GRE Math Review

GRE MATH REVIEW #1

Basic Arithmetic The whole numbers, or counting numbers, are 0,1,2,3,4,5,6… The integers are …-4, -3, -2, -1,0,1,2,3,4… Recall the number line in deciding whether one negative number is larger than another. Positive integers increase as they move away from 0; negative integers decrease as they move away from 0. For example, 4 is greater than 3, but -4 is less than –3. Consecutive integers are in increasing order without any integers missing between them. For example, 0,1,2,3,4… are consecutive integers; -2,0,2,4…are consecutive even integers; -3, -1,1,3… are consecutive odd integers. Zero is both a whole number and an even integer, but it is neither positive nor negative. The sum of 0 and any number is that number; the product of 0 and any number is 0. There are 10 digits in our number system: 0,1,2,3,4,5,6,7,8,9. An integer greater than 9 is made up of several digits. For example, in the number 5234, 4 is the ones digit, 3 is the tens digit, 2 is the hundreds digit, 5 is the thousands digit. When multiplying positive and negative numbers, a positive times a positive is a positive; a negative times a negative is a positive; a positive times a negative is a negative: neg + neg = neg pos + pos = pos neg + pos = !

(neg) x (neg) = pos (pos) x (pos) = pos (neg) x (pos) = neg

An even number is any number that can be divided evenly by 2 with no remainder left over. An odd number cannot be divided evenly by 2. Any integer is even if its ones digit is even; an integer is odd if its ones digit is odd. Don’t confuse even/odd with positive/negative. On the GRE, remember the following: even + even = even odd + odd = even even + odd = odd

(even) x (even) = even (odd) x (odd) = odd (even) x (odd) = even

An integer is divisible by 2 if its ones digit is divisible by 2. An integer is divisible by 3 if the sum of its digits is divisible by 3. An integer is divisible by 5 if its ones digit is either 0 or 5. An integer is divisible by 10 if its ones digit is 0. A prime number is a number that can be divided evenly by only itself and 1. 0 and 1 are not prime numbers; 2 is the only even prime number. Some examples of prime numbers are 2, 3, 5, 7, 11, 13, 17… A number x is a factor of a number y if y is divisible by x. For example, 2, 3, 4, and 6 are all factors of 12 since they all divide evenly into 12. A multiple of a number x is x multiplied by any integer except 0. For example, 10, 20, 30, 40, etc. are all multiples of 10. The following is a list of symbols and their meanings you need to know on the GRE: = ≠ > < ≥ ≤

equal to not equal to greater than less than greater than or equal to less than or equal to

The following is a list of terms and their definitions you need to know on the GRE: sum difference product quotient numerator denominator

the result of addition the result of subtraction the result of multiplication the result of division the “top” number in a fraction the “bottom” number in a fraction

The six basic operations you will need to perform on the GRE are as follows. Let a and b be any numbers 1. 2. 3. 4. 5. 6.

addition: a + b subtraction: a – b multiplication: ab = a • b = a × b = (a)(b) division: a/b = a ÷ b raising a number to an exponent: a2 finding square roots and cube roots: a , 3 b

In order to find an answer when more than one operation is involved, you must know the correct order of operations. Remember the following mnemonic: Please Excuse My Dear Aunt Sally. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. This is the order in which the operations must be performed. For example, 10(2) + (11-1) ÷ 5 – 4 = 18. NOTE: MULTIPLICATION AND DIVISION GOES LEFT TO RIGHT, i.e., 10 ÷ 2 × 5 =25 NOT 1; ADDITION AND SUBTRACTION ALSO GOES LEFT TO RIGHT, i.e., 10 – 2 + 5 =13 NOT 3. The associative laws of addition and multiplication allow you to regroup numbers in any order when adding or multiplying. If a, b, and c are any numbers, a + (b + c) = (a + b) + c and a (b c) = (a b) c For example, 2 + (4 + 6) = 4 + (2 + 6) = (6 + 4) + 2 = 2 + 4 + 6 and 2(4 • 6) = (4 • 2)6 = 6(2 • 4) = 6 • 4 • 2. The distributive laws are very important on the GRE. Apply them every chance you get. If a, b, and c are any numbers, then

a(b + c) = ab + ac a(b – c) = ab - ac

and and

(b + c)a = ba + ca = ab + ac (b – c)a = ba – ca = ab –ac

When you use the distributive law to go from ab + ac to a(b + c), you are actually factoring the expression ab + ac by finding the common term to ab and ac which is “a”.

EXERCISE 1

1.

List five consecutive negative even integers.

2.

List five consecutive integers, one of which is 0.

3.

List all the prime numbers less than 30.

4.

What is the least integer greater than –5.8?

5.

What is the greatest integer less than –3.6?

6.

What is a 3-digit number whose digits add up to 14?

7.

Without performing division, what is the remainder when: (a) 99 is divided by 5? (b) 12,345,671 is divided by 10?

8.

A multiple of both 3 and 7 is also a multiple of what number?

9.

Is the product of 34,569 and 227 odd or even?

10.

Is 5223 divisible by (a) 3? (b) 2? (c) 5? (d) 10?

11.

If 2 even numbers are multiplied together and then the product is multiplied by 37, will the result be even or odd?

12.

Express 56 as the product of prime numbers.

13.

If –3 is multiplied by –345, is the result positive or negative? odd or even?

14.

2(4 + 10) – 6 ÷ 3 =

15.

5[3(2 - 4)] ÷ 15 – 2 =

16.

Use the distributive law to rewrite 3(x + 2y) – 5x(y – 4).

17.

Factor the expression 4xyz – 12 xy + 2yz.

EXERCISE 1 SOLUTIONS 1.

For example, -10, -8, -6, -4, -2.

2.

For example, -2, -1, 0, 1, 2.

3.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

4.

–5

5.

–4

6.

For example, 275.

7.

(a) Find the greatest multiple of 5 that is still less than 99 and find the difference between that number and 99. That multiple is 95, so the remainder is 4. (b) The greatest multiple of 10 that is still less than 12,345,671 is 12,345,670, so the remainder is 1.

8.

21

9.

odd

10.

(a) yes

11.

even

12.

56 = 2 • 2 • 2 • 7

13.

positive, odd

14.

26

15.

–4

16.

The distributive law results in 3x + 6y – 5xy + 20x. Although simplifying algebraic expressions has not yet been discussed, this expression can be simplified to 23x + 6y – 5xy.

17.

2y(2xz – 6x + z)

(b) no

(c) no

(d) no

Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference: Robinson, Adam, and John Katzman. The Princeton Review – Cracking the System; The GRE 1992 Edition. New York: Villard, 1991. 105 – 201.

GRE MATH REVIEW #2 Fractions A fraction is just a shorthand way of expressing a division problem. In other words, 5/2 = 5 ÷ 2. The Numerator is the ‘top’ number in a fraction and the Denominator is the ‘bottom’ number. For example: 5/2, the 5 is the numerator and the 2 is the denominator. When adding and subtracting fractions, the denominators of the fractions must be the same. If the denominators of the fractions are already alike, just add or subtract the numerators and put the result over the common denominator. For example, 11 5 1 11 + 5 − 1 15 + − = = 12 12 12 12 12

If the denominators of the fractions are different, multiply the fraction(s) by the fractional form(s) of the number 1 which makes the denominators the same. For example, 3 5 3  3  5  4  9 20 29 + = + =  +   = 4 3 4  3  3  4  12 12 12 Since 12 is the lowest common denominator of 3 and 4, then we had to multiply by 3/3 and 4/4 to get 12 in both denominators. Since 3/3 and 4/4 both equal 1, we are not changing the value of the fractions. 1 is the only number you can multiply by without changing the values of the fractions. When multiplying fractions, simply multiply the numerators and multiply the denominators; then reduce. Reducing fractions is discussed below. For example, (5/7)*(2/3) = 10/21. When one fraction is multiplied by another fraction, the product is smaller than either of the original fractions. When dividing fractions, invert (put the denominator over the numerator) the second fraction and multiply. For example, (5/7) ÷ (1/3) = (5/7)*(3/1) = (5*3) /(7*1) = 15/7. The GRE may include problems in which the numerator or denominator itself is a fraction such as 5/(2/3). Just remember this is just another notation for division. For example, 5 ÷ (2/3) = 5*(3/2) = (5/1)*(3/2) = 15/2. When adding, subtracting, multiplying, and dividing fractions, you must make sure the final fraction is in reduced form, especially if your final answer does not appear in the answer choices. Consider the example we used when discussing adding and subtracting fractions. The answer we got was 15/12. This answer would probably not appear as one of the answer choices for that question on the GRE because this fraction is not reduced. To reduce a fraction, simply factor the numerator and denominator into prime factors and cancel out like factors: 15/12 = (3*5)/ 2*2*3) = 5/(2 * 2) = 5/4

Another way to tackle this problem is to just divide the numerator and the denominator by the largest factor that is common to both which in this example would be 3. The result would be 5/4. A mixed number is a fractional number greater than 1 which is expressed as a whole 3 number and a fraction such as 5 . In most cases on the GRE, you will need to convert 4 mixed numbers to fractions by multiplying the denominator by the whole number, adding this product to the numerator, and putting this sum above the denominator. For example, 3 23 5 = 4 4 To convert a fraction greater than 1 to a mixed number, divide the denominator into the numerator to get the whole number and put the remainder over the denominator to get 1 the fractional part. For example, 17/8 = 2 8 The GRE may ask you to compare fractions and decide which one is larger or smaller. There are two ways to compare fractions. One way is to find a common denominator and multiply the fractions by the fractional forms of 1 that give them the same denominator. For example, suppose we want to know which fraction is larger, 3/7 or 7/14. Since 14 is a common denominator, multiply 3/7 by 2/2 to get 6/14. Clearly 7/14 is larger than 6/14. Another method for comparing fractions is cross-multiplication. Consider the two fractions in the example above, 3/7 and 7/14. In cross-multiplication, multiply the denominator of each fraction by the numerator of the opposite fraction; then compare the two products as follows. 7 3 14 7 14*3 is 42 and 7*7 is 42. Since 49 is bigger than 42, 7/14 is bigger than 3/7. When crossmultiplying you must be careful to work from bottom to top in the direction of the arrows. Had you multiplied numerator by denominator working from top to bottom, you would have gotten the wrong answer. The following is an example of a GRE question asking you to compare more than two fractions. Example: Which of the following fractions is smaller? 4/7

5/8

8/11

To solve this type of problem, just compare the fractions two at a time. First, compare 4/7 and 5/8 and eliminate the largest of the two. The smallest of these two is 4/7. Now compare 4/7 to 8/11 and eliminate the largest. The smallest fraction is 4/7. One reminder about fractions: if the denominators are alike and the numerators are different, the one with the largest numerator is the largest fraction; if the numerators are alike and the denominators are different, the one with the largest denominator is the

smallest fraction. For example, 11/13 is larger than 11/14. If both the numerators and the denominators are different, you must use one of the methods described above to determine which is larger.

