Q1. When a person reduces his speed from 105 km/hr to 90 km/hr he takes 20 minutes more than his usual time to reach his destination. Find the usual time taken by him to cover the distance.Correct Option 2 … Explanation: Let the distance be d km. The extra time is 20 minutes = 1/3 hour, so d/90 − d/105 = 1/3. The LCM of 90 and 105 is 630, so this becomes (7d − 6d)/630 = 1/3, which gives d = 210 km. Usual time = 210/105 = 2 hours. Option 1 (140 minutes) is the time at the reduced speed, not the usual time. Hence, option 2.Q2. A vendor initially earns a 20% profit by selling rice. Driven by greed, he aims to increase his profit margin to 50% while keeping the selling price unchanged. To achieve this, he switches to a lower-quality variety of rice. By what percentage is the cost price of the new rice lower than that of the original variety?Correct Option 2 … Explanation: Let the original cost price be 100. At 20% profit, SP = 120. To earn 50% profit at the same SP, the new cost price C' must satisfy 1.5C' = 120, so C' = 80. The new cost price is lower by (100 − 80)/100 × 100 = 20%. Option 1 (16.67%) would wrongly compare the SP with the cost price. Hence, option 2.Q3. 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 2 … Explanation: The SP of 5 sheets is Re 1 at a 20% loss, so their CP = 1/0.8 = Rs 1.25. CP per sheet = 1.25/5 = Rs 0.25. For a 20% gain, the SP per sheet = 0.25 × 1.2 = Rs 0.30. Number of sheets for Rs 12 = 12/0.30 = 40. Hence, option 2.Q4. 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. 43000 to the bank while receiving 45580 as interest from his friend. What is the rate of interest?Correct Option 4 … Explanation: The SI for 2 years is 43000, so the SI for one year is 21500. For 2 years, CI − SI equals the interest on one year's SI, which is 21500 × r. So 45580 − 43000 = 2580 = 21500 × r, which gives r = 2580/21500 = 0.12 = 12%. Hence, option 4.Q5. 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 3 … Explanation: D(1 + 3D/100) = 272. Testing D = 80 gives 80 × (1 + 2.4) = 80 × 3.4 = 272, so D = 80. Decreasing 80 by 80% gives 80 × 0.2 = 16. Hence, option 3.Q6. A shopkeeper manufactures a product. Raw materials account for 30% of its total cost, manufacturing accounts for 35%, and the remaining cost is attributable to packaging, transportation, and other expenses. The shopkeeper marks up the product's cost price by 20% to determine its selling price. If the cost of raw materials decreases by 8% and the manufacturing cost increases by 5%, while all other costs and the selling price remain unchanged, what is the approximate profit percentage earned by the shopkeeper?Correct Option 2 … Explanation: Let the total cost be 100: raw materials 30, manufacturing 35 and others 35, with SP = 120. New raw material cost = 30 × 0.92 = 27.6. New manufacturing cost = 35 × 1.05 = 36.75. Other costs stay at 35. New total cost = 99.35. Profit % = (120 − 99.35)/99.35 × 100 = 20.65/99.35 × 100 ≈ 20.78%. Hence, option 2.Q7. After two successive increases, the population of a town becomes 48/25 times its initial population. If the second percentage increase is three times the first percentage increase, what is the percentage increase in the first instance?Correct Option 2 … Explanation: Let the first increase be x (as a fraction), so the second is 3x. Then (1 + x)(1 + 3x) = 48/25 = 1.92. Testing x = 0.2 gives 1.2 × 1.6 = 1.92. So the first increase is 20%, and the second is 60%. Hence, option 2.Q8. The daily earnings of three workers, A, B, and C, are constant. The total earnings of A and B over 4 days equal C's earnings over 11 days. Similarly, the total earnings of A and C over 5 days equal B's earnings over 7 days. What is the ratio of the highest daily earning to the lowest daily earning?Correct Option 1 … Explanation: The conditions give 4(A + B) = 11C and 5(A + C) = 7B. Multiply the first by 5: 20A + 20B = 55C. Multiply the second by 4: 20A − 28B = −20C. Subtracting gives 48B = 75C, so B = 25C/16. Then A = (11C − 4B)/4 = 19C/16. So A : B : C = 19 : 25 : 16. Highest : lowest = B : C = 25 : 16. Option 3 (25:19) compares B with A instead of the lowest earner, C. Hence, option 1.Q9. Sheetal invested a total of 65000 