Q1. Divide 1870 into three parts such that half of the first part, one-third of the second part, and one-sixth of the third part are equal. What is the difference between the largest and the smallest parts?Correct Option A … Explanation: Let a/2 = b/3 = c/6 = k. Then a = 2k, b = 3k and c = 6k. Total = 2k + 3k + 6k = 11k = 1870, so k = 170. The parts are 340, 510 and 1020. Difference between the largest and smallest = 1020 − 340 = 680. Hence, option A.Q2. Tank A is initially 2/5th full of water. Pipe P can fill the entire tank in 10 minutes, while pipe Q can empty the entire tank in 6 minutes. If both pipes are opened simultaneously, how long will it take to empty the tank completely?Correct Option E … Explanation: Pipe P fills 1/10 of the tank per minute and pipe Q empties 1/6 of the tank per minute. Net emptying rate = 1/6 − 1/10 = (5 − 3)/30 = 1/15 of the tank per minute. Water to be emptied = 2/5 of the tank. Time = (2/5) ÷ (1/15) = 6 minutes. Hence, option E.Q3. A shopkeeper marked up the price of an item by 27.27%. After giving two successive discounts of 18.18% and x%, he finds that he makes neither profit nor loss. Find the value of x.Correct Option D … Explanation: 27.27% = 3/11 and 18.18% = 2/11. Let CP = 1. MP = 1 × 14/11 = 14/11. After the first discount, price = 14/11 × 9/11 = 126/121. For no profit no loss, the second discount must bring the price down to 1: (126/121) × (1 − x/100) = 1, so 1 − x/100 = 121/126 and x/100 = 5/126. Therefore x = 500/126 ≈ 3.97%. Hence, option D.Q4. Container A contains liquids X and Y in the ratio 3:5, and Container B contains liquids Y and Z in the ratio 4:6. 40 litres of mixture from Container A is mixed with 50 litres of mixture from Container B in Container C. Then, 45 litres of the mixture in Container C is removed and replaced by pure liquid Z. Find the ratio of liquids Y and Z in Container C now.Correct Option E … Explanation: From A (40 L, X:Y = 3:5): X = 15 L, Y = 25 L. From B (50 L, Y:Z = 4:6): Y = 20 L, Z = 30 L. Container C has X = 15, Y = 45, Z = 30 (total 90 L). Removing 45 L (half the mixture) leaves X = 7.5, Y = 22.5, Z = 15. Adding 45 L of pure Z makes Z = 60. Required ratio Y:Z = 22.5:60 = 3:8. Hence, option E.Q5. Rajesh invested ₹75,000 across three schemes A, B, and C offering simple interest at 5%, 8%, and 10% per annum respectively. The total interest earned after 2 years was ₹11720. If the amount invested in scheme A was 80% of the amount invested in scheme C, how much was invested in scheme B?Correct Option A … Explanation: Annual interest = 11720/2 = ₹5860. Let the amount in C be C, so A = 0.8C and B = 75000 − 1.8C. Then 0.05(0.8C) + 0.08(75000 − 1.8C) + 0.10C = 5860, which gives 0.04C + 6000 − 0.144C + 0.10C = 5860. So −0.004C = −140 and C = ₹35,000. A = ₹28,000 and B = 75000 − 63000 = ₹12,000. Check: 1400 + 960 + 3500 = 5860 per year. Hence, option A.Q6. A, B, and C work on a project in rotation: A and B on Day 1, B and C on Day 2, and A and C on Day 3, repeating this 3-day cycle until completion. Working alone, A can finish the job in 24 days. B is 1.5 times as efficient as A, and C is twice as efficient as A. By what percentage is the time taken under this rotational scheme more than the time taken if all three worked together every day?Correct Option C … Explanation: Let the efficiency of A = 1 unit/day, so total work = 24 units, B = 1.5 units/day and C = 2 units/day. All three together do 4.5 units/day, so they need 24/4.5 = 16/3 days. In rotation, Day 1 (A+B) = 2.5, Day 2 (B+C) = 3.5 and Day 3 (A+C) = 3, giving 9 units per 3-day cycle. After 2 cycles (6 days), 18 units are done and 6 remain. Day 7 does 2.5 units, leaving 3.5. Day 8 (B+C) does exactly 3.5 units. Total = 8 days. Percentage more = (8 − 16/3)/(16/3) × 100 = (8/3)/(16/3) × 100 = 50%. Hence, option C.Q7. Two trains start simultaneously from stations P and Q toward each other. After crossing each other, Train 1 takes 9 hours to reach Q and Train 2 takes 16 hours to reach P. If Train 1 travels at 80 km/h, what is the distance between P and Q?Correct Option D … Explanation: For two trains after crossing, S1/S2 = √(t2/t1) = √(16/9) = 4/3. So Train 2's speed = 80 × 3/4 = 60 km/h. After meeting, Train 1 covers in 9 hours the distance Train 2 covered before meeting: 80 × 9 = 720 km. Therefore time before meeting = 720/60 = 12 hours. Total distance = (80 + 60) × 12 = 1680 km. Hence, option D.Q8. 15 men and 15 women can complete a piece of work in 24 days. 15 men and 15 children can complete the same work in 30 days. 30 women and 30 children can complete the same work in 20 days. How many days will 12 men and 18 women take to complete the same work?Correct Option C … Explanation: Let one man, woman and child do m, w and c of the work per day. Then m + w = 1/360, m + c = 1/450 and w + c = 1/600. Adding: 2(m + w + c) = (5 + 4 + 3)/1800 = 1/150, so m + w + c = 1/300. Hence m = 1/300 − 1/600 = 1/600 and w = 1/360 − 1/600 = 1/900. Work done by 12 men and 18 women per day = 12/600 + 18/900 = 1/50 + 1/50 = 1/25. Time required = 25 days. Hence, option C.Q9. A's income is 25% more than B's income, and A's expenditure is 15% more than B's expenditure. If A's savings are 75% more than B's savings, then B's expenditure is what percentage of B's income?Correct Option C … Explanation: Let B's income = I and expenditure = E, so B's savings = I − E. A's savings = 1.25I − 1.15E. Given 1.25I − 1.15E = 1.75(I − E), we get 1.25I − 1.15E = 1.75I − 1.75E. So 0.6E = 0.5I and E/I = 5/6. B's expenditure is 83.33% of B's income. Hence, option C.Q10. CP of 75 kg of sugar is equal to SP of x kg of sugar. The profit is 5%. What is the value of x?Correct Option E … Explanation: Let the CP of 1 kg = 1, so CP of 75 kg = 75 = SP of x kg. The CP of x kg is x, so profit % = (75 − x)/x × 100 = 5. This gives 75 = 1.05x, so x = 75/1.05 ≈ 71.43. Hence, option E.Q11. If Suraj travels at 8/9th of his usual speed, he misses a train by 10 minutes. But if he travels at 5/4th of his usual speed, he reaches the station 5 minutes before the departure of the train. What is the approximate normal time taken by Suraj to reach the railway station?Correct Option D … Explanation: Let the usual time be T minutes. Time is inversely proportional to speed, so at 8/9 speed the time is 9T/8 and at 5/4 speed the time is 4T/5. The difference between the two times is 10 + 5 = 15 minutes. So 9T/8 − 4T/5 = 13T/40 = 15, which gives T = 600/13 ≈ 46 minutes. Hence, option D.Q12. A person borrows some amount from a bank at simple interest and lends it to his friend at the same rate but compounded annually. After 2 years, he is liable to pay an interest of Rs. 75000 to the bank while receiving 79500 as interest from his friend. What is the rate of interest?Correct Option D … Explanation: SI for 2 years = 75000, so SI for 1 year = PR/100 = 37500. For 2 years, CI − SI = P(R/100)² = 79500 − 75000 = 4500. This can be written as (PR/100) × (R/100) = 4500, so 37500 × R/100 = 4500 and R = 12%. Hence, option D.Q13. A certain number of men were supposed to complete a task in 11 days. However, after the first day, 3 men dropped out every day. Eventually, they completed the work on the 16th day. How many men started the work in the beginning?Correct Option E … Explanation: Let M men start, so total work = 11M man-days. The workforce on the 16 days is M, M − 3, M − 6, …, M − 45. Total man-days = 16M − 3(0 + 1 + 2 + … + 15) = 16M − 3 × 120 = 16M − 360. Setting 16M − 360 = 11M gives 5M = 360, so