BITS Pilani Goa Test 3 Answers

BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers 1.36 men together can build a wall 140 m long in 21 days ; the number of men working at the same rate required to build the same wall in 14 days is 1) 54 2) 48 3) 36 4) 18 2.The perimeters of a square and a rectangle are equal . If their area be 'A' m2 and 'B' m2 then correct statement is 1) A < B 2) A ≤ B 3) A > B 4) A ≥ B 3.A man bought a watch for 10% discount.If he had bought for 20% discount he would have got the watch for ₹ 125 less . The marked price of the watch is 1) rs. 2500 2) rs. 1250 3) rs. 3750 4) rs. 1000 4.The compound ratio of the inverse ratios of the ratios x : yz , y : zx , z : xy is : 1) 1 : xyz 2) xyz : 1 3) 1: 1 4) x : yz 5.What will be the percentage of increase in the area of a square when each of the its sides is increased by 10% ? 1) 20` 2) 11 3) 121 4) 21 6.If p3 - q3 = ( p - q ) { ( p + q )2 - x p q } then the value of x is 1) 1 2) -1 3) 2 4) -2 7.On simplification th value of 1) 2√2 - 1

2) 1 - 2√2

1−

1 1 is + 1+ 2 1− 2

3) 1 - √2

4) -2√2

8.If sec 15θ = cosec 15θ (0° < θ < 10°) then value of θ is 1) 9° 2) 5° 3) 8° 4) 3° 1/2

2

1/3,

9.If x=(0.25) , y=(0.4) , z=(0.216) then 1) Y > X > Z 2) x > y > z

3) Z > X > Y

4) X > Z > Y

10.Average runs scored by 11 players of a cricket team is 23 runs . If the first player scored 113 runs . Find the average runs of the remaining players 1) 8 runs 2) 12 runs 3) 14 runs 4) 27 runs 11.Find the highest power of 7 present in (60! – 20!)? (1) 7 (2) 9 (3) 2 (4) 5 (5) 8 Solution: Corrected Explanation: 60! – 20! = 20! × [(60!/20!) – 1]. Now, 60!/20! can be written as 7k where k is an integer because 60!/20! = 21*22*…*60 = 7 *3*22*…60. Any integer of form (7k – 1) will not be divisible by 7. And therefore, only 20! is divisible by 7, which has as highest power 2. Hence, the correct answer is option 3. 12.How many distinct pairs of integral values of x and y exist such that they satisfy the equation: 3x + 5y = 1, where |x| < 400 and |y| < 200? (1) 160 (2) 134 (3) 120 (4) 133 (5) None of these Solution: Given: 3x + 5y = 1. So, x = (1 – 5y)/3. Also, |x| < 400 and |y| < 200. i) Consider y = –199 or |y| = 199, then x = (1 – 5(-199))/3 = 332. When y = -198 or -197, x has a non-integral value. Again when y = -196, x = 327. Thus, it can be observed that when y = -199, -196, -193, … , -4, -1; x has an integral value. There are a total of [(199 – 1)/3] + 1 = 67 such values for y (for y < 0). © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers ii) Now, consider y = 2, then x = (1 – 5(2))/3 = -3. When y = 3 or 4, x has a non-integral value. Again when y = 5, x = -8. Thus, it can be observed that when y = 2, 5, 8, … , 194, 197; x has an integral value. There are a total of [(197 – 2)/3] + 1 = 66 such values for y (for y < 0). Thus, in all the number of integral set of values for x and y that satisfy the given equation = 67 + 66 = 133. Hence, the correct answer is option 4. 13.What is the Highest Common Factor of (11 20 (1) 10 (2) 10

500

20

– 1, 11 – 1)? 10 10 (3) 11 (4) 11 – 1

(5) 11

20

–1

Solution: 500 20 Highest Common Factor of (11 – 1, 11 – 1). 500 20 25 n 20 11 – 1 = (11 ) – 1 = a – 1, where a = 11 . n It is known that, (a – 1) is exactly divisible by (a – 1), when n is odd. 20 25 20 Similarly, when (11 ) – 1 is divided by (11 – 1), n = 25 (odd), hence it leaves no remainder as it is exactly divisible. 500 20 20 Thus, HCF of (11 – 1, 11 – 1) = 11 – 1. Hence, the correct answer is option 5. th

