IBPS PO Prelims 2022 Quantitative Aptitude Questions with Solutions

Explanation

Let the cost price of the shirt be C. Marked price = 1.6C. Selling price = 1.6C × (1 - 0.25) = 1.2C. Profit on shirt = 1.2C - C = 0.2C = Rs. 950. Cost price of shirt C = 950 / 0.2 = Rs. 4750.

Cost price of pant = 1.4 × 4750 = Rs. 6650. Since the profit on both items is the same (Rs. 950): Selling price of pant = 6650 + 950 = Rs. 7600.

Explanation

Interest rate in scheme A = 12.5% = 1/8 per annum compound interest for 2 years. Amount after 2 years from scheme A = P × (1 + 1/8)² = P × (9/8)² = 81P / 64.

This amount is invested in scheme B at 20% simple interest per annum for 3 years. Interest from scheme B = (81P / 64) × (20 / 100) × 3 = 243P / 320. Given interest = Rs. 2430. 243P / 320 = 2430 P = 3200.

Explanation

Dev travels from 10:00 AM to 2:30 PM, total time = 4.5 hours. Distance covered by Dev = 42 × 4.5 = 189 km.

Harsh travels from 12:00 PM to 2:30 PM, total time = 2.5 hours. Distance covered by Harsh = 28 × 2.5 = 70 km.

Total distance between A and B = 189 + 70 = 259 km.

Explanation

Investment-time product over 24 months: For A: Months 1-6: 32500 × 6 = 195,000 Months 7-12: 35000 × 6 = 210,000 Months 13-24: 45000 × 12 = 540,000 Total for A = 945,000

For B: Months 1-6: 40000 × 6 = 240,000 Total for B = 240,000

For C: Months 13-24: 60000 × 12 = 720,000 Total for C = 720,000

Profit sharing ratio A : B : C = 945,000 : 240,000 : 720,000 = 63 : 16 : 48. Given C's profit share = Rs. 9600. 1 unit = 9600 / 48 = Rs. 200. Profit share of B = 16 × 200 = Rs. 3200.

Explanation

Efficiency ratio of Q : P = 160 : 100 = 8 : 5. Time ratio Q : P = 5 : 8. Let Q take 5x days and P take 8x days. Difference = 8x - 5x = 3x = 15 days, so x = 5. Time taken by Q = 25 days, P = 40 days.

P, Q, and R together complete the work in 8 days. 1/P + 1/Q + 1/R = 1/8 1/40 + 1/25 + 1/R = 1/8 1/R = 1/8 - 1/40 - 1/25 = (25 - 5 - 8) / 200 = 12 / 200 = 3 / 50.

Time taken by R alone = 50 / 3 = 16 2/3 days.

Explanation

Since the value of ₹1 coins is ₹25, the number of ₹1 coins is 25 / 1 = 25. Given the ratio of ₹1, ₹2, and ₹5 coins is 5 : 6 : 7, the common multiplier is 25 / 5 = 5. The number of ₹5 coins is 7 × 5 = 35. Total value of ₹5 coins = 35 × ₹5 = ₹175.

Explanation

Let the speed of the stream be 5x km/h, then the speed of the boat in still water is 1.4 × 5x = 7x km/h. Downstream speed = 7x + 5x = 12x km/h, and upstream speed = 7x - 5x = 2x km/h. Total time = 96 / (12x) + 40 / (2x) = 8 / x + 20 / x = 28 / x = 14 hours, which gives x = 2. Downstream speed = 12 × 2 = 24 km/h. Time required to cover 288 km downstream = 288 / 24 = 12 hours.

Explanation

Sum of present ages of P, Q, and R = 35 × 3 + 3 × 5 = 120 years. Sum of present ages of P and R = 40 × 2 - 2 × 5 = 70 years. Present age of Q = 120 - 70 = 50 years. Present age of P = 50 - 10 = 40 years. Present age of R = 70 - 40 = 30 years.

Explanation

Let the train length be L and its speed be v. L = 8v and L + 300 = 20v. 20v - 8v = 300, so 12v = 300, giving v = 25 m/s and L = 200 m. Car speed = 54 × (5 / 18) = 15 m/s. Relative speed moving in opposite directions = 25 + 15 = 40 m/s. Time to cross the car = 200 / 40 = 5 seconds.

Explanation

Milk in Jar P = 70% of 20 = 14 liters. Milk in Jar Q = (3 / 5) × 30 = 18 liters. Total milk in Jar Z = 14 + 18 = 32 liters. Total volume in Jar Z = 20 + 30 = 50 liters. Percentage of milk in the mixture = (32 / 50) × 100 = 64%.

