BITS PILANI Test 8 – Answer Key & Explanatory Answers

BITS PILANI Test - 8 – Answer Key & Explanatory Answers 1.If sinθ+ cosecθ=2, then the value of sin-7θ + cosec7θ is 1) 27 2) 2-7 3) 2

4) 2-1

2.A man has some hens and some cows. If the total number of heads of hens and cows together is 50 and the nunber of feet of hens and cows together is 142, then the number of cows is 1) 21 2) 25 3) 27 4) 29 3.In a class, average height of all students is 'a' cms. Among them, average height of 10 students is 'b' cms and the average height of the remaining students is 'c' cms. Find the number of students in the class. (Here a>c and b>c ) 1) (a(b-c))/(a-c) 2) ((b-c))/(a-c)) 3) (b-c))/(a-c)) 4) (10(b-c))/(a-c) 4.If (a+b-6)2 + a2+b2+1+2b = 2ab+2a, then the value of a is 1) 7 2) 6 3) 3.5

4) 2.5

5.Two circles of radii 17 cm and 8 cm are concentric. The length of a chord of greater circle which touches the smaller circle is 1) 15 cm 2) 16 cm 3) 30 cm 4) 34 cm 6.The least six digit number which is a perfect square is 1) 100489 2) 100000 3) 100256

4) 100225

7.The percentage change of a number when it is first decreased by 10% and then increased by 10% is 1) 0.1 % increase 2) 1 % decrease 3) 0.1 % decrease 4)No changes 8.If 2y cosθ = x sinθ and 2x secθ - y cosecθ = 3 then what is the value of x2 + 4y2 ? 1) 4 2) 1 3) 2 4) 5 9.The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. The rate of interest per annum is? 1) 6% 2) 7% 3) 8% 4) 9% 10.If 20 men working 8 hours per day can complete a piece of work in 21 days. How many hours per day must 48 men work to complete the same job in 7 days? 1) 12 2) 20 3) 10 4) 15 2

11.The equation x + 5x + 3 = 0 has its roots in the ratio of m:n. Find the value of [√(m/n)] + [√(n/m)]? (1) 5/√3 (2) 3/√5 (3) √(5/3) (4) √(3/5)

(5) None of these

Solution: 2 Given equation: x + 5x + 3 = 0. Let the roots of the equation be mk and nk. Sum of the roots = (mk + nk) = k(m + n) = –5. 2 → {k(m + n)} = 25. 2 Product of the roots = (mk × nk) = k (mn) = 3. 2 2 2 {k(m + n)} / k (mn) = (m + n) /mn = 25/3. → (m + n) /√(mn) = 5/√3 → [√(m/n)] + [√(n/m)]= 5/√3. Hence, the correct answer is option 1. st

rd

th

12.An arithmetic progression has an even number of terms. The sum of all the terms in the odd places (1 , 3 , 5 terms and nd th th so on) is 3100 and the sum of all the terms in the even places (2 , 4 , 6 terms and so on) is 3225. The difference of the last and the first term is 245. How many terms are there in this progression? (1) 48 (2) 64 (3) 50 (4) 60 (5) 56 Solution: Let the first term, common difference and the number of terms of the A.P be denoted by a, d and 2n respectively. The terms in the odd places are represented by: a, a + 2d, a + 4d, … a + (2n – 2)d; and there are ‘n’ such terms. Thus, a + a + 2d + a + 4d + …+ a + (2n – 2)d = n(a + (n – 1)d) = 3100. (1) © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers The terms in the even places are represented by: a + d, a + 3d, a + 5d, … a + (2n – 1)d; and there are ‘n’ such terms. Thus, a + d + a + 3d + a + 5d + …+ a + (2n – 1)d = n(a + d + (n – 1)d) = 3225. (2) (2) – (1) gives (dn) = 3225 – 3100 = 125 (3) Also, it is given that the difference of the last and the first term is 245. → a + (2n – 1)d – a = (2n – 1)(d) = 2(dn) – d = 245 (4) Using (3) in (4), we get d = 5 and n = 25. Hence, the total number of terms of the A.P = 2n = 50. Hence, the correct answer is option 3. 13. Find the sum of all the terms in the following series: (41 × 70) + (42 × 69) + (43 × 68) + ... (70 × 41) (1) 109070 (2) 80240 (3) 85200 (4) 90160

