Test 2 Quant & DI - 40 Qs - Explanatory Answers

Explanatory Answers 41

1. What is the remainder when 15 + 7 (1) 0 (2) 1 n

41

is divided by 11? (3) 4

(4) 8

(5) 10

n

Solution:It is known that a + b is exactly divisible by (a + n), when n is odd. For example, 3 3 2 2 i) a + b = (a + b)(a – ab + b ) 5 5 4 3 2 2 3 4 ii) a + b = (a + b)(a – a b + a b – ab + b ). 41 41 n n Let N = 15 + 7 = a + b , where a = 15, b = 7, n = 41 (odd). 41 41 Thus, (15 + 7 ) is exactly divisible by (15 + 7 = 22). 41 41 Since 22 is a multiple of 11, (15 + 7 ) is also exactly divisible by 11. Thus, the remainder is zero. Hence, the correct answer is option 1. 2. Consider Set S = {N1, N2... Nn}. The set S contains all those natural numbers that are less than 840, such that the highest number that can divide any of the number in set S and 840 is 1. What is the sum of all the numbers in the set S? (1) 840 (2) 8640 (3) 80640 (4) 806400 (5) None of these Solution: By definition, set S contains all the co-primes of the number 840. The number 840 expressed in terms of its factors: p q r s 3 2 1 1 Let N = a × b × c × d = 840 = 2 × 3 × 5 × 7 . Number of co-primes of 840 = Nc = N × [1 – (1/a)] × [1 – (1/b)] × [1 – (1/c)] × [1 – (1/d)] = 840 × [1 – (1/2)] × [1 – (1/3)] × [1 – (1/5)] × [1 – (1/7)] = 840 × (1/2) × (2/3) × (4/5) × (6/7) = 192. Sum of co-primes of a number less than the number itself is given by: M = (N × Nc)/2 = (840 × 192)/2 Thus, M = 80640. Hence, the correct answer is option 3. 3.

st

nd

Two brothers, Makaanwala and Gharwala sell both of their houses. They sell the 1 one at a loss of 25%. On the 2 one, they mark its price up by 80% and then offer a discount of 40%. What is the overall loss percentage on the two houses if the selling prices of both the houses are the same? (1) 15.0 (2) 12.6 (3) 10.2 (4) 11.5 (5) None of these

Solution: Let the selling price of both the houses be Rs 1000 each. st i) Loss % on the 1 house = 25% st Cost price of the 1 house = [100/(100 – 25)] × (1000) = Rs 4000/3. nd ii) A discount of 40% on the marked price leads to the selling price, thus Marked Price of the 2 house = [100/(100 – 40)] × (1000) = Rs 5000/3. nd Cost price of the 2 house = (5000/3) × [100/(100 + 80)] = Rs 25000/27 Total Cost Price = (4000/3) + (25000/27) = Rs 61000/27 = Rs 2259.26. Total Selling Price = Rs 2000. Thus, Overall Loss = 100 × (2259.26 – 2000)/2259.26 = 11.5%. Hence, the correct answer is option 4.

4. India and Sri Lanka are playing a 5 match ODI cricket series. Each win earns four points for the winner and none to the loser, while a tie earns two points for both the teams. The team which has earned more points at the end of this 5 match series wins the tournament. In how many ways India can win the tournament? (1) 31 (2) 96 (3) 100 (4) 64 (5) None of these Solution: The total number of ways of winning the series is given as follows: 5 i) India won all 5 = P0 = 1 way. © 2017 ETHNUS

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Explanatory Answers 5

ii) India won 4 and lost 1 = P1 = 5 ways. 5 iii) India won 4 and drew 1 = P1 = 5 ways. 5 iv) India won 3 and lost 2 = P2/2! = 10 ways. 5 v) India won 3 and drew 2 = P2/2! = 10 ways. 5 vi) India won 3, drew 1 and lost 1 = P5/3! = 20 ways. 5 vii) India won 2 and drew 3 = P2/2! = 10 ways. 5 viii) India won 2, drew 2 and lost 1 = P5/(2! × 2! ×1!) = 30 ways. 5 ix) India won 1 and drew 4 = P1 = 5 ways. Hence, the total number of ways in which India can win the tournament = 1 + 5 + 5 + 10 + 10 + 20 + 10 + 30 + 5 = 96 ways. Hence, the correct answer is option 2. 5. Two swimmers, Philips and Thorps are on the two opposite sides of a rectangular pool and the distance between the swimmers is 100 m. If Philips can swim at twice the rate at which Thorps can, then how much distance (in km) would st Philips have covered by the time they both meet for the 101 time, if it is known that both start swimming at the same time and they keep on swimming from one end to the other end and back at a constant speed? (1) 5.0 (2) 13.3 (3) 13333 (4) 13.4 (5) None of these Solution:Since Philips swims at twice the speed of Thorps and they start simultaneously from the opposite ends of the pool, Philips always covers twice the distance that Thorps covers in the same time or in other words, Philips covers rd rd 2/3 and Thorps covers 1/3 of the total distance covered by both together. Let the distance between the two ends of the pool be denoted by ‘D’ m. Meeting Point Distance travelled by Philips and Distance that is (MP) Thorps together covered by Philips st

Till 1 MP

(2D/3) + (D/3) = D

2D/3

MP

(D) + [(D/3) + D] + (2D/3) = 3D

3D × 2/3 = 6D/3 = 2D

Till 3 MP

3D + [(D/3) + D] + (2D/3) = 5D

5D × 2/3 = 10D/3

Till 2

nd

rd

From the table above, it can be seen that Philips has to travel a distance of (4D/3) to get to the every successive MP st after the 1 . st Thus, the distance travelled by Philips till 101 MP = (2D/3) + (100) × (4D/3) = 402D/3 = 134D. st Since we know that the length of the pool = D = 100 m, the distance that Philips has covered until 101 MP = 134 × 100 m = 13400 m = 13.4 km. Hence, the correct answer is option 4. 2

3

4

6. Find the sum of all the terms of the series F: (3), (4/13), (44/13 ), (444/13 ), (4444/13 )...? (1) 40/9 (2) 44/13 (3) 564/169 (4) 43/11

(5) None of these

Solution:Let the sum of the series F be denoted by: 2 3 4 T = 3 + (4/13) + (44/13 ) + (444/13 ) + (4444/13 )… (1) 2 3 4 T/13 = (3/13) + (4/13 ) + (44/13 ) + (444/13 ) … (2) (1) – (2) gives, 2 3 4 [T – (T/13)] = 3 + (1/13) + (40/13 ) + (400/13 ) + (4000/13 )… 2 3 4 → [(12T – 40)/13] = (40/13 ) + (400/13 ) + (4000/13 )… st 2 The RHS of the above equation is a Geometric Progression with the 1 term = a = (40/13 ) and the common ratio = r = (10/13) < 1. Sum to infinite terms of this series is given by: S = a/(1 – r) 2 Thus, S = (40/13 )/ [1 – (10/13)] = 40/(3 × 13). → [(12T – 40)/13] = 40/(3 × 13) → 3 × (12T – 40) = 40 © 2017 ETHNUS

