 => f (1) = (a − 1n ) m
1 m => f (1) = p = (a − 1)
Let us express the value of f p
1 m n f p = a−p
1 n 1 1 m => f p = a − (a − 1) m ; by plugging in the value of p = (a − 1) m
1 ! n m = a − (a − 1) m . . . (i)
From statement 1: n=m Thus, from (i), we have: 1 ! m m f p = a − (a − 1) m
1 n o 1 m = a − (a − 1)
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196
Algebra Guide – Solutions 1 m =1
= 1 – Sufficient From statement 2: a=1 Thus, from (i), we have: 1 ! n m f p = 1 − (1 − 1) m
1 m = {1 − 0}
1 m =1
= 1 – Sufficient The correct answer is option D.
124.
1 In the sequence a, b, x, y, . . . , each term of the sequence above is 9 more than of the previous 3 term. Thus, we have: b =9+
a . . . (i) 3
x =9+
b . . . (ii) 3
y =9+
x . . . (iii) 3
From statement 1: a = 54 Thus, from (i), we have: => b = 9 +
54 = 27 3
Thus, from (ii), we have:
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Algebra Guide – Solutions
=> x = 9 +
197
27 = 18 3
Thus, from (iii), we have: => y = 9 +
18 = 15 3
=> x − y = 18 − 15 = 3 – Sufficient From statement 2: y=
5 x 6
From (iii), we have: y =9+
x 3
=>
x 5 x =9+ 6 3
=>
x =9 2
=> x = 18 => y =
5 × 18 = 15 6
=> x − y = 3 – Sufficient The correct answer is option D. Alternate Approach: The question can be solved through Logical deduction approach too. We know that the sequence a, b, x, y, . . . follows a common logic: “Each term of the above 1 of the previous term.” This follows that if the value of any term or sequence is 9 more than 3 the relationship between any two terms of the sequence: a, b, x, y is given, the whole sequence can be deduced. We see that statement 1 provides us with the value of the first term of the sequence: a = 54 and statement 2 provides us with a relationship between two terms of the sequence: x & y, thus each statement alone is sufficient to reach the answer. 125.
f(n+1) (x) = f(n) (x) + 1 . . . (i) From statement 1: f(1) (x) = 0
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198
Algebra Guide – Solutions Thus, from (i), we have: For n = 1 For n = 2 For n = 3
f(1+1) (x) = f(1) (x) + 1 => f(2) (x) = 0 + 1 = 1 f(2+1) (x) = f(2) (x) + 1 => f(3) (x) = 1 + 1 = 2 f(3+1) (x) = f(3) (x) + 1 => f(4) (x) = 2 + 1 = 3
Thus, we have: f(4) (x) = 3 – Sufficient From statement 2: f(5) (x) = 4 Thus, from (i), we have: For n = 4: f(4+1) (x) = f(4) (x) + 1 => f(4) (x) = f(5) (x) − 1 => f(4) (x) = 4 − 1 = 3 – Sufficient The correct answer is option D. 126.
f (n + 3) = f (n + 2) × f (n) From statement 1: f (1) = f (2) = 1 Thus, we have: For n = 1 For n = 2
f (1 + 3) = f (1 + 2) × f (1) => f (4) = f (3) × f (1) => f (4) = f (3) f (2 + 3) = f (2 + 2) × f (2) => f (5) = f (4) × f (2) => f (5) = f (4)
Thus, we have: f (3) = f (4) = f (5) => f (5) − f (3) = 0 – Sufficient From statement 2: f (3) = −1 Thus, we have:
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199
f (1 + 3) = f (1 + 2) × f (1) => f (4) = f (3) × f (1) => f (4) = −f (1) f (2 + 3) = f (2 + 2) × f (2) => f (5) = f (4) × f (2)
Thus, we have: f (5) − f (3) = f (4) × f (2) − (−1) = −f (1) × f (2) + 1 However, the values of f (1) and f (2) are unknown. – Insufficient Thus, the correct answer is option A. 127.
