IBPS PO Prelims 2022 Quantitative Aptitude Questions with Solutions

Explanation

Curved surface area = 2 × π × r × h = 10692. Given r : h = 3 : 7, let r = 3k and h = 7k. 2 × (22 / 7) × 3k × 7k = 10692 132 k² = 10692 k² = 81, so k = 9.

Radius r = 27 cm, height h = 63 cm. Volume = π × r² × h = (22 / 7) × 27² × 63 = 144342 cm³. Time required to empty vessel = 144342 / 1782 = 81 minutes.

Explanation

Let daily work rates of a man and a woman be m and w. Total work W = 15 × (8m + 7w) = 120m + 105w. Also, (4/5)W = 10 × (10m + 7w) = 100m + 70w, so W = 125m + 87.5w.

120m + 105w = 125m + 87.5w 5m = 17.5w m = 3.5w.

Total work W = 120(3.5w) + 105w = 525w. Time required for 21 women = 525w / (21w) = 25 days.

Explanation

Delhi Express speed = 144 × (5 / 18) = 40 m/s. Length of Delhi Express = 40 × 12.5 = 500 m.

Patna Express speed = 162 × (5 / 18) = 45 m/s. Relative speed in opposite directions = 40 + 45 = 85 m/s. Total distance covered in 12 seconds = 85 × 12 = 1020 m. Length of Patna Express = 1020 - 500 = 520 m.

Explanation

Let total balls be N. Probability of orange ball = (N - 36) / N = 1 / 5. 5N - 180 = N, giving N = 45. Orange balls = 9, Pink balls = 36.

Probability of picking one pink and one orange ball when drawing two balls = (36C1 × 9C1) / 45C2 = (36 × 9) / 990 = 324 / 990 = 18 / 55.

Explanation

Distance = Speed × Time. D = (x + 13) × 10 D = (4x - 12) × 12.5

10(x + 13) = 12.5(4x - 12) 10x + 130 = 50x - 150 40x = 280 x = 7.

Distance = (7 + 13) × 10 = 200 km.

Explanation

Items sold of type A = 35% of 2000 = 700. Items sold of type C = 40% of 1600 = 640. Total sold items of A and C = 700 + 640 = 1340.

Items unsold of type B = (100 - 65)% of 1500 = 35% of 1500 = 525. Items unsold of type E = (100 - 70)% of 2800 = 30% of 2800 = 840. Total unsold items of B and E = 525 + 840 = 1365.

Ratio = 1340 : 1365 = 268 : 273.

Explanation

Unsold items of type B = 35% of 1500 = 525. Unsold defective items of type B = (7/15) × 525 = 245. Unsold non-defective items of type B = 525 - 245 = 280.

Unsold items of type D = 37% of 600 = 222. Unsold defective items of type D = (5/6) × 222 = 185. Unsold non-defective items of type D = 222 - 185 = 37.

Total unsold defective items = 245 + 185 = 430. Total unsold non-defective items = 280 + 37 = 317. Difference = 430 - 317 = 113.

Explanation

Non-defective items of type A = (100 - 60)% of 2000 = 40% of 2000 = 800.

Unsold items of type C = 60% of 1600 = 960. Unsold items of type E = 30% of 2800 = 840. Average unsold items of C and E = (960 + 840) / 2 = 900.

Percentage = (800 / 900) × 100 = 88.88% ≈ 89%.

Explanation

Items sold of type B = 65% of 1500 = 975. Selling cost of type B = 975 × 50 = ₹48,750.

Items sold of type E = 70% of 2800 = 1960. Selling cost of type E = 1960 × 70 = ₹1,37,200.

Average total selling cost = (48750 + 137200) / 2 = 185950 / 2 = ₹92,975.

Explanation

Sold items of type A and D = (35% of 2000) + (63% of 600) = 700 + 378 = 1078. Sold items of type C and B = (40% of 1600) + (65% of 1500) = 640 + 975 = 1615.