th

nd

14.In an Arithmetic Progression, the difference of 9 times the 5 term and the 13 term is 200 and the product of the 2 and th th the 7 terms is 629. What is the 10 term of this series, given that all the terms of this series are positive numbers? (1) 49 (2) 26 (3) 21 (4) 36 (5) Cannot be determined uniquely Solution: Let the terms of the A.P be denoted by T1, T2, T3 and so on. Also, let the first term and the common difference of the A.P be denoted by a and d respectively. Given that: i) (9 × T5) – T13 = 200. → 9 × (a + 4d) – (a + 12d) = 200 → 8a + 24d = 200 → a + 3d = 25 = T4 (1) ii) T2 × T7 = 629. → (a + d) × (a + 6d) = 629 → (a + 3d – 2d) × (a + 3d + 3d) = 629 (2) Using (1) in (2), we get → (25 – 2d) × (25 + 3d) = 629 2 → 6d – 25d + 4 = 0 Solving the above equation, we get d = 4 or 1/6. Using d = 4 in (1), we get a = 13, thus T10 = a + 9d = 13 + 9 × 4 = 49. Using d = 1/6 in (1), we get a = 24.5, thus T10 = a + 9d = 24.5 + 9 × (1/6) = 26. Hence, the correct answer is option 5. 15.Find the remainder when 28! is divided by 29? (1) 1 (2) 27 (3) 28

(4) 2

(5) None of these

Solution: By Wilson’s Theorem, for a prime number n, (n – 1)! = -1 mod n. → [(n – 1)! + 1] is exactly divisible by n. Here, let n = 29 (prime number). Thus, [(29 – 1)! + 1] = (28! + 1) is exactly divisible by 29. Thus, when 28! is divided by 29, a remainder of -1 or (29 – 1) = 28 is left behind. Hence, the correct answer is option 3. 16.Find the range of x which satisfies the in-equation: 4^[log64(48|x| + 32)] < 8, where x is a real number and |x| ≠ (2/3)? (1) -10 < x < 10 (2) -10 < x < 0 (3) 0 < x < 10 (4) 0 < x (5) None of these Solution: © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers Given expression: 4^[log64(48|x| + 32)] < 32. Corrected: Given expression: 4^[log64(48|x| + 32)] < 8. → 4^[log64(48|x| + 32)] < 4^(3/2) → [log64(48|x| + 32)] < (3/2) → 64^(3/2) > (48|x| + 32) → 512 > (48|x| + 32) → 480 > 48|x|. When x = x, then 480 > 48x → x < 10. When x = -x, then 480 > -48x → x > -10. Hence, the range of values for x is given by: -10 < x < 10 Hence, the correct answer is option 1. 17.If m is a positive integer greater than 5, the highest number by which the following expression: 4 3 2 3 (m – 5m + 6m )(3m – 3) is always divisible by is? (1) 972 (2) 648 (3) 256 (4) 108 (5) 72 Solution: The terms of the given expression could be re-arranged to form: 4 3 2 3 (m – 5m + 6m )(3m – 3) 3 2 2 3 = 3 × m × (m – 5m + 6) × (m – 1) 2 2 = (27) × [(m – 3) × (m – 2) × (m – 1)] × [m × (m – 1) ] It is known that: [(m – 3) × (m – 2) × (m – 1)] is always divisible by 6 (Since 3 Consecutive integers will have a number divisible by 3 and since an even number would also be present, hence divisible by 2 as well). 2 2 Also [m × (m – 1) ] is always divisible by 4 (Since, either m or (m – 1) is even, and its square would be divisible by 4). Hence the given expression: 2 2 (27) × [(m – 3) × (m – 2) × (m – 1)] × [m × (m – 1) ] is always divisible by a number = 27 × 6 × 4 = 648. Hence, the correct answer is option 2. 2

