Q1. A single discount equivalent to three successive discounts of 20%, 10%, and 5% is:Correct Option A … Explanation: Successive discounts multiply, they do not add. Taking the marked price as 100, the price after each discount is 100 × 0.80 × 0.90 × 0.95. First discount: 100 × 0.80 = 80. Second: 80 × 0.90 = 72. Third: 72 × 0.95 = 68.4. The final price is 68.4, so the total discount is 100 − 68.4 = 31.6. Hence the single equivalent discount is 31.6%. Options B, C, D and E result from wrongly adding the discounts (20 + 10 + 5 = 35%) or from partially adding and partially multiplying them. Hence, option A.Q2. A batsman has a certain average run rate after 15 innings. In his 16th inning, he scores 88 runs, thereby increasing his average by 3 runs. What is his new average after the 16th inning?Correct Option C … Explanation: Let the average after 15 innings be x. Total runs after 15 innings = 15x. After the 16th inning the total becomes 15x + 88 and the new average is x + 3, so 15x + 88 = 16(x + 3). Expanding gives 15x + 88 = 16x + 48, therefore x = 40. The new average is x + 3 = 43. Shortcut: the extra 88 runs must cover the old average (40) plus 3 extra runs for each of the 16 innings, i.e. 88 − 3 × 16 = 40 is the old average. Option A is the old average, not the new one. Hence, option C.Q3. A train 150 meters long passes a stationary pole in 15 seconds. Travelling at the same speed, how many seconds will it take to completely cross a 300-meter-long platform?Correct Option D … Explanation: While crossing a pole the train covers only its own length, so speed = 150 ÷ 15 = 10 m/s. While crossing a platform the train must cover its own length plus the platform length, i.e. 150 + 300 = 450 metres. Time required = 450 ÷ 10 = 45 seconds. Option A (30 s) arises from using only the platform length of 300 m and ignoring the length of the train. Hence, option D.Q4. A, B, and C started a business. A invested 3 times as much as B, and B invested two-thirds of what C invested. A left the business at the end of 7 months, and B left after 9 months. If the total profit at the end of the year was Rs. 132000, what is B's share of the profit?Correct Option B … Explanation: Let C's investment be 3 units. Then B = (2/3) × 3 = 2 units and A = 3 × 2 = 6 units. Profit is shared in the ratio of capital × time. A stayed 7 months: 6 × 7 = 42. B stayed 9 months: 2 × 9 = 18. C stayed the full 12 months: 3 × 12 = 36. The ratio is 42 : 18 : 36, and the total is 96 parts. B's share = 132000 × 18/96 = 132000 × 0.1875 = Rs. 24750. Option C (Rs. 27000) results from dividing by capital alone and ignoring the differing time periods. Hence, option B.Q5. A car travels from City A to City B at an average speed of 34 km/hr and returns from City B to City A along the same route at an average speed of 51 km/hr. What is the average speed of the car for the entire journey?Correct Option A … Explanation: When equal distances are covered at two different speeds, the average speed is the harmonic mean, not the arithmetic mean. Average speed = 2xy/(x + y) = (2 × 34 × 51)/(34 + 51) = 3468/85 = 40.8 km/hr. Verification: if the one-way distance is 102 km, the outward trip takes 3 hours and the return takes 2 hours, giving 204 km in 5 hours = 40.8 km/hr. Option C (42.5 km/hr) is the arithmetic mean of 34 and 51, which is the classic trap here. Hence, option A.Q6. A town's population increases by 10% in the first year, decreases by 20% in the second year, and increases by 30% in the third year. If the current population is 24310, what was the population 3 years ago?Correct Option D … Explanation: Let the population 3 years ago be P. Then P × 1.10 × 0.80 × 1.30 = 24310. The combined multiplier is 1.10 × 0.80 = 0.88, and 0.88 × 1.30 = 1.144. So P = 24310 ÷ 1.144 = 21250. Check: 21250 × 1.1 = 23375; 23375 × 0.8 = 18700; 18700 × 1.3 = 24310. The other options fail this verification because they assume the percentages cancel out (+10 − 20 + 30 = +20%), which is invalid for successive percentage change. Hence, option D.Q7. Arjun and Bhim each have the same quantity of a mixture of water and milk. In Arjun's mixture, the ratio of water to milk is 3:7, while in Bhim's mixture, the ratio is 2:8. They combine both mixtures and set out to sell the resulting mixture in the city. On the way, 10% of the mixture is spilt due to several speed breakers. The remaining mixture is sold at the cost price of pure milk. Find the overall percentage profit they earn.Correct Option C … Explanation: Take 10 litres each. Arjun's mixture has 3 L water and 7 L milk; Bhim's has 2 L water and 8 L milk. Combined: 5 L water and 15 L milk, total 20 L. Water is free, so the cost incurred is only for 15 L of milk. Let the cost of milk be Rs. 1 per litre, so cost price = Rs. 15. After 10% spillage, 20 × 0.9 = 18 L remain, sold at the cost price of pure milk, i.e. Rs. 1 per litre, giving revenue = Rs. 18. Profit = 18 − 15 = Rs. 3, so profit percentage = 3/15 × 100 = 20%. Option D (33.33%) is what the profit would have been with no spillage. Hence, option C.Q8. In an election between two candidates, the winner secures 55% of the valid votes. 