Percentage difference = [(1615 - 1078) / 1615] × 100 = (537 / 1615) × 100 ≈ 33.25% ≈ 33%.

Explanation

Let the remaining quantity of mixture after removing 180 ml be 30k. Alcohol = 13k and Water = 17k.

After adding 12 ml alcohol and 24 ml water: (13k + 12) / (17k + 24) = 5 / 7 7 × (13k + 12) = 5 × (17k + 24) 91k + 84 = 85k + 120 6k = 36 ⇒ k = 6.

Remaining mixture = 30 × 6 = 180 ml. Original quantity = 180 + 180 = 360 ml.

Explanation

Equation I: x² + 13x - 114 = 0 (x + 19)(x - 6) = 0 ⇒ x = -19, 6.

Equation II: y³ = 216 y = 6.

Comparing x and y: For x = -19, x < y. For x = 6, x = y. Hence, x ≤ y.

Explanation

Equation I: 3x² + 15x + 18 = 0 x² + 5x + 6 = 0 ⇒ (x + 2)(x + 3) = 0 ⇒ x = -2, -3.

Equation II: y² + 14y + 48 = 0 (y + 6)(y + 8) = 0 ⇒ y = -6, -8.

Comparing each value of x with y: -2 > -6 and -2 > -8 -3 > -6 and -3 > -8 Hence, x > y.

Explanation

Equation I: x² - 6x + 8 = 0 (x - 2)(x - 4) = 0 ⇒ x = 2, 4.

Equation II: y² + 2y - 8 = 0 (y + 4)(y - 2) = 0 ⇒ y = -4, 2.

Comparing x and y: 4 > 2, 4 > -4 2 = 2, 2 > -4 Hence, x ≥ y.

Explanation

Equation I: 10x² - 9x + 2 = 0 (5x - 2)(2x - 1) = 0 ⇒ x = 2/5 = 0.4, x = 1/2 = 0.5.

Equation II: 15y² - 11y + 2 = 0 (3y - 1)(5y - 2) = 0 ⇒ y = 1/3 ≈ 0.33, y = 2/5 = 0.4.

Comparing x and y: 0.5 > 0.4 and 0.5 > 0.33 0.4 = 0.4 and 0.4 > 0.33 Hence, x ≥ y.

Explanation

The differences between consecutive terms are: 7.5 - 4 = 3.5 14.5 - 7.5 = 7.0 25 - 14.5 = 10.5 39 - 25 = 14.0

The differences increase by 3.5 at each step. Next difference = 14.0 + 3.5 = 17.5 Missing term = 39 + 17.5 = 56.5 Checking the next term: 56.5 + 21.0 = 77.5.

Explanation

Working backward from the last term: 18900 ÷ 3780 = 5 3780 ÷ 840 = 4.5 840 ÷ 210 = 4 210 ÷ 60 = 3.5 60 ÷ 20 = 3

The multiplier decreases by 0.5 for each step to the left, so the first multiplication step is: ? × 2.5 = 20 ? = 20 ÷ 2.5 = 8

Explanation

The differences between consecutive terms are: 6 - 2 = 4 12 - 6 = 6 20 - 12 = 8 30 - 20 = 10 42 - 30 = 12

The difference increases by 2 each step. Next difference = 14 Missing term = 42 + 14 = 56

Explanation

The consecutive differences follow alternating signs with consecutive prime numbers: 57 - 7 = 50 50 + 11 = 61 61 - 13 = 48 48 + 17 = 65 65 - 19 = 46 46 + 23 = 69 69 - 29 = 40

Missing term = 46

Explanation

The pattern combines multiplication by n and alternating addition/subtraction of n + 1: 4 × 1 + 2 = 6 6 × 2 - 3 = 9 9 × 3 + 4 = 31 31 × 4 - 5 = 119 119 × 5 + 6 = 601 601 × 6 - 7 = 3599

Explanation

Solving equation I: x² + 7x + 10 = 0 (x + 5)(x + 2) = 0 x = -5, -2

Solving equation II: y² + 6y + 8 = 0 (y + 4)(y + 2) = 0 y = -4, -2

Comparing x and y: When x = -2 and y = -4, x > y. When x = -5 and y = -4, x < y. Since x can be both greater than and less than y, no relation can be established.