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Explanatory Answers → T = 160/36 = 40/9. Hence, the correct answer is option 1. 7. Pipes A and B alone can fill an empty tank in 15 and 10 hours respectively. Initially the tank was empty and pipe A was opened. After some time, pipe B was also opened. In this way, the time taken for the tank to fill up completely was 9 hours less than thrice the number of hours it would have taken for the tank to fill up if both the pipes had been opened simultaneously. Find out how much time (in hrs) after the opening of pipe A was pipe B opened? (1) 4 (2) 5 (3) 6 (4) 8 (5) 9 Solution: th th In 1 hour, pipe A would have filled 1/15 of the tank while pipe B would have filled 1/10 of the tank. th Together, in 1 hour, pipes A and B would have filled (1/15) + (1/10) = 1/6 of the tank. Hence, when open together the pipes A and B would take 6 hours to completely fill the empty tank. Now consider pipe A being opened first and then pipe B after some time: The time taken to fill the tank this way = 3 × 6 – 9 = 9 hrs. Let the time for which only pipe A was opened be t hours. Then, the time the pipe B was open = 9 – t hrs. th Thus, in 9 hours, pipe A fills (9/15 ) of the tank and pipe B fills th (9 – t)/(10 ) of the tank which together lead to filling the tank completely. Thus, (9/15) + [(9 – t)/(10)] = 1 → t = 5 hrs. Hence, the correct answer is option 2. 8. A particular grade of steel contains 95% iron and 5% carbon. Another grade of steel contains 92% iron and 2% carbon. If both these alloys are mixed in such a way that the resultant steel alloy contains 4% carbon, then what percentage of iron would the resultant steel alloy contain? (1) 93.0 (2) 93.5 (3) 94 (4) 94.5 (5) None of these Given that: Grade 1 steel: 95% iron and 5% carbon Grade 2 steel: 92% iron and 2% carbon Mixture of 2 grades: 4% carbon Thus, the alligation rule for carbon % is as listed below: 5 2 4 2 1 Thus, to get 4% carbon, the alloys (grade 1 and 2 steel) need to be mixed in the ratio of 2:1. So, if there is 2 kgs of grade 1 steel, then the alloy mixture should have 1 kg of grade 2 steel. Now, 2 kgs of grade 1 steel would have 95% iron = 0.95 × 2 = 1.9 kgs. 1 kg of grade 2 steel would have 92% iron = 0.92 kg. The mixture would weigh 2 + 1 = 3 kgs in all. The total amount of iron in this mixture = 1.9 + 0.92 = 2.82 kgs. Thus, the percentage of iron in the mixture = (2.82/3) × 100 = 94%. Hence, the correct answer is option 3. 2

9. If the roots of the quadratic equation: ax + bx + c = 0 are denoted by p and q respectively (p, q ≠ 0), then the expression: [(p/q) + (q/p) + 2{(1/p) + (1/q)} – pq] would be equal to which of the following expressions in terms of a, b and c? 2 2 2 2 (1) (1/a) ×{[(b – a )/c] – 2(a + b)} (2) (1/a) ×{[(b – a )/a] – 2(b + c)} 2 2 2 2 (3) (1/b) ×{[(c – b )/b] – 2(a + c)} (4) (1/c) ×{[(b – c )/a] – 2(b + c)} (5) None of these 2

Solution:Given equation: ax + bx + c = 0, where p and q are its roots. Sum of the roots = p + q = -b/a. Product of the roots = pq = c/a. Let given expression = E = [(p/q) + (q/p) + 2{(1/p) + (1/q)} – pq] 2 2 E = [(p + q )/(pq) + 2{(p + q)/(pq)} – pq] © 2017 ETHNUS

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Explanatory Answers 2

2

2 2

→ E = [{p + q + 2(p + q) – p q }/(pq)] 2 2 2 2 → E = [{p + q + 2pq + 2(p + q) – p q }/(pq)] – 2 2 2 → E = [{(p + q) + 2(p + q) – (pq) }/(pq)] – 2 2 2 → E = [{(-b/a) + 2(-b/a) – (c/a) }/(c/a)] – 2 2 2 2 2 → E = [a ×{(b /a ) + 2(-b/a) – (c /a )}/(c)] – 2 2 2 → E = [{(b /a) + 2(-b) – (c /a)}/(c)] – 2 2 2 → E = [[{(b – c )/(a)} – 2b]/(c)] – 2 2 2 → E = [[{(b – c )/(a)} – 2b – 2c]/(c)] 2 2 → E = [[{(b – c )/(a)} – 2(b + c)]/(c)] 2 2 → E = (1/c) × [{(b – c )/(a)} – 2(b + c)] Hence, the correct answer is option 4. 10. If P(n) denotes the product of all the digits of the number n, then find the value of P(52) + P(53) + ... + P(151)? (1) 1570 (2) 2025 (3) 2375 (4) 3025 (5) None of these Solution: Let X = P(52) + P(53) + ... + P(151). Let X1 = P(52) + P(53) + … + P(99) and X2 = P(100) + P(101) + … + P(151) X1 = P(52) + P(53) + … + P(99) X1 = (5 × 2) + (5 × 3) + (5 × 4) + … + (5 × 9) + (6 × 1) + (6 × 2) … + (6 × 9) + (7 × 1) + (7 × 2) … + (7 × 9) + (8 × 1) + (8 × 2) … + (8 × 9) + (9 × 1) + (9 × 2) … + (9 × 9) → X1 = (5 + 6 + 7 + 8 + 9) × (1 + 2 + … + 9) – (5 × 1) X2 = P(101) + P(102) + … + P(151) = P(111) + P(112) + … + P(151) {Since P(101) = P(102) = 0} X2 = [(1 × 1) × (1 + 2 + … + 9)] + [(1 × 2) × (1 + 2 + … + 9)] + [(1 × 3) × (1 + 2 + …+ 9)] + [(1 × 4) × (1 + 2 + … + 9)] + [(1 × 5)] → X2 = (1 + 2 + 3 + 4) × (1 + 2 + … + 9) + (5 × 1) → X = X1 + X2 = (1 + 2 + … + 9) × (1 + 2 + … + 9) → X = (45 × 45) = 2025. Hence, the correct answer is option 2. 11. How many distinct sets, of positive integral values of x and y, satisfy the equation: (2/x) + (3/y) = 1/12, where x, y < 100? (1) 6 (2) 8 (3) 12 (4) 14 (5) 9 Solution: x 9 6 7 8 7 2 6 0 5 6 5 1 4 8 4 2 4 0