Since 10 kilograms of alloy K consists of x kilograms of aluminum and y kilograms of copper, we have: x + y = 10 . . . (i) Cost of x kilograms of aluminum = $2x. Cost of y kilograms of copper = $4y. We need to determine whether x > y From statement 1: y >3 x + y = 10 . . . from (i) x = 10 − y => x < 7 Thus, we may have: (1)
y = 5, x = 5 => x ≯ y
(2)
y = 4, x = 6 => x > y
Thus, a unique answer cannot be determined. – Insufficient From statement 2: Total cost of 10 kilograms of alloy K, which consists of x kilograms of aluminum and y kilograms of copper = $ 2x + 4y
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200
Algebra Guide – Solutions Thus, we have: 2x + 4y < 30 => 2 x + y + 2y < 30 => 20 + 2y < 30 . . . from (i) => 2y < 10 => y < 5 => x > 10 − 5 = 5 . . . from (i) => x > y – Sufficient The correct answer is option B.
128.
Let the ages of P, Q and R be p, q and r respectively. We have: p+q+r = 24 3 => p + q + r = 72 . . . (i) From statement 1: Since R is 6 years elder to Q, we have: r = q + 6 . . . (ii) Since the difference between the ages of P and Q is 6 years, we can have the following possibilities (depending on who between P and Q is older): (a)
p−q =6 => p = q + 6 => p + q + r = q + 6 + q + q + 6 , using (ii) above = 3q + 12 => 3q + 12 = 72, using (i) above => q = 20 => p = 20 + 6 = 26
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201
OR (b)
q−p =6 => p = q − 6 => p + q + r = q − 6 + q + q + 6 , using (ii) above = 3q => 3q = 72, using (i) above => q = 24 => p = 24 − 6 = 18
Thus, the value of p cannot be uniquely determined. – Insufficient From statement 2: The actual ages of any of the persons is not known. – Insufficient Thus, from statements 1 and 2 together: We know that: p p = q + 6 => p = r , using (ii) above Thus, this situation is not possible, since p 6= q 6= r OR
(b)
q−p =6 => p = q − 6 => p + q + r = q − 6 + q + q + 6 , using (ii) above = 3q => 3q = 72, using (i) above => q = 24 => p = 24 − 6 = 18 – Sufficient
From statement 2: The actual ages of any of the persons is not known. – Insufficient The correct answer is option A. 130.
Let the amounts with A, B and C be $a, $b and $c respectively. Thus, we have: a + b + c = 60 . . . (i) From statement 1: a + b = (100 + 40) % of c
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Algebra Guide – Solutions
=> a + b =
140 ×c 100
=> a + b =
7c 5
=> a + b + c =
203
7c 12c +c = 5 5
Since a + b + c = 60, we have: =>
12c = 60 5
=> c = 25 => a + b = 60 − c, using (i) above => a + b = 35 It may be possible that: (1)
a = 34, b = 1 => a has the highest amount
(2)
a = 1, b = 34 => a does not have the highest amount
Thus, the answer cannot be uniquely determined. – Insufficient From statement 2: Since B has more than what A and C together have, it is clear that B must have the highest amount. Thus, A does not have the highest amount. – Sufficient The correct answer is option B. 131.
Let the present ages of A and B be a years and b years, respectively. We need to determine whether: b > 2a From statement 1: Four years back, A’s age = (a − 4) years. Four years back, B’s age = (b − 4) years. Thus, we have:
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204
Algebra Guide – Solutions b − 4 = 2 (a − 4) => b = 2a − 4 => b < 2a; the answer is No. – Sufficient From statement 2: b = a + 20 We can apply a logical deduction approach to reach to a conclusion on whether b > 2a. If a is a relatively smaller value compared to 20, for example, say a = 5, then b = a + 20 = 25, thus b = 5a => b > 2a. If a is a relatively larger value compared to 20, for example, say a = 40, then b = a + 20 = 60, thus b = 1.5a => b < 2a. Thus, there is no unique answer. – Insufficient The correct answer is option A.
132.
