Dear Aspirants,

### Quantitative Aptitude Quiz For SBI PO

Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

Q1. The ratio between two numbers is 3:5. If the smaller number is increased by 20% and the bigger one is decreased by 25%, the new ratio of numbers (smaller: bigger) will be –
25 : 24
24 : 25
23 : 24
24 : 23
11 : 13
Q2. The difference between C.I. and S.I. on a certain sum at the rate of 10% per annum for 2 years is Rs. 122. Find that sum (in Rs).
12,000
12,400
12,200
13,400
14,600
Q3. The average of 5 consecutive odd numbers is 25. What is the product of smallest and second largest numbers?
567
657
562
576
476
Q4. In how many ways the letter of the word ‘TWEET’ can be arranged?
120
30
240
70
60
Q5. A train crosses a tunnel and a standing man in 36 and 6 seconds respectively. If the speed of train be 48 kmph, what is the length of tunnel?
420 m
380 m
400 m
410 m
450 m
Directions (6-10): What should come in place of the question mark (?) in the following questions?

Q6. 2115 ÷ ? = 94 × 15
1.25
2.75
1.5
3
1.75
Q7. Q8. 748 × ? × 9 = 861696
122
132
128
124
136
Q9. 6573 ÷ 21 × (0.2)² =?
7825
62.5
1565
12.52
125.2
Q10. 74156 – ? – 321 – 20 + 520 = 69894
3451
4441
5401
4531
4414
Solution:
? = 74676 – 69894 – 341
= 4441

Directions (11-15): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer accordingly.

Q11. I. a² – 11a + 24 = 0
II. b² + 3b – 18 = 0
if a > b
if a ≥ b
if a = b or no relation can be established between a and b.
if a ≤ b
if a < b
Solution:
I. a² – 11a + 24 = 0
a² – 8a – 3a + 24 = 0
a (a– 8) – 3 (a– 8) = 0
(a – 8) (a– 3) = 0
a = 8, 3
II. b² + 3b – 18 = 0
b² +6b – 3b – 18 = 0
b (b + 6) – 3 (b + 6) = 0
(b – 3) (b+ 6) = 0
b = 3, –6
a ≥ b

Q12. I. 2a² + 17a + 35 = 0
II. 3b² + 17b + 24 = 0
if a > b
if a ≥ b
if a = b or no relation can be established between a and b.
if a ≤ b
if a < b
Solution:
I. 2a² + 17a + 35 = 0
2a² + 10a+ 7a + 35 = 0
2a (a+ 5) +7 (a+ 5) = 0
(2a+ 7) (a+ 5) =0
a = -7/2, -5
II. 3b² + 17b + 24 = 0
3b² + 9b + 8b + 24 = 0
3b (b + 3) + 8 (b+ 3) = 0
(b+ 3) (3b + 8) = 0
b= -3,-8/3
b > a

Q13. I. a² + 72 = 108
II. b³ + 581 = 365
if a > b
if a ≥ b
if a = b or no relation can be established between a and b.
if a ≤ b
if a < b
Solution:
I. a² + 72 = 108
a² =108 – 72 = 36
a = ±6
II. b³ + 581 = 365
b³ = –216
b = –6
a ≥ b

Q14. I. 2a + 3b = 72
II. a + 2b = 42
if a > b
if a ≥ b
if a = b or no relation can be established between a and b.
if a ≤ b
if a < b
Solution:
I. 2a + 3b = 72 …(i)
II. a + 2b = 42 …(ii)
On solving (i) & (ii)
a = 18 b = 12
a > b

Q15. I. a² – 14a + 48 = 0
II. b² – 10b + 24 = 0
if a > b
if a ≥ b
if a = b or no relation can be established between a and b.
if a ≤ b
if a < b
Solution:
I. a² – 14a + 48 = 0
a² – 8a – 6a + 48 = 0
a(a–8) –6 (a– 8) = 0
(a– 8) (a–6) = 0
a = 8, 6
II. b² – 10b + 24 = 0
b² – 6b– 4b + 24 = 0
b (b– 6) – 4 (b –6) = 0
(b–4) (b– 6) = 0
b= 4, 6
a ≥ b

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