y 4 8 5 2 5 4 6 0 6 3 6 8 7 2 8 4 9 0

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Explanatory Answers Given equation: (2/x) + (3/y) = 1/12, where x, y < 100. → 12 × (3x + 2y) = xy → 36x + 24y = xy. → x = 24y/(y – 36); → y = 36x/(x – 24). Thus, it can be seen that x > 24 and y > 36. The set of positive values for x and y which satisfy the given equation are listed in the table. This table is obtained by considering values for y (greater than 36 but less than 100) such that the corresponding value for x is greater than 24 but less than 100. Thus, it can be seen that there are 9 pairs of positive integral values for x and y which satisfy the given equation. Hence, the correct answer is option 5. 12. PQRS is a square of side 2 cms. T is a point on side PS such that the lines PT: RT is 1:5. If RT and SQ intersect at point U, then find the measure (in cm) of line segment TU? (1) 6/7 (2) 13/14 (3) 15/14 (4) 8/7 (5) None of these Solution: Let the following diagram represent the given data: Here, PQ = QR = RS = SP = 2 cms. Let PT = x, then TS = 2 – x. Since PT: RT = 1:5, RT = 5x. In triangle TSR, it can be seen that: 2 2 2 RS + TS = RT 2 2 2 2 → 2 + (2 – x) = (5x) → 6x + x – 2 = 0 → x = (1/2) or (-2/3), where only x = 1/2 is a valid value. Thus, PT = 0.5, TS = 1.5, RT = 2.5 cms. Here, it can be seen that triangle SUT is similar to triangle QUR. ST/QR = TU/RU = SU/QU = 3/4. RT/TU = (RU + TU)/TU = 1 + 4/3 = 7/3. Also, RT/RS = 2.5/2 = 5/4. RT/TU = (RT/RS) × (RS/TU) → 7/3 = (5/4) × (2/TU) → TU = 15/14 cms. Hence, the correct answer is option 3. 13. A group of families, each family consisting of 2 adults and 1 child or more went on a fun trip to Six Flags Amusement Park. The price of an entrance ticket for an adult was different from that for a child. They paid the total entrance cost and entered the park. But soon the ticket seller realised that he got Rs 600 less than what he should have actually got. This happened because while he made the bill, he had interchanged the number of adults and children. Now, how many of the given statements are not valid? Statement 1: The price for an adult ticket is Rs 30 more than that of one ticket for a child. Statement 2: The price for a child ticket is Rs 40 more than that of one ticket for an adult. Statement 3: There were 80 adults less than the total number of children. Statement 4: There were 25 children more than the number of adults. (1) 4 (2) 2 (3) 3 (4) 1 (5) 0 Solution: Let the price of 1 ticket for an adult and a child be represented by p and q respectively. Also, let the number of adults and children be represented by a and b respectively. Let the total entrance cost paid be denoted by E. Then, i) pa + qb = E (1) ii) qa + pb = E + 600 (2) (2) – (1) gives, (q – p) × (a – b) = 600. Now, consider the given statements: Statement 1: If (p – q) = 30, then (b – a) = 600/30 = 20. Thus, it is a valid statement. Statement 2: If (q – p) = 40, then (a – b) = 600/40 = 15. Thus, it is a valid statement. Statement 3: If (b – a) = 80, then (p – q) = 600/80 = 7.5. Thus, it is not a valid statement. © 2017 ETHNUS

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Explanatory Answers Statement 4: If (b – a) = 25, then (p – q) = 600/25 = 24. Thus, it is a valid statement. Hence, only one statement is invalid. Hence, the correct answer is option 4. 14. How many of the following statements are definitely true? Statement 1: (1/3) > log303 > (1/4). Statement 2: (1/2) < log205 < (1). Statement 3: (1/5) > log302 > (1/6). Statement 4: (2/3) > log74 > (2/5). (1) 2 (2) 4 (3) 0

(4) 1

(5) 3

Solution: Consider the 4 given statements: Statement 1: (1/3) > log303 > (1/4) → 3 < (log 30/ log 3) < 4 → 3 log 3 < (log 30) < 4 log 3 3 4 → log 3 < (log 30) < log 3 → 27 < 30 < 81. Thus, statement 1 is true. Statement 2: (1/2) < log205 < (1) → 2 > (log 20/ log 5) > 1 → 2 log 5 > (log 20) > log 5 2 → log 5 > (log 20) > log 5 → 25 > 20 > 5. Thus, statement 2 is true. Statement 3: (1/5) > log302 > (1/6) → 5 < (log 30/ log 2) < 6 → 5 log 2 < (log 30) < 6 log 2 5 6 → log 2 < (log 30) < log 2 → 32 < 30 < 64, which is a false statement Thus, statement 3 is false. Statement 4: (2/3) > log74 > (2/5). → 3/2 < (log 7/ log 4) < 5/2 → (3/2) log 4 < (log 7) < (5/2) log 4 1.5 2.5 → log 4 < (log 7) < log 4 → 8 < 7 < 32, which is a false statement Thus, statement 4 is false. Hence, 2 statements are true. Hence, the correct answer is option 1. 15. The volume of a particular cube is numerically equal to the sum of the measures (in cm) of all its edges. What is the total 3 volume (in cm ) of another cube whose surface area is equal to thrice that of the original cube? (1) 54√2 (2) 64 (3) 216 (4) 27 (5) None of these Solution: Given: Volume of a cube is equal to the measure of the sum of its edges. Let the side of the cube be denoted by a. Number of edges of a cube = 12. Thus, the sum of the edges = 12a. 3 Volume of the cube = a . 3 → a = 12a → a = √12 2 Surface area of the cube = 6a = 6 × 12 = 72 sq.cms. Thus, surface area of another cube = 3 × 72 = 216 sq.cms. Hence, side of another cube = b = √(216/6) = 6. © 2017 ETHNUS

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Explanatory Answers 3

Thus, volume of another cube = b = 216 cm Hence, the correct answer is option 3.