across three different schemes, A, B, and C, that offered simple interest of 6%, 8%, and 9%, respectively. The total interest received after 2 years was 10180. If the amount invested in scheme A was 72% of the amount invested in scheme C, how much was invested in scheme B?Correct Option 2 … Explanation: Annual interest = 10180/2 = 5090. Let C = c, so A = 0.72c and B = 65000 − 1.72c. Then 0.06(0.72c) + 0.08(65000 − 1.72c) + 0.09c = 5090. This simplifies to 0.0432c + 5200 − 0.1376c + 0.09c = 5090, so −0.0044c = −110 and c = 25000. A = 18000 and B = 65000 − 18000 − 25000 = 22000. Check: 1080 + 1760 + 2250 = 5090. Option 1 (18000) is the amount in scheme A, not B. Hence, option 2.Q10. In the last three years, Ajit has received a salary hike of 10%, 20% and 25% respectively. If the hike in the third year was Rs. 66000, what was the hike received in the second year?Correct Option 4 … Explanation: Let the initial salary be S. After two hikes, the salary is S × 1.1 × 1.2 = 1.32S. The third-year hike = 25% of 1.32S = 0.33S = 66000, so S = 200000. Second-year hike = 20% of 1.1S = 0.22 × 200000 = Rs. 44000. Hence, option 4.Q11. Rakesh invests a certain sum at 20% per annum compound interest, compounded annually, for 12 years. Sooraj invests a certain sum at 40% per annum compound interest, compounded half-yearly, for 5 years. At the end of 5 years, Sooraj withdraws the entire amount and invests it in another scheme offering 15% per annum simple interest for 2 years. If the final amounts received by Rakesh and Sooraj after 12 years and 7 years, respectively, are in the ratio 12:13, by what percentage is Rakesh's initial investment less than Sooraj's initial investment?Correct Option 4 … Explanation: Rakesh's amount = R(1.2)^12. Sooraj earns 20% per half-year for 10 half-years, giving S(1.2)^10. The 2 years of SI at 15% multiply this by 1.3, so his final amount is S(1.2)^10 × 1.3. The ratio is R(1.2)^12 / [S(1.2)^10 × 1.3] = 12/13. This gives 1.44R/(1.3S) = 12/13, so 1.44R = 1.2S and R/S = 5/6. Rakesh's investment is less by 1/6 ≈ 16.67%. Option 5 (20%) is how much Sooraj's investment exceeds Rakesh's, which is the reverse comparison. Hence, option 4.Q12. Working at their usual efficiency levels, Risha and Swati together can complete a task in 21 days. After working together for 9 days, Swati reduces her efficiency to half of her usual rate, while Risha increases her efficiency to three times her usual rate. Together, they complete the remaining task in 9 days. How many days would Swati take to complete the entire task alone at her usual efficiency?Correct Option 3 … Explanation: Let the daily rates be r (Risha) and s (Swati), so the total work = 21(r + s). After 9 days, 12(r + s) remains. This is done in 9 days at the new rate (3r + s/2), so 9(3r + 0.5s) = 12r + 12s. That gives 27r + 4.5s = 12r + 12s, so 15r = 7.5s and s = 2r. Since r + s = 3r = 1/21, r = 1/63 and s = 2/63. Swati alone takes 63/2 = 31.5 days. Hence, option 3.Q13. The price of a laddoo is x% higher than that of a barfi. A box can hold 33.33% more barfis than laddoos. These boxes are then packed into separate cartons, with the number of laddoo boxes per carton and the number of barfi boxes per carton being in a ratio of 7:5. If the total price of a carton of laddoos is 26% more than the total price of a carton of barfis, what is the value of x?Correct Option 3 … Explanation: Let a barfi cost b and a box hold L laddoos, so a box holds 4L/3 barfis. A laddoo carton is worth 7 × L × b(1 + x/100). A barfi carton is worth 5 × (4L/3) × b = 20Lb/3. The ratio is 7(1 + x/100) ÷ (20/3) = 21(1 + x/100)/20 = 1.26. So 1 + x/100 = 1.2, and x = 20. Hence, option 3.Q14. Saurabh works for two companies, A and B. His hourly wage at company B is T% higher than his hourly wage at company A, while the number of hours he works per day at company B is 3T/8 % less than the hours he works at company A. If his daily earnings at company B are 19/40% higher than his daily earnings at company A, and his total daily working time across both companies is 11 hours and 6 minutes, how many hours does he work per day at company A?Correct Option 3 … Explanation: This solution reads the earnings increase as 