# SBI PO Quantitative Aptitude For Prelims: 12th March 2019

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

= 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

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

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

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

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

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

**Quantitative Aptitude Study Notes for Bank Exams****100 MCQs Data Interpretation | Download Free PDF's of DI****Quantitative Aptitude Questions for all Competitive Exams**

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