Latest Banking jobs   »   NIACL AO Quantitative Aptitude For Phase...

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

Dear Aspirants,


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

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
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_3.1

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
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_4.1

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

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
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_5.1

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
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_6.1

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
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_7.1

Q7. I. x²-5x-14=0 
II. y²+7y+10 = 0

 if x<y
if x >y
if x ≤y
if x ≥y
if x = y or no relation can be established
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_8.1

Q8. I. 2x² + 9x + 9 = 0 
II. 2y² + 17y + 36 = 0

 if x<y
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

Q9. I. 5x² + 29x + 20 = 0 
II. 25y² + 25y + 6 = 0

 if x<y
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

Q10. I. 3x² – 16x + 21 = 0 
II. 3y² – 28y + 65 = 0

 if x<y
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

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


Q11.NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_9.1

72
70
68
65
85
Solution:

⇒ 249 + 1149 + 2 = ? + 1335
⇒ ? = 1400 - 1335
= 65

Q12.NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_10.1

37.5
3.75
375
35.7
32.5
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_11.1

Q13. 36.5% of 140 ÷ 12.5% of 80 =?

6.12
4.71
5.11
5.91
8.11
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_12.1

Q14.NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_13.1

7005
7.005
70.05
700.5
600.5
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_14.1

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

49
56
64
66
42
Solution:

NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_15.1

               



NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_16.1




You May also like to Read:

                         
   NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_17.1           NIACL AO Quantitative Aptitude For Phase II: 25th February 2019 |_18.1
     
Check the Video Course of Quantitative Aptitude  


Print Friendly and PDF

Leave a comment

Your email address will not be published. Required fields are marked *