EXERCISE 2 1.

2.

Reduce each of the following fractions: (a) 5/20 (c) 51/39 (b) 18/24 (d) 45/56

(e) (f)

78/48 12/63

Perform the following operations: (a) 5/6 + 9/10 (c) 6/25 ÷ 15/2 (b) (2/3)/6 (d) 5 + 5/6

(e) (f)

15/32 – 7/8 (3/10)/(2/15)

3.

Determine the larger fraction in each pair of the following fractions: (a) 28/53, 27/53 (b) 14/21, 14/22 (c) 7/5, 11/12

4.

Convert the mixed number 5

5.

Convert the fraction 25/7 to a mixed number.

6.

How many halves are in the number 6? How many thirds?

4 to a fraction. 7

EXERCISE 2 SOLUTIONS 1. (a) (b)

1/4 3/4

(c) (d)

17/13 45/56

(e) (f)

13/8 4/21

2. (a) (b)

26/15 1/9

(c) (d)

4/125 35/6

(e) (f)

-13/32 9/4

3. (a)

28/53

(b)

14/21

(c)

7/5

4.

39/7

5.

3

6.

12, 18

4 7

Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference:Robinson, Adam, and John Katzman. The Princeton Review – Cracking the System: The GRE 1992 Edition. New York: Villard, 1991. 105-201.

GRE MATH REVIEW #3 Decimals Decimal numbers are just another way of expressing fractions. Every fraction can be written as a decimal, and every decimal can be written as a fraction. Sometimes it is necessary to convert decimals to fractions because the fractional form of the number may be easier to work with than the decimal form. For instance, it is easier to find the square root of 1/4 than 0.25. However, if the answer choices are in decimal form, work with the decimal form in your computations. In a decimal number, such as 0.357, the first digit to the right of the decimal point, which is 3 in this example, is the tenths digit (i.e. 1/10), the 5 is the 100ths digit (1/100), the 7 is the 1000ths digit (1/1000), etc. To convert a decimal to a fraction, use the decimal number (without the decimal point) as the numerator, and put it over the appropriate denominator (How?). For example, 0.357 = 357/1000. Then reduce if possible. To convert a fraction into its decimal equivalent, divide the numerator by the denominator. For example, .75 3/4 = 3 ÷ 4 = 4 3.00 = 0.75

28 20 20 0 When adding and subtracting decimals, you must realize that all numbers can be written with a decimal point. For example, 12 can be written as 12.0. You must know where the decimal point should go in order to add and subtract decimal numbers. To add and subtract decimals, line up the decimal points in a column and add the digits as you would in any addition problem. If decimal points are missing, put them in at least mentally. You may even want to put zeros in the columns to the right of the decimal point to make the columns line up evenly. For example, 34.5 87 123.456 + 0.98 245.936

34.500 87.000 123.456 + 0.980 245.936

When multiplying decimals, first multiply the numbers as you would integers. Next, count the total number of digits to the right of the decimal points in the two numbers you are multiplying. Finally, count that number of digits from right to left in your product and put the decimal point there. For example:

3.4517 × 80.9 279.24253 Since 3.4517 has four decimal places and 80.9 has one, the product has five decimal places. When dividing decimals, convert the divisor into a whole number by moving the decimal point to the right. Hence, you must move the decimal point in the dividend the same number of places to the right. After performing the division, the decimal point in the quotient goes directly above the decimal point in the dividend. In the division problem 24 ÷ 1.25, 1.25 is the divisor and 24 is the dividend. Before dividing, change the problem to 2400 ÷ 125: 19.2 24/1.25= 2400/125 = 125 2400.0 = 19.2

125 1150 1125 250 250 0 To determine which of two decimal numbers, such as 0.00099 and 0.001, is larger or smaller, (1) vertically align the decimal points, and (2) fill in the missing zeros. For example, 0.00099 = 0.00099 0.001 = 0.00100 Now it is obvious that 100 is larger than 99, so 0.001 is larger than 0.00099. It is often helpful on the GRE to remember that money is based on the decimal system. For example, $.98 is 98/100 of a dollar. Hence, $5.98 means 5 dollars plus 98/100 of a dollar. If you are stuck on a decimal problem, reminding yourself of this fact may help you.

EXERCISE 3 1.

Add 101.054 to 5.12.

2.

Subtract 10.31 from 125.823.

3.

Multply 22.65 by 0.5.

4.

Divide 22.65 by 0.5.

5.

Reduce 5.76/.3.

6.

Add four one-thousandths and three tenths.

7.

Which is larger, .002 or .0015?

8.

Convert 13/2 to a decimal.

9.

Convert .125 to a fraction.

10. Approximate 2.00465/3.98136 without using your pencil.

EXERCISE 3 SOLUTIONS 1. 106.174 2. 115.513 3. 11.325 4. 45.3 5. 19.2 6. 0.304 7. .002 8. 6.5 9. 1/8 10. approximately 1/2

Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference: Robinson, Adam and John Katzman. The Princeton Review - Cracking the System: the GRE 1992 Edition. New York: Villard, 1991. 105 - 201.

GRE MATH REVIEW #4 Percentages A percent is just a shorthand way of expressing a fraction whose denominator is 100. Percent means “per 100”, “out of 100”, or “divided by 100”. For example, 25% = 25/100 = 0.25 and 0.3% = 0.3/100 = 0.003. In terms of money, 50 cents out of a dollar is 50 cents out of 100, which is 50/100 of a dollar or 50% of a dollar. To find a percentage of something, the percents must be converted to decimals and then multiplied by some number. Never directly add and/or subtract percents; you must first multiply them by something. When finding percentages, convert the sentence into a mathematical equation. For example, “5 is what percent of 100” can be converted into “5 = x(100)” where x is the percent you are looking for. When working with percents, it is usually necessary to convert percents to decimals before performing computations with them. Since a percent is just a fraction with denominator 100, you convert a percent to a decimal by moving the decimal point two places from right to left. For example, 6% is equivalent to (.06). In the following example, it is necessary to convert the percent to a decimal. Example 1:

What is 30 percent of 200?

To find 30% of 200, convert 30% to .30. Then multiply 200 by .30, which results in 60. Hence, 60 is 30% of 200. To convert a decimal to a percent, move the decimal point two places from left to right and add a % sign. For example, 0.8 = 80% and 0.02 = 2%. To convert a percent to a fraction, just make the percent the numerator of a fraction with denominator 100 and reduce the fraction. For example, 40% = 40/100 = 4/10 = 2/5. To convert a fraction to a percent, divide the numerator by the denominator and move the decimal point two places to the right. Example 2:

Express 4/5 as a percent.

To do so, use long division and move the decimal point. 5 divided into 4 is 0.8 which is equivalent to 80%. Hence, 4/5 = 0.8 = 80%. The following is a list of percents and their fraction and decimal equivalents which should be committed to memory: 0.01= 0.1= 0.2= 0.25=

1/100= 1/10= 1/5= 1/4=

1% 10% 20% 25%

0.5= 0.75=

1/2= 3/4=

50% 75%

Ratios Ratios, like fractions, decimals, and percents, are just another way of expressing division. Every fraction is a ratio and every ratio is a fraction. A fraction is just the ratio of the numerator to the denominator. The ratio 1: 2 (read “1 to 2”) is equivalent to the fraction 1/2 or the decimal 0.5 or 50% or just 1 divided by 2. On the GRE, ratios may be expressed in any of the following ways: (1) (2) (3) (4)

x/y the ratio of x to y x is to y x:y

Anytime you see a ratio, treat it just like a fraction. Anything you can do to a fraction, you can also do to a ratio, including cross-multiplying, reducing, finding common denominators, etc. Here are some examples of how to work ratio problems on the GRE. Example 3: If you have 3 coins and the ratio of pennies to nickels is 2 : 1, how many pennies and how many nickels are there? Clearly, there are 2 pennies and 1 nickel since you only have 3 coins altogether and there are 3 parts in the ratio (2 and 1 = 3). Example 4: If you have 24 coins and the ratio of pennies to nickels is 2 : 1, how many of each type of coin do you have? Since the ratio again contains 3 parts (2 and 1), divide 24 by 3 to get the number of coins in each part; then multiply each part of the ratio by the result . Dividing 24 by 3 yields 8 which means each of the 3 parts in the ratio consists of 8 coins. Two of the parts are pennies at 8 coins per part, so there are 16 pennies. One part is nickels which makes 8 nickels. Example 5: At a camp for boys and girls, the ratio of girls to boys is 5 : 3. If the camp’s enrollment is 160, how many of the children are boys? (a) 20

(b) 36

(c) 45

(d) 60

(e) 100

A ratio of 5 : 3 means 8 total parts. To find out how many children are in each part, divide the total enrollment by the total number of parts. (160 divided by 8 is 20); that means each part is 20 children. Three parts are boys and 3 multiplied by 20 is 60 which is the answer (d). Note that there is no need to find out how many girls there are. Also, be careful and use the correct part of the ratio that answers the question.