3

16. What number should be added to D = 484848...4848 (120 digits in all), such that it would be exactly divisible by 198? (1) 6 (2) 48 (3) 98 (4) 150 (5) 192 Solution: Given: D = 484848...4848 (120 digits). Dividing the numerator and denominator by 2, we get that 4848…4848/ 198 = 242424...2424/99. Now, dividing the numerator and denominator by 3, we get 242424...2424/99 = 080808…0808/33. Adding all the digits of the numerator, we get the sum to be equal to (0 + 8) × 120/2 = 480, which is a multiple of 3; hence the numerator is also a multiple of 3. Thus, 080808…080808/33 = 026936…026936 (20 sets of 6 numbers)/11. Here, the number set 026936 when divided by 11 leaves a remainder of (2 + 9 + 6) – (0 + 6 + 3) = 8. Thus, for a number consisting 20 such sets of 026936, the remainder should be equal to 20 × [(2 + 9 + 6) – (0 + 6 + 3)] = 20 × 8 = 160, which when divided by 11 leaves a final remainder of 6. Thus, (198 – 6) = 192 needs to added to 484848…4848 so that it is divisible by 198. Hence, the correct answer is option 5.

17. Courier delivery business consists of 2 processes: One is sorting of the packages at the distribution centre and the other is delivering the courier package to the home/office of the customer from the distribution centre. Sorting any kind of package takes the same time and it is equal to the time taken to travel 4 kms. The delivery schedule for the day was as follows: Courier delivery was required for four places (A, B, C and D), which were at a distance of 3, 5, 8 and 12 kms respectively from the distribution centre. After each delivery, it was required to come back to distribution centre and then go for another delivery. If the total time spent on the courier delivery needs to be reduced by 25%, then by what percentage must the travel speed per km be increased if the sorting time remains the same? (1) 64.7 (2) 60.7 (3) 154.5 (4) 35.5 (5) None of these Solution: Let the time taken to travel 1 km be denoted by t. Then, the time taken to sort 1 package = 4t. Since there are a total of 4 packages, the time taken to sort all 4 would be equal to (4 × 4t) = 16t. The total time taken to transport all the 4 packages = 3t + 5t + 8t + 12t = 28t. Hence, the total time taken to complete the entire courier delivery business = 16t + 28t = 44t. The total travel distance = 3 + 5 + 8 + 12 = 28 kms. Thus, the average travel speed = 28/28t = 1t. Since the total time has to be reduced by 25%, the new total courier delivery time should be 0.75 × 44t = 33t. Since sorting time remains constant = 16t, the new total transport time should be = 33t – 16t = 17t. Thus, the average travel speed = 28/17t. Hence, the percentage increase in travel speed = 100 × [(28/17t) – (t)]/(t) = 1100/17 = 64.7% Hence, the correct answer is option 1. Solutions for questions 18 and 19: Let the prize money in the 5 suitcases be denoted by p, q, r, s and t respectively where p < q < r < s < t. 5 Since 5 suitcases are checked in sets of 3, a total of C3 = 10 prize money set counts (pqr, pqs, pqt, prs, prt, pst, qrs, qrt, qst, rst) are present. Here, in the 10 sets, it can be seen that each suitcase is counted 6 times. Thus, sum of all the prize money set counts = 13 + 16 + 18 + 19 + 20 + 22 + 23 + 25 + 26 + 28 = 210 = 6 × (p + q + r + s + t) → (p + q + r + s + t) = 210/6 = 35 (1) The least prize money set count = 13 = p + q + r (2) The highest prize money set count = 28 = r + s + t (3) (1) – (2) gives, © 2017 ETHNUS

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Explanatory Answers s + t = 22 (4) (1) – (3) gives, p+q=7 (5) nd The 2 least prize money set count = 16 = p + q + s (6) Using (5) and (6) and then (4), we get s = 9 and t = 13. nd The 2 highest prize money set count = 26 = q + s + t Using (4) and (7), then (5) and finally (1), we get q = 4, p = 3 and r = 6.

(7)

Directions for questions 18 and 19: Refer the following data to answer the questions given below. In a new TV reality game show, “Pick and Take” (ala Deal or No Deal), there are five suitcases, each of which contains a cash price (in multiples of 1000s of Rs). The total prize money summed up when three suitcases are grouped in each set is revealed to be as follows: 13, 16, 18, 19, 20, 22, 23, 25, 26 and 28. 18. What is the average of the prize money (in multiples of 1000 Rs) present in all the 5 suitcases? (1) 5 (2) 8 (3) 6 (4) 9 Solution: Average of the prize money in all the 5 suitcases = 35/5 = 7. Hence, the correct answer is option 5.

(5) 7

19. What is the difference between the maximum and the minimum prize money (in multiples of 1000 Rs) in those 5 suitcases? (1) 8 (2) 10 (3) 9 (4) 11 (5) 12 Solution: The difference between the maximum and the minimum prize money in the suitcases = t – p = 13 – 3 = 10. Hence, the correct answer is option 2. 20. A shop is selling mints (wrapper covered ones) at 1 Re. each. It also gives a free mint if 3 empty wrappers are returned back to the shop. With Rs 149, a maximum of how many mints could be obtained from the shop? (1) 198 (2) 220 (3) 224 (4) 222 (5) 223 Solution: The maximum number of mints could be obtained as shown: i) With Rs 149, at a price of Rs 1 per mint, a total of 149 mints could be purchased. ii) These 149 mints generate 149 wrappers which in turn could be exchanged for 49 mints leaving behind 2 extra wrappers. iii) The 49 mints generate 49 wrappers which in turn could be exchanged for 16 mints leaving behind 1 extra wrapper. iv) Now there are a total of 16 + 1 + 2 = 19 wrappers which could be exchanged for 6 mints leaving behind 1 extra wrapper. v) The 6 mints generate 6 wrappers which in turn could be exchanged for 2 mints leaving behind no extra wrappers. vi) Now there are a total of 1 + 2 = 3 wrappers which could be exchanged for 1 mint leaving behind no extra wrappers. Hence, a total of 149 + 49 + 16 + 6 + 2 + 1 = 223 mints could be obtained from the shop with Rs. 149. Hence, the correct answer is option 5. Directions for questions 21 to 24: Refer the following data to answer the questions given below. The world’s top twelve tennis players, Roger Federer (RF), Rafael Nadal (RN), Novak Djokovic (ND), Pete Sampras (PS), Andy Roddick (AR), Andre Aggassi (AA), Tommy Rick (TR), James Blake (JB), Robby Ginepri (RG), Tim Henman (TH), Roy Emerson (RE) and Bobby Riggs (BR) respectively ranked 1 through 12, participate in a tournament called WIMBLORE consisting two rounds held in the city of Bangalore. In the first round, every tennis player has to play exactly once with every other tennis player. In the table given below the names (in the short form) represent the winners of the matches played in the first round. For example, the winner of the match played between Tommy Rick and Andy Roddick, is AR who is nothing but Andy Roddick. For every win in the © 2017 ETHNUS

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Explanatory Answers first round, the number of points that a tennis player gets is either three or the number by which the losers rank is better than his rank, whichever is greater. The top four performers (in terms of total points) advance to the second round. The players those are qualified for the second round plays once with every other player in the second round. For every win in the second round, the number of points that a tennis player gets is either six or double the number by which the losers rank is better than his rank, whichever is greater. Considering both the first and the second rounds the player with the highest total number of points is announced as the winner of the tournament.

