IBPS PO2019Quantitative AptitudePage 2 of 2

IBPS PO Prelims 2019 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 51–85 of 85, each with the correct answer and a step-by-step explanation.

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51Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I.x³ = 2744 II.y² = 324
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Answer: (A) x = y or no relation.

Explanation

Solving equation I: x³ = 2744 yields x = ∛2744 = 14. Solving equation II: y² = 324 yields y = √324 = 18 or y = -18.

Comparing each possible pair of roots: When y = 18, x = 14 < 18. When y = -18, x = 14 > -18.

Because x is greater than y in one case and less than y in another case, no definite relation can be established between x and y.

52Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I.7x² - 17x + 6 = 0 II.5y² - 24y + 16 = 0
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Answer: (A) x = y or no relation.

Explanation

Solving quadratic equation I: 7x² - 17x + 6 = 0 Factoring gives 7x² - 14x - 3x + 6 = 0, which simplifies to (7x - 3)(x - 2) = 0. Roots are x = 2 or x = 3/7 ≈ 0.43.

Solving quadratic equation II: 5y² - 24y + 16 = 0 Factoring gives 5y² - 20y - 4y + 16 = 0, which simplifies to (5y - 4)(y - 4) = 0. Roots are y = 4 or y = 4/5 = 0.8.

Comparing roots: When x = 2 and y = 4, x < y. When x = 2 and y = 0.8, x > y. Since x is smaller than y in one pair and larger in another, no relation can be established between x and y.

53Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I.13x² + 9x - 4 = 0 II.2y² + y - 3 = 0
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Answer: (A) x = y or no relation.

Explanation

Solving equation I: 13x² + 9x - 4 = 0 Factoring gives 13x² + 13x - 4x - 4 = 0, leading to (13x - 4)(x + 1) = 0. Roots are x = 4/13 ≈ 0.31 and x = -1.

Solving equation II: 2y² + y - 3 = 0 Factoring gives 2y² - 2y + 3y - 3 = 0, leading to (2y + 3)(y - 1) = 0. Roots are y = 1 and y = -1.5.

Comparing roots: 0.31 < 1, but 0.31 > -1.5. Since x is less than y for one root combination and greater than y for another, no relationship can be established between x and y.

54Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I.x² + 13x + 42 = 0 II.y² + 8y + 12 = 0
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Answer: (C) x≤ y

Explanation

Solving equation I: x² + 13x + 42 = 0 Factoring gives (x + 6)(x + 7) = 0. Roots are x = -6 and x = -7.

Solving equation II: y² + 8y + 12 = 0 Factoring gives (y + 6)(y + 2) = 0. Roots are y = -6 and y = -2.

Comparing roots: For x = -6 and y = -6, x = y. For x = -6 and y = -2, x < y. For x = -7 and y = -6, x < y. For x = -7 and y = -2, x < y. In all cases, x is less than or equal to y (x ≤ y).

55Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I.x² + 21x + 108 = 0 II.y² + 14y + 48 = 0
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Answer: (B) x< y

Explanation

Solving equation I: x² + 21x + 108 = 0 Factoring gives (x + 12)(x + 9) = 0. Roots are x = -12 and x = -9.

Solving equation II: y² + 14y + 48 = 0 Factoring gives (y + 8)(y + 6) = 0. Roots are y = -8 and y = -6.

Comparing roots: -12 < -8 -12 < -6 -9 < -8 -9 < -6 Every value of x is strictly less than every value of y, so x < y.

56Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I.x² + 24x = -119 II.3y² + 7 = -10y
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Answer: (B) x< y

Explanation

Rearranging equation I: x² + 24x + 119 = 0 Factoring gives (x + 17)(x + 7) = 0. Roots are x = -17 and x = -7.

Rearranging equation II: 3y² + 10y + 7 = 0 Factoring gives 3y² + 3y + 7y + 7 = 0, leading to (3y + 7)(y + 1) = 0. Roots are y = -7/3 ≈ -2.33 and y = -1.

Comparing roots: -17 < -2.33 -17 < -1 -7 < -2.33 -7 < -1 Every value of x is strictly smaller than every value of y, so x < y.

57Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and answer the following questions. I. 2x² + 7x - 60 = 0 II. 3y² - 28y + 64 = 0
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Answer: (C) x≤ y

Explanation

Solving equation I, 2x² + 7x - 60 = 0: Factorizing by splitting the middle term gives 2x² + 15x - 8x - 60 = 0, which simplifies to (2x + 15)(x - 4) = 0. This yields roots x = -15/2 = -7.5 and x = 4.