18.When written in decimal representation, (ABA) = (CCDCC), where A, B, C and D are different natural numbers. Find the sum of A, B, C and D? (1) 12 (2) 14 (3) 16 (4) 17 (5) None of these Solution: 2 Given: (ABA) = (CCDCC), where A, B, C and D are different natural numbers. Range of ABA could be from 100 to 316, square of which would yield a 5 digit number. As last digit of a perfect square, C could take the values of 1, 4, 5, 6 or 9 only. Among this, since last digit of ABA = A is 2 not equal to C, i.e., xx1 end with 1 and so on. C could take only the values of 1, 4, 6 or 9. Correspondingly, A could take the values of 2 or 3. For A = 2, ABA could be 202, 212, 222 and so on. 2 First consider ABA = 202, then (ABA) = 40804, which is not of the form CCDCC. 2 Next consider ABA = 212, then (ABA) = 44944, which is exactly of the form CCDCC. 2 Note that upon further trials for values of A, no set of values satisfy the given equation (ABA) = (CCDCC). Thus, ABA = 212 and CCDCC = 44944. Hence, A + B + C + D = 2 + 1 + 4 + 9 = 16. Hence, the correct answer is option 3. 19.Given that x, y, z are non-zero real numbers, the expression {x ^ [log (y/z)]} × {y ^ [log (z/x)]} × {z ^ [log (x/y)]} could be equal to which of the following? (1) xyz (2) log (xyz) (3) x + y + z (4) 1 (5) 0 Solution: Let given expression: © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers L = {x ^ [log (y/z)]} × {y ^ [log (z/x)]} × {z ^ [log (x/y)]}. → L = {x ^ [log y – log z]} × {y ^ [log z – log x]} × {z ^ [log x – log y]}. Let log x = X, log y = Y, log z = Z , then L = {x ^ [Y – Z]} × {y ^ [Z – X]} × {z ^ [X – Y]}. Log L = {[Y – Z] × X} + {[Z – X] × Y} + {[X – Y] × Z} → Log L = 0. → L = 1. 20.The co-ordinates of four points are defined as follows: A (-5, 2), B (-3, -4), C (3, 4), D (x, -6). Find the value of x if it is given that the area of the triangle ABD is twice the area of the triangle ACD? (1) 81 (2) 67 (3) 5 (4) 1 (5) None of these Solution: Given: A (-5, 2), B (-3, -4), C (3, 4), D (x, -6). Area (A) of a triangle with co-ordinates (x1, y1), (x2, y2), (x3, y3) is given by: A = (1/2) × determinant of |(x1 – x2), (y1 – y2)| |(x1 – x3), (y1 – y3)| Thus, area of triangle ABD (A1) is given by: A (-5, 2), B (-3, -4), D (x, -6) A1 = (1/2) × determinant of |(-5 + 3), (2 + 4)| |(-5 – x), (2 + 6)| → A1 = (1/2) × (14 + 6x) = 3x + 7. Similarly, area of triangle ACD (A2) is given by: A (-5, 2), C (3, 4), D (x, -6) A2 = (1/2) × determinant of |(-5 – x), (2 + 6)| |(-5 – 3), (2 – 4)| → A2 = (1/2) × (74 + 2x). It is known that A1 = 2A2 → 3x + 7 = (74 + 2x) → x = 74 – 7 = 67. Hence, the correct answer is option 2. 21.Select the related word/letters/number from the given alternatives. EARTH : PLANET :: MOON : ? 1) SUN 2) UNIVERSE 3) VENUS

4) SATELLITE

22.Select the related word/letters/number from the given alternatives.EJOT : ________::YDIN : VAFK 1) LQGB 2) BGLQ 3) QBGL 4) BGQL 23.Select the related word/letters/number from the given alternatives.100 : 102 : : 100000 : ? 1) 105 2) 104 3) 1003 4) 1004 24.For the following questions Find the odd word/letters/number pair from the given alternatives 1) Temple 2) Worship 3) Mosque 4) Church 25.For the following questions Find the odd word/letters/number pair from the given alternatives 1) M N O M 2) B D C B 3) X Z Y X 4) P R Q P 26.For the following questions Find the odd word/letters/number pair from the given alternatives 1) 70, 80 2) 54, 62 3) 28, 32 4) 21, 24 © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers 27.Arrange the following words as per order in the dicitonary 1) Pearl 2) Peasant 3) Pea 4) Peanut 1) 3, 1, 4, 2 2) 3, 4, 1, 2 3) 1, 3, 2, 4 4) 1, 2, 3, 4 28.Which one set of letters when sequentially placed at the gaps in the given letter series shall complete it? g f e _ i g_e i i _f e i_g f _i i 1) i f g i e 2) f i g i e 3) e i f g I 4) i f i g e 29.Which one number when placed at the sign of interrogation shall complete the series 3, 6, 18, 72, ...?..., 2160 1) 144 2) 216 3) 288 4) 360 30.Introducing a lady, a lady said, "She is the only daughter of Mohan's grandfather who is my husband's father". How does the lady relate herself with the introduced lady? 1) Aunt 2) Mother 3) Mother-in-law 4) Sister-in-law Directions for questions 31 to 35: Refer the following data to answer the questions given below. During the practice sessions at Olympics 2008, eight players had to be allotted in two different rooms such that at least three persons are allotted in each room. The players in each room must be able to converse directly with atleast one person in their room. Player Edward and Haley can converse with each other in German. Player Aaron can speak three languages, French, Spanish and German. Among the eight, the players Barak and Ford wanted to be the roommates because they are from the same place and they can speak only German. Player Ford objects to have Player Daniel to be his roommate. Player Caleb can speak only French and Spanish. Player Gyula who speaks only Spanish, objects to have Player Barak to be his roommate. Player Edward can speak two languages, one among them is French. Player Haley can speak two languages, one among them is Spanish. Player Daniel knows only one language and it is also known that the only three men with whom he cannot converse is Barak, Ford and Edward. Solutions for questions 31 to 35 Let us represent the names of the players with their first letters. From the conditions given, B and F should be together in the same room; D and G should be together and in the other room. So, for the purpose of answering different questions we can use the table below and fill the other names in the two rooms. Let us first find out the total number of possible combinations for the two rooms. No. of No. of Combinati persons persons on In Room 1 In Room 2 I 3 5 II 4 4 III 5 3 Let us first assume that B and F are in room 1. So, D and G should be in room 2. Room – 1 Room – 2 BF D, G (German) (Spanish) 31.How many different possible ways can the rooms be allotted? (1) 26 (2) 24 (3) 28 (4) 20

