# NIACL AO Quantitative Aptitude For Phase II: 25th February 2019

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**Quantitative Aptitude Quiz For NIACL AO**

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. A shopkeeper sells goods on 20% profit. When he purchases on 10% more and sells it on 25% profit then he gets Rs. 14 more. What is the initial cost price of goods?**

Rs. 80

Rs. 100

Rs. 125

Rs. 105

Rs. 400

**Q2. By selling an article for Rs. 160, a dealer makes 20% loss. Next time, the dealer sold the same article in such a way that he gained 25%. What is the difference between the percentage increase in his selling price and percentage loss made by him earlier?**

26.75%

30%

36.25%

34.25%

None of these

**Q3. A playground is in the shape of a rectangle. A sum of Rs. 1000 was spent to make the ground usable at the rate of 25 paise per sq. m. The breadth of the ground is 50m. If the length of the ground is increased by 20 m, what will be the expenditure in making this ground usable at the same rate?**

Rs.1500

Rs.2250

Rs.1250

Rs.1000

Rs. 1800

Solution:

Area of ground = 1000/0.25 = 4000 m²

Breadth = 50 m

Length = 4000/50 = 80 m

New length = 80 + 20 = 100 m

New area = 100 × 50 = 5000 m²

So, expenditure = 5000 × 0.25 = Rs 1250

Breadth = 50 m

Length = 4000/50 = 80 m

New length = 80 + 20 = 100 m

New area = 100 × 50 = 5000 m²

So, expenditure = 5000 × 0.25 = Rs 1250

**Q4. The probabilities of solving a problem by three students. A, B and C are 1/2,1/3 and 1/4, respectively. The probability that the problem will be solved, is**

1/4

1/2

3/4

1/3

3/5

**Q5. Two trains A and B are running on parallel tracks in the same direction. Train A which is coming from behind takes 50 seconds to cross B completely. One person in train A observes that he crosses train B in 30 seconds. If the speed of the train A and train B is in the ratio of 2: 1, what is the ratio of their lengths?**

1 : 2

2 : 3

4 : 3

5 : 3

5 : 4

**Directions (6 -10): Two equations I and II are given below. You have to solve these equations and give answer**

**Q6. I. x²-5 =0**

**II. 4y²-24y+35 = 0**

if x<y

if x>y

if x ≤y

if x ≥y

if x = y or no relation can be established

**Q7. I. x²-5x-14=0**

**II. y²+7y+10 = 0**

if x >y

if x ≤y

if x ≥y

if x = y or no relation can be established

**Q8. I. 2x² + 9x + 9 = 0**

**II. 2y² + 17y + 36 = 0**

if x >y

if x ≤y

if x ≥y

if x = y or no relation can be established

Solution:

I. 2x² + 9x + 9 = 0

⇒ 2x² + 6x + 3x + 9 = 0

⇒ (x + 3) (2x + 3) = 0

⇒ x = –3, –3/2

II. 2y² + 17y + 36 = 0

⇒ 2y² + 8y + 9y + 36 = 0

⇒ (y + 4) (2y + 9) = 0

y = – 4, –9/2

x > y

⇒ 2x² + 6x + 3x + 9 = 0

⇒ (x + 3) (2x + 3) = 0

⇒ x = –3, –3/2

II. 2y² + 17y + 36 = 0

⇒ 2y² + 8y + 9y + 36 = 0

⇒ (y + 4) (2y + 9) = 0

y = – 4, –9/2

x > y

**Q9. I. 5x² + 29x + 20 = 0**

**II. 25y² + 25y + 6 = 0**

if x >y

if x ≤y

if x ≥y

if x = y or no relation can be established

Solution:

I. 5x² + 29x + 20 = 0

⇒ 5x² + 25x + 4x + 20 = 0

⇒ (x + 5) (5x + 4) = 0

⇒ x = –5, –4/5

II. 25y² + 25y + 6 = 0

⇒ 25y² + 15y + 10y + 6 = 0

⇒ (5y + 3) (5y + 2) = 0

⇒ y = – 3/5, –2/5

y > x

⇒ 5x² + 25x + 4x + 20 = 0

⇒ (x + 5) (5x + 4) = 0

⇒ x = –5, –4/5

II. 25y² + 25y + 6 = 0

⇒ 25y² + 15y + 10y + 6 = 0

⇒ (5y + 3) (5y + 2) = 0

⇒ y = – 3/5, –2/5

y > x

**Q10. I. 3x² – 16x + 21 = 0**

**II. 3y² – 28y + 65 = 0**

if x >y

if x ≤y

if x ≥y

if x = y or no relation can be established

Solution:

I. 3x² – 16x + 21 = 0

⇒ 3x² – 9x – 7x + 21 = 0

⇒ (x – 3) (3x – 7) = 0

⇒ x = 3, 7/3

II. 3y² – 28y + 65 = 0

⇒ 3y² – 15y – 13y + 65 = 0

⇒ (y – 5)(3y – 13) = 0

⇒ y = 5, 13/3

y > x

⇒ 3x² – 9x – 7x + 21 = 0

⇒ (x – 3) (3x – 7) = 0

⇒ x = 3, 7/3

II. 3y² – 28y + 65 = 0

⇒ 3y² – 15y – 13y + 65 = 0

⇒ (y – 5)(3y – 13) = 0

⇒ y = 5, 13/3

y > x

**Directions (11-15): What should come in place of question mark (?) in the following question?**

**Q11.**

72

70

68

65

85

Solution:

⇒ 249 + 1149 + 2 = ? + 1335

⇒ ? = 1400 - 1335

= 65

⇒ ? = 1400 - 1335

= 65

**Q12.**

37.5

3.75

375

35.7

32.5

**Q13. 36.5% of 140 ÷ 12.5% of 80 =?**

6.12

4.71

5.11

5.91

8.11

**Q14.**

7005

7.005

70.05

700.5

600.5

**Q15. 3.5% of 40 + 3.5% of 80 =? % of 10**

49

56

64

66

42

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