Quantitative Aptitude For NIACL AO Phase II: 7th February 2019

Dear Aspirants,

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, B and C can complete a piece of work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work was:

Solution:

Q2. Two trains R and S having their lengths in the ratio of 9 : 11 are running towards each-other from two opposite stations. They cross a platform which length is 50% more than the average length of trains in 16 and 24 seconds respectively. Find the ratio of speed of train R to that of train S.

18:13

13:18

17:13

13:17

14:13

Solution:

Q3. How many numbers of five digits can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits?

150

160

120

140

130

Solution:

Total no. formed = 1 × 5 × 4 × 3 × 2 = 120

Q4. The ratio of efficiencies of two filling pipes is 4 : 5. There is a third emptying pipe which efficiency is two third of the average efficiency of first two filling pipes can empty a filled tank in 36 minutes. In how much time both the filling pipes can fill the tank when it is empty?

12 min.

14 min.

10 min.

16 min.

18 min.

Solution:

Q5. Mr. Harshit can make 40 questions in 5 hours and Mr. Akash can make 45 questions in 9 hours. If Mr. Harshit takes help of Mr. Brijendra then both can complete 40 questions in 2 hours and if Mr. Akash takes help of Mr. Prabhat then both can complete 45 questions in 5 hours. Find in how much time Mr. Brijendra and Mr. Prabhat can complete a set of 24 questions together?

3 h

2.5 h

2 h

1.5 h

4 h

Solution:

Directions (6-10): In each of the following questions two equations are given. You have to solve the equations and Give answer –

Q6. I. 3x² + 16x + 21 = 0
II. 6y² + 17y + 12 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 3x² + 16x + 21 = 0
⇒ 3x² + 9x + 7x + 21 = 0
⇒ (x + 3) (3x + 7) = 0
⇒ x = –3, –7/3
II. 6y² + 17y + 12 = 0
⇒ 6y² + 9y + 8y + 12 = 0
⇒ 3y (2y + 3) + 4 (2y + 3) = 0
⇒ y = – 3/2, –4/3
y > x

Q7. I. 16x² + 20x + 6 = 0
II. 10y² + 38y + 24 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 16x² + 20x + 6 = 0
⇒ 8x² + 10x + 3 = 0
⇒ 8x² + 4x + 6x + 3 = 0
⇒ (2x + 1) (4x + 3) = 0