(5) None of these

Solution: Given series: (41 × 70) + (42 × 69) + (43 × 68) + ... (70 × 41) Here, it can be observed that the terms of the series follow the pattern as shown below. th 2 n term = (n + 40)(71 – n) = 2840 + 31n – n . Also, the number of terms of the series = 70 – 41 + 1 = 30 terms. 2 Summation of this series (S) = Sum of all ‘n’ terms = Summation of {(n + 40)(71 – n) = (2840 + 31n – n )} where n varies from 1 to 30. S = [2840(n) + 31(n)(n + 1)/2 – (n)(n + 1)(2n + 1)/6] → S = [{2840(30)} + {31(30)(31)/2} – {(30)(31)(61)/6}] Thus, S = [85200 + 14415 – 9455] = 90160. Hence, the correct answer is option 4. 2

14. A box with a square base has no top surface and its total surface area is 1440 cm . What is the maximum possible 3 volume (in cm ) of such a box? (1) 5258.1 (2) 8095.4 (3) 2629.0 (4) 10516.3 (5) None of these Solution: Let the side of the square base and the height of the square box be denoted by s and h respectively. 2 Then, volume = V = s × s × h = s h (1) 2 Also, surface area = area of 4 lateral surfaces + area of bottom surface = 4sh + s = 1440. 2 → h = (1440 – s )/4s = (360/s) – (s/4) (2) 2 3 Using (2) in (1), we get V = s × {(360/s) – (s/4)} = 360s – s /4 To find the maxima, set dV/ds to be equal to zero. 2 → dV/ds = 360 – (3s /4) = 0 2 → s = 480 → s = 4√30. Using this value in (2), we get h = 2√30. 2 3 Thus, V = s h = (16 × 30) × 2√30 = 5258.1 cm . Hence, the correct answer is option 1. 15. If f(x) = sin(log x) and g(x) = cos(log x), then what is the value of [f(x).f(y) – (1/2) ×{g(x/y) – g(xy)}]? (1) log xy (2) tan (log xy) (3) 0 (4) 1

(5) xy – (1/xy)

Solution: Given: f(x) = sin(log x) and g(x) = cos(log x) Let F = [f(x).f(y) – (1/2) × {g(x/y) – g(xy)}] F = [sin(log x).sin(log y) – (1/2) ×{cos(log (x/y)) – cos(log(xy) ) } ] F = [sin(log x).sin(log y) – (1/2) ×{cos(log x – log y) – cos(log x + log y)}] Now, let log x = A, log y = B. F = [sin A. sin B – (1/2) ×{cos(A – B) – cos(A + B)}] By basic trigonometric identities: i) cos(A – B) = cosA. cosB – sinA. sinB ii) cos(A + B) = cosA. cosB + sinA. sinB Hence, cos(A – B) – cos(A + B) = 2 sinA. sinB © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers → F = [sin A. sin B – (1/2) × 2 sinA. sinB] → F = [sin A. sin B – sinA. sinB] = 0. Hence, the correct answer is option 3. 4