Proportions The GRE often contains questions in which you must compare two ratios which are proportional. These questions take a given ratio, or relationship, and project it onto a larger or smaller scale while leaving out one piece of information. Example 6:

If 10 baskets contain a total of 50 eggs, how many eggs would 7 baskets contain?

(a) 10

(b) 17

(c) 35

(d) 40

(e) 50

To solve a problem like this, set up the two proportional ratios, one of which will have a missing piece of information. Think of the ratios like this: 10 baskets is to 50 eggs as 7 baskets is to x eggs. Then set up these ratios as proportional fractions and cross-multiply: 10 50

7 × 10x = 350 x = 35

OR

10 : 50 7:x 10x = 350 x = 35

Hence, the answer is (c). Note that we could have reduced the fraction 10/50 to 1/5 and made the cross-multiplication easier. Averages The average, or arithmetic mean, of a set of numbers is the sum of all the numbers in the set divided by the total number of numbers in the set. The formula to remember is average = the sum of the numbers being averaged the number of elements For example, the average of the numbers 1, 2, 3, 4, 5, 6, 7 is 28 divided by 7 which is 4. The GRE always refers to an average as an “average (arithmetic mean)”. Just ignore the parenthetical remark so it doesn’t confuse you. In an averaging problem, you may be asked to find the total first. For example, suppose a problem states that the average of 4 test scores is 80 find the sum of the tests. Recall the formula above. In this case we already know the average and the number of elements; we need to find the sum of the tests. Hence, just cross-multiply and the total of the test scores is the product of the two: 80 × 4 = 320. Suppose you are told that two of these scores are 90 and 95 and you want to find the average of the other two scores. The sum of 90 and 95 is 185. So, the total of the other two scores is 320 – 185 = 135. Hence, the average of the remaining two scores is 135 divided by 2 which is (67.5).

Do not get confused if two or more of the elements being averaged are the same. For example, the average of 5, 5, 5, and 20 is 5 + 5 + 5 + 20 divided by 4 which is 35 divided by 4 which is (8.75). You do not add 5 and 20 and divide by 2, nor do you add 5 and 20 and divide by 4. Another situation which may confuse you is when a new element is added to a set that has already been averaged. Suppose you take two tests and earn scores of 70 and 80. The average of your two tests is 75. Now, suppose you take a third test and earn another score of 70. Does your average remain 75? No. Your new average is (70 + 80 + 70)/3 = 73.33. Note that your new average is NOT (75 + 70)/2 = 72.5. Suppose your third test score was 75. Then your average over all three tests is still 75 (Why?). Another common error occurs when averaging a set of numbers that includes 0. For example, what is the average of 0, 0, 0, and 4? A careless person would say 4, but the answer is 1! An easy averaging problem can be made difficult on the GRE if certain information is left out. Example 7: The average test score earned by a group of students is 80. If 40% of the students have an average score of 70, what is the average score of the remaining 60%? (a) 70.33

(b) 80

(c) 86.67

(d) 90

(e) 95

An important piece of information is missing from this question. Do not try to solve this problem by setting up a complex algebraic equation. Since the problem is dealing with percentages, the actual number in the group is irrelevant. Just pick a number that’s easy to work with. Since all of our percents are multiples of 10, choose 10 as the number of students in the group. 40%, or 4, of these students have an average score of 70. We want to know the average score of the remaining 6 (60% of 10). Now that we’ve gotten rid of the percents, it is just an averaging problem. So first find the totals. If the average score of 10 students is 80, then the total sum of their test scores is 800. Just remember the averaging formula and cross-multiply the average and the number of elements. Since 4 of the students have an average score of 70, their total score is 280. Again, just cross-multiply. To find the total score of the remaining 6 students, just subtract: 800 – 280 = 520. Hence, the average score must be 520 divided by 6 which is 86.67 or choice (c).

EXERCISE 4 1. What is 40% of 350? 2. 2 is what percent of 16? 3. 2.2 is 20% of what number? 4. 10% of 24 equals 20% of what number? 5. 20% of 25% of x is 10. What is x? 6. The following chart illustrates fraction, decimal, and percent equivalents. Fill in the blanks.

Fraction 1/2 1/3

Decimal 0.5

Percent 50%

Ratio 1:2 2:3

25% 0.75 1/5 40% 0.6 4/5 1:6 12.5% 7. What is the average of the numbers 24, 24, 26, 28, and 40? 8. If the average of 5 numbers is 20, what is their total? 9. If the average of 5 numbers is 20, what is the largest that any of the numbers could be? 10. If the average of 11, 17, 15, 28, and x is 19.6, what is the value of x? 11. Jim’s average score on 4 math tests was 80 out of a possible 100. If his scores on 2 of the tests were 65 and 70, what is the lowest that either of his other scores could have been?

EXERCISE 4 SOLUTIONS 1. 140

7. 28.4

2. 12.5%

8. 100

3. 11

9. 100

4. 12

10. 27

5. 200

11. 85

6. Fraction 1/2 1/3 2/3 1/4 3/4 1/5 2/5 3/5 4/5 1/6 1/8

Decimal 0.5 0.3333 0.6666 0.25 0.75 0.2 0.4 0.6 0.8 0.1666 0.125

Percent 50% 33.33% 66.66% 25% 75% 20% 40% 60% 80% 16.67% 12.5%

Ratio 1:2 1:3 2:3 1:4 3:4 1:5 2:5 3:5 4:5 1:6 1:8

Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference: Robinson, Adam and John Katzman. The Princeton Review – Cracking the System: The GRE 1992 Edition. New York: Villard, 1991. 105 – 201.

GRE MATH REVIEW #5 Exponents and Radicals Many numbers can be expressed as the product of a number multiplied by itself a number of times. For example, 16 can be expressed as 2 ⋅ 2 ⋅ 2 ⋅ 2 . Another way to write this is 24. The 4 is called the exponent and the 2 is called the base. The expression 24 is read “2 to the fourth power”. Here is a list of some definitions you need to know in order to follow this discussion of algebra. However, you will not need to know the definitions on the GRE. 1. Variable: A letter that represents an unknown number. 2. Term: A product of any combination of variables and numbers. For example, 2x, 3xy2, 10y3, etc. are all terms. 3. Coefficient: A number or variable in a term. For example, in the term 2xy2, 2 is a coefficient of xy2, x is a coefficient of 2y2, y2 is a coefficient of 2x, etc. 4. Expression: Any number of terms combined by addition or subtraction signs. For example, 3xyz + 6x2y – 10z is an algebraic expression. 5. Equation: Two expressions or terms set equal to each other. Do not get an expression and an equation confused. An equation has an equal sign in it; and an expression does not. An easy way to remember the definitions of term, equation, and expression is to think of an equation as the algebraic equivalent of a sentence and an expression as the algebraic equivalent of a phrase. Then a term is just one word in the sentence or phrase. When multiplying terms with the same base, just add the exponents. For example, x5 • x 3 = x8. However, x5 • y3 ≠ (xy)8 because x5 and y3 do not have the same base. Do not make the mistake of adding the exponents when multiplying numbers with different bases. Another common error is adding the exponents when adding two terms with like bases. This rule does not apply to addition. For example, 24 + 22 ≠ 26. When dividing terms with the same base, just subtract the exponents. For example, 35 ÷ 33 = 32. Again, this rule does not apply to division of terms with different bases or to subtraction of terms with like bases. For example, x5 ÷ y3 ≠ (x/y)2and 24 – 22 ≠ 22. When adding or subtracting two or more terms with exponents the terms must have like bases and like exponents. If the bases and exponents are just alike, simply add the

numerical coefficients. Remember that if there is no numerical coefficient, it is understood to be 1. For example, x2 + 5x2 – 2x2 = 4x2. When raising an exponent to another exponent, simply multiply the exponents. For example, (x5)2 = x10. If there is a term with several coefficients in the parentheses, you must distribute the exponent to every coefficient. For example, (2xy2)3 = 8x3y6 and (3/2)2 = 9/4. This rule does not apply if there is an addition or subtraction sign inside the parentheses. For example, (2x + 3y)2 ≠ 4x2 + 9y2 , this is a very common mistake, so be careful not to make it! The following is a list of characteristics of exponents that you should commit to memory for the GRE: 1.

Any number raised to 0 is always 1. For example, 50 = 1, x0 = 1, etc.

2.

Any number without an exponent is understood to have an exponent of 1. For example, x = x1, 2 = 21, etc.

3.

Raising a number greater than 1 to a power greater than 1 results in a bigger number. For example, 22 = 4.

4.

Raising a fraction between 0 and 1to a power greater than 1 results in a smaller fraction. For example, (1/2)2 = 1/4. Recall from our discussion on fractions that we said multiplying a fraction by another fraction results in a smaller fraction. Raising a number to a power is equivalent to multiplying that number by itself.