Solving equation II, 3y² - 28y + 64 = 0: Factorizing by splitting the middle term gives 3y² - 12y - 16y + 64 = 0, which simplifies to (3y - 16)(y - 4) = 0. This yields roots y = 16/3 ≈ 5.33 and y = 4.

Comparing all pairs of roots: When x = -7.5, x < 4 and x < 5.33. When x = 4, x = 4 and x < 5.33. Therefore, x ≤ y holds true for all root pairs.

58Quantitative Aptitude
Ratio of speed of a bus to a car is 6 : 7. They start from the same point and move towards the same direction. After four hours distance between them is 28 km. Find the time in which car will cover 196 km.
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Answer: (B) 4 hours

Explanation

Let the speed of the bus be 6s and the speed of the car be 7s. Since both start from the same point in the same direction, the relative speed of the car with respect to the bus is 7s - 6s = s. The distance between them after 4 hours is given as 28 km. Distance = Relative speed × Time 28 = s × 4 ⇒ s = 7 km/h.

Speed of the car = 7 × 7 = 49 km/h. Time required by the car to cover 196 km = Distance ÷ Speed = 196 ÷ 49 = 4 hours.

59Quantitative Aptitude
The average weight of a class of 45 girls is 53 kg. It was latter found that weight of two girls was read as 49 kg and 57 kg instead of 45 kg and 52 kg. Find the actual average weight of the class.
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Answer: (D) 52.80 kg

Explanation

Original average weight of 45 girls = 53 kg. Sum of wrong weights read = 49 + 57 = 106 kg. Sum of correct actual weights = 45 + 52 = 97 kg. Net change in total weight = Correct sum - Wrong sum = 97 - 106 = -9 kg.

Change in class average weight = Net change in total weight ÷ Number of girls Change in average = -9 ÷ 45 = -0.20 kg.

Actual average weight = Original average + Change in average Actual average weight = 53 - 0.20 = 52.80 kg.

60Quantitative Aptitude
A rectangular field was fenced at 4 Rs./meter and ratio of breadth to length of field is 9 : 4. If total cost of fencing was Rs. 208, then find area of the rectangular field?
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Answer: (A) 144m²

Explanation

Perimeter of a rectangular field = Total cost of fencing ÷ Rate per meter Perimeter = 208 ÷ 4 = 52 meters.

Let the breadth be 9x and the length be 4x according to the given ratio 9 : 4. Perimeter formula = 2 × (Length + Breadth) 2 × (9x + 4x) = 52 2 × 13x = 52 26x = 52 ⇒ x = 2.

Breadth = 9 × 2 = 18 meters. Length = 4 × 2 = 8 meters. Area of the rectangular field = Length × Breadth = 18 × 8 = 144 m².

61Quantitative Aptitude
The average age of Megha and Ritu is 18 years. Six years hence, Megha's age will be two times of Ritu's age. Find present age of Megha.
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Answer: (B) 26 years

Explanation

Let M and R be the present ages of Megha and Ritu respectively. Sum of present ages: M + R = 18 × 2 = 36 Six years hence: M + 6 = 2 × (R + 6) M - 2R = 6

Subtracting the second equation from the first: 3R = 30 ⇒ R = 10 M = 36 - 10 = 26

Megha's present age is 26 years.

62Quantitative Aptitude
Sidhu purchased 100 articles at rate Rs 10 per article. If he sold 50% articles at 20% profit while 50% of remaining at 10% profit, 20% of remaining at 10% loss and rest at 20% loss. Find his loss or profit percent in entire transaction.
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Answer: (A) 8%

Explanation

Using the weighted percentage change formula: Portion 1: 50% of articles sold at +20% profit = 0.50 × 20% = +10% Portion 2: 50% of remaining (25%) sold at +10% profit = 0.25 × 10% = +2.5% Portion 3: 20% of remaining (5%) sold at -10% loss = 0.05 × (-10%) = -0.5% Portion 4: Rest (20%) sold at -20% loss = 0.20 × (-20%) = -4%

Net profit percentage = 10 + 2.5 - 0.5 - 4 = 8%

63Quantitative Aptitude
What is probability of drawing 2 red balls from a bag containing 4 red, 5 yellow and some green balls such that green balls are twice the average of red and yellow color balls?
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Answer: (D) 2 / 51

Explanation

Number of red balls = 4 Number of yellow balls = 5 Average of red and yellow balls = (4 + 5) / 2 = 4.5 Number of green balls = 2 × 4.5 = 9 Total balls = 4 + 5 + 9 = 18