3

2

16. Given: f(x) = x + (n – 3)x – (3n – 5)x – (2n – 9)x – 6, such that f(x) is exactly divisible by (x – 2), which of the following is a possible integral value of n? (1) 0 (2) 2 (3) 1 (4) 3 (5) –2 Solution: 4 3 2 Given: f(x) = x + (n – 3)x – (3n – 5)x – (2n – 9)x – 6. Since (x – 2) exactly divides f(x), setting x = 2 in the above expression should satisfy the equation and equate it to zero. 4 3 2 → f(x) = 2 + (n – 3)2 – (3n – 5)2 – (2n – 9)2 – 6 = 0 → f(x) = 16 + 8(n – 3) – 4(3n – 5) – 2(2n – 9) – 6 = 0 → n = 3. Hence, the correct answer is option 4. 17. The vertices of a triangle ABC with sides 17 cms, 18 cms and 25 cms are positioned such that they are also the centres 2 of three circles, each touch the other two circles externally, and are off different radii. What is the total area (in cm ) of the three circles put together? (1) 2827 (2) 1062 (3) 707 (4) 1257 (5) 1018 Solution: Let the radii of the three circles be denoted by x, y and z respectively such that x < y < z. It is given that the three circles touch each other externally. Thus, the distance between the centres of the circles is equal to the sum of the two radii of the corresponding circles which are the sides of the given triangle. Thus, x + y = 17; x + z = 18; y + z = 25. (1) Adding all the expression in (1), we get x + y + x + z + y + z = 2(x + y + z) = 60 → (x + y + z) = 30 (2) Using (1) and (2), we get x = 5; y = 12; z = 13. Hence, the total area of the three circles 2 2 2 = π(5 + 12 + 13 ) 2 = 338 π ≈ 1062 cm . Hence, the correct answer is option 2. 18. Ashwin takes 10 secs more than Appu to complete one round around a circular running track. When they start running at the same time from the same start point, they meet each other again for the first time after 2 minutes. How many times would they meet each other in 2 minutes if they start running from the same start point but in opposite directions? (1) 4 (2) 6 (3) 5 (4) 8 (5) 7 Solution: Let the speeds of Appu and Ashwin be denoted by p and s respectively and the track length be denoted by L. Since Ashwin takes 10 secs more than Appu to complete one round, it can be written that: (L/s) – (L/p) = 10 → L = 10sp/(p – s) (1) Consider Appu and Ashwin running in the same direction: Relative speed = (p – s) L/(p – s) = 2 minutes = 120 seconds. → L = 120(p – s) (2) Using (2) in (1), we get 120(p – s) = 10sp/(p – s) 2 → 12(p – s) = sp 2 2 → 12p + 12s – 25sp = 0 2 Dividing throughout by s , we get 2 → 12(p/s) + 12 – 25(p/s) = 0 2 Let (p/s) = r, then 12r – 25r + 12 = 0 Solving this equation, we get r = 4/3 or 3/4 (this value is not valid since p > s). Also, L = 120(p – s) = 120s(r – 1) = 40s = 30p. © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers Consider Appu and Ashwin running in the opposite direction: Relative speed = (p + s) st To meet for the 1 time, the distance covered would be equal to the length of the track = L. st Thus, the time taken to meet for the 1 time = L/(p + s) = 40s/(7/3s) = 120/7 seconds. Hence, in 2 minutes = 120 seconds, they would meet each other 7 times. Hence, the correct answer is option 5. 19. Kumar and Singh are two painters, who are hired to paint 3 identical houses. They both paint the first house together st and then Kumar paints the second house while Singh paints the third house. The ratio of the times taken to paint the 1 nd st rd and the 2 house is 3:4. Find the ratio of the time taken to paint the 1 and the 3 house? (1) 1:2 (2) 1:4 (3) 3:8 (4) 3:4 (5) 1:3 Solution: Let the rate of painting of Kumar and Singh be denoted by k and s per hour respectively. st Then, the time taken to paint the 1 house = 1/(k + s). nd Also, the time taken to paint the 2 house = 1/(k). st nd The ratio of the times taken to paint the 1 and the 2 house = 3/4 = [1/(k + s)]/[1/(k)] = k/(k + s) = 3/4. → 4k = 3k + 3s → k = 3s st rd Now, the ratio of the times taken to paint the 1 and the 3 house = [1/(k + s)]/[1/(s)] = s/(k + s) = 1/4. Hence, the correct answer is option 2. 3