M = 72. Hence, option E.Q14. A number D is increased by 3D%, and the result is 272. What is the result if the number is decreased by D%?Correct Option C … Explanation: D(1 + 3D/100) = 272 gives 3D² + 100D − 27200 = 0. So D = [−100 + √(10000 + 326400)]/6 = (−100 + 580)/6 = 80. Check: 80 increased by 240% = 80 × 3.4 = 272. Decreasing 80 by 80% gives 80 × 0.2 = 16. Hence, option C.Q15. Wardhan borrowed 43000 at 5% simple interest payable at 4 equal annual instalments. What is the annual instalment paid by him?Correct Option B … Explanation: Amount due after 4 years = 43000 + 43000 × 5 × 4/100 = 51600. Let each instalment be x. The instalments paid at the end of years 1, 2 and 3 earn simple interest for 3, 2 and 1 years respectively, and the last earns none. So the total value is 4x + x × 5/100 × (3 + 2 + 1) = 4x + 0.3x = 4.3x. Setting 4.3x = 51600 gives x = Rs 12000. Hence, option B.Q16. P, Q, R are in continued proportion and Q, R, S are also in continued proportion such that P:Q = 3:5. If all four numbers are positive integers, find the minimum difference between P and S.Correct Option A … Explanation: Since P:Q = Q:R = R:S = 3:5, the four numbers form a geometric progression with ratio 5/3. So P:Q:R:S = 3 : 5 : 25/3 : 125/9. Multiplying by 27 to get the smallest positive integers gives 27 : 45 : 75 : 125. Minimum difference S − P = 125 − 27 = 98. Hence, option A.Q17. If a seller loses 20% by selling 5 sheets of paper for a rupee, how many sheets should he sell for Rs 12 in order to gain 20%?Correct Option B … Explanation: SP of 1 sheet = 1/5 = Rs 0.20, and this is at a 20% loss, so CP = 0.20/0.8 = Rs 0.25. For a 20% gain, SP = 0.25 × 1.2 = Rs 0.30 per sheet. Number of sheets for Rs 12 = 12/0.30 = 40. Hence, option B.Q18. A trader marks up his goods by 50% over the cost price and gives a discount of 20% on the marked price. He also cheats his customers by using a weight of 900 g in place of 1 kg. What is his overall profit percentage?Correct Option D … Explanation: Let the CP of 1 kg = 100. MP = 150, and SP after the 20% discount = 120 for what the customer believes is 1 kg. He actually gives only 900 g, which costs him 90. Profit % = (120 − 90)/90 × 100 = 33.33%. Hence, option D.Q19. In a group of students, adding one student weighing 75 kg increases the average weight of the group by 5 kg. If one more student weighing 75 kg is then added, the average weight becomes 8 kg more than the original average. If yet another student weighing 75 kg is added after that, what would be the average weight of the group (in kg)?Correct Option C … Explanation: Let there be n students with average A. From the first addition, nA + 75 = (n + 1)(A + 5), which gives A + 5n = 70. From the second addition, nA + 150 = (n + 2)(A + 8), which gives 2A + 8n = 134, or A + 4n = 67. Subtracting gives n = 3, so A = 55. After the third addition, total weight = 165 + 225 = 390 for 6 students. New average = 390/6 = 65 kg. Hence, option C.Q20. A and B sell water-diluted milk: A adds water equal to 25% of the milk, and B adds water equal to 40% of the milk. A buyer purchases 80 litres from A and 70 litres from B and mixes them in a container. He then replaces 20% of this mixture with pure milk twice sequentially. What is the total volume of pure milk in the container after both replacements? (rounded off to the nearest integer)Correct Option B … Explanation: A's milk:water = 4:1, so 80 L contains 64 L milk and 16 L water. B's milk:water = 5:2, so 70 L contains 50 L milk and 20 L water. The mixture has 150 L in total, with 114 L milk and 36 L water. Each replacement with pure milk removes 20% of the water, so water left = 36 × (0.8)² = 23.04 L. Pure milk = 150 − 23.04 = 126.96 ≈ 127 litres. Hence, option B.
