# Free Quant quiz for IBPS PO & IBPS CLERK : Part – 6 ## Free Quant Quiz for IBPS PO & IBPS CLERK : Part – 6

In this quiz we will be discussing 10 questions based on the concept of approximation and quadratic equations and are important from bank exams point of view. Try attempting these questions:

Free Quant quiz for IBPS PO & IBPS CLERK Questions

Directions (1-5): Approximation based questions:

Question 1. (421.98 + 478.21) ÷ ? = 60.029

(a) 6

(b) 8

(c) 9

(d) 12

(e) 15

Question 2. √??? × 19.17 + 8.15 × 13.78 = ?

(a) 406

(b) 450

(c) 432

(d) 416

(e) 392

Question 3. 16.217 × 23.88 + ? = 18.98 × 32.12

(a) 216

(b) 224

(c) 200

(d) 228

(e) 250

Question 4. 27.897 × 16.21 = ? × 13.98 + 69.87

(a) 15

(b) 22

(c) 27

(d) 32

(e) 39

Question 5. 272.112 + 189.98 + 84.101 = ? × 12.89 × 6.11

(a) 5

(b) 7

(c) 9

(d) 11

(e) 13

Directions (6-10): In each of these questions, two equations (i) and (ii) are given. You have to solve both the equations and give answer as

(a) if x > y

(b) if x ≥ y

(c) if x < y

(d) if x ≤y

(e) if x = y or no relation can be established between x and y.

Question 6. (i) 2x² – 5x + 3 = 0

(ii) 3y² – 4y + 1 = 0

Question 7. (i) x² – 17x + 72 = 0

(ii) y² – 27y + 180 = 0

Question 8. (i) (x – 12)² = 0

(ii) y² – 21y + 108 = 0

Question 9. (i) 2x² + 7x + 5 = 0

(ii) 3y² + 12y + 9 = 0

Question 10. (i) x² + 2x – 35 = 0

(ii) y² + 15y + 56 = 0

Free Quant quiz for IBPS PO & IBPS CLERK Solutions

Solution 1. Ans. (e)

Sol. (422 + 478) ÷ ? ≃ 60

? ≃ 900 / 60 = 15

Solution 2. Ans. (d)

Sol. ? ≃ √256 × 19 + 8 × 14

? ≃ 16 × 19 + 8 × 14 = 416

Solution 3. Ans. (b)

Sol. 16 × 24 + ? ≃ 19 × 32

? = 608 – 384 = 224

Solution 4. Ans. (c)

Sol. 28 × 16 ≃ ? × 14 + 70

448 ≃ ? × 14 + 70

? = 378 / 14 = 27

Solution 5. Ans. (b)

Sol. 272 + 190 + 84 ≃ ? × 13 × 6

⇒ ? ≃ 546 / 13×6 = 7

Solution 6. Ans. (b)

Sol. (i) 2x² – 5x + 3 = 0

2x² – 2x – 3x + 3 = 0

2x (x – 1) – 3(x – 1) = 0

(x – 1) (2x – 3) = 0

x = 1, 3 / 2

(ii) 3y² – 4y + 1 = 0

3y² – 3y – y + 1 = 0

3y(y – 1) –1 (y – 1) = 0

(3y – 1) (y – 1) = 0

? = 1 / 3, 1

So, x ≥ y

Solution 7. Ans. (c)

Sol. (i) x² – 17x + 72 = 0

x² – 9x – 8x + 72 = 0

x(x – 9) – 8 (x – 9) = 0

(x – 8) (x – 9) = 0

x = 8, 9

(ii) y² – 27y + 180 = 0

y² – 12y – 15y + 180 = 0

y(y – 12) – 15 (y – 12) = 0

(y – 15) (y – 12) = 0

y = 15, 12

So, y > x

Solution 8. Ans. (b)

Sol. (i) (x – 12)² = 0

x – 12 = 0

x = 12

(ii) y² – 21y + 108 = 0

y² – 12y – 9y + 108 = 0

y (y – 12) – 9 (y – 12) = 0

(y – 9) (y – 12) = 0

y = 9, 12

x ≥ y

Solution 9. Ans. (e)

Sol. (i) 2x² + 7x + 5 = 0

2x² + 2x + 5x + 5 = 0

2x (x + 1) + 5 (x + 1) = 0

(2x + 5) (x + 1) = 0

? = – 5 / 2, – 1

(ii) 3y² + 12y + 9 = 0

3y² + 9y + 3y + 9 = 0

3y (y + 3) +3 (y + 3) = 0

(3y + 3) (y + 3) = 0

y = –1, – 3

No relation can be established.

S10. Ans. (b)

Sol. (i) x² + 2x – 35 = 0

x² + 7x – 5x – 35 = 0

x (x + 7) – 5 (x + 7) = 0

(x – 5) (x + 7) = 0

x = 5, –7

(ii) y² + 15y + 56 = 0

y² + 7y + 8y + 56 = 0

y (y + 7) + 6 (y + 7) = 0

(y + 8) (y + 7) = 0

y = – 8, – 7

x ≥ y

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December 13, 2018

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