20. A pyramid shaped container box has a volume of 10240 cm and a square base of 32 cm by 32 cm. Smaller toy 3 pyramids have a volume of 160 cm and 8 cm by 8 cm base. How many such toys can be packed in the container such that is completely filled? (1) 32 (2) 48 (3) 36 (4) 64 (5) None of these Solution: Let base of the box pyramid be denoted by b1. 3 2 Volume of the box pyramid = 10240 cm = (1/3)b1 h1 2 2 → h1 = (10240 × 3)/b1 = (10240 × 3)/32 → h1 = 30 cm. Let base of the toy pyramid be denoted by b2. 3 2 Volume of the box pyramid = 160 cm = (1/3)b2 h2 2 2 → h2= (160 × 3)/b2 = (160 × 3)/8 → h2= 7.5 cm. The box and the toy pyramid are similar since the height to base length is equal for both. Also, the height ratios of the box and the toy pyramid = 30/7.5 = 4:1. This means that 4 levels or layers of the toys can be fitted into the box. In the box, the toys are placed either facing up or down. st In the 1 or the bottom layer, in one line, 4 toys can be fitted facing up and 3 toys facing down can be fitted in between the toys facing up. There would be 4 such lines, hence 4 × 4 = 16 toys facing up and 3 × 3 = 9 toys facing down could be fitted in the bottom layer. nd In the 2 layer, there would be 3 such lines, hence 3 × 3 = 9 toys facing up and 2 × 2 = 4 toys facing down could be fitted. rd In the 3 layer, there would be 2 such lines, hence 2 × 2 = 4 toys facing up and 1 × 1 = 1 toy facing down could be fitted. th In the 4 or top layer, there would be 1 such line, hence 1 × 1 = 1 toy facing up and no toys facing down could be fitted. Thus, in all (16 + 9 + 4 + 1) + (9 + 4 + 1) = 44 toy pyramids could be fitted completely in the box pyramid. Hence, the correct answer is option 5. 21.Ramesh is richer than Satish but Jaya is less rich than Ramesh. Ram is less rich than Jaya but richer than Satish, but is not as rich as Ramesh. Ramesh is less rich than Navin. The richest amongst them is? 1) Ramesh 2) Satish 3) Navin 4) Jaya 22.From the given alternative words, select the word which cannot be formed using the letters of the given word: CARPENTER © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers 1) NECTAR

2) CARPET

3) PAINTER

4) REPENT

23.If FRIEND is coded as HUMJTK, how can CANDLE be written in that code? 1) EDRIRL 2) ESJFME 3) DCQHQK 4) DEQJQM 24.If A denotes −, C denotes ×, D denotes ÷, E denotes +, then 14C3A12E4D2= ? 1) 6 2) 17 3) 28 4) 32 25.If 879= 8, 625= 1, 586= 9, then 785 = ? 1) 6

2) 7

3) 8

4) 9

26.Sham goes to his friend's house that is straight 10 Km from his house. On his way back, he takes a right turn and walks 2 Km and again takes a right turn and walks for 10 Km before he takes a right turn again. How much distance is Sham still away from his house? 1) 10 Km 2) 8 Km 3) 12 Km 4) 2 Km 27.One or two statements are given followed by two conclusions, I and II. You have to consider the statement to be true, even if it seems to be at variance from commonly known facts. You are to decide which of the given conclusions/assumptions can definitely be drawn from the given statement. Indicate your answer. Statement: All boys are tall. Rajiv is a boy. Conclusion I: Rajiv is tall. II: Rajiv is not tall. 1) Only conclusion I follows 2) Only conclusion II follows 3) Both conclusion I and conclusion II follow 4) Neither conclusion I nor conclusion II follows 28.Select the related word/letters/numbers from the given alternatives: Owl : Hoots :: Hen : ? 1) Chirps

2) Clucks

3) Coos

4) Cackles

29.Select the related word/letters/numbers from the given alternatives: AKU : ? :: CMW : DNX 1) BGL

2) BLQ

3) BGQ

4) BLV

30.Select the related word/letters/numbers from the given alternatives: 5 : 100 :: 7 : ? 1) 49

2) 196

3) 91

4) 135

Directions for questions 31: Refer the following data to answer the questions given below. Two people Sripriya and Srividhya play a card game each of them starting with 50 points. There are 10 cards kept on the table, numbered from 1 to 10. The player who plays first says a number between 1 to 10 and picks up a card. If the face value of the card is greater than or equal to that number, then she gains as many points as the face value of the card. Else the opponent gains points equal to twice the face value of the card. Srividhya challenges Sripriya and starts playing the game. In the first round, Srividhya said ‘3’ and picked the card numbered 10. Sripriya said ‘1’ and picked the card with the number 8. In the second round, Srividhya said ‘6’ and picked the card numbered 3, while Sripriya said ‘6’ and picked the card with the number 10. 31.In the third round when Srividhya plays first, she gets the maximum number of points possible. If Sripriya plays next, in how many ways can Srividhya have more points than Sripriya after the third round? © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers (1) 36