(5) None of these

Solution: If we take 3 and 5 combinations for the first and second rooms respectively, the third person in the first room can be A or E or H. If there are three people in the second room the third person can be C or A or H. So, we have only six possibilities for one room having 3 persons and the other having 5 persons. Similarly, the 4 and 4 combination can be done in six ways. Combi Persons in Room Persons in Room 2 nation 1 (BFA)/(BFE)/(BFH (DGCEH)/ (DGACH)/ I ) (DGAEC) © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers (3 ways) II (BFAH)/(BFAC)/ (DGCE)/ (DGEH)/ (6 (BFEH)/(BFEC)/ (DGAC)/ (DGAH) ways) (BFAE)/ (BFHC) (DGHC)/ (DGAE) III (BFCEH)/ (3 (BFAEH)/ (DGA)/ (DGC)/ (DGH) ways) (BFACE) Since there are two rooms, the players can be allotted in (12 × 2), i.e., 24 different ways. Hence, the correct answer is option 2. 32.If Player Edward objects to have Player Aaron as his roommate, then how many possible ways can the rooms be allotted? (1) 18 (2) 16 (3) 14 (4) 10 (5) None of these Solution: Combi Persons in Persons in Room 2 nation Room 1 I (2 (BFA)/ (BFE) (DGCEH)/ (DGACH) ways) II (BFAH)/ (DGCE)/ (DGEH)/ (4 (BFAC)/ (DGAC)/ (DGAH) ways) (BFEH)/ (BFEC) III (BFCEH) (DGA) (1 way) Since there are two rooms, the players can be allotted in (7 × 2), i.e., 14 different ways. Hence, the correct answer is option 3. 33.If Player Edward knows Spanish, then how many additional new combinations are possible in allotting the rooms? (1) 5 (2) 4 (3) 3 (4) 2 (5) 1 Solution: Combi nation I (3 ways) II (6 ways) III (4 ways)

Persons in Room 1

Persons in Room 2

(BFA)/(BFE)/(BF H)

(DGCEH)/ (DGACH)/ (DGAEC)

(BFAH)/(BFAC)/ (BFEH)/(BFEC)/ (BFAE)/ (BFHC) (BFCEH)/ (BFAEH)/ (BFACE)/(BFAC H)

(DGCE)/ (DGEH)/ (DGAC)/ (DGAH) (DGHC)/ (DGAE) (DGA)/ (DGC)/ (DGH)/(DGE)