Let the number of apples purchased be x and oranges purchased be y. From statement 1: Worth of apples = $(0.30x). Worth of oranges = $(0.58y). Thus, total worth of all fruits = $(0.30x + 0.58y). Thus, we have: 0.30x + 0.58y = 8.80 => 0.15x + 0.29y = 4.40 (Dividing by 2 throughout) => 15x + 29y = 440 (Multiplying by 10 throughout) We know that x and y are non-negative integers. Thus, we have: 15x = 440 − 29y
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Algebra Guide – Solutions
=> x =
205
440 − 29y 15
Since x is an integer, 440 − 29y must be divisible by 15. We separate out the part from 440 − 29y which is divisible by 15. Thus, we have: => x =
435 − 30y + 5 + y 15
5+y => x = 29 − 2y + 15 Thus, the value of y should be such that 5 + y is divisible by 15 => y = 10, 25, 40 . . . etc. Thus, we have: If y = 10: 5+y x = 29 − 2y + 15 = (29 − 2 × 10) +
5 + 10 15
= 10 Working with the next value of y = 25, we have: 5+y x = 29 − 2y + = −19, i.e.not possible 15 Thus, the only solution is: x = y = 10 – Sufficient From statement 2: We have: x=y However, the information provided is not sufficient to determine the value of x. – Insufficient The correct answer is option A. 133.
Let the number of paperback and hardcover books be x and y, respectively. We know that:
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206
Algebra Guide – Solutions x > 10 Thus, the cost of paperback books = 8x > 80 Since the minimum value of x = 11, we have => 8x ≥ 88 . . . (i) From statement 1: Cost of hardcover books = $ 25y
Thus, we have: 25y ≥ 150 => y ≥ 6 . . . (ii) However, the unique value of y cannot be determined. (y could be 6, 7, etc.) – Insufficient From statement 2: The total cost of all the books was less than $260 => 8x + 25y < 260 However, from (i), we know that: 8x ≥ 88 => 25y < 260 − 88 = 172 => y <
172 25
y < 6.48 . . . (iii) Thus, possible values of y cannot be uniquely determined (y could be 6, 5, etc.). – Insufficient Thus, from statements 1 and 2 together: From (ii) and (iii), we have: 6 ≤ y < 7.2 => y = 6 – Sufficient The correct answer is option C. 134.
Let the number of Brand X computers and Brand Y computers be x and y, respectively.
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207
Thus, we have: x + y = 880 . . . (i) From statement 1: The ratio of the number of Brand Y computers to the number of Brand X computers at the company is 5 to 6. Thus, we have: 5 y = x 6 Using (i), we have: => y =
5 × 880 = 400 – Sufficient 5+6
From statement 2: Since the number of Brand X computers is greater than the number of Brand Y computers at the company by 80, we have: x − y = 80 Using (i), we have, by adding the equations: 2x = 960 => x = 480 => y = 400 – Sufficient The correct answer is option D. 135.
From the table, we have: m=d+z n=e+y Thus, we have: m + n = (d + z) + e + y
= d + y + (e + z) From statement 1: The value of (e + z) is not known. – Insufficient
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208
Algebra Guide – Solutions From statement 2: The value of d + y is not known. – Insufficient Thus, from statements 1 and 2 together: m + n = d + y + (e + z) => m + n = (−3) + 12 = 9 – Sufficient The correct answer is option C.
136.
Total charge to rent a car from Company J = $ (15 + 0.2x) Total charge to rent a car from Company K = $ (20 + 0.1x) We need to determine whether: 15 + 0.2x < 25 From statement 1: 20 + 0.1x < 25 => 0.1x < 5 => x < 50 => 0.2x < 50 × 0.2 => 0.2x < 10 => 15 + 0.2x < 15 + 10 => 15 + 0.2x < 25 – Sufficient From statement 2: x < 50 This is the same condition as obtained from Statement 1. – Sufficient The correct answer is option D.
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Algebra Guide – Solutions 137.