Probability of choosing 2 red balls = ⁴C₂ / ¹⁸C₂ ⁴C₂ = 6 ¹⁸C₂ = 153 P = 6 / 153 = 2 / 51

64Quantitative Aptitude
In how many ways 10 girls be arranged in a row all facing north such that 4 girls always sit together?
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Answer: (E) 120960

Explanation

Treat the 4 girls who must sit together as a single block. Total entities to arrange = (10 - 4 + 1) = 7 Arrangements of 7 entities = 7! = 5040 Internal arrangements of the 4 girls within the block = 4! = 24

Total number of arrangements = 7! × 4! = 5040 × 24 = 120960

65Quantitative Aptitude
A boat covers 'x+20' km downstream or 'x- 5' km upstream in 5 hours. If speed of boat in still water is 300% more than speed of stream. find x.
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Answer: (A) 42.5

Explanation

Let the speed of the stream be y km/h. Speed of boat in still water = y + 300% of y = 4y km/h. Downstream speed = 4y + y = 5y km/h. Upstream speed = 4y - y = 3y km/h.

From downstream motion: (x + 20) / 5y = 5 ⇒ x + 20 = 25y From upstream motion: (x - 5) / 3y = 5 ⇒ x - 5 = 15y

Subtracting the two equations: 25 = 10y ⇒ y = 2.5

Substituting y into x - 5 = 15y: x - 5 = 15 × 2.5 = 37.5 x = 42.5

66Quantitative Aptitude
Solve the given quadratic equations and mark the correct option based on your answer— I. 2x² + 3x - 27 = 0 II. 3y² - 5y - 42 = 0
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Answer: (E) x = y or no relation can be established between x and y .

Explanation

For the first equation, factorizing gives (2x + 9)(x - 3) = 0, yielding x = -4.5 and x = 3. For the second equation, factorizing gives (3y - 14)(y + 3) = 0, yielding y = 14/3 ≈ 4.67 and y = -3. Comparing the roots, x = 3 is greater than y = -3, but x = -4.5 is smaller than y = -3. Therefore, no relation can be established between x and y.

67Quantitative Aptitude
Solve the given quadratic equations and mark the correct option based on your answer— I. x² - 2x - 15 = 0 II. y² - 4y - 12 = 0
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Answer: (E) x = y or no relation can be established between x and y .

Explanation

Solving x² - 2x - 15 = 0 gives (x - 5)(x + 3) = 0, so x = 5 and x = -3. Solving y² - 4y - 12 = 0 gives (y - 6)(y + 2) = 0, so y = 6 and y = -2. Comparing x = 5 with y = -2 gives x > y, but comparing x = 5 with y = 6 gives x < y. Since the relationship varies depending on the chosen roots, no relationship can be established between x and y.

68Quantitative Aptitude
Solve the given quadratic equations and mark the correct option based on your answer— I. x² - 20x + 96 = 0 II. y² - 10y + 24 = 0
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Answer: (A) x > y

Explanation

Solving x² - 20x + 96 = 0 gives (x - 12)(x - 8) = 0, so x = 12 and x = 8. Solving y² - 10y + 24 = 0 gives (y - 6)(y - 4) = 0, so y = 6 and y = 4. Comparing all pairs of roots: 12 > 6, 12 > 4, 8 > 6, and 8 > 4. Since x is strictly greater than y in all cases, x > y.

69Quantitative Aptitude
Solve the given quadratic equations and mark the correct option based on your answer— I. 6x² + 13x + 5 = 0 II. 3y² + 11y + 10 = 0
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Answer: (B) x ≥ y

Explanation

Solving 6x² + 13x + 5 = 0 gives roots x = -5/3 ≈ -1.67 and x = -1/2 = -0.5. Solving 3y² + 11y + 10 = 0 gives roots y = -2 and y = -5/3 ≈ -1.67. Comparing these values shows that -0.5 is greater than both -2 and -1.67, while -1.67 is equal to -1.67 and greater than -2. Thus, x ≥ y.

70Quantitative Aptitude
Solve the given quadratic equations and mark the correct option based on your answer— I. 6x + 7y = 52 II. 14x + 4y = 35
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Answer: (C) x < y

Explanation

Multiplying 6x + 7y = 52 by 2 gives 12x + 14y = 104. Multiplying 14x + 4y = 35 by 3.5 gives 49x + 14y = 122.5. Subtracting the first resulting equation from the second gives 37x = 18.5, so x = 0.5. Substituting x = 0.5 into 6x + 7y = 52 gives 3 + 7y = 52, which yields y = 7. Since 0.5 < 7, x < y.