Explanation

Solving equation I: 2x² - 15x + 28 = 0 (2x - 7)(x - 4) = 0 x = 3.5, 4

Solving equation II: 2y² + 3y - 20 = 0 (2y - 5)(y + 4) = 0 y = 2.5, -4

Comparing all values: 3.5 > 2.5 and 3.5 > -4 4 > 2.5 and 4 > -4 In all cases, x > y.

Explanation

Solving equation I: x² - 7x - 98 = 0 (x - 14)(x + 7) = 0 x = 14, -7

Solving equation II: y² + 19y + 84 = 0 (y + 12)(y + 7) = 0 y = -12, -7

Comparing all values: 14 > -12 and 14 > -7 -7 > -12 and -7 = -7 In all cases, x ≥ y.

Explanation

Solving equation I: x² - 11x + 30 = 0 (x - 5)(x - 6) = 0 x = 5, 6

Solving equation II: y² - 13y + 42 = 0 (y - 6)(y - 7) = 0 y = 6, 7

Comparing all values: 5 < 6 and 5 < 7 6 = 6 and 6 < 7 In all cases, x ≤ y.

Explanation

Solving the system of linear equations: 7x + 4y = 55 5x + 6y = 55

Multiplying the first equation by 3 and the second equation by 2: 21x + 12y = 165 10x + 12y = 110

Subtracting the second equation from the first gives: 11x = 55 → x = 5

Substituting x = 5 into the first equation: 35 + 4y = 55 → 4y = 20 → y = 5

Since x = 5 and y = 5, x = y.

Explanation

For equation I, x² - 12x + 35 = 0 factors as (x - 5)(x - 7) = 0, giving roots x = 5 and x = 7. For equation II, y² + 8y + 15 = 0 factors as (y + 3)(y + 5) = 0, giving roots y = -3 and y = -5. Comparing all values, x = 5 and x = 7 are both strictly greater than y = -3 and y = -5.

Explanation

Cost price of mixture = Selling Price / (1 + Profit %) = 77 / (1 + 1/6) = 77 × (6/7) = ₹66 per kg. Let the cost per kg of rice A be 5x and rice B be 4x. Weighted average cost price = (40 × 5x + 60 × 4x) / (40 + 60) = 4.4x. Equating to the mixture cost price: 4.4x = 66, which gives x = 15. The difference between the cost per kg of rice A and rice B is 5x - 4x = x = ₹15.

Explanation

Profit share of Naman = 14500 - 4000 = ₹10500. Ratio of profit share = Naitik / Naman = 4000 / 10500 = 8/21. Investment-time product ratio = (P × T) / [(P + 3000) × 1.5T] = P / [1.5 × (P + 3000)]. Equating ratios: P / (1.5P + 4500) = 8/21. 21P = 12P + 36000 9P = 36000 ⇒ P = 4000. Investment of Naman = P + 3000 = 4000 + 3000 = ₹7000.

Explanation

34 ⅔% is equivalent to 240/700 = 12/35. Let the second number be x. First number = x + (12/35)x = (47/35)x. Average of first and second number = [(47/35)x + x] / 2 = (41/35)x. Third number = (41/35)x + 4. Sum of all three numbers = (47/35)x + x + (41/35)x + 4 = (123/35)x + 4 = 496. (123/35)x = 492 ⇒ x = (492 × 35) / 123 = 140.

Explanation

Downstream speed = 432 / 9 = 48 km/h. Upstream speed = 288 / 16 = 18 km/h. Speed of boat in still water = (48 + 18) / 2 = 33 km/h. Time taken to cover 132 km in still water = 132 / 33 = 4 hours.