209
From statement 1: The call rates are not known. – Insufficient From statement 2: The call duration is not known. – Insufficient Thus, from statements 1 and 2 together: Cost of the first minute of the call = $0.32 0.32 Cost per minute after the first minute = $ 2
= $0.16
Total call duration = 8 minutes. Thus, cost of the remaining 8 – 1 = 7 minutes of the call = $(0.16 × 7) = $1.12 Thus, total cost of the call = $(0.32 + 1.12) = $1.44 – Sufficient The correct answer is option C. 138.
Let the total hours the car was parked be t. Let the charge for the first hour be $x and for each additional hour (or fraction of an hour), for (t − 1) hours be $y. Thus, the total parking charge = $ x + (t − 1) y
Thus, we have: x + (t − 1) y = 3.75 . . . (i) We need to determine the value of t. From statement 1: x = 0.75, y = 0.50 Thus, we have, from (i): 0.75 + (t − 1) × 0.5 = 3.75 => (t − 1) × 0.5 = 3 => t − 1 = 6
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210
Algebra Guide – Solutions => t = 7 – Sufficient From statement 2: If the charge for the first hour had been $1.00, Nan’s total parking charge would have been $4.00 Thus, we have: 1 + (t − 1) × y = 4 => (t − 1) y = 3 Since the value of y is not know, we cannot determine the value of t. – Insufficient The correct answer is option A.
139.
Let the lengths of the 3 pieces, in cm, be a, b and c, where a > b > c (Since the lengths are distinct integers, we can arrange them in a pre-defined order) Thus, we have: a + b + c = 8 . . . (i) We need to determine the value of (a × b × c). From statement 1: a=b+c From (i): a + (b + c) = 8 => a + a = 8 => a = 4 => b + c = 4 => b = 3, c = 1 (since b > c) Thus, the product of the lengths = a × b × c = 4 × 3 × 1 = 12 – Sufficient
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211
From statement 2: c=1 From (i), we have: a + b + 1 = 8 => a + b = 7 => a = 4, b = 3 OR a = 5, b = 2 (since a > b > c) Thus, the product of the lengths =a×b×c = 4 × 3 × 1 = 12 OR 5 × 2 × 1 = 10 Thus, the answer cannot be uniquely determined. – Insufficient The correct answer is option A. 140.
Let the number of units of products A and B sold be a and b, respectively. Thus, we have: => 12a + 5b = 300 We need to determine the values of a and b, given that they are positive integers. The obvious set of values is: a =
300 = 25, b = 0 12
To generate the other integer values, we subtract the coefficient of b from the value of a obtained in previous set and simultaneously, add the coefficient of a to the value of b obtained in previous set. a 25 25 – 5 = 20 20 – 5 = 15 15 – 5 = 10 10 – 5 = 5 5–5=0
b 0 0 + 12 = 12 12 + 12 = 24 24 + 12 = 36 36 + 12 = 48 48 + 12 = 60
From statement 1: 10a ≥ 120 => a ≥ 12 Thus, the possible values of a and b are:
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Algebra Guide – Solutions a 25 20 15
b 0 12 24
Thus, there is no unique solution. – Insufficient From statement 2: a=b+8 Thus, the only set of values is: a = 20, b = 12 – Sufficient The correct answer is option B. 141.
Let the price of 1 apple, 1 orange and 1 lemon be a, r and l respectively. We need to determine the value of (a + r + l) From statement 1: 5a + 4r + 3l = 130 . . . (i) We cannot determine the values of a, r and l from a single equation. Hence, the answer cannot be determined. – Insufficient From statement 2: 3a + 4r + 5l = 110 . . . (ii) We cannot determine the values of a, r and l from a single equation. Hence, the answer cannot be determined. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we cannot determine the individual values of a, r and l. However, we need to check, if by combining both equations, we can find the value of (a + r + l). We can see that if we add (i) and (ii): 8a + 8r + 8l = 240 => a + r + l =
240 = 30 – Sufficient 8
The correct answer is option C. www.manhattanreview.com
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Algebra Guide – Solutions 142.