Since there are two rooms, the players can be allotted in (13 × 2), i.e., 26 different ways. Hence, the correct answer is option 4. 34.If Player Haley does not know Spanish but the players Daniel and Gyula know French, then how many possible ways can the rooms be allotted? (1) 24 (2) 26 (3) 28 (4) 20 (5) None of these Solution: Combi Persons in Persons in Room 2 nation Room 1 © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers I (BFA)/(BFE)/(BF (DGCEH)/ (DGACH)/ (3 H) (DGAEC) ways) II (BFAH)/(BFAC)/ (DGCE)/ (DGEH)/ (6 (BFEH)/(BFEC)/ (DGAC)/ (DGAH) ways) (BFAE)/ (BFHC) (DGHC)/ (DGAE) III (BFCEH)/ (3 (BFAEH)/ (DGA)/ (DGC)/ (DGE) ways) (BFAHC) Since there are two rooms, the players can be allotted in (12 × 2), i.e., 24 different ways. Hence, the correct answer is option 1. 35.If a new player Ifan, who can speak only Spanish, joins the group, in how many possible ways can the rooms be allotted if each room has to have more than three members? (1) 28 (2) 30 (3) 36 (4) 34 (5) None of these Solution: Combinations 4 persons in room 8 ways 1 5 persons in room 9 ways 1 Since there are two rooms, the players can be allotted in (17 × 2), i.e., 34 different ways. Hence, the correct answer is option 4. Directions for questions 36 to 40: Refer the following data to answer the questions given below. Each Mala, Malavika and Malathi goes to buy hair bands of three varieties A, B and C. The cost of each hair band of variety A, B and C is 50 paise, Re. 1, and Rs. 2 respectively. It is known that Mala, Malavika and Malathi together have spent Rs. 42. Each one of the three has bought atleast one earring of each variety. The total number of hair bands bought by any of the three is equal to the numerical value of the amount spent by that person. It is also known that each of them spent equal amount in buying hair bands. Solutions to questions 36 to 40: It is given that the amount and the total number of hair bands receive by each is the same. Amount spent and number of hair bands with each = 42/3 = 14. The possible number of hair bands of the three varieties is as follows.

Possible combinati ons I II III IV

Number of hair bands 50 Rs. 2 pai Re. 1 se Variety Variety Variety C A B 1 2 11 2 4 8 3 6 5 4 8 2

36.The number of hair bands of variety A bought by Malathi is equal to the number of hair bands of variety C bought by Malavika is the same. The only girl who bought odd number of hair bands of variety B is Mala. Which of the following is the least possible number of hair bands of variety A bought by these three girls? (1) 12 (2) 18 (3) 16 (4) 14 (5) (1) or (2) Solution: © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers

Name

Number of hair bands 50 Rs. 2 pai Re. 1 se Variety Variety Variety C A B

Malathi/Malavi 2 4 8 ka Malathi/Malavi 4 8 2 ka Mala 1 2 11 The total number of variety A hair bands is 14. Hence, the correct answer is option 4. 37.The number of hair bands of variety A bought by Mala is equal to the number of hair bands of variety C bought by Malathi is the same. If each Malathi and Malavika buys odd number hair bands of variety B, then what is the total number of hair bands of variety B bought by these three girls? (1) 6 (2) 12 (3) 15 (4) 21 (5) 24 Solution: The two possible combinations are: Number of hair bands 50 Rs. 2 pai Re. 1 Name se Variety Variety Variety C A B Mala 1 2 11 Malathi 2 4 8 Malavik 3 6 5 a The total number of variety C hair bands is 24. Hence, the correct answer is option 5. 38.What is the difference between the highest possible number of hair bands of variety A and the highest possible number of hair bands of variety B, could have been bought by these three girls? (1) 6 (2) 8 (3) 4 (4) 10 (5) None of these Solution: The highest possible number of hair bands of variety A could have been bought by these 3 girls is = 8 + 6 + 4 = 18 The highest possible number of hair bands of variety B could have been bought by these 3 girls is = 11+ 8 + 5 = 24. The required difference is 24 – 18 = 6. Hence, the correct answer is option 1. 39.The number of hair bands of variety A bought by Malathi is equal to the number of hair bands of variety B bought by Malavika is the same. The only girl who bought odd number of hair bands of variety B is Mala. If Mala buys more than 5 hair bands of variety B, then what is the total number of hair bands of variety A bought by these three girls? (1) 18 (2) 12 (3) 14 (4) 10 (5) 16 Solution: The only possible combination is: Number of hair bands 50 Name Rs. 2 pai Re. 1 se © 2017 ETHNUS

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BITS PILANI Goa Test 3 – Answer Key & Explanatory Answers Variety C

Variety A

Variety B

Malathi/Malavi 2 4 8 ka Malathi/Malavi 4 8 2 ka Mala 1 2 11 The total number of variety A hair bands is 14. Hence, the correct answer is option 3. 40.The number of hair bands of variety A bought by Mala is equal to the number of hair bands of variety B bought by Malathi is the same. If each Malathi and Malavika buys odd number hair bands of variety B, then what is the total number of hair bands of variety C bought by these three girls? (1) 6 (2) 8 (3) 7 (4) 9 (5) None of these Solution: The only possible combination is: Number of hair bands 50 Rs. 2 pai Re. 1 Name se Variety Variety Variety C A B Malathi 1 2 11 Mala 4 8 2 Malavik 3 6 5 a The total number of variety C hair bands is 8. Hence, the correct answer is option 2.

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