5.

A negative number raised to an even power results in a positive number. For example, (-3)2 = 9.

6.

A negative number raised to an odd power results in a negative number. For example, (-2)3 = -8.

Although negative exponents, such as 2-3, are a very important concept in your algebra class, you will NOT see them on the GRE, so don’t worry about them. You should have a feel for the relative size of exponents. Remember that they are just shorthand notation for multiplication. So, 25 is twice as large as 24. And 210 is more than 10 times as large as 102 (Why?).

The radical sign

indicates the square root of the number under the radical. Similarly,

the sign 3 indicates the cube root of the number under the radical. If x2 = 16, then x = + 4 and + 4 are called the positive and negative square roots of the number 16. However, 16 = +4. In other words, the radical sign only refers to the positive root of the number under the radical. Hence, if the GRE asks for 25 , the answer is +5, not –5. Unlike for an algebra class, there are only two radical rules you need to know for the GRE: 1.

x

y=

xy

For example,

2.

x y

=

2

3=

6 and

5 16

=

5 16

32 = 16

2= 4 2

x y

For example,

=

5 and 4

32 = 2

32 = 16 = 4 2

The following values should be committed to memory for the GRE in order to be able to work radical problems more quickly:

1 2 3 4 5

= = = = =

1 1.4 1.7 2 2.2

Algebra You are required to know very little real algebra for the GRE. Algebra methods learned in your algebra classes will often mislead you and will usually take up way too much time. For instance, never try to set up an algebraic and work through to an answer. The GRE only cares which space you blacken on the answer sheet, not how you work an algebra problem. In this section, we will discuss the small subset of algebra rules that you actually need to know to do well on the GRE.

There are a few more definitions that you will need to know in order to follow this section. However, again, you do not need to know these for the GRE. 1. Binomial:

An algebraic expression containing 2 terms.

2. Trinomial:

An algebraic expression containing 3 terms.

3. Polynomial:

A binomial, trinomial, or any other algebraic expression containing two or more terms.

The following are some helpful hints to remember for the GRE. See Review #1 if you need to review factoring and “unfactoring” (i.e., the distributive laws). 1. When you encounter a problem containing an expression that can be factored, you should always factor that expression. For example, if you see an expression such as 4x + 4y, you should immediately factor it into 4(x + y). 2. Similarly, whenever you see an expression that has been factored, you should immediately “unfactor” it, i.e. “multiply it out”. For example, if a problem contains the expression 4(x + y), multiply it out, to get 4x + 4y. When multiplying polynomials, remember the distributive law and multiply every term in the first polynomial by every term in the second polynomial (FOIL). For example, (x + 4)(2x – 1) = x(2x) + x(-1) + 4(2x) + 4(-1) = 2x2+ (-x) + 8x + (-4) = 2x2 + 7x – 4 There are three expressions that you need to commit to memory for the GRE in both their factored and unfactored forms. They are: 1.

x2 – y2 = (x + y)(x – y) x2 – y2 is the unfactored form; (x + y)(x – y) is the factored form.

2.

x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2 x2 + 2xy + y2 is the unfactored form; (x + y)(x + y) and (x + y)2 are equivalent factored forms. x2 – 2xy + y2 = (x – y)(x – y) = (x – y)2 x2 – 2xy + y2 is the unfactored form; (x – y)(x – y) and (x – y)2 are equivalent factored forms.

3.

Here is an example of a problem in which you need to factor and know one of these expressions. Example: Simplify the following expression: 4x2 – 4

x–1 Your should immediately recognize that you need to factor the numerator which results in the expression 4(x2 – 1) for the numerator. Next, you should recognize that x2 – 1 is of the form x2 – y2 and therefore can be written in its factored form (x – 1)(x + 1). Hence, the original fraction can be written as 4(x + 1)(x – 1) (x – 1) Now you should recognize that the fraction can be reduced by canceling out the common factor (x – 1) in the numerator and the denominator. The final simplified, factored form is 4(x + 1). Whenever you see a complicated-looking algebraic expression, simplify it if possible by combining similar terms, i.e. terms with like bases and like exponents. For example, (4x2 + 4x + 2) + (3 – 7x) – (5 – 3x) = 4x2 + 4x + 2 + 3 – 7x – 5 + 3x = 4x2 + (4x – 7x + 3x) + (2 + 3 – 5) = 4x2 When given two simultaneous equations on the GRE and asked to solve them, don’t solve them using the techniques you learned in algebra class. Instead, look for shortcuts which usually involve adding or subtracting the two equations. And you’ll never have to worry about solving a system with more than two equations. Let’s see an example. If 5x + 4y = 6 and 4x + 3y = 5, then x + y =? Instead of using substitution or elimination like you learned in algebra, just add or subtract the two equations. First, let’s add them and see what we get. Remember, we are looking for x + y, so we really don’t need to know what x and y are individually. 5x + 4y = 6 + 4x + 3y = 5 9x + 7y = 11 Obviously, that didn’t help us much. So let’s subtract them.

5x + 4y = 6 -(4x + 3y = 5) x+y=1 Since we’re looking for the value of the expression x + y, the answer would be 1. We never even had to find out the values of x and y. Whenever you encounter simultaneous equations, try adding or subtracting them, factoring something out, or multiplying by

something. You will never need to use the methods you learned in your algebra classes to solve simultaneous equations on the GRE. The following is an example of the type of problem on the GRE that involves factoring and the distributive law. Example: If y + 3 = 2x, then 3y – 6x = (a) –9

(b) –3

(c) 0

(d) 3

(e) 9

Instead of solving for y and plugging –3 + 2x in for y in the second expression, just factor out a three in the second expression: 3(y – 2x). This is just the distributive law working in reverse. Notice that the first equation can be written y – 2x = -3 (why?). By substituting –3 in for y – 2x, we get 3(-3) = -9, so the answer is (a). Always be on the lookout for chances to factor and use the distributive laws. The GRE loves equations set equal to zero because of the unique properties of 0. One of the most important properties of 0, which was mentioned in Review #1, is the fact that the product of anything and 0 is 0. Hence, if a product is equal to 0, one of the factors in that product must be 0. In other words, if ab = 0, then either a or b must be 0 or both are 0. This fact can be used to solve some equations on the GRE. Here’s an example. What are all the values of y for which y(y + 5) = 0? In order for the product y(y +5) to equal 0, either y must be 0 or y + 5 must be 0 or both of them must be 0. In order for y +5 to be 0, y would have to be –5. Hence, the values of y for which y(y + 5) is 0 are y = 0 and y = -5. In an equation, one expression is equal to another. In an inequality, one expression is not equal to another expression. (See Review #1, page 3, for a list of symbols and their meanings.) However, inequalities are solved just like equations. You can factor, unfactor, simplify, multiply/divide both sides by a constant, add/subtract terms from both sides, etc. The one primary difference is that if you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality symbol. It’s easy to see why. For instance, we know that 2 < 4. Multiplying both sides by –2 results in –4 on the left side and –8 on the right side. Clearly, -4 is greater than –8, so the inequality symbol must be reversed: -4 > -8. From your algebra classes, you probably remember hearing about functions even if didn’t you understand them. Usually, the symbol f(x), read “f of x”, was used to represent a function. The GRE contains function problems, but instead of using the f(x), they are disguised by funny-looking symbols such as #, *, @, etc. If you remember how functions work, just think of functions when you work these problems and perform the operations. However, you can still work these problems even if you don’t remember functions. Think of a funny-looking symbol as representing a set of operations or instructions. Here’s an example: If x @ y = (x – y )/2, what is the value of 3 @ 5? To find the answer, just substitute 3 in for x and 5 in for y. Since x @ y = (x – y)/2, then 3 @ 5 = (3 – 5)/2 = -1.

The GRE also contains your favorite type of problems: word problems. You have to learn how to translate them into mathematical equations. The following is list of words found in word problems and their mathematical translations. These are the same translations that are used in percentage problems (see Review #4). Word is of, times, product what (or the unknown value) more, sum less, difference ratio, quotient

Symbol = × any variable + ÷

The following formulas frequently appear in GRE word problems, so you should commit them to memory: 1. distance = (rate)(time) This formula can also be expressed as rate = distance/time or time = distance/rate. 2. total price = (number of items)(cost per item) 3. sale price = (original price) – (% discount)(original price)

Exercise 5 1. 34 • 32 = ? 2. If x = 3, what is (2x)3? 3. If x = 4, what is (x2)3? 4. If 42 + 32 = x2, what is x? 5. Approximate 3 3 . 6. Simplify:

(a)

56

(b) 14 / 98

7. (3x + 4)(8x – 3) = ? 8. Factor or unfactor the following expressions: (a) (b) (c) (d) (e)

4x2 – 9y2 (2x + 3y)2 16x2 – 24xy + 9y2 (4x + y)(4x – y) x2 – 10x + 9

9. Simplify the following expression by combining like terms: (6x2 + 7x –7) – 2x(3x2y + 3x +2) + 9 10. If x – y = 7 and –x + 2y = 3, then y = ? 11. If 4ab = 0 and a > 1, then b = ? 12. If x + 3 < 2x + 4, then x > ? 13. If x & y = x2 + 3y, what is the value of x2 & 3y?

EXERCISE 5 SOLUTIONS 1. 36 = 729 2. 63 = 216 3. 163 = 4096 4. +5 or –5 5. Approximately 5 6. (a) 2 14

(b)

7 /7

7. 24x2 + 23x –12 8.