(2) 28

(3) 21

(4) 15

(5) None of these

Solution: Not considering the starting points into account Points at the Srividhya Sripriya end of Round 1 10 8 10 + 0 = 8 + 6 + 10 = 2 10 24 10 + 10 = 3 20 In the third round Sripriya can say any number from 1 to 10 and we should see in how many cases Sripriya looses more than 4 points to Srividhya. This can happen when Sripriya picks a card of face value 3 or more while he said a larger number. To calculate the number of ways when Sripriya will give Srividhya more than 4 points The number Unfavourable card that is Number of said picked up ways by Sripriya Any card from with face 10 7 value 3 to 9 Any card from with face 9 6 value 3 to 8 Any card from with face 8 5 value 3 to 7 Any card from with face 7 4 value 3 to 6 Any card from with face 6 3 value 3 to 5 Any card from with face 5 2 value 3 to 4 The card with face value 4 1 3 Total number of ways 28 Hence, the correct answer is option 2.

Directions for questions 32 to 34: Refer the following data to answer the questions given below. The table below gives the analysis of quantitative ability section of CAT in its topics and subtopics from the year 2002 to 2007. ANALYSIS OF QUANTITATIVE ABILITY SECTION OF CAT 2002 – 2007 Number of questions CAT 2007 CAT 2006 CAT 2005 CAT 2004 CAT 2003 CAT 2002 Time, speed, distance & Work 2 1 4 3 1 5 Ratio, Proportion & Variation 0 1 0 2 1 2 Arithmetic Number systems 1 7 5 3 11 19 Percentages 5 1 0 0 1 0 Profit and Loss 0 0 0 0 1 0 Mordent Math Permutations & Combinations 0 0 0 4 3 5 Maxima and Minima 1 2 2 0 1 0 Algebra Equations & Inequalities 5 4 2 3 7 6 Sequences and Series 2 2 2 3 2 1 TOPICS

SUB-TOPICS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers

Geometry Others

Functions Logarithms Coordinate Geometry Trigonometry Mensuration Logic

1 0 0 2 1 5

1 0 0 4 0 2

2 0 1 6 3 3

3 1 1 8 2 2

3 0 1 4 9 5

2 0 0 5 3 1

MM

Arithmetic

Topics

Solutions to questions 32 to 34 Number of questions Sub200 200 topics 2007 2005 2003 6 4 TSD & 2 1 4 3 1 W RPV 0 1 0 2 1 NS 1 7 5 3 11 Per. 5 1 0 0 1 P&L 0 0 0 0 1 PC

Algebra

MM E&I S&S Function s Logarith ms CG Geomet Trig. ry Mens. Others Logic Total

Total 200 No. of Qs 2 5

16

2 6 76 19 46 0 7 0 1

0

0

0

4

3

5

12 12

1 5 2

2 4 2

2 2 2

0 3 3

1 7 2

0 6 1

6 27 12

1

1

2

3

3

2

12

0

0

0

1

0

0

1

0 2 1 5 25

0 4 0 2 25

1 6 3 3 30

1 8 2 2 35

1 4 9 5 50

0 5 3 1 49

58

3 29 50 18 18 18 214 214

32. When the given six years taken together, the number of subtopics that have the same number of questions is utmost (1) 2 (2) 3 (3) 4 (4) 5 (5) None of these Solution: There are 18 questions in each of the subtopics Logic and Mensuration. There is one question in each of the subtopics Profit & Loss and Logarithms. There are 12 questions in each of the subtopics Permutations & Combinations, Functions and Sequence & Series. The number of subtopics that had the same number of questions is utmost 3. Hence, the correct answer is option 2. 33. When the given six years taken together, what are the following topics, put together will have the same number of questions as that of the Arithmetic topic? (1) Algebra & Others (2) Geometry & Others (3) Modern Math & Algebra (4) Modern Math & Geometry (5) None of these Solution: The total number of questions in Arithmetic is 76. The total number of questions in Algebra is 58 The total number of questions in Geometry is 50 The total number of questions in Others is 18 The topics ‘Others’ and ‘Algebra’ put together will have 76 questions. Hence, the correct answer is option 1.