Explanation

For city B, non-male population = 60% of 84000 = 50400. Females in city B = (7 / 12) × 50400 = 29400. For city A, non-male population = 75% of 45000 = 33750. Children in city A = (7 / 15) × 33750 = 15750. Required percentage = (29400 / 15750) × 100 = 186.67% ≈ 187%.

Explanation

For city D, non-male population = 45% of 66000 = 29700. Children in city D = (3 / 11) × 29700 = 8100. Males in city E = 60% of 72000 = 43200. Ratio of children in city D to males in city E = 8100 : 43200 = 3 : 16.

Explanation

Children in city A = (7 / 15) × (75% of 45000) = 15750. Children in city E = (2 / 9) × (40% of 72000) = 6400. Total children in city E and A together = 6400 + 15750 = 22150. Males in city D = 55% of 66000 = 36300. Males in city B = 40% of 84000 = 33600. Total males in city D and B together = 36300 + 33600 = 69900. Difference = 69900 - 22150 = 47750.

Explanation

Females in city A = (8 / 15) × 33750 = 18000. Females in city B = (7 / 12) × 50400 = 29400. Females in city C = (14 / 25) × (55% of 50000) = 15400. Females in city D = (8 / 11) × 29700 = 21600. Females in city E = (7 / 9) × 28800 = 22400. Total females = 18000 + 29400 + 15400 + 21600 + 22400 = 106800. Average number of females = 106800 / 5 = 21360.

Explanation

Children in city A = (7 / 15) × (75% of 45000) = 15750. Males in city E = 60% of 72000 = 43200. Percentage less = [(43200 - 15750) / 43200] × 100 = (27450 / 43200) × 100 = 63.54% ≈ 63.5%.

Explanation

Total time for the journey is equal to total distance divided by average speed, which is 100 / 37.5 = 8/3 hours. Let the distance between A and B be d km. Then the distance between B and C is (100 - d) km. Using time = distance / speed, d / 45 + (100 - d) / 30 = 8/3. Multiplying the equation by 90 gives 2d + 3(100 - d) = 240, which simplifies to 300 - d = 240, giving d = 60 km.

Explanation

Let the initial quantity of milk be M liters. Since water is 20% of milk, initial water is 0.2M liters. When 10 liters of water are added, the new quantity of water becomes 0.2M + 10 liters, which is equal to 40% of milk, or 0.4M. Solving 0.2M + 10 = 0.4M yields 0.2M = 10, so M = 50 liters. The initial water is 0.2 × 50 = 10 liters. The difference between the initial quantities of milk and water is 50 - 10 = 40 liters.

Explanation

Let A's investment be 8 units. B invested 37.5% less, so B's investment is 5 units. C invested 20% more than B, so C's investment is 6 units. The ratio of their investments multiplied by time period gives the profit share ratio A : B : C = (8 × 4) : (5 × 8) : (6 × 6) = 32 : 40 : 36 = 8 : 10 : 9. The combined share of B and C is 10 + 9 = 19 units, which corresponds to Rs. 17100. Thus, 1 unit = 900 Rs. A's profit share is 8 × 900 = 7200 Rs.

Explanation

Train A crosses a 600 m platform in 36 seconds and a pole in 16 seconds, meaning it covers 600 m in 36 - 16 = 20 seconds. The speed of Train A is 600 / 20 = 30 m/s, and its length is 30 × 16 = 480 m. Train B's speed is 72 km/hr = 20 m/s and its length is 640 m. Running in the same direction, relative speed is 30 - 20 = 10 m/s, and total distance to cover is 480 + 640 = 1120 m. The time required is 1120 / 10 = 112 seconds.

Explanation

The area of the square is 784 m², so its side length is √784 = 28 m, and its perimeter is 4 × 28 = 112 m. The perimeter of the rectangle is 2(L + B) = 112, so L + B = 56 m. Since B = 0.4L, L + 0.4L = 56, giving 1.4L = 56, so L = 40 m and B = 16 m. The area of the rectangle is L × B = 40 × 16 = 640 m².