71Quantitative Aptitude
Solve the given quadratic equations and mark the correct option based on your answer— I. x² - 12x + 32 = 0 II. y² - 18y + 77 = 0
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Answer: (E) x = y or no relation can be established between x and y .

Explanation

Solving x² - 12x + 32 = 0 yields x = 8 and x = 4. Solving y² - 18y + 77 = 0 yields y = 11 and y = 7. Comparing the roots shows that x = 8 is greater than y = 7, but x = 8 is less than y = 11. Because x can be both greater than and less than y, no relationship can be established between x and y.

72Quantitative Aptitude
Find the volume of a cylinder having radius 7 cm while its height is equal to diagonal of square of area 64 sq.cm.
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Answer: (A) 1232√2 cm³

Explanation

The area of the square is 64 sq.cm, so its side length is √64 = 8 cm. The diagonal of the square, which equals the height of the cylinder, is 8√2 cm. Using the volume formula V = π r² h with radius r = 7 cm and height h = 8√2 cm gives V = (22/7) × 7² × 8√2 = 22 × 7 × 8√2 = 1232√2 cm³.

73Quantitative Aptitude
An amount becomes twice when invested for 3 years at simple interest. What would be the interest received if Rs 5000 invested at same rate for 2 years at compound interest.(approx.)
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Answer: (D) Rs 3889

Explanation

Since the principal doubles in 3 years at simple interest, the interest equals the principal, so SI = P × R × 3 / 100 = P, which gives R = 100/3% per annum. For compound interest, the total amount after 2 years on Rs 5000 is A = P(1 + R/100)² = 5000 × (1 + 1/3)² = 5000 × (4/3)² = 80000/9 ≈ Rs 8888.89. The compound interest is 8888.89 - 5000 = Rs 3888.89, which rounds to Rs 3889.

74Quantitative Aptitude
A work can be completed by 7 men or 10 women in 10 days. If the work is to be completed in 4 days then how many men will be required to assist 15 women?
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Answer: (D) 7

Explanation

Since 7 men or 10 women can complete the work in 10 days, 7 men = 10 women, and total work = 10 × 10 = 100 woman-days. To complete the work in 4 days, the daily requirement is 100 / 4 = 25 women per day. With 15 women already working, an additional workforce equivalent to 10 women per day is required. Since 10 women are equivalent to 7 men, 7 men are needed to assist the 15 women.

75Quantitative Aptitude
Dinesh's monthly salary is 50% more than that of Chappu. Chappu's salary is 20% less than Guthi. Guthi earns 30% more than Rajesh. Salary of Dinesh is what percent more or less than that of Rajesh?
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Answer: (B) 56%

Explanation

Let Rajesh's salary be 100. Guthi earns 30% more than Rajesh, so Guthi's salary is 130. Chappu earns 20% less than Guthi, so Chappu's salary is 130 × 0.8 = 104. Dinesh earns 50% more than Chappu, so Dinesh's salary is 104 × 1.5 = 156. Comparing Dinesh's salary of 156 to Rajesh's salary of 100 gives a difference of 56, meaning Dinesh's salary is 56% more than Rajesh's.

76Quantitative Aptitude
2, 3, 7, 25, 121, ? What will come in the place of the question mark (?) in the following number series?
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Answer: (B) 721

Explanation

The pattern in the series is: 2 × 2 - 1 = 3 3 × 3 - 2 = 7 7 × 4 - 3 = 25 25 × 5 - 4 = 121 121 × 6 - 5 = 721

77Quantitative Aptitude
5,19,45,95,?,387 What will come in the place of the question mark (?) in the following number series?
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Answer: (D) 193

Explanation

The pattern in the given series is: 5 × 2 + 9 = 19 19 × 2 + 7 = 45 45 × 2 + 5 = 95 95 × 2 + 3 = 193 193 × 2 + 1 = 387

78Quantitative Aptitude
2,8,20,56,200,? What will come in the place of the question mark (?) in the following number series?
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Answer: (E) 920

Explanation

The differences between successive terms are: 8 - 2 = 6 20 - 8 = 12 = 6 × 2 56 - 20 = 36 = 12 × 3 200 - 56 = 144 = 36 × 4

The next difference will be 144 × 5 = 720. The missing term is 200 + 720 = 920.