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers 34. Considering the given six years together, which of the following is approximately equal to the percentage of “Time, speed, distance & work” questions in Arithmetic topic? (1) The percentage of questions in “Number systems” in all six years taken together. (2) The percentage of “Sequence and Series” questions in Algebra topic in all six years taken together. (3) The percentage of “Functions” questions in Algebra in all six years taken together. (4) The percentage of questions in “Trigonometry and Mensuration” in all six years taken together. (5) All the above. Solution: The percentage of “Time, Speed and Work” questions in Arithmetic topic = (16 × 100)/76 = 21.05% ≅ 21% Option 1: The percentage of Number systems questions = (46 × 100)/224 = 20.54% ≅ 21% Option 2: The percentage of ‘sequence & series’ questions in Algebra: = (12 × 100)/58 = 20.70% ≅ 21% Option 3: The percentage of ‘Functions’ questions in Algebra: = (12 × 100)/58 = 20.70% ≅ 21% Option 5: The percentage of ‘Trigonometry and Mensuration’ questions in Algebra: = (47 × 100)/224 = 21% Hence, the correct answer is option 5. Directions for questions 35 to 37: Refer the following data to answer the questions given below. The table below gives the number of Airfone Mobile connection users per thousand of Vodatel Mobile connection of eighteen different cities of India in different census years. City

2003

2004

2005

2006

2007

A B C D E F G H I J K L M N O P Q R

899 875 780 976 963 937 917 912 845 968 1002 983 920 945 960 1005 850 973

923 912 743 960 948 952 920 924 837 970 1008 975 918 950 980 1010 860 986

950 936 828 948 976 968 950 940 858 965 1010 970 927 953 990 980 888 980

981 948 869 932 989 978 980 948 850 963 1012 960 925 960 980 975 900 960

1000 976 1001 897 998 999 940 960 864 976 1009 965 930 963 990 1008 930 972

35. For how many of the given cities, the number of Airfone users as a percentage of Vodatel users is the at least two of the given census years? (1) 2 (2) 4 (3) 3 (4) 0 (5) 1