Table of outcomes of all the matches between 12 tennis players in the first round

Pete Sampra s James Blake Novak Djokovi c Roger Federer Andy Roddick Andre Aggassi Tommy Rick Rafael Nadal Robby Ginepri Tim Henma n Roy Emerso n Bobby Riggs

Pete Sampr as

Jam es Blak e

Novak Djokovi c

Roger Feder er

Andy Roddic k

Andre Aggas si

Tomm y Rick

Rafa el Nada l

Robby Ginepr i

Tim Henma n

Roy Emerso n

Bobb y Riggs

-

PS

ND

RF

PS

PS

PS

RN

PS

PS

PS

PS

PS

-

ND

RF

JB

JB

JB

RN

JB

TH

JB

JB

ND

ND

-

ND

AR

ND

ND

RN

ND

ND

ND

ND

RF

RF

ND

-

RF

RF

RF

RF

RF

RF

RF

RF

PS

JB

AR

RF

-

AR

AR

RN

RG

AR

AR

BR

PS

JB

ND

RF

AR

-

AA

AA

AA

AA

RE

AA

PS

JB

ND

RF

AR

AA

-

RN

RG

TR

TR

TR

RN

RN

RN

RF

RN

AA

RN

-

RN

RN

RN

RN

PS

JB

ND

RF

RG

AA

RG

RN

-

RG

RG

RG

PS

TH

ND

RF

AR

AA

TR

RN

RG

-

TH

BR

PS

JB

ND

RF

AR

RE

TR

RN

RG

TH

-

RE

PS

JB

ND

RF

BR

AA

TR

RN

RG

BR

RE

-

21. Whom among the following gets the maximum number of points in the first round of this tournament? (1) Roger Federer (2) Rafael Nadal (3) Novak Djokovic (4) Andy Roddick (5) James Blake Solution: Total points after the tournament: RF è 10 × 3 = 30 points RN è 9 × 3 = 27 points ND è 8 × 3 + 3 = 27 points AR è 5 × 3 = 15 points JB è (3× 3) + 3 + 3 + 3 = 18 points © 2017 ETHNUS

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Explanatory Answers Hence, the correct answer is option 1. 22. Which of the following players has defeated the maximum number of players ranked better than himself in the first round? (1) James Blake (2) Robby Ginepri (3) Tim Henman (4) Andre Aggassi (5) Pete Sampras Solution: By observation, we see that JB has defeated tennis players AR, AA and TR and no one else has defeated (3 or more) better ranked tennis players. Hence, the correct answer is option 1. 23. It is known that the Rafael Nadal fails a doping test and is disqualified because of which the next best performer advances to the second round. Which of the following groups of four players go to the second round from the first to last in the order, in-terms of their total points in the first round? (1) Roger Federer, Novak Djokovi, James Blake, Robby Ginepri (2) Roger Federer, Novak Djokovi, Robby Ginepri, James Blake (3) Roger Federer, Pete Sampras, Novak Djokovi, James Blake (4) Roger Federer, Pete Sampras, Novak Djokovi, James Blake (5) Roger Federer, Novak Djokovi, Pete Sampras, James Blake Solution: Points scored by tennis players PS, RG, RF, RN, ND and JB: PS è 8 × 3 = 24 points RG è 3 × 3 + 4 + 3 = 16 points RF è 10 × 3 = 30 points RN è 9 × 3 = 27 points ND è 8 × 3 + 3 = 27 points JB è (3× 3) + 3 + 3 + 3 = 18 points The four tennis players that go through are RF, ND, PS and JB in that order. Hence, the correct answer is option 5. 24. It is known that the Novak Djokovic fails a doping test and is disqualified because of which the next best performer advances to the second round. Considering both the first and the second rounds, what is the highest possible total number of points that any tennis player can score? (1) 56 (2) 52 (3) 57 (4) 55 (5) None of these Solution: The only contestants into the second round are RF, RN, PS and JB. If RF won all the matches, then The net total of RF is = 30 + 6 + 6 + 6 = 48. If RN won all the matches, then The net total of RN is = 27 + 6 + 6 + 6 = 45 If PS won all the matches, then The net total of PS is = 24 + 6 + 6 + 6 = 42 If JB won all the matches, then The net total of JB is = 18 + 14 + 12 + 8 = 52 Hence, the correct answer is option 2. Directions for questions 25 to 28: Refer the following data to answer the questions given below. Given below is the pipeline network of ‘Nandhini Water Works’ that manages the drinking water supply for Bangalore city across 10 hubs – Jayanagar, Basavangudi, Lalbagh, Richmond Road, Residency Road, M.G.Road, Manipal Centre, Shavaji Nagar, R.T.Nagar and Banaswadi. The directions of the flow through some of the pipelines are indicated on the top of the lines joining two hubs by an arrow-mark. Every hub except Jayanagar, has a certain minimum requirement, which for some of the hubs, is indicated in the circles next to the respective hubs. Each of the pipeline segments has a maximum capacity of 1000 litres per minute. The slack in any pipeline is defined as the extra flow required in it to bring it to full capacity. The magnitudes of the flow in some of the pipelines along with the directions and the requirements at some of the hubs are given (both in litres per minute). The magnitude and the direction of flow in certain pipelines and the requirements at certain hubs are to be found. The Jayanagar hub is the only source of ‘Nandhini Water Works’ and has no minimum requirement. The slack in the pipeline connecting Residency Road and Manipal Centre is 800 litres. © 2017 ETHNUS

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Explanatory Answers

Manipal Centre 100

40

Basavangudi

100

20 0 Shivaji Nagar

R.T.Nagar

10 0

Residency Road

30 0 M.G. Road

60 100

100

Lalbagh Jayanagar

100 30 0

100 Richmond Road

Banaswadi

30 0 25. What is the requirement at the hubs in Shivaji Nagar and Basavangudi put together? (1) 600 litres (2) 500 litres (3) 200 litres (4) 400 litres

(5) 300 litres

Solution:Let us indicate the hubs with the following alphabets. Jà Jayanagar BASVà Basavangudi LàLalbagh RIàRichmond Road REàResidency Road MGàM.G.Road MCàManipal Centre SNàShavaji Nagar B àBanaswadi RTà R.T.Nagar MC 1

20

40 0

RE

100

10

20

60

30 10 0

SN

R

BASV

MG

L J

10

10 3 0

10 B

RI

30 0

Consider the hub in R.T. Nagar, Total inflow at R.T. Nagar hub = 300 + 100 = 400 litres Since R.T. Nagar hub’s requirement is only 100 litres, the remaining 300 litres will be an inflow to Shivaji nagar. Now consider the hub at Shivaji nagar, The 300 litres inflow from R.T. Nagar will be channelled as outflow from Shivaji nagar to Manipal Centre and Banaswadi. So the minimum requirement at Shivaji nagar hub must be supplied from MG Road. If the minimum requirement at hub SN is ‘x’, then the inflow from Basavangudi to M.G. Road is = 300 + x Now consider the hub at Basavangudi, Requirement at Basavangudi hub = (J-BASV) – (B-RE) – (B-MG) = 1000 – 400 – (300 + x) = 300 – x Total requirement at Basavangudi and Shivaji Nagar = 300 – x + x = 300 litres. Hence, the correct answer is option 5.