Explanation

Perimeter of square = 88 cm, so side = 88 / 4 = 22 cm. Cylinder radius r = 22 / 2 = 11 cm. Circle area = π × R² = 616 cm², so R² = (616 × 7) / 22 = 196, giving R = 14 cm. Cylinder height h = 14 cm. Volume of cylinder = π × r² × h = (22 / 7) × 11² × 14 = 22 × 121 × 2 = 5324 cm³.

Explanation

Let A's initial share be A. B's share = 2A, and C's share = 1.2A. Total toffees = A + 2A + 1.2A = 4.2A = 84, so A = 20 and B = 40. B gives 25% of 40 = 10 toffees to A. Final number of toffees A possesses = 20 + 10 = 30.

Explanation

Total number of outcomes when 3 coins are tossed = 2³ = 8. Number of outcomes with exactly two tails = 3C2 = 3. Number of outcomes with three tails = 3C3 = 1. Favorable outcomes for at least two tails = 3 + 1 = 4. Probability = 4 / 8 = 0.5.

Explanation

For equation I: x² - 16x + 55 = 0 gives (x - 11)(x - 5) = 0, so x = 11 or x = 5. For equation II: y² + 16y + 55 = 0 gives (y + 11)(y + 5) = 0, so y = -11 or y = -5. Comparing the roots, 11 > -5, 11 > -11, 5 > -5, and 5 > -11, so x > y.

Explanation

For equation I: x² - 20x + 75 = 0 gives (x - 15)(x - 5) = 0, so x = 15 or x = 5. For equation II: y² - 13y + 42 = 0 gives (y - 7)(y - 6) = 0, so y = 7 or y = 6. Comparing the values, 15 is greater than both 7 and 6, but 5 is less than both 7 and 6. Therefore, no definite relationship can be established between x and y.

Explanation

Solving 8x² − 2x − 45 = 0: 8x² − 20x + 18x − 45 = 0 4x(2x − 5) + 9(2x − 5) = 0 (4x + 9)(2x − 5) = 0 x = 2.5 or x = −2.25

Solving y² − 10y + 21 = 0: (y − 7)(y − 3) = 0 y = 7 or y = 3

Comparing the roots: 2.5 < 3 and 2.5 < 7 −2.25 < 3 and −2.25 < 7

In every case, x < y.

Explanation

Solving x² + 8x + 15 = 0: (x + 5)(x + 3) = 0 x = −5 or x = −3

Solving y² + 13y + 40 = 0: (y + 8)(y + 5) = 0 y = −8 or y = −5

Comparing the roots: −3 > −5 and −3 > −8 −5 = −5 and −5 > −8

Thus, x ≥ y.

Explanation

Solving x² − 9x + 20 = 0: (x − 4)(x − 5) = 0 x = 4 or x = 5

Solving y² − 22y + 85 = 0: (y − 17)(y − 5) = 0 y = 17 or y = 5

Comparing the roots: 4 < 5 and 4 < 17 5 = 5 and 5 < 17

Thus, x ≤ y.

Explanation

Solving x² − 7x + 6 = 0: (x − 1)(x − 6) = 0 x = 1 or x = 6

Solving y² − 14y + 48 = 0: (y − 6)(y − 8) = 0 y = 6 or y = 8

Comparing the roots: 1 < 6 and 1 < 8 6 = 6 and 6 < 8

Thus, x ≤ y.

Explanation

Let the work rate of pipes A, B, and C be A, B, and C respectively. A + B = 1/12 A + B + C = 1/15 B + C = 1/30

Subtracting (A + B) from (A + B + C): C = 1/15 − 1/12 = (4 − 5)/60 = −1/60

Substituting C into (B + C): B − 1/60 = 1/30 B = 1/30 + 1/60 = 3/60 = 1/20

Substituting B into (A + B): A + 1/20 = 1/12 A = 1/12 − 1/20 = (5 − 3)/60 = 2/60 = 1/30

Pipe A alone fills the tank in 30 minutes.

Explanation

Solving x² - 15x + 56 = 0 gives x = 7 and x = 8. Solving y² - 8y + 15 = 0 gives y = 3 and y = 5. Comparing the roots, both values of x (7 and 8) are strictly greater than both values of y (3 and 5).

Explanation

Solving x² + 12x + 35 = 0 gives x = -5 and x = -7. Solving y² + 7y + 10 = 0 gives y = -2 and y = -5. Comparing the values, x = -5 equals y = -5, while for all other pairs x is less than y. Thus, x ≤ y.