(a) (b) (c) (d) (e)

(2x + 3y)(2x – 3y) 4x2 + 12xy + 9y2 (4x – 3y)(4x – 3y) 16x2 – y2 (x – 9)(x – 1)

9. 6x2 + 7x – 7 – 6x3y – 6x2 – 4x + 9 = -6x3y + 3x + 2 10. 10 11. 0 12. –1 13. x4 + 9y

Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference: Robinson, Adam, and John Katzman. The Princeton Review – Cracking the System: The GRE 1992 Edition. New York: Villard, 1991.105 – 201.

GRE MATH REVIEW #6 Geometry As in the case of algebra, you don’t need to know much of the actual geometry you learned in your geometry class for the GRE. Here is a list of facts about degrees and angles that you should know for the GRE: 1. A circle contains 360 degrees (360o). 2. A line is a 180o angle, i.e. a perfectly flat angle. 3. When two lines intersect, four angles are formed, and the sum of these angles is 3600. 4. When two lines are perpendicular to each other, their intersection forms four 90o angles. The symbol used on the GRE to indicate perpendicularity is ⊥ . 5. Ninety-degree angles are also called right angles. A right angle on the GRE is identified by the following symbol:

6. When two parallel lines, indicated by the symbol // , are cut by a third line, angles that look equal are equal. This is easier to remember than the definitions of corresponding exterior and interior angles that you learned in geometry class. However, be careful when applying this rule since diagrams in the GRE are often not drawn to scale. The following diagram is an example to which this rule applies.

xo

xo

7. When two lines intersect, the angles across from each other, which are called vertical angles, are equal. See the diagram below.

x a

b y

x=y a=b

a + b + x + y = 360o

There are also many facts about triangles that you need to know for the GRE. For instance, every triangle contains three interior angles which add up to 180o no matter what the shape of the triangle. For example:

b

b b

a

c

a + b + c = 180o

a

c a + b + c = 180o

a

c a + b + c = 1800

An equilateral triangle is one in which all three sides are equal in length. Since the sides are equal, all the angles are equal too. Hence, each angle in an equilateral triangle is 60o. The following is a diagram of an equilateral triangle: 60

60

60

An isosceles triangle is one in which two of the three sides are equal in length. Hence, the two angles opposite these sides are also equal in length. If you know the degrees of any angle in an isosceles triangle, you can easily figure out the degrees of the other two. For example, if one of the two equal angles equals 40o, then the other angle is also 40 degrees. Hence, the third angle is 180o – 80o = 100o degrees. This same strategy also applies to the length of the sides. The following is a diagram of an isosceles triangle. Remember though: you can’t just assume two sides or angles are equal because they look equal except in the case described in #6 above. Be careful; the diagrams on the GRE are NOT always drawn to scale.

A a

b B

AB = AC b=c

c C

A right triangle is a triangle in which one of the angles is a right angle, i.e. a 90o angle. The symbol mentioned earlier will be used to indicate the right angle in a right triangle. The longest side of a right triangle, which is the side opposite the 90-degree angle, is called the hypotenuse.

In any triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. If two interior angles are equal, the sides opposite them are also equal as in the case of an isosceles triangle and an equilateral triangle.

A

a

c=b AB = AC

b

c

B

C

The perimeter of a triangle is the sum of the lengths of the sides of the triangle.

10

4

12

perimeter = 26

The area of a triangle is the product of the base and the height of the triangle multiplied by 1/2: area = (1/2) • base • height This formula is for the area of any triangle even though the height of some triangles may not be immediately obvious. For example:

4

4

4

6

6

area = 4 × 6 = 12 2

area = 4 × 6 = 12 2

6 area = 4 × 6 = 12 2

The Pythagorean theorem is probably the most difficult geometry concept on the GRE, but you must know it. The Pythagorean theorem applies only to right triangles. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the triangle below, c2 = a2 + b2.

b

c

a Pythagorean problems on the GRE often involve right triangles with side lengths 3, 4, and 5 or multiples of these numbers. This particular triangle is the smallest one in which the side lengths are all integers. Here are three examples of 3-4-5 right triangles that you could see on the GRE. Learn to recognize multiples of the lengths 3, 4, and 5.

10 4

5

8

3

1

6

5/4

3/4

The Pythagorean theorem can also be used to solve other problems in which the application isn’t as obvious. For example, every square or rectangle is made up of two right triangles. Hence, if you know the length and width of any rectangle or square, you can find the length of the diagonal by using the Pythagorean theorem. For example, if the lengths of the sides of the rectangle below are 3 and 4, then using the Pythagorean theorem, 32 + 42 = 9 + 16 = 25. Hence, the length of the diagonal is 5.

3

4

There are two special right triangles that you will see on the GRE. You need to commit these triangles and their characteristics to memory.

1.

The first of these triangles is called the 30:60:90 right triangle, referring to the degree measurements of the angles: 60o

30o 5 The ratio between the lengths of the sides in a 30:60:90 degree triangle is constant. So, if you know the length of any side, you can find the lengths of the other two sides. The constant ratio is 1:2: 3 . In other words, if the shortest side has length 5, then the hypotenuse has length 10 and the other side has length 5 3 . Here are two examples of 30:60:90 triangles:

4

8

4 3

2.

4

24

12

12 3

The second special right triangle is the 45:45:90 right triangle in which the two sides opposite the hypotenuse are equal. The ratio between the length of either of these sides and the length of the hypotenuse is 1: 2 . In other words , if the length of each short leg is 5 , then the length of the hypotenuse is 5 2 . Here are two examples: 4 2

4

2

2

2 2

In algebra class, you learned that the value of pi ( π ) is 3.14 or 3.14159. However, on the GRE you can approximate π as 3. Three is a close enough approximation to find the correct answer on the GRE and it is much easier to work with than 3.14. Some problems on the GRE involving π can be solved simply by plugging in 3 for each π in the answer choices and comparing the results. The circumference of a circle is like the perimeter of a triangle: it is the distance around the outside. The circumference of a circle is the product of pi and the radius r multiplied by 2 or pi times the diameter d: circumference = 2 π r = π d You also need to remember the relationship that π expressed. Pi is the ratio between the circumference of a circle and its diameter. Since π is approximately 3, then every circle is approximately three times as far around as it is across. If the diameter of a circle is 4, then its radius is 2 and its circumference is 4 π , or approximately 12. If the circumference is 10, then its diameter is 10/ π , or a little more than 3, and its radius is 5/ π , or a little more than 1.5.

5 •

circumference = 2 π r = 2 π 5 = 2 × 3 × 5 = 30 The area of a circle is the product of pi and the square of the radius: area = π r2 For example, if the radius of a circle is 5, then the area is approximately 3 times 25 or approximately 75: 5 •

area = π r2 = π × 25 = 3 × 25 = 75

The perimeter of a rectangle is just like the perimeter of a triangle: the sum of the lengths of the sides. For example:

8

4

4

8 perimeter = 4 + 8 + 4 + 8 = 24 The area of rectangle is the length of the rectangle times its width: area = length • width. In the example just above, the area is (8)(4) = 32. A square is just a rectangle with 4 equal sides, s. Therefore, the perimeter is just 4 times the length of a side, i.e. perimeter = 4s. The area of a square is just the length of a side s times itself or the square of the length of a side, i.e. area = s2. The only geometric figure for which you will have to calculate the volume is a rectangular solid, or a box. The volume of a box is just the length times the width times the height: volume = length • width • height You will also need to be familiar with the Cartesian coordinate system shown below. The horizontal axis is the x-axis; the vertical axis is the y-axis. The four areas formed by these axes are called quadrants. The point where the two axes intersect is called the origin. 5 4 3 2 1

•B

y

•A

1 2 3 4 5 6

-6 -5 -4 -3 -2 -1

•C

-1 -2 -3 -4 -5

x

The Cartesian coordinate system is a method of describing the location of any point on the plane formed by the two axes. In the diagram above, the point A can be described by the coordinates (2, 4) where 2 is the distance to the right of the origin on the x-axis and 4 is the distance above the origin on the y-axis. The coordinates (-6, 1) describe the point B which is 6 spaces to the left of the origin on the x-axis and 1 space above the origin on the y-axis.

EXERCISE 6 1. In the figure below, if l 1 // l 2 , what is the measure of angle b? 40o

11

12

b

2. In the triangle below, what is the measure of angle c? 25o 30o

c

3. In the triangle below, what is the length of side AB? A 60O 60O

60O

C

B

4. In the triangle below, what is the measure of angle c? b

a

c

5. What are the perimeters of the triangle and rectangle below?

3

3 5

3 4

6. What are the areas of the triangle and rectangle below?

8

4

4

6

7. What is the length of the third side in the right triangle below?

6

8 8. What is the approximate circumference of the circle below?

•

3

9. What is the approximate area of the circle in problem 8? 10. If a box is 5 inches wide, 10 inches long, and 4 inches deep, what is its volume in cubic inches? 11. Determine the coordinates of points A, B, C, and D in the graph below.

y 5 4 3 2 1

D • -6 -5 -4 -3 -2 -1

•C

•A

1 2 3 4 5 6 -1 -2 -3 -4 -5

x •B

EXERCISE 6 SOLUTIONS 1. b = 140o 2. c = 125o 3. AB = 20 4 c = 90o 5. triangle perimeter = 11 rectangle perimeter = 14 6. triangle area = 16 rectangle area = 24 7. 10 8. approximately 18 9. approximately 27 10. volume = 200 cubic inches 11. A = (3, 4) B = (4, -4) C = (-6, -5) D = (-4, 0)

Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference: Robinson, Adam and John Katzman. The Princeton Review – Cracking the System: The GRE 1992 Edition. New York: Villard, 1991. 105 – 201.