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Explanatory Answers 26. If the pipeline connecting Richmond Road to Banaswadi has a slack of 400 litres, then what is the total supply from Jayanagar? (1) 2600 litres (2) 2400 litres (3) 2500 litres (4) 2700 litres (5) 2300 litres Solution:Let us indicate the hubs with the following alphabets. Jà Jayanagar BASVà Basavangudi LàLalbagh RIàRichmond Road REàResidency Road MGàM.G.Road MCàManipal Centre SNàShavaji Nagar B àBanaswadi RTà R.T.Nagar MC 1

20

40 0

RE

100

10

20

60

30 10 0

SN

R

BASV

L

MG

J

10

10 10

3 0

RI

B

30 0

The flow in the pipeline connecting Richmond Road and Banaswadi = 1000 – 400 = 600 litres/min. If there is no pipeline connecting Lalbagh and Richmond Road, all the required water to Richmond Road should be supplied by the pipeline connecting Jayanagar and Richmond Road. Let us assume that there is no pipeline connecting Lalbagh and Richmond Road. From Jayanagr, the total amount of water required to be supplied to (i) Richmond Road = RI + RI-B = 300 + 600 = 900 (ii) Lalbagh = L + L-MG + L-B = 600 +100 +100 = 800 (iii) Basavangudi = 1000. Total supply from Jayanagar = 900 + 800 + 1000 = 2700. Hence, the correct answer is option 4. 27. What is the slack in the pipeline joining Jayangar and Richmond Road? (1) 0 litres (2) 400 litres (3) 300 litres Solution:Let us indicate the hubs with the following alphabets. Jà Jayanagar BASVà Basavangudi LàLalbagh RIàRichmond Road REàResidency Road MGàM.G.Road MCàManipal Centre SNàShavaji Nagar B àBanaswadi RTà R.T.Nagar MC 1

20

40 0

RE

100 60

30 10 0

SN

R

MG

(5) Indeterminate

BASV

10

20

(4) 200 litres

L J

10

10 3 0

10 B

RI

30 0

The slack in J-RI cannot be determined without knowing the requirement at B. Hence, the correct answer is option 5. © 2017 ETHNUS

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Explanatory Answers 28. What is maximum possible requirement that can be met at hubs in Manipal centre, Residency Road, Basavangudi, Shivaji Nagar and Banaswadi put together? (1) 1800 litres (2) 1900 litres (3) 1700 litres (4) 1600 litres (5) Indeterminate Solution:Let us indicate the hubs with the following alphabets. Jà Jayanagar BASVà Basavangudi LàLalbagh RIàRichmond Road REàResidency Road MGàM.G.Road MCàManipal Centre SNàShavaji Nagar B àBanaswadi RTà R.T.Nagar MC 1

20

40 0

RE

100

10

20

60

30 10 0

SN

R

BASV

MG

L J

10

10 3 0

10 B

RI

30 0

The maximum water supply from the source hub to all the other hubs = 1000 × 3 = 3000 litres. The maximum possible requirement that can be met in G, E, B, H, and I is = 3000 – 100 – 300 – 600 – 300 = 1700 litres. Hence, the correct answer is option 3. Directions for questions 29 and 30: Refer the following data to answer the questions given below. Each of three men – Saif, Surya and Abhishek got married to exactly one woman amongst Aishwarya, Kareena and Jothika, not necessarily in that order. Each couple has only one daughter among Nagma, Jaya and Karishma. No girl was born within one year of her parents’ marriage. Abhishek was blessed with a daughter after sixteen months of his marriage. Jaya was born in January, 2009. Nagma was born sixteen months after her parents’ marriage. Karishma is the not youngest. The first one to get married among the men was Saif, who got married in February, 2007. Aishwarya got married 6 months after Jothika got married. The last man to get married was Abhishek, who got married in September, 2007. Kareena gave birth to a child two years after her marriage. No one among Nagama, Jaya and Karishma was born in any month between August to December, including both these months. Nagma was not born to Aishwarya. 29. The wife and daughter of Saif is respectively (1) Jothika and Nagma (2) Jothika and Jaya (4) Jothika and Karishma (5) Indeterminate

(3) Kareena and Karishma

Solution: 1. The first one to get married among the men was Saif, who got married in February, 2007. 2. Aishwarya got married 6 months after Jothika got married. 3. Karishma is not the youngest. 4. Nagma was born 16 months after her parents’ marriage, but not to Aishwarya. 5. The last man to get married was Abhishek, who got married in September, 2007 and was blessed with a daughter after 16 months of marriage. 6. Kareena gave birth to a child two years after her marriage. 7. Jaya was born in January, 2009. 8. No girl was born within one year of her parents’ marriage, and definitely not in any month between August to December, including both these months.

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Explanatory Answers

From the given information, we get the following initial arrangement. Husband Saif

Wife Not Aishwary a (1) & (2)

Married

Child

Born on

Jaya (5) & (7)

Jan. 2009

Feb.200 7 From (1)

Surya Abhishek

Not Jothika (2) & (5)

Sep. 2007

Now Aishwarya could not have got married in August, 2007, which is six months from February, 2007. But 16 months from August is December and two years from august is august that is when her daughter would be born. But this violates condition 8. Hence, Aishwarya must have married in September, 2007, i.e., 6 months from March, 2007 in which Surya got married to Jothika. Hence, Kareena got married to Saif and gave birth to a child after 2 years, who is not the youngest, i.e. Karishma then. Nagma must have been born to Surya & Jothika after 16 months of their marriage, i.e., in July, 2009. Hence the final arrangement is as given below. Husban d

Wife

Saif

Kareena

Surya

Jothika

Abhishe k

Aishwary a

Married

Child

Born on

Feb. 2007 Mar. 2007 Sep. 2007

Karish ma

Feb. 2009 July 2009 Jan. 2009

Nagma Jaya

Hence, the correct answer is option 3. 30. When was Nagma born? (1) January, 2009

