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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