GRE MATH REVIEW #7 Specific Techniques for Attacking Arithmetic, Algebra, and Geometry Problems on the GRE Now that we have reviewed the basic arithmetic, algebra, and geometry required for the GRE, we can discuss the specific techniques used to attack these types of problems. Following these suggestions rather than your algebra class methods will make you much more efficient on the GRE. Many people find it easier to improve their math scores than their verbal scores just by practicing and trusting these techniques that follow. Before taking the GRE, make sure you are familiar with the instructions for each type of math question. Do not waste any of your test time reading the instructions or the examples. Here is a list of information contained in the math instructions on the GRE. Once you understand what it means here, do not bother reading it during the test. 1. All numbers used are real numbers. 2. Position of points, angles, regions, etc., can be assumed to be in the order shown; angle measures can be assumed to be positive. These two statements are for math whizzes who try to argue about their answers. Don’t worry about them. 3. Lines shown as straight can be assumed to be straight. 4. Figures can be assumed to lie in a plane unless otherwise indicated. 5. Figures accompanying questions are intended to provide information useful in answering the questions. However, unless a note states that a figure is drawn to scale, do NOT solve these problems by estimating sizes by sight or measurement, but by your knowledge of mathematics. This information simply means you can’t look at a drawing and figure out if one side of a triangle is longer than another or about how large an angle is. Everything about drawings on the GRE is accurate except the dimensions. Example 1: How many different positions can there be for a square that must have corners at both (0, 1) and (0, 0)? (a) One

(b) Two

(c) Three

(d) Four

(e) Five

Solution: The most obvious, easy solution to this question is the two squares shown below with corners at each of the given points.

This question would be at the difficult level on the GRE; therefore, since two is the first answer that comes to mind for an average person, eliminate choice (b). We can also eliminate (a) since we’ve already found two squares. So, we’ve improved our odds of choosing the correct answer to 1 in 3, and this question is the difficult one of its section. Only about 8% are expected to answer this question correctly! Now, we need to look for another way in which a square could have corners at the two given points. In the diagram below, we see that there is one more square, so the answer is three. Careless students might choose four by assuming there could be a square in each quadrant.

Average students like to find answers by using easy arithmetic that they understand. On difficult questions, these choices can be eliminated. Example 2: A suit is selling for $100 after a 20% discount. What was the original selling price? (a) $200

(b) $125

(c) $120

(d) $80

(e) $75

Solution: A careless student will choose (c) since $120 is 20% more than $100. But that’s not what the question asked. So, eliminate choice (c). You can also eliminate $80

since that is 20% less than $100. A careless person will choose (c) and (d) first since those answers are arrived at by easy manipulation of the numbers in the problem. Choice (e) can be eliminated because we’re looking for an amount greater than $100. In addition, $200 is much too high since 20% of $200 is $40. Hence, the answer must be (b). A careless student also likes answers that remind him/her of the problem. On difficult questions, do not be attracted to answers that simply repeat the numbers in the problem or are easily derived from the numbers in the problem. So, eliminate these types of answer choices. Example 3: After 6 gallons of water are transferred from container A to container B, there are 10 gallons more water in container A than in container B. Container A originally had how many more gallons than container B? (a) 0

(b) 6

(c) 10

(d) 16

(e) 22

Solution: Careless students are immediately attracted to choices (b) and (c) because they repeat numbers that are in the problem. So, eliminate these two choices. Eliminate (d) as well since it is easily derived from numbers in the problem: 6 + 10 = 16. Obviously, choice (a) makes no sense, so the answer is (e). On some difficult questions, you will be asked to find the least or greatest possible value that satisfies certain conditions. On such problems, immediately eliminate the least or greatest value among the choices. When “It cannot be determined from the information given” is an answer choice on easy or medium questions, it could possibly be the correct answer. However, when this is an answer choice on difficult questions, it is almost never the answer. Keep in mind that we are not talking about quantitative comparison questions where this is an answer choice on every question! Working backward is a powerful technique on the GRE although it is not recommended in an algebra class. Instead of setting up an equation and solving it, simply plug in the answer choices until you find the one that works. At most, you will have to try four of the choices, but you can usually find the answer in one or two tries. This technique always works when all of the answer choices are numbers. Example 4: Which of the following values of x does not satisfy: 5x – 3 < 3x + 5? (a) –2

(b) 0

(c) 2

(d) 3

(e) 4

Solution: When the answer choices are numbers, start plugging in with the middle answer and work outward because the answer choices on the GRE are usually arranged in order of size. By starting in the middle, you may save time by eliminating choices that are

too large or too small. If you are quick at solving an inequality in your head and it is not a difficult question, it may be faster for you to do so. But be careful. Otherwise, plug in: (c) 5(2) – 3 < 3(2) + 5, or 7 < 11. True? Yes. Eliminate. (d) 5(3) – 3 < 3(3) + 5, or 12 < 14. True? Yes. Eliminate. (e) 5(4) – 3 < 3(4) + 5, or 17 < 17. True? No. The answer. Example 5: The units digit of a 2- digit number is 3 times the tens digit. If the digits are reversed, the resulting number is 36 more than the original number. What is the original number? (a) 13

(b) 26

(c) 36

(d) 62

(e) 93

Solution: There are ninety 2-digit numbers. Eighty-five of these have already been eliminated because they are not answer choices. We need to eliminate four more by trying out the answer choices. Don’t try to set up some kind of complicated equation; it would take too much time. The two conditions that the answer must satisfy are (1) the ones digit is 3 times the tens digit, and (2) the number resulting from reversing the digits must be 36 more than the original number. If any of the answer choices fails to satisfy either of these conditions, we can eliminate it. Just consider one condition at a time.

(c) Is six 3 times 3? No. Eliminate. (b) Is six 3 times 2? Yes. A possible answer. (d) Is two 3 times 6? No. Eliminate. (a) Is three 3 times 1? Yes. A possible answer. (e) Is three 3 times 9? No. Eliminate. So, the only two possibilities are (a) and (b). Now try the second condition. (a) Is 31 = 13 + 36? No. Eliminate. Hence, the correct answer is (b) since 62 = 26 + 36. The technique of working backward from the answers can also be used on word problems instead of setting up an equation. On the GRE, you must be aware of extra information in a problem that makes the problem a little more difficult. Example 6: A restaurant owner sold 2 dishes to each of his customers at $4 per dish. At the end of the day he had taken in $180, which included $20 in tips. How many customers did he serve? (a) 18

(b) 20

(c) 22

(d) 40

(e) 44

Solution: The information about tips is just extra information to make the problem a little more difficult. Subtract the $20 in tips from the daily total to find out how much was

earned on just the dishes. Since each customer bought two 4-dollar dishes, each customer spent $8 on food. The day’s total, $160, divided by $8 is 20 people, answer choice (b). If a problem has variables in the answer choices instead of just numbers, you can often find the answer by making up numbers and plugging them in. (1) First, pick a different number for each variable in the problem. (2) Solve the problem using your numbers. (3) Plug your numbers into the answer choices to see which one equals the solution in step 2. (4) Plug in working from the outside in, i.e. a, e, b, d, c, in that order. Example 7: If x + y = z and x = y, then all of the following are true EXCEPT (a) (b) (c) (d) (e)

2x + 2y = 2z x–y=0 x–z=y–z x = z/2 x – y = 2z

Solution: If you are pretty comfortable with algebra, you should recognize immediately that the first three are true. In that case, only use the following procedure on (d) and (e). Pick values for x, y, and z that are consistent with the two given equations. Since x and y are equal, let’s pick them both to be 2. Since the sum of x and y is z, z must be 4. Remember we are looking for the answer choice which is false. (a) 2(2) + 2(2) = 2(4), or 4 + 4 = 8. True. Eliminate. (e) 2 –2 = 2(4), or 0 = 8. False. If you have enough time, you may want to verify that the other answer choices are true just to make sure you didn’t make a mistake. 2 – 2 = 0, or 0 = 0. True. (d) 2 = 4/2, or 2 = 2. True. (b) 2 – 4 = 2 –4, or –2 = -2. True. When plugging numbers into a problem, choose numbers that are easy to work with. This depends, however, on the problem. Usually small numbers are easier to work with than large numbers, especially if there are exponents involved. When exponents are involved, choose numbers like 2 or 3; 0 and 1 have special properties that may make it more difficult to recognize the answer. Small numbers are not always the best though. When you’re working with percentages, 10 and 100 are easy numbers to work with. In a problem involving minutes or seconds, 60 may be the easiest number to work with. Look for clues in the problem to determine which numbers to use. Example 8: A street vendor has just purchased a carton containing 250 hot dogs. If the carton cost x dollars, what is the cost in dollars of 10 of the hot dogs? (a) x/25

(b) x/10

(c) 10x

(d) 10/x

(e) 25/x

Solution: An easy number to pick for x would be 250 so that the hot dogs cost $1 each. Hence, 10 of the hot dogs cost $10. Now we look for the answer choice that yields 10 when we plug 250 in for x. (a) x/25 = 250/25 =10. The answer. If you need more proof that you will get this answer with any number, try a few more numbers for x and see if you get the same answer. But don’t do this during the test! You can also solve this problem using ratios and proportions. Sometimes you may have to plug in more than once to find the correct answer. Here’s an example. Example 9: The positive difference between the squares of any two consecutive integers is always: (a) (b) (c) (d) (e)

the square of an integer a multiple of 5 an even integer an odd number a prime number

Solution: The word always in the question means we only have to find one instance in which the condition is false. Let’s choose 2 and 3 as our two consecutive integers. Their squares are 4 and 9 and the difference is 5. (a) (b) (c) (d) (e)

5 is NOT the square of an integer. Eliminate. 5 IS a multiple of 5. A possibility. 5 is NOT an even integer. Eliminate. 5 IS an odd integer. A possibility. 5 IS a prime number. A possibility.