(2) February, 2009

(3) September, 2009

(4) July, 2009

(5) None of these

Solution: 1. The first one to get married among the men was Saif, who got married in February, 2007. 2. Aishwarya got married 6 months after Jothika got married. 3. Karishma is not the youngest. 4. Nagma was born 16 months after her parents’ marriage, but not to Aishwarya. 5. The last man to get married was Abhishek, who got married in September, 2007 and was blessed with a daughter after 16 months of marriage. 6. Kareena gave birth to a child two years after her marriage. 7. Jaya was born in January, 2009. 8. No girl was born within one year of her parents’ marriage, and definitely not in any month between August to December, including both these months. From the given information, we get the following initial arrangement. Husband Saif

Wife Not Aishwary a (1) & (2)

Married

Not Jothika

Sep. 2007

Child

Born on

Jaya (5) & (7)

Jan. 2009

Feb.200 7 From (1)

Surya Abhishek

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Explanatory Answers (2) & (5) Now Aishwarya could not have got married in August, 2007, which is six months from February, 2007. But 16 months from August is December and two years from august is august that is when her daughter would be born. But this violates condition 8. Hence, Aishwarya must have married in September, 2007, i.e., 6 months from March, 2007 in which Surya got married to Jothika. Hence, Kareena got married to Saif and gave birth to a child after 2 years, who is not the youngest, i.e. Karishma then. Nagma must have been born to Surya & Jothika after 16 months of their marriage, i.e., in July, 2009. Hence the final arrangement is as given below. Husban d

Wife

Saif

Kareena

Surya

Jothika

Abhishe k

Aishwary a

Married

Child

Born on

Feb. 2007 Mar. 2007 Sep. 2007

Karish ma

Feb. 2009 July 2009 Jan. 2009

Nagma Jaya

Hence, the correct answer is option 4. Directions for questions 31 and 32: Refer the following data to answer the questions given below. Five friends Kabir, Kadir, Kadin, Kael and Kabil are studying MBA, MCA, MSc, BSc and MCom, not necessarily in the same order. They are arranged in terms of their weight & height and ranked separately with respect to height and weight in such a way that • The heaviest and tallest person is ranked 1 in weight and height respectively. • Similarly the lightest and shortest person is ranked 5 in weight and height respectively and so on. It is known that no two persons are of the same height and weight. The following details are respect to their arrangement in terms of their height and weight: The second heaviest person studies MBA. The person who is the third heaviest studies MSc. Kabir studies BSc. The heaviest person studies MCom. Kadir studies MCA. Kael is not the third heaviest. The person who is second in height studies Msc. Kabil is not the heaviest. The shortest person studies MCom. The person who is fourth in height studies MBA. Kabil is not the second in height. 31. Who among the following is definitely true? (1) Kael is the heaviest and shortest and studies MCom. (2) Kadir studies MCA and is the heaviest and the shortest. (3) Kadin is the third in height, the heaviest and MCom. (4) Kabil is the second heaviest, fourth in height and studies MBA. (5) None of these Solution: 1. The person who is the third heaviest and second in height studies MSc. Kabir studies BSc. 2. The heaviest person is the shortest and studies MCom. Kadir studies MCA. 3. Kabil is neither the heaviest nor the second in height and Kael is not the third heaviest. 4. The person that studies MBA is the second heaviest and fourth in height. From (1) and (3), Kabil does not study MSc. From (2) and (3), Kabil does not study MCom. We know that Kabir studies BSc (from 1) and Kadir studies MCA (from 2), hence Kabil should be studying MCom. Pers on Kabir Kadir Kadin Kael

Course BSc (From 1) MCA (From 2) MCom.(From 1, 2, 3) MBA (From 1, 3)

Heig ht 3/1 1/3

Weig ht 4/5 5/4

5

1

4

2

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Explanatory Answers Kabil

MSc

2

3

Hence, the correct answer is option 4. 32.

Which of the following course could the third tallest person study? (1) BSc (2) MSc (3) MBA (4) MCom

Solution: 1. The person who is the third heaviest and second in height 2. The heaviest person is the shortest and studies MCom. Kadir 3. Kabil is neither the heaviest nor the second in height and Kael 4. The person that studies MBA is the second heaviest and

(5) Indeterminate

studies MSc. Kabir studies BSc. studies MCA. is not the third heaviest. fourth in height.

From (1) and (3), Kabil does not study MSc. From (2) and (3), Kabil does not study MCom. We know that Kabir studies BSc (from 1) and Kadir studies MCA (from 2), hence Kabil should be studying MCom. Pers on Kabir Kadir Kadin Kael Kabil

Course BSc (From 1) MCA (From 2) MCom.(From 1, 2, 3) MBA (From 1, 3) MSc

Heig ht 3/1 1/3

Weig ht 4/5 5/4

5

1

4 2

2 3

Hence, the correct answer is option 1. Directions for questions 33 to 35: Refer the following data to answer the questions given below. Table below gives the details of engineering graduates at XYZ Institute for the batch 2009 – 2010. Please note that all the students have graduated in all the disciplines in the year 2009 – 2010. The average salaries of students from Civil, Electrical, Mechanical and Computer Science disciplines are 4.7, 2.1, 2.6 and 1.7 lakhs respectively. The average salary of students from the other disciplines is 1.4 lakhs. The tables below give the discipline-wise and region-wise breakup of students at XYZ Institute for the batch 2009 – 2010.

Discipline-wise breakup Percentag Disciplines e Civil Engineering 34% Mechanical Engineering 26% Electrical Engineering 18% Computer Science 14% Engineering Other disciplines 8%

Region-wise breakup Students from Percentag different parts of India e North 5% South 37% East 28% West 16% Central

14%

33. The sum of the salaries of the students belonging to which of the discipline is the lowest? (1) Electrical Engineering (2) Computer Science Engineering (3) Mechanical Engineering (4) Other disciplines (5) Indeterminate Solution:The average salary of ‘other disciplines’, as well as the percentage of total students from ‘other disciplines’ are the least. Therefore ‘other disciplines’ is the lowest. © 2017 ETHNUS