213
From statement 1: We have no information about the value of the ages. – Insufficient From statement 2: We have no information about the value of the ages. – Insufficient Thus, from statements 1 and 2 together: Let the age of B x years ago was b years. Thus, the present age of B = (x + b) years. Since A, at present, is twice as old as B was x years ago, we have: A’s present age = 2b years. Thus, A’s age x years ago = (2b − x) years. Since B, at present, has the same age as A had x years ago, we have: x + b = 2b − x => b = 2x Thus, the present age of B = (x + b) years = 3x years. Also, the present age of A = 2b years = 4x years. Since the sum of the present ages of A and B is 70 years, we have: 3x + 4x = 70 => x = 10 Thus, A’s present age = 4x years = 40 years. – Sufficient The correct answer is option C.
143.
From statement 1: 2x+2y = 8 => 2x+2y = 23
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Algebra Guide – Solutions => x + 2y = 3 Since x and y are positive integers, the only possible values of x and y are: x = 1, y = 1 => xy = 1 – Sufficient From statement 2: 23x+2y = 32 => 23x+2y = 25 => 3x + 2y = 5 Since x and y are positive integers, the only possible values of x and y are: x = 1, y = 1 => xy = 1 – Sufficient The correct answer is option D.
144.
Let the number of cakes and the number of biscuit packets purchased be x and y, respectively. We need to determine the value of x + y . From statement 1: 13x + 7y = 33 The number of items purchased must be positive integers. Let us try with a few positive integer values of x: x 1 2 3
y 13 + 7y = 33 => y = which is not an integer – Not a valid solution 26 + 7y = 33 => y = 1, which is an integer – A valid solution 39 + 7y = 33 => y is negative – Not a valid solution (Thus, for all other higher values of x, the value of y would be negative, and hence, can be ignored) 20 7 ,
Thus, the only solution is: x = 2, y = 1
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=> x + y = 3 – Sufficient From statement 2: The number of cakes and the number of biscuit packets purchased are y and x, respectively. 7x + 13y = 27 The number of items purchased must be positive integers. Let us try with a few positive integer values of x: x 1 2 3 4
y 7 + 13y = 27 => y = which is not an integer – Not a valid solution 14 + 13y = 27 => y = 1, which is an integer – A valid solution 6 21 + 13y = 27 => y = 13 , which is not an integer – Not a valid solution 28 + 13y = 27 => y is negative – Not a valid solution (Thus, for all other higher values of x, the value of y would be negative, and hence, can be ignored) 20 13 ,
Thus, the only solution is: x = 2, y = 1 => x + y = 3 – Sufficient Note: It apparently seems that we have two equations: 13x + 7y = 33 . . . (i) 7x + 13y = 27 . . . (ii) By adding them, we have: 20x + 20y = 60 => x + y = 3, implying the answer as ‘C’; which is not correct. The correct answer is option D. 145.
Let the price of each bottle of beer = $x. From statement 1: Total bottles purchased = 4 + 6 + 2 = 12. Thus, total cost of all beer bottles = $12x.
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Algebra Guide – Solutions
Number of bottles of beer consumed by each friend =
12 = 4 bottles. 3
However, C had purchased only 2 bottles of beer. Thus, he had to pay to A and B the price of 4 – 2 = 2 bottles of beer. Thus, the amount C paid to A and B = $2x. However, we cannot determine the value of x. – Insufficient From statement 2: There is no information about the number of bottles of beer purchased. – Insufficient Thus from statement 1 and 2 together: We have: 2x = 16 => x = 8 – Sufficient The correct answer is option C. 146.
Each of the initial n bacteria weighed 10−12 grams. Thus, total weight of the initial n bacteria = 10−12 n grams. Each of the n bacteria gave birth to n more bacteria weighing 10−12 grams each. Thus, total number of bacteria born = n2 Thus, total weight of the bacteria born = 10−12 n2 grams. From statement 1: 1 of the total weight of all bacteria. 16 1 15 Thus, the weight of the new bacteria born was 1 − = of the total weight of all bacteria. 16 16 The first n bacteria weighed
Thus, we have: Ratio of the weight of the initial n bacteria and the new bacteria born 1 1 16 = = 15 15 16 www.manhattanreview.com
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Algebra Guide – Solutions
=>
10−12 n 10−12 n2
=
217
1 15
=> n = 15 – Sufficient From statement 2: Total weight of all bacteria = 10−12 n + 10−12 n2 grams. Thus, we have: 10−12 n + 10−12 n2 = 24 10−11
=> n + n2 =
24 10−11 10−12
=
240 10−12 10−12
= 240
=> n2 + n − 240 = 0 => n2 + 16n − 15n − 240 = 0 => (n + 16) (n − 15) = 0 => n = 15 OR −16 Since n must be positive, we have: n = 15. – Sufficient The correct answer is option D. 147.