We have improved our odds to 1 in 3. But we need to eliminate 2 more choices, Let’s choose 0 and 1 as our consecutive integers. The difference between the squares is 1. Now we need to check (b), (d), and (e). (b) 1 is NOT a multiple of 5. Eliminate. (d) 1 IS an odd integer. A possibility. (e) 1 is NOT a prime number. Eliminate. So, the answer is (d). Plugging in more than once is very often necessary on inequalities. This is especially true on inequalities which have squared variables, such as the following example, or

fractions with variables in the denominators. Do not try to solve for x on these types of inequalities. Example 10: What are all the values of x such that x2 – 3x – 4 is negative? (a) (b) (c) (d) (e)

x < -1 or x > 4 x < -4 or x > 4 1 18, which of the following is true? (a) (b) (c) (d) (e)

x < -4 x< 6 x > -4 x > -6 x=2

Solution: Clearly, this inequality is simple to solve. Solving it would be faster than plugging in the answer choices on a problem like this. -3x + 6 > 18 -3x > 12 -x > 4 x < -4

Hence, the answer is (a). Some problems in the GRE will give you the value of one of the terms in an expression, and then ask you for the value of that expression. Never bother to solve for the variable on a problem like this. It is much faster to plug in and there is less room for error. Example 12: If 3x = - 2, then (3x – 3)2 = ? (a) –25

(b) 13

(c) –5

(d) 25

(e) 9

Solution: Don’t bother to solve for x and mess with fractions. Since the value of 3x is –2, just plug –2 into the expression (3x – 3): (3x – 3) = (-2 –3) = (-5) = 25 Plugging in simpler numbers for the numbers in the problem can really simplify a problem. Example 13: Which if the following numbers is the closest approximation to the value of 0.507(507 ) ? 5.07 (a) 1

(b) 5

(c) 10

(d) 50

(e) 100

Solution: Rather than spend time manipulating these numbers, just approximate each number since we are only looking for an approximate value anyway. Replace the last two digits of each number with zeros: 0.5(500 ) 5 Now we can see that this number is just half of 500 divided by 5. Half of 500 is 250, and 250 divided by 5 is 50. The answer is (d). Example 14: 28 – 27 = (a) 2

(b) 27/8

(c) 27

(d) 28

(e) 215

Solution: Do not make the mistake of subtracting the exponents on this problem. Remember that you can only add and subtract exponents when multiplying and dividing. Instead of figuring out the values of 28 and 27, let’s try this problem first with smaller numbers: 24 – 23 = 23

Since 24 is 16 and 23 is 8, then 24 – 23 = 16 – 8 = 8 = 23. Note that the solution is the same as the second number in the problem. This makes sense since 24 is 2 times 23 or 23 + 23 = 24. Using this same principle on the original problem, the answer must be 27 or answer (c). Had you not known what to do on this problem, you could have immediately eliminated (e), since the difference can’t be greater than either of the two numbers, and (d), since the difference can’t be the same as the larger number unless the second number was zero. Choices (a) and (b) are attractive to careless people who think they can use the division rule for exponents. Watch out for tricky wording or changes in wording on the GRE. If there is an incorrect answer that would result from misreading the problem, you can bet it will be an answer choice. Example 15: (PQ)(PQ) = 144 If P and Q represent the digits in the two-digit numbers above, the ratio of Q to P is (a) 1 : 2

(b) 1 : 1

(c) 2 : 1

(d) 3 : 1

(e) 12 : 1

Solution: Since the square root of 144 is 12, then PQ must be 12. Hence, P = 1 and Q = 2. So, the ratio must be 1 : 2, right? Wrong! The question asked for the ratio of Q to P, so the answer is 2 : 1, or (c). Be careful! On geometry problems, you can plug in values for angles or lengths as long as the values do not contradict the problem statement or geometry laws. So, you must remember the geometry laws we learned.

A

B

D

C

(0, 0)

(b, 0)

Example 16: The area of the rectangle ABCD is 3b2. The coordinates of C and D are given. In terms of b, BD = (a) b

(b) 2b

(c)

5b

(d) 3b

(e) 10 b

Solution: The distance from D to C is b (the width). If the area is 3b2, then the length is 3b (Why)? Looking at triangle ABD the length of the hypotenuse is

(b) 2 + (3b) 2 =

10 b (e). Remember that c2 = a2 + b2. Another way to approach this problem is to eliminate answer choices that don’t make common sense based on the geometry rules we’ve learned. Suppose again that we plug in 2 for b. The length of DC is 2, and the length of BC is 6 (area = length × width). Recall from our geometry review that the longest side is opposite the largest angle and that the hypotenuse of a right triangle is always longer than either of its other two sides. Hence, the length of the hypotenuse BD must be more than 6. The hypotenuse must also be shorter than the sum of the other two sides. So, the length of the hypotenuse must be less than 8. Now look at the answer choices. Plugging in 2 for b, we see immediately that (a), (b), and (d) are all too small. Remember that the square root of 5 is approximately 2.2 (you were supposed to commit this to memory!). Hence, 2 5 is approximately 4.4 which is also too small. The answer must be (e). Example 17: In square ABCD below, what is the value of (AD)(AB) ? (AC)(BD)

(a) 1/2

A

B

D

C

(b)

2/2

(c) 1

(d)

2

(e) 2

Solution: Since the diagonal of a square (or rectangle like in the previous problem) is always larger than the sides, then (AC)(BD) must be larger than (AD)(AB). Since the numerator is smaller than the denominator, we are looking for a fraction between 0 and 1. That eliminates (c), (d), and (e). Now recall the side relationships of a 45:45:90 triangle. If one side of the 45:45:90 triangle is 1, then the hypotenuse is 2 . In other words, if AD and AB have length 1, then AC and BD must have length 2 . Hence: (AD)(AB) = (1)(1) = 1/2; which is (a). (AC)(BD) = 2 2 Some math problems on the GRE don’t seem to fit any of the shortcuts that we’ve discussed. However, rather than solving it by some lengthy operation you learned in algebra class, look for a shortcut. Virtually all GRE math problems can be solved without time-consuming mathematical techniques. Improve your score by teaching yourself to look for the easy way out.

Example 18: If x2 – 4 = (18)(14), then x could be (a) 14

(b) 16

(c) 18

(d) 26

(e) 32

Solution: Solving for x or plugging in would take way too much time. Remember the expression x2 – y2 from our algebra review? Notice that the left side of the equation in this problem fits this form. Also, remember that x2 – y2 factors into (x + y)(x – y). Hence, x2 – 4 factors into (x + 2)(x – 2). Since (x + 2)(x – 2) = (18)(14), then x + 2 = 18 and x – 2 = 14. Since 16 + 2 = 18 and 16 – 2 = 14, then x could be 16 which is answer choice (b). You can also find shortcuts on geometry problems on the GRE. You must remember however that diagrams are not drawn to scale, so you must not try to guess the answer by eyeballing or measuring a diagram. However, diagrams on the GRE do provide a lot of useful information which may provide a shortcut to the answer. X 135o

W

Z

Y Example 19: In the circle above, if segment WZ and segment XY are diameters of the circle with length 12, what is the area of the two right triangles? (a) 36

(b) 33

(c) 30

(d) 18

(e) 12

Solution: Since the diameter of the circle is 12, the radius must be 6. Therefore, the hypotenuse of each of the two right triangles is 6. The small angle (either one) next to the 135o = 180o – 135o = 45o. Hence, the two right triangles are 45:45:90 triangles, so the other two sides of the triangles must be equal. Recalling the ratio of the sides of a 45:45:90 triangle (1: 2 ), the other sides must have length 6/ 2 . The area of one triangle = 1 / 2 • base • height = 1 / 2 • 6 / 2 • 6 / 2 = 36 / 2( 4 ) = 36 / 2 • 2 = 36 / 4 = 9

The total area = 2 • 9 = 18 (d). Compiled by Robyn Wright, 1992 Revised by Mosbah Dannaoui, 1992; Ziad Diab, 1993; John Everett, 1999 Reference: Robinson, Adam and John Katzman. The Princeton Review – Cracking the System: The GRE 1992 Edition. New York: Villard, 1991. 105-201.