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Explanatory Answers Hence, the correct answer is option 4. 34. Assuming that the distribution by regions of the students for each discipline is exactly similar to that of the total students, the students who are Civil Engineers from the North form what percent of the Computer Science Engineering students from the East? (1) 38% (2) 42.5% (3) 43.4% (4) 34% (5) Indeterminate Solution:Since the overall region-wise distribution applies to individual disciplines as well, percentage of civil engineers from North part of India is = 5% of 34% of total students, i.e., 1.7% of total students [0.05 × 0.34 = 0.017]. Similarly, the computer science Engineering students from the East part of India is = 14% of 28% of total students. i.e., 3.92 % of total students [0.14 × 0.28 = 0.0392] The required percentage = (1.7/3.92) × 100 = 43.36% Hence, the correct answer is option 3. 35. It is known that the distribution by regions of the students for each discipline is exactly similar to that of the total students. If the total number of students is 400, then what is average salary of the Computer Science Engineering students from the East? (1) 0.51 lakhs (2) 3.61 lakhs (3) 5.1 lakhs (4) 4.3 lakhs (5) Indeterminate Solution:Since we do not know the distribution of the salaries within the computer science discipline by regions, the question cannot be answered. Hence, the correct answer is option 5. Directions for questions 36 to 40: Each question is followed by two statements A and B. Indicate your responses based on the following directives: Mark (1) if the question can be answered using A alone but not using B alone. Mark (2) if the question can be answered using B alone but not using A alone. Mark (3) if the question can be answered using A and B together, but not using either A or B alone Mark (4) if the question can be answered using either statement alone. Mark (5) if the question cannot be answered even using A and B together 36. After how many years from now will Madhu Chitra be twice as old as her daughter Madhumita? A. Eight years ago, Madhu Chitra’s age was six years less than thrice her daughter Madhumita’s age B. Six years hence, Madhu Chitra’s age will be four years less than twice her daughter Madhumita’s age. 1. if the question can be answered using A alone but not using B alone. 2. if the question can be answered using B alone but not using A alone. 3. if the question can be answered using A and B together, but not using either A or B alone 4. if the question can be answered using either statement alone. 5. if the question cannot be answered even using A and B together Solution:Let the present age of Madhu Chitra and her daughter’s age be M and D respectively. From Statement A, M – 8 = 3D – 24 – 6 M = 3D – 22 Hence, statement A alone is not sufficient. From statement B, M + 6 = 2D + 12 – 4 M = 2D + 2 Adding 2 both the sides, M + 2 = 2D + 4 = 2 (D + 2) So, in two years time Madhu Chitra will be twice as old as her daughter. Statement B alone is sufficient to answer the question. Hence, the correct answer is option 2. 2

2

37. If (2D – 1)A + (2D + 1)A + C = 0 and (E + 1)B + (4E + 1)B + 3C = 0, then what is the sum of D and E? A. ‘C’ is a natural number and E is greater than D 2 2 B. The quadratic equations (2D – 1)A + (2D + 1)A + C = 0 and (E + 1)B + (4E + 1)B + 3C = 0 have the same pair of roots © 2017 ETHNUS

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Explanatory Answers 1. if the question can be answered using A alone but not using B alone. 2. if the question can be answered using B alone but not using A alone. 3. if the question can be answered using A and B together, but not using either A or B alone 4. if the question can be answered using either statement alone. 5. if the question cannot be answered even using A and B together Solution:From B, Because the quadratic equations have the same roots, we have (2D – 1)/(E + 1) = (2D + 1)/(4E + 1) = C/3C Here we need to know whether C is non-zero or not, otherwise we cannot find the answer uniquely. Statement A tells us C is non-zero. Now we can find out the value of D and E. Hence, the correct answer is option 3. 3

3

38. If the numbers A and B are in base 6, then what is the value of A + B in the same number 2 A. The quadratic equation x – 9x + 20 = 0 is written base 6. 2 B. A and B are the roots of the quadratic equation x – 9x + 20 = 0

system with base 6?

1. if the question can be answered using A alone but not using B alone. 2. if the question can be answered using B alone but not using A alone. 3. if the question can be answered using A and B together, but not using either A or B alone 4. if the question can be answered using either statement alone. 5. if the question cannot be answered even using A and B together 2

Solution:From B, it is known that the roots of the quadratic equation x – 9x + 20 = 0 are real and they are 5 and 4. Since the roots are less than 6, statement A does not help in finding the solution. Hence, the correct answer is option 2. 39. There were only three different varieties of chocolates in a shop. If total number of chocolates bought by Manish is a prime number, then how many chocolates did Manish buy for Rs. 107? A. Manish bought at least one chocolate costing Rs. 50 and at least 15 chocolates costing less than Rs. 50. B. The cost of a first variety chocolate is Rs. 1, the cost of a second variety chocolate is Rs. 10, and the cost of the third variety chocolate is Rs. 50. 1. if the question can be answered using A alone but not using B alone. 2. if the question can be answered using B alone but not using A alone. 3. if the question can be answered using A and B together, but not using either A or B alone 4. if the question can be answered using either statement alone. 5. if the question cannot be answered even using A and B together Solution:By observation, we can say that either statement alone is not sufficient to find out the answer. Let the number of first, second and third variety of chocolates bought by Manish be x, y and z. From statement A, Manish buys at least 16 chocolates. From statement B, ∴ x + 10y + 50z = 107 (1) From statement A and B together, Since he buys more than 1 chocolate in third variety, z ≥ 1 Case 1: z = 1 Equation (1) becomes x + 10y = 57 y = (57 – x)/10 ∴ The minimum value of y is 0 and the maximum value of y is 6. ∴ There are 5 possibilities when z = 1 They are (17, 4, 1), (27, 3, 1), (37, 2, 1), (47, 1, 1) and (57, 0, 1). Case 2: z = 2 Equation (1) becomes x + 10y = 7 Clearly x = 7 and y = 0 is the only solution when z =2. This possibility of (x, y, z) is (7, 0, 2), since he buys less than 16 chocolates, this cannot be included. © 2017 ETHNUS

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Explanatory Answers Hence, the total number of buys Manish could buy the chocolates is 5. Out of 5 possibilities, there is only one way where the total number of chocolates bought buy Manish can be a prime number. That is (27, 3, 1). Hence, the correct answer is option 3. 40. How many pieces of pants, shirts and T-shrits did the dealer sell at the end of the day? A. The cost of a pant, a shirt and a T-shrit is Rs. 100, Rs. 120 and Rs. 25 respectively B. The dealer made a total collection of Rs. 560 by selling atleast one of each pant, shirt and T-shrit. 1. if the question can be answered using A alone but not using B alone. 2. if the question can be answered using B alone but not using A alone. 3. if the question can be answered using A and B together, but not using either A or B alone 4. if the question can be answered using either statement alone. 5. if the question cannot be answered even using A and B together Solution: Either statement alone will not be sufficient to answer the question. Now let us combine them together. Let the number of pants, shirts and T-shrits sold are P, S and T respectively. There is only one combination that we can get for the sale Rs. 560, i.e. (1P, 3S, 4T). Hence, the correct answer is option 3.

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