From statement 1: pq = 4 . . . (i) The value of p cannot be determined from a single equation with two unknowns. – Insufficient From statement 2: q − 2p = 7 . . . (ii) The value of p cannot be determined from a single equation with two unknowns. – Insufficient Thus, from statements 1 and 2 together: From (ii), we have: q − 2p = 7 => q = 2p + 7
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Algebra Guide – Solutions Substituting the value of q in (i): p 2p + 7 = 4 => 2p 2 + 7p − 4 = 0 => p + 4 2p − 1 = 0 => p = −4 OR
1 2
Since p is an integer, we have: p = −4 – Sufficient The correct answer is option C. 148.
a2 − b2 = 20 . . . (i) From statement 1: We know that a and b are positive integers. a2 − b2 = 20 => (a + b) (a − b) = 20 Since a and b are positive integers, (a + b) and (a − b) are also integers. Thus, (a + b) and (a − b) are factors of 20. Let us assume that: a + b = k and a − b = l, where k × l = 20 Thus, by adding the above two equations, we have: 2a = k + l => a =
(k + l) 2
Since a is an integer, (k + l) must be even, which is possible if:
•
k and l are both odd – Not possible, since 20 is an even number
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Algebra Guide – Solutions •
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k and l are both even – The only possible scenario => (a + b) and (a − b) are both even
Since 20 = 22 × 5, the possible cases are: •
(a + b) = 2 × 5 = 10 and (a − b) = 2 OR
•
(a + b) = 2 and (a − b) = 2 × 5 = 10
For any two positive integers, their sum is always greater than their difference. Since a and b are positive integers, we have: a+b >a−b Thus, we finally have: a + b = 10 AND a−b =2 Adding the above two equations, we have: 2a = 12 => a = 6 – Sufficient From statement 2: a + b = 10 . . . (ii) Dividing (i) by (ii), we have: 20 a2 − b 2 = a+b 10 => a − b = 2 This is the same situation as obtained from statement 1. – Sufficient The correct answer is option D. © 1999–2016 Manhattan Review
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220 149.
Algebra Guide – Solutions From statement 1: a2 + b = 2a b => a2 + b2 = 2ab => a2 − 2ab + b2 = 0 => (a − b)2 = 0 => a − b = 0 => a = b =>
a = 1 – Sufficient b
From statement 2: (a − 2)2 + |b − 2| = 0 . . . (i) We can observe that (a − 2)2 is a ‘perfect square’ term => (a − 2)2 ≥ 0 Again, |b − 2| is an ‘absolute value’ term => |b − 2| ≥ 0 Thus, from (i), we observe that two ‘non-negative’ terms add up to ‘0’, which is only possible if both terms are individually ‘0’. Thus, we have: (a − 2)2 = 0 => a − 2 = 0 => a = 2 . . . (ii) Also, we have: |b − 2| = 0 => b − 2 = 0 => b = 2 . . . (iii)
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Thus, from (ii) and (iii), we have: a 2 = = 1 – Sufficient b 2 The correct answer is option D. 150.
From statement 1: 1−a 1−b + =2 a b 1 1 => −1 + −1 =2 a b =>
1 1 + = 4 – Sufficient a b
From statement 2: 1 ab = a+b 4 Taking reciprocal on both sides: =>
a+b =4 ab
=>
b a + =4 ab ab
=>
1 1 + = 4 – Sufficient b a
The correct answer is option D.
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Chapter 6
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