79Quantitative Aptitude
32,32,16,48,12,? What will come in the place of the question mark (?) in the following number series?
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Answer: (A) 60

Explanation

The pattern of operations in the series is: 32 × 1 = 32 32 ÷ 2 = 16 16 × 3 = 48 48 ÷ 4 = 12 12 × 5 = 60

80Quantitative Aptitude
6,10,19,27,54,? What will come in the place of the question mark (?) in the following number series?
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Answer: (C) 70

Explanation

The pattern of differences between successive terms is: 6 + 2² = 6 + 4 = 10 10 + 3² = 10 + 9 = 19 19 + 2³ = 19 + 8 = 27 27 + 3³ = 27 + 27 = 54 54 + 2⁴ = 54 + 16 = 70

81Quantitative Aptitude
1,0,1,16,81,?,625 What will come in the place of the question mark (?) in the following number series?
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Answer: (C) 256

Explanation

The nth term of the series is given by Tₙ = (n - 2)⁴: T₁ = (1 - 2)⁴ = 1 T₂ = (2 - 2)⁴ = 0 T₃ = (3 - 2)⁴ = 1 T₄ = (4 - 2)⁴ = 16 T₅ = (5 - 2)⁴ = 81 T₆ = (6 - 2)⁴ = 256 T₇ = (7 - 2)⁴ = 625

82Quantitative Aptitude
A certain amount was distributed among A, B and C in the ratio 1:3:5 but it was actually distributed in ratio 2 : 3 : 6 due to which B got Rs 42 less. Find the amount actually to be received by A.
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Answer: (B) Rs 77

Explanation

Let the total amount be T. Intended ratio = 1 : 3 : 5 (total parts = 9) Actual ratio = 2 : 3 : 6 (total parts = 11)

B's intended share = T / 3 B's actual share = 3T / 11

Difference in B's share = T / 3 - 3T / 11 = 2T / 33 = 42 T = (42 × 33) / 2 = 693

The amount intended to be received by A = (1 / 9) × 693 = 77

83Quantitative Aptitude
A train crosses a man, who is running in the same direction of train at the speed of 2m / sec. in 10 seconds. The same train crosses a tunnel in 54 seconds. If speed of train is 72km / h then what is the length of tunnel?
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Answer: (C) 900 m

Explanation

Train speed = 72 km/h = 72 × (5 / 18) = 20 m/s Relative speed of train with respect to the man = 20 - 2 = 18 m/s

Length of train = Relative speed × Time = 18 × 10 = 180 m

When crossing the tunnel of length L: Total distance = Length of train + Length of tunnel = 180 + L Speed = 20 m/s, Time = 54 s

180 + L = 20 × 54 = 1080 L = 1080 - 180 = 900 m

84Quantitative Aptitude
Neeraj and Arun started a business by making investment in the ratio of 3 : 5. After four months, Arun withdraws one fifth of his initial investment. If total annual profit was Rs. 880, find profit share of Neeraj.
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Answer: (B) Rs.360

Explanation

Let Neeraj's initial investment be 3x and Arun's initial investment be 5x.

Neeraj's investment for 12 months = 3x × 12 = 36x Arun's investment for first 4 months = 5x × 4 = 20x Arun's reduced investment for remaining 8 months = (5x - x) × 8 = 32x Total investment units for Arun = 20x + 32x = 52x

Ratio of profit shares (Neeraj : Arun) = 36 : 52 = 9 : 13 Neeraj's profit share = (9 / 22) × 880 = 360

85Quantitative Aptitude
A boat can travel with the speed of 17 kmph in upstream. If the speed of river is 3 kmph, then find the speed of boat in downstream in the same river.
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Answer: (A) 23 kmph

Explanation

Upstream speed = Boat speed - Stream speed = 17 kmph Stream speed = 3 kmph

Boat speed in still water = 17 + 3 = 20 kmph Downstream speed = Boat speed + Stream speed = 20 + 3 = 23 kmph

Answer key

Tap a number to jump to that question.

  1. 51A
  2. 52A
  3. 53A
  4. 54C
  5. 55B
  6. 56B
  7. 57C
  8. 58B
  9. 59D
  10. 60A
  11. 61B
  12. 62A
  13. 63D
  14. 64E
  15. 65A
  16. 66E
  17. 67E
  18. 68A
  19. 69B
  20. 70C
  21. 71E
  22. 72A
  23. 73D
  24. 74D
  25. 75B
  26. 76B
  27. 77D
  28. 78E
  29. 79A
  30. 80C
  31. 81C
  32. 82B
  33. 83C
  34. 84B
  35. 85A

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