GRE MATH REVIEW #8

Charts and Tables These questions test your ability to read and interpret charts and graphs. The mathematics involved in these questions is nothing more than percentages, ratios, averages, etc. So, if you remember our review on these topics, this section should not be difficult for you. Most students, however, are careless in reading the questions and interpreting the charts rather than with the calculations. In the following discussion, we’ll review the types of graphs and charts used on the GRE before working examples. A grid graph is very similar to the Cartesian coordinate system. The following grid graph depicts the relationship between a person’s age and his/her height: Height in Inches 90 80 70 60 50 40 30 20 10 2 4 6 8 10 12 14 16 Age in Years The vertical axis represents height; the horizontal represents age; the plotted curve represents the relationship between them. This graph tells a person’s age at any height or a person’s height at any age. For example, the point on the curve shows us that at age 15, the person was 70 inches tall. A bar graph is a lot like a grid graph. The following example shows the range in temperature during four consecutive days:

90 80 70 60 50 40 30 20 10 M

T

W

Th

The bottom of each bar represents the low temperature for each day; the top of each bar represents the high temperature; the entire bar represents the range of temperatures. For example, on Monday the temperature ranged from a low of about 45 degrees to a high of about 75 degrees. The bars in any bar graph are like thermometers; they convey information as you move along one axis. A pie chart is a circular chart that depicts fractional parts of whole as wedge-shaped slices. The following example shows how much of a blueberry pie was eaten by each member of a family:

DAD 45%

SALLY 17%

JOE 25% MOM 10%

FIDO 3%

The entire pie stands for 100% of whatever is being represented. In this example, the pie represents 100% of a blueberry pie. Pie charts will always be proportionally accurate on the GRE, so you may be able to get rough estimates of answers by looking at the pie chart. Pie charts could be made more difficult by using a second graph, chart, or table to elaborate on information contained in the pie chart. For example, we could elaborate on the above pie chart with a second pie chart that represents the 45% of the pie eaten by Dad:

2

DAD

DESSERT 48%

BREAKFAST 50%

LEFT ON PLATE 2%

The second pie chart shows when and how Dad consumed his 45% share of the pie. If you were asked what part of the entire pie Dad ate for breakfast, you would multiply 45% by 50% to get 22.5% of the total pie consumed by Dad at breakfast. A chart depicts numerical information as a picture. Instead of just listing what percentage of the pie each family member ate, a chart shows the information graphically. A table on the other hand conveys the same information without the picture. A table is just an organized list of numbers broken into categories. Some tables on the GRE will refer to information in other tables or charts. The following table conveys the same information as the first pie chart above: The Blueberry Pie: Who Ate What Dad 45% Joe 25% Sally 17% Mom 10% Fido 3% Your first concern on a chart problem is understanding what the question is asking you to find, and then understanding how to extract that information from the chart or table. After succeeding at this, the questions are the same as the arithmetic questions we covered earlier. A percentage problem is worked the same way as any percentage problem. The only difference is that some of the information needed to work the problem may be embedded in a chart or table. Here is a sample chart section: Questions 21-25 refer to the following graphs:

3

(Fictional) Nationwide survey of people’s ice cream preference in 1975 and in 1985 by flavor. Vanilla 26.53%

Other 15.73%

Chocolate Chip 8.23%

Strawberry 11.93%

Chocolate 28.77%

Chocolate Chip 9.88%

Other Vanilla 21.06% 19.08%

Strawberry 12.54% Chocolate 25.63% Butter Pecan 8.81%

Butter Pecan 11.81%

1975

21.

To the nearest one percent, what percentage decrease in popularity occurred for chocolate from 1975 to 1985? (a) 9%

22.

(b) 10%

(b) 1:1

(e) 90%

(c) 3:2

(d) 2:1

(e) 5:1

(b) 10,000

(c) 42,120

(d) 100,000

(e) 1,000,000

If a percentage point shift results in annual additional sales of $50,000, how much, in dollars, did combined annual Butter Pecan and Chocolate Chip sales increase between 1975 and 1985? (a) $2335

25.

(d) 89%

In 1985, if 20 percent of the “other” category is Lemon flavor, and 4,212 people surveyed preferred Lemon, then how many people were surveyed? (a) 1000

24.

(c) 11%

What is the ratio of categories that increased in popularity to those that decreased? (a) 1:2

23.

1985

(b) $4650

(c) $232, 500 (d) $465,000 (e) $23,250,000

Which of the following statements can be deduced from the pie graphs? I.

Both the Butter Pecan and Vanilla percentages increased by more than 33 percent between 1975 and 1985.

II.

A higher percentage of people chose Butter Pecan and Strawberry in 1975 than chose Butter Pecan and Chocolate Chip in 1985.

III.

The total share of Vanilla, Chocolate, and Strawberry decreased by less than 20 percent from 1975 to 1985.

4

(a) I only

(b) II only

(c) III only

(d) II and III (e) I, II, and III

Solution to 21: Simply find the difference between 28.77 and 25.63 and then determine what percentage of 28.77 that difference is. The difference is 3.14. Now we want to know what percent of 28.77 is 3.14. Obviously, we can eliminate (d) and (e) because they are much too large. Notice that 10% of 28.77 is 2.877 which is smaller than 3.14. Hence, 3.14 must be 11% of 28.77, which is answer (c). Solution to 22: The first step in a ratio problem is to count the parts. There are six wedges in each of the pies, so there are six parts. Four of the flavors increased in popularity, and two of the flavors decreased. The ratio of increases to decreases, therefore, is 4:2 which can be reduced to 2:1, answer choice (d). Solution to 23: Based on the first piece of information in the question, the percentage of people who preferred Lemon in 1985 is 20% of 21.06% which is roughly 45 (Notice that our calculations can be rough since the answer choices are so far apart). This means that 4% of the total number of people surveyed is 4212. So, (d) is the correct answer. Solution to 24: The first thing to remember is that a percentage-point decrease is calculated simply by subtracting percentages. For example, the percentage-point decrease between 120% and 100% is 20%, but the percentage decrease between 120% and 100% is 16.67%. The percentage-point change in Butter Pecan from 1975 to 1985 is a 3 percentage-point increase. The percentage-point change in Chocolate Chip from 1975 to 1985 is 1.65 percentage-point increase. The total increase in percentage points is 4.65. Notice again that the answers are far apart, so we don’t have to be exact in our calculations. Simply multiply 4.65 by 50,000, or just using 4 and 5 would result in a number between 200,000 and 250,000. Hence, the answer must be (c). Solution to 25: This is a very time-consuming problem. To solve it, first check out Statement I. Did the Butter Pecan percentage increase by more than 33% or one-third? It increased from 8.81 to 11.81, which is a 3-point increase. Since 3 is more than one-third of 8.81, the first part of Statement I is true. Vanilla went from 26.53 to 19.08, which is a decrease, not an increase. Therefore, Statement I is not true. We can now eliminate (a) and (e). Looking at Statement II, the combined percentage for Butter Pecan and Strawberry in 1975 is 8.81 + 11.93 or 20.74. The combined percentage for Butter Pecan and Chocolate Chip in 1985 is 11.81 + 9.88 or 21.69. Statement II is false which eliminates (b) and (d). Hence, the answer is (c).

5

GRE MATH REVIEW 9 Quantitative Comparisons Quantitative comparisons are basic arithmetic, algebra, and geometry problems, which use the same concepts we reviewed earlier. We will learn a few special techniques for quantitative comparisons in this section. Here are the directions for quantitative comparisons as they appear on the GRE: Directions: Each of the Questions consists of two quantities, one in Column A and one in Column B. You are to compare the two quantities and choose A if the quantity in Column A is greater; B if the quantity in Column B is greater; C if the quantities are equal; D if the relationship cannot be determined from the information given. Note: Since there are only four choices, NEVER MARK (E). Common Information: In a question, information concerning one or both of the quantities to be compared is centered above the two columns. A symbol that appears in both columns represents the same thing in Column A as it does in Column B.

Do not bother to read these directions. Here are the directions worded more explicitly that you should memorize before taking the GRE: Directions: Each of the Questions consists of two quantities, one in Column A and one in Column B. You are to compare the two quantities and choose A if the quantity in Column A is always greater; B is the quantity in Column B is always greater; C if the quantities are always equal; D if none of the other choices is always correct.

The only difference between these directions and the GRE’s is the fact that the quantity in Column A must always be greater than the quantity in Column B in order to choose answer A. The following example illustrates this point.

Example 1:

Column A x+1

Column B 1–x

Solution: Plug in the number 1 for x. Then A is greater than B. But what if we plug in (– 1). Then B is greater than A. So, the answer would be D since neither A nor B is always true. Be very careful on quantitative comparisons to never mark E. There are only four answer choices. If a quantitative comparison problem contains only numbers, there will be an exact answer. Therefore, always eliminate choice D on these problems.

Example 2:

Column A 2/7 – 1

Column B 1/3 – 1

Solution: Immediately eliminate choice D because there are only numbers involved. Also, since (–1) appears in both expressions, we can ignore it. Now we only have to decide which is bigger, 2/7 or 1/3. Since 1/3 can also be written as 2/6, 2/7 is obviously smaller. Therefore, the answer is B. On some problems, you will be able to visualize the problem and avoid computations.

Example 3:

Column A Area of circle with diameter 12

Column B Surface area of a sphere with diameter 12

Solution: Just picture a soccer ball and a paper plate. The answer is B. Quantitative comparisons are supposed to be fast. If you find yourself setting up an elaborate calculation or equation, you are on the wrong track. Look for a shortcut.

Example 4:

Column A 9(3 + 24)

Column B (9 × 3) + (9 × 24)

Solution: Notice first that D cannot be the answer. Now notice that A is simply the factored form of B. The answer is C. You are not expected to multiply out numbers like these. Treat the two columns as if they were the two sides of an equation. Anything you can do to both sides of an equation, you can also do to both columns. You can add or subtract numbers from both columns; you can multiply or divide both columns by a positive number; you can multiply one side by some form of 1. Do not, however, multiply or divide both columns by a negative number. The reason is that we don’t know if the two columns represent an equation or an inequality. If they represent an inequality, the direction of the inequality will change if you multiply or divide by a negative number. You should always simplify the terms in a quantitative comparison by reducing, factoring, unfactoring, etc.

Example 5:

Column A 25 × 7.39

Column B 739/4

Solution: Notice that D cannot be the answer. Do not do the division or multiplication in this problem. Try to simplify it. Multiplying both sides by 4, we get: Column A 100 × 7.39

Column B 739

2

Now it’s obvious that the two quantities are equal. The answer is C. For quantitative comparisons involving variables, it is usually easier to just plug in numbers. Column A Example 6:

[(k × 1/2) ÷ 3] × 6

Column B k