same

for

Solution: Observing the given data. It is the same for only city O. Hence, the correct answer is option 5. © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers 36. If the number of Airfone users as a percentage of Vodatel users is between 94.9% and 97.5%, then the city is graded as A. If the number of Airfone users as a percentage of Vodatel users is between 89.9% and 95%, then the city is graded as B. During which of the census years mentioned above did the number of cities graded A and B, respectively increase and decrease compared to the previous census year? (1) 2007 (2) 2004 (3) 2006 (4) 2005 (5) None of these Solution: The number of cities graded as A for the given census years are as follows: 4, 4, 6, 4, 4. The number of cities graded as B for the given census years are as follows: 5, 6, 4, 5, 3. Only during 2005, the number of cities graded as A increased but the number of cities graded as B decreased. Hence, the correct answer is option 4. 37. If the number of Airfone users as a percentage of Vodatel users is between 69.9% and 90%, then the city is graded as C. If the number of Airfone users as a percentage of Vodatel users is between 94.9% and 97.5%, then the city is graded as A. If the number of Airfone users as a percentage of Vodatel users is between 97.4% and 100.1%, then the city is graded as D. How many cities was graded C in 2003, but became a city with grade A in 2005 and became a city with grade D in 2007? (1) 2 (2) 1 (3) 3 (4) 0 (5) 4 Solution: By observation only city A, which was a city with grade C in 2003, became a city with grade A in 2005 and became a city with grade D in 2007. Hence, the correct answer is option 2. Directions for questions 38 to 40: Refer the following data to answer the questions given below. Four friends – Veeru, Bajji, Yuvi and Sachhu – gave me the following information about the comparisons between their marks in Physics, Chemistry, Biology and Mathematics. Veeru : I have scored more marks in Physics than Bajji. Yuvi has scored more marks in Chemistry than Sachhu. Bajji : Yuvi has scored more marks in Biology than Sachhu. Veeru is the top scorer in Mathematics. Yuvi : Bajji has scored more marks in Chemistry than me. I am not the top scorer in Physics. Sachhu : Veeru is not the top scorer in Physics. I have not scored lowest in Mathematics. Each person among them always alternates between truth and lie, in any order. Each person must be ranked first in only one of the four given subjects, in such a way that the highest scorer in any of the given subjects is ranked first and the lowest scorer in any of the given subjects is ranked fourth and so on. No person gets the same rank in any two subjects, and no two persons are ranked the same in any one subject. Also, I was provided with the following facts: Veeru is the top scorer in Mathematics, but has not scored the lowest in Physics. Bajji is the lowest scorer in Biology, but not the highest scorer in Chemistry. The lowest scorer in Mathematics among them has not secured the third rank in Physics. The second highest ranker in Mathematics has scored the second lowest in Biology. Solutions to questions 38 to 40: Let us not the information as shown below: (1) Four friends, Veeru (A), Bajji (B), Yuvi (C) and Sachhu (D), are compared in terms of their marks in Physics, Chemistry, Mathematics and Biology. (2) All of them are alternators (TF or FT) (3) A is the top scorer in Mathematics but A has not scored the lowest in Physics. (4) B has scored the lowest in Biology but B is not the top scorer in Chemistry. (5) The lowest scorer in Mathematics has not secured the third rank in Physics. (6) The second ranker in Mathematics has scored the second lowest in Biology. From (3), B is second statement is true and the first statement is false. st st A is 1 ranker in Mathematics; therefore A cannot be the 1 ranker in Physics (For he has to get a different rank). th From (3), A is not 4 ranker in Physics. th Hence, D’s first statement is true and the second statement is false which means D is the 4 ranker in Mathematics. © 2017 ETHNUS

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BITS PILANI Test - 8 – Answer Key & Explanatory Answers nd

rd

From (4) & (6), the 2 ranker in Mathematics is the 3 ranker in Biology. nd rd rd B cannot be the 2 ranker in Mathematics, because he is not the 3 ranker in Biology. So, B is the 3 ranker in nd rd Mathematics and C is the 2 ranker in Mathematics and 3 ranker in Biology. st st nd From (4) B is not the 1 ranker in Chemistry, which means B must be the 1 ranker in Physics and 2 ranker in Chemistry. rd th th From (5) D is not the 3 ranker in Physics, D cannot also be the 4 ranker in Physics for he is the 4 ranker in nd Mathematics. So, D has to the 2 ranker in Physics. th rd th From (3) A is not the 4 ranker in Physics, so A has to be 3 ranker in Physics and C is the 4 ranker in Physics and the st 1 ranker in Chemistry. st st nd Since A cannot be the 1 ranker in Biology for A is already the 1 ranker in Mathematics, A has to be the 2 ranker in th Biology and 4 ranker in Chemistry. The left out ranks are the third rank in Chemistry and first rank in Biology will be filled by D. Ran Physi Chemist Biolog Mathemat ks cs ry y ics 1 B C D A 2 D B A C 3 A D C B 4 C A B D 38. Who among the following has secured the second rank Biology? (1) Veeru (2) Yuvi (3) Sachhu (4) Bajji

(5) Indeterminate

Solution: nd A is the 2 ranker in Biology Hence, the correct answer is option 1. 39. Which among the following is the correct order from first ranker to the fourth ranker in Chemistry? (1) Veeru, Bajji, Yuvi, Sachhu (2) Bajji, Yuvi, Sachhu, Veeru (3) Sachhu, Bajji, Veeru, Yuvi (4) Yuvi, Bajji, Sachhu, Veeru (5) None of these Solution: Hence, the correct answer is option 4. 40. Which of the following statements is true? (1) Yuvi’s first statement is true. (2) Bajji is the lowest scorer in Chemistry (3) Sachhu and Veeru have secured the second rank in Chemistry and Physics respectively (4) Out of the first statement given by each of Bajji and Yuvi, only one is true. (5) None of these Solution: Hence, the correct answer is option 3.

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