Explanation

Let the present ages of Veer and Ayush be 16x and 7x. Twelve years hence, (16x + 12) = (20/11) × (7x + 12). Multiplying across gives 176x + 132 = 140x + 240, so 36x = 108, yielding x = 3. Present age of Veer is 48 years and Ayush is 21 years. The sum of present ages of Veer, Ayush, and Shivam is 35 × 3 = 105 years. Therefore, Shivam's present age is 105 - (48 + 21) = 36 years.

Explanation

Simple interest from scheme P is P × 15 × 2 / 100 = 0.30P = 3900, giving P = 13000 Rs. For scheme Q, principal is 13000 + X, rate is 10% p.a. compound interest for 2 years, which equals an effective interest rate of 21%. Thus, 0.21 × (13000 + X) = 3360. Solving for 13000 + X gives 16000, so X = 16000 - 13000 = 3000.

Explanation

The class has 60 girls and 30 boys, making the ratio of girls to boys 2 : 1. Total average weight is 146/3 kg, and boys' average weight is 56 kg. Let G be the average weight of girls. (2G + 56) / 3 = 146 / 3, which simplifies to 2G + 56 = 146, giving 2G = 90 and G = 45 kg. The difference between the average weight of boys and girls is 56 - 45 = 11 kg.

Explanation

Ankit buys the table from Rahul at a 20% profit for Rs. 1296. Let C be Rahul's cost price for the table. Since Rahul sold it at a 20% profit, 1.20 × C = 1296. Dividing 1296 by 1.20 gives C = 1080 Rs.

Explanation

Let total work be 40 units. Bholu's efficiency is 1 unit/day. Happy is 25% more efficient, so Happy's efficiency is 1.25 units/day. In 12 days, Happy completes 12 × 1.25 = 15 units of work. The remaining 25 units are completed by Shivam in 30 days, so Shivam's efficiency is 25/30 = 5/6 units/day. The time taken by Shivam alone to complete 40 units is 40 / (5/6) = 48 days.

Explanation

The given series follows a double difference pattern: First differences: 7 - 4 = 3, 13 - 7 = 6, 24 - 13 = 11, 42 - 24 = 18, 69 - 42 = 27. Second differences: 6 - 3 = 3, 11 - 6 = 5, 18 - 11 = 7, 27 - 18 = 9.

The second differences increase by 2 each time. The next second difference is 11. Adding 11 to the last first difference gives 27 + 11 = 38. The next term in the series is 69 + 38 = 107.

Explanation

The prime numbers starting from 7 are 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59... The series picks alternate prime numbers: 7 (skip 11) → 13 13 (skip 17) → 19 19 (skip 23) → 29 29 (skip 31) → 37 37 (skip 41) → 43 43 (skip 47) → 53

The missing term is 53.

Explanation

The differences between consecutive terms follow the pattern n × (n + 1): 1 × 2 = 2 2 × 3 = 6 3 × 4 = 12 4 × 5 = 20 5 × 6 = 30 6 × 7 = 42

Working backwards from 5 using the first difference of 2: Missing first term = 5 - 2 = 3.

Checking the entire sequence: 3 + 2 = 5 5 + 6 = 11 11 + 12 = 23 23 + 20 = 43 43 + 30 = 73 73 + 42 = 115

Explanation

The multiplication pattern between consecutive terms is: 24 × 2.5 = 60 60 × 3.5 = 210 210 × 4.5 = 945 945 × 5.5 = 5197.5

The multipliers increase by 1 at each step: ×0.5, ×1.5, ×2.5, ×3.5, ×4.5, ×5.5. Applying this to the first step: 32 × 0.5 = 16 16 × 1.5 = 24

Thus, the missing number is 16.

Explanation

The series pattern is: 5 × 1 - 2 = 3 3 × 2 - 3 = 3 3 × 3 - 4 = 5 5 × 4 - 5 = 15 15 × 5 - 6 = 69 69 × 6 - 7 = 414 - 7 = 407

The required next term in the series is 407.