19T/40 %. Let t = T/100. Then (1 + t)(1 − 3t/8) = 1 + 19t/40. Expanding the left side gives 1 + 5t/8 − 3t²/8. Setting this equal to the right side gives 5t/8 − 19t/40 = 3t²/8, so 6t/40 = 3t²/8 and t = 0.4, i.e. T = 40. Hours at B are therefore 15% less than at A. Let h be the hours at A; then h + 0.85h = 1.85h = 11.1 hours (11 h 6 min), so h = 6. Check: earnings ratio = 1.4 × 0.85 = 1.19, a 19% (= 19 × 40/40 %) increase. Hence, option 3.Q15. Rajesh invests equal amounts of money at the same fixed annual rate of interest under both simple interest and compound interest, compounded annually. The simple interest earned at the end of 3 years is ₹45,000, while the compound interest earned over the same period is ₹54,600. Find the annual rate of interest.Correct Option 5 … Explanation: The SI per year is Pr = 45000/3 = 15000. For 3 years, CI − SI = Pr²(3 + r), so 15000 × r(3 + r) = 54600 − 45000 = 9600. This gives r(3 + r) = 0.64, and r = 0.2 satisfies it (0.2 × 3.2 = 0.64). So the rate is 20%. Hence, option 5.Q16. 15 men, 18 women, or 45 children can independently paint a house in 44 days. Assuming all workers of a type work at constant rates, in how many days can 25 men, 12 women, and 27 children paint 5 such identical houses?Correct Option 4 … Explanation: Since 15 men = 18 women = 45 children, 1 man = 3 children and 1 woman = 2.5 children. The group 25 men + 12 women + 27 children equals 75 + 30 + 27 = 132 children. One house needs 45 × 44 = 1980 child-days, so 5 houses need 9900 child-days. Days = 9900/132 = 75. Hence, option 4.Q17. At a municipal clinic, staff members are classified into three groups: doctors, nurses, and attendants. Doctors supervise both nurses and attendants, while nurses supervise attendants and report to doctors. A doctor states that the ratio of her peers to her subordinates is 1:6. A nurse states that the ratio of her superiors to her subordinates is 1:3. What is the minimum possible total number of staff members at the clinic?Correct Option 1 … Explanation: Let there be D doctors, N nurses and A attendants. A doctor's peers are the other doctors and her subordinates are all nurses and attendants, so (D − 1) : (N + A) = 1 : 6. A nurse's superiors are the doctors and her subordinates are the attendants, so D : A = 1 : 3, i.e. A = 3D. Then N + 3D = 6(D − 1), so N = 3D − 6. For N ≥ 1 we need D ≥ 3. With D = 3: N = 3 and A = 9, so the total is 15. Hence, option 1.Q18. Train A leaves Delhi at 9:00 a.m. and reaches Agra at 1:00 p.m. Train B, which leaves Agra at 8:30 a.m. reaches Delhi at 12:00 p.m. At what time do they pass each other?Correct Option 1 … Explanation: Let the distance be d. Train A takes 4 hours, so its speed is d/4. Train B takes 3.5 hours, so its speed is 2d/7. By 9:00 a.m., B has run for 30 minutes and covered d/7, leaving 6d/7. Their relative speed is d/4 + 2d/7 = 15d/28. Time to meet = (6/7) ÷ (15/28) = 1.6 hours = 1 hour 36 minutes after 9:00 a.m., i.e. 10:36 a.m. Hence, option 1.Q19. A vendor typically earns a 25% profit on each electric bulb sold. However, from his current stock, he sold 200 bulbs at a 30% loss due to their approaching expiry date. Despite this, he managed an overall profit of 20%. What was the total number of bulbs sold?Correct Option 3 … Explanation: Let the CP of each bulb be 1, and let n bulbs be sold at 25% profit. Then 0.25n − 0.30 × 200 = 0.20(n + 200). This gives 0.25n − 60 = 0.2n + 40, so 0.05n = 100 and n = 2000. Total bulbs sold = 2000 + 200 = 2200. Option 2 (2000) counts only the bulbs sold at a profit. Hence, option 3.Q20. A work can be done with three types of machines: A, B, and C. It can be done in 55 hours by: machine A or 2 units of machine B, or 3 units of machine C. If instead of machine A alone, a unit each of machines B and C are also used, how many fewer hours would it take to complete the work?Correct Option 5 … Explanation: Let machine A's rate be a. Then B's rate = a/2 and C's rate = a/3. Together, A + B + C = a(1 + 1/2 + 1/3) = 11a/6. The new time = 55 × 6/11 = 30 hours. That is 55 − 30 = 25 hours fewer. Option 1 (22) is not the difference, and option 4 (20) would follow from adding only one extra machine's rate incorrectly. Hence, option 5.