20% of the total votes polled were declared invalid. If the total number of votes polled was 12700, how many valid votes did the losing candidate receive?Correct Option A … Explanation: Invalid votes are 20% of the total polled, so valid votes = 80% of 12700 = 10160. The winner takes 55% of the valid votes, so the loser takes the remaining 45% of the valid votes. Loser's votes = 10160 × 0.45 = 4572. A common error is to apply 45% to the total polled (12700 × 0.45 = 5715) or to deduct the invalid share twice, which produces the other options. Hence, option A.Q9. A shopkeeper bought 170 articles for Rs. 4590. During transit, 17 articles were damaged and had to be thrown away. At what price per article must he sell the remaining inventory to make an overall profit of 35%?Correct Option E … Explanation: The total cost of Rs. 4590 was incurred on all 170 articles, and the damaged ones still count as a cost. The required total selling price = 4590 × 1.35 = Rs. 6196.50. Only 170 − 17 = 153 articles remain to be sold, so the price per article = 6196.50 ÷ 153 = Rs. 40.50. Option A (Rs. 38.75) comes from spreading the marked-up amount over all 170 articles, i.e. forgetting that the 17 damaged pieces earn nothing. Hence, option E.Q10. The cost of 3 pens, 4 pencils, and 5 erasers is Rs. 75. The unit costs of a pen, a pencil, and an eraser are in the ratio 4:2:1. What is the total cost of 4 pens, 7 pencils and 3 erasers?Correct Option A … Explanation: Let the unit costs be 4k, 2k and k for a pen, a pencil and an eraser respectively. Then 3(4k) + 4(2k) + 5(k) = 75, i.e. 12k + 8k + 5k = 25k = 75, so k = 3. Therefore a pen costs Rs. 12, a pencil Rs. 6 and an eraser Rs. 3. The required cost = 4(12) + 7(6) + 3(3) = 48 + 42 + 9 = Rs. 99. Hence, option A.Q11. The average age of 30 students in a class is 15 years. If the teacher's age is included, the average age of the class increases by 1 year. If the principal's age is also included, the average age becomes 16.75 years. What is the difference between the age of the teacher and the principal?Correct Option A … Explanation: Total age of 30 students = 30 × 15 = 450 years. With the teacher included, there are 31 people with an average of 16 years, so the total = 31 × 16 = 496, which makes the teacher's age 496 − 450 = 46 years. With the principal also included, there are 32 people with an average of 16.75 years, so the total = 32 × 16.75 = 536, which makes the principal's age 536 − 496 = 40 years. The difference = 46 − 40 = 6 years. Hence, option A.Q12. 10 men can complete a project in 15 days, while 15 women can complete the same project in 20 days. If 5 men and 10 women work together for 6 days, what fraction of the total work remains to be completed?Correct Option D … Explanation: The whole job needs 10 × 15 = 150 man-days, so one man does 1/150 of the work per day. It also needs 15 × 20 = 300 woman-days, so one woman does 1/300 per day. In one day, 5 men and 10 women together do 5/150 + 10/300 = 1/30 + 1/30 = 1/15 of the work. In 6 days they complete 6 × 1/15 = 2/5 of the work. The fraction remaining = 1 − 2/5 = 3/5. Option B (2/5) is the work completed, not the work remaining. Hence, option D.Q13. A and B started a business by investing ₹7,500 and ₹10,000, respectively. After 4 months, A withdrew ₹2,000 from the business. Six months later, B withdrew ₹3,000. Find the ratio of their profits at the end of the year.Correct Option A … Explanation: Profits are divided in the ratio of capital multiplied by the time for which it was employed. A invested ₹7,500 for the first 4 months, then ₹5,500 for the remaining 8 months: (7500 × 4) + (5500 × 8) = 30000 + 44000 = 74000. B invested ₹10,000 for the first 10 months (4 months plus the six that followed), then ₹7,000 for the last 2 months: (10000 × 10) + (7000 × 2) = 100000 + 14000 = 114000. The ratio is 74000 : 114000 = 74 : 114 = 37 : 57. Note the timing trap in the question — B withdrew six months after A's withdrawal, i.e. at the end of month 10, not month 6. Hence, option A.Q14. A and B can do a piece of work in 12 days, B and C in 15 days, and C and A in 20 days. If A works alone, how many days will he take to finish the work?Correct Option C … Explanation: Write the daily rates: A + B = 1/12, B + C = 1/15, C + A = 1/20. Adding all three gives 2(A + B + C) = 1/12 + 1/15 + 1/20 = (5 + 4 + 3)/60 = 12/60 = 1/5, so A + B + C = 1/10. A's individual rate = (A + B + C) − (B + C) = 1/10 − 1/15 = (3 − 2)/30 = 1/30. Working alone, A therefore takes 30 days. Hence, option C.Q15. The difference between the compound interest and simple interest on a certain sum of money for 2 years, at an annual interest rate of 8%, is Rs. 768. What interest would be earned on the principal over 2 years if the interest were compounded half-yearly?Correct Option E … Explanation: For 2 years, the difference between CI and SI equals P(r/100)². So P × (0.08)² = 768, i.e. 0.0064P = 768, giving P = Rs. 1,20,000. With half-yearly compounding the rate becomes 4% per half-year for 4 periods: Amount = 120000 × (1.04)⁴ = 120000 × 1.16985856 = Rs. 1,40,383.03. Interest earned = 140383 − 120000 ≈ Rs. 20,383. Option A (Rs. 19,968) is the interest under annual compounding, which is lower because compounding is less frequent. Hence, option E.Q16. A person invests a total of Rs. 20,000 across two different schemes offering simple interest at 8% and 10% per annum, respectively. If the total interest earned from both schemes after 2 years is Rs. 3,520, how much money was invested in the 10% scheme?Correct Option A … Explanation: Let Rs. x be invested at 8%, so Rs. (20000 − x) is invested at 10%. Over 2 years of simple interest: 2[0.08x + 0.10(20000 − x)] = 3520, so 0.08x + 2000 − 0.10x = 1760, giving −0.02x = −240 and x = Rs. 12,000. That is the amount at 8%, so the amount in the 10% scheme = 20000 − 12000 = Rs. 8,000. Option C (Rs. 12,000) is the amount in the 8% scheme and is the trap for candidates who stop one step early. Hence, option A.Q17. A specific task can be completed by 2 units of Machine X, or 3 units of Machine Y, or 4 units of Machine Z in exactly 26 hours. If one unit of each machine (X, Y, and Z) is used simultaneously, how many hours will it take to finish the task?Correct Option D … Explanation: One unit of X alone would take 2 × 26 = 52 hours, so its rate is 1/52 per hour. One unit of Y alone would take 3 × 26 = 78 hours, rate 1/78. One unit of Z alone would take 4 × 26 = 104 hours, rate 1/104. Working together, the combined rate = 1/52 + 1/78 + 1/104. Using the LCM 312: 6/312 + 4/312 + 3/312 = 13/312 = 1/24. The task therefore takes 24 hours. Hence, option D.Q18. A 60-liter mixture contains milk and water in the ratio of 2:1. How many liters of pure water must be added to the mixture to reverse the ratio of milk to water to 1:2?Correct Option E … Explanation: In 60 L with milk to water as 2:1, milk = 40 L and water = 20 L. Only water is added, so the milk stays fixed at 40 L. Let x litres of water be added. The new ratio requires 40/(20 + x) = 1/2, so 20 + x = 80 and x = 60 L. Check: the mixture becomes 40 L milk and 80 L water, a ratio of 1:2. Option A (20 L) is the amount of water originally present, not the amount to be added. Hence, option E.Q19. Swarndeep drives from his home to the airport to catch a flight. He drives 35 km in the first hour, but realizes that he will be 1 hour late if he continues at this speed. He increases his speed by 15 km per hour for the rest of the way to the airport and arrives 30 minutes early. How many km is the airport from his home?Correct Option C … Explanation: The first hour covers 35 km, so the original speed is 35 km/hr and the increased speed is 50 km/hr. Let the total distance be D, so the remaining distance after the first hour is (D − 35). Continuing at 35 km/hr means arriving 1 hour late; covering it at 50 km/hr means arriving 30 minutes early. The two scenarios differ by 1 hour 30 minutes, i.e. 1.5 hours. So (D − 35)/35 − (D − 35)/50 = 1.5. Since 1/35 − 1/50 = (10 − 7)/350 = 3/350, we get (D − 35) × 3/350 = 1.5, so D − 35 = 175 and D = 210 km. Option A (175 km) is the remaining distance, not the total. Hence, option C.Q20. The sum of the cost prices of two articles, A and B, is ₹ 6580. Article A is sold at a 10% loss, while Article B is sold at a 20% profit. If the selling prices of both articles are equal, and Article B is instead sold at a y% profit while Article A is sold at a y/2% loss, their selling prices are again equal. Find the value of y.Correct Option D … Explanation: From the first condition, 0.9A = 1.2B, so A : B = 1.2 : 0.9 = 4 : 3. With A + B = 6580, the parts give A = 6580 × 4/7 = ₹3760 and B = 6580 × 3/7 = ₹2820. Now apply the second condition: 3760(1 − y/200) = 2820(1 + y/100). Dividing both sides by 940 gives 4(1 − y/200) = 3(1 + y/100), i.e. 4 − y/50 = 3 + 3y/100. Rearranging, 1 = 3y/100 + 2y/100 = 5y/100 = y/20, so y = 20. Check: A sold at a 10% loss gives 3760 × 0.9 = 3384, and B sold at a 20% profit gives 2820 × 1.2 = 3384 — equal, as required. Hence, option D.