Explanation

Solving x² - 3x - 40 = 0 gives x = 8 and x = -5. Solving y² + 11y + 30 = 0 gives y = -5 and y = -6. Comparing the values, x = 8 is greater than both roots of y, and x = -5 equals y = -5 while being greater than y = -6. Thus, x ≥ y.

Explanation

Simplifying 2x² + 14x - 16 = 0 gives x² + 7x - 8 = 0, which yields x = 1 and x = -8. Solving y² - y - 12 = 0 gives y = 4 and y = -3. Comparing x = 1 with y = 4 gives x < y, but comparing x = 1 with y = -3 gives x > y. Since opposite comparison relationships exist, no relationship can be established.

Explanation

Solving 2x² - x - 3 = 0 gives x = 1.5 and x = -1. Simplifying 2y² - 6y + 4 = 0 gives y² - 3y + 2 = 0, yielding y = 1 and y = 2. Comparing x = 1.5 with y = 1 gives x > y, whereas comparing x = 1.5 with y = 2 gives x < y. Because contradictory relations exist, no relation can be established.

Explanation

Let the total students in only A, only B, and only C be 2k, 3k, and 5k respectively. The students in all three categories are 0.5 × 5k = 2.5k. Students in only A & B = 3k - 30. Sum of students in (only B & C) and (only A & C) = 155. Total students = 2k + 3k + 5k + 2.5k + (3k - 30) + 155 = 900. 15.5k + 125 = 900, which gives k = 50. Only A = 100, Only B = 150, Only C = 250, All three = 125, Only A & B = 120. Because the individual number of students in only A & C is not uniquely specified, the total number of students in category A cannot be calculated.

Explanation

Using k = 50, Only B = 150, Only C = 250, All three = 125, Only A & B = 120, and the sum of (only B & C) and (only A & C) is 155. Total in category C = Only C + (Only B & C + Only A & C) + All three = 250 + 155 + 125 = 530. Total in category B = Only B + Only A & B + Only B & C + All three = 150 + 120 + (Only B & C) + 125 = 395 + (Only B & C). Since the individual value for students in only B & C is unknown, category B cannot be uniquely determined, making it impossible to find the ratio.

Explanation

From the college Venn diagram analysis, the value of k = 50. The number of students in only category A is 2k = 2 × 50 = 100. Given that 30% are boys, the percentage of girls is 100% - 30% = 70%. Number of girls in only category A = 70% of 100 = 70.

Explanation

From the Venn diagram calculations, k = 50 and the number of students in all three categories is 2.5k = 2.5 × 50 = 125. The total number of students in the college is 900. Students in at most two categories = Total students - Students in all three categories = 900 - 125 = 775.

Explanation

From the Venn diagram analysis, k = 50. Only category A = 2k = 100. Only category B = 3k = 150. Only category C = 5k = 250. Total students in only one category = 100 + 150 + 250 = 500.

Explanation

The differences between consecutive terms in the series are 4, 6, 8, and 10. The difference increases by 2 at each step, so the next difference is 12. Adding 12 to 30 gives 42.

Explanation

The difference between the first two terms is 5² = 25. The next differences follow the squares of prime numbers: 7² = 49, 11² = 121, 13² = 169, and 17² = 289. Adding 49 to 49 yields 98, and verifying 98 + 121 gives 219.

Explanation

The first difference is 7 (7 × 1) and the second is 21 (7 × 3). Continuing this pattern of adding consecutive odd multiples of 7, the third difference is 35 (7 × 5). Adding 35 to 76 gives 111, which is confirmed by 111 + 49 = 160 and 160 + 63 = 223.

Explanation

The difference between terms increases by 1.5 at each step: +1.5, +3.0, +4.5, +6.0, and +7.5. Adding 6.0 to 24 results in 30, and adding 7.5 to 30 equals 37.5.

Explanation

The ratio between consecutive terms forms the sequence of odd numbers 3, 5, 7, 9, 11. Multiplying 210 by 9 gives 1890, and multiplying 1890 by 11 yields 20790.

Explanation

Let the present age of B be B years. Since C is 3 years younger than B, C = B - 3. A is 12 years older than C, so A = B - 3 + 12 = B + 9. Given the ratio A : B = 6 : 5, (B + 9) / B = 6 / 5, which simplifies to 5B + 45 = 6B, so B = 45. Thus, C = 42. Ten years hence, B will be 55 and C will be 52, making their sum 107 years.

Explanation

The sum of profit shares is 5800 and the difference is 1000, which gives profit shares of 2400 for A and 3400 for B. The ratio of profit share of A to B is 2400 : 3400 = 12 : 17. Profit share ratio equals investment product ratio: (12 × X) / [8 × (X + 900)] = 12 / 17. Simplifying gives 17X = 8X + 7200, which yields 9X = 7200, so X = 800.

Explanation

Let initial milk be 4m liters and water be m liters. Taking out 10 liters of mixture removes 8 liters of milk and 2 liters of water. Adding 25 liters of milk makes the final milk quantity 4m + 17 liters and water quantity m - 2 liters. Since final milk is 400% more than final water, milk is 5 times water: 4m + 17 = 5(m - 2). Solving gives m = 27, so initial milk quantity is 4 × 27 = 108 liters.

Explanation

CP_A + CP_B = 350. Markup is 60% on both, so MP_B - MP_A = 1.6 x (CP_B - CP_A) = 80, giving CP_B - CP_A = 50. Solving: CP_B = 200, CP_A = 150.

Explanation

The breadth B of the rectangle is found using perimeter = 2(L + B) = 108, which gives 2(40 + B) = 108, so B = 14 cm. The maximum radius R of a semicircle that can fit inside a 40 cm by 14 cm rectangle is 14 cm, requiring a base diameter of 28 cm and height of 14 cm. Because the condition regarding whether semicircles can overlap or must remain disjoint is not specified, the exact count cannot be determined.

Explanation

Let upstream speed be 3x and downstream speed be 5x. Time equation for the total distance covered gives 60/5x + 60/3x = 16 + 10/60, which simplifies to 32/x = 97/6, yielding x ≈ 1.979 km/h. Thus, upstream speed = 3x = 5.938 km/h. Time taken to cover 37 km upstream = 37 / 5.938 ≈ 6.23 hours, which is approximately 6 hours.

Explanation

Time taken to cover distance A to B = 400 / 40 = 10 hours. Since time taken for section B to C is the same, speed for B to C = 375 / 10 = 37.5 km/hr. Total distance = 400 + 375 = 775 km and total time = 10 + 10 = 20 hours. Average speed = 775 / 20 = 38.75 km/hr. The required difference is 38.75 - 37.5 = 1.25 km/hr.

Explanation

Efficiency ratio of Ritu to Rakesh = 160 : 100 = 8 : 5. Time ratio = 5 : 8. Given the difference in time is 3 units = 18 days, 1 unit = 6 days. Thus, Ritu takes 30 days and Rakesh takes 48 days. Total work = 8 × 30 = 240 units. Rahul's efficiency = 8 × (1 - 0.125) = 7 units/day. Total efficiency of all three = 8 + 5 + 7 = 20 units/day. Time taken together = 240 / 20 = 12 days.

Explanation

Simple Interest on Rs. X at 15% for 2 years = (X × 15 × 2) / 100 = 0.30X. Compound Interest on Rs. 2X at 8% for 2 years = 2X × [(1 + 8/100)² - 1] = 2X × 0.1664 = 0.3328X. The given difference is 0.3328X - 0.30X = 0.0328X = 820. Therefore, X = 820 / 0.0328 = Rs. 25000.

Explanation

While the savings percentage for 2017 allows calculating the income for 2017 as 45000 / 0.15 = 300000 Rs., there is no information provided regarding whether the income in 2018 remained the same, increased, or decreased. Consequently, the savings amount for 2018 cannot be evaluated.

Explanation

Invalid votes in block A for 2012 = 750 - 200 = 550. Female invalid votes in block A = (5 / 11) × 550 = 250. Invalid votes in block B for 2015 = 1300 - 1000 = 300. The required difference = 300 - 250 = 50.

Explanation

Total invalid votes in 2015 = (900 - 720) + (1300 - 1000) = 180 + 300 = 480. Total valid votes in 2012 = 200 + 1000 = 1200. Percentage less = [(1200 - 480) / 1200] × 100 = (720 / 1200) × 100 = 60%.

Explanation

Invalid votes in block B in 2012 = 1200 - 1000 = 200. Valid votes obtained after recasting = 25% of 200 = 50. Total valid votes in block B after recasting = 1000 + 50 = 1050.

Explanation

Invalid votes in 2012: Block A = 750 - 200 = 550, Block B = 1200 - 1000 = 200. Average invalid votes in 2012 = (550 + 200) / 2 = 375. Invalid votes in 2015: Block A = 900 - 720 = 180, Block B = 1300 - 1000 = 300. Average invalid votes in 2015 = (180 + 300) / 2 = 240. Required difference = 375 - 240 = 135.

Explanation

Total valid votes of block A in 2012 = 200. Total registered votes of block B in 2015 = 1300. The required ratio = 200 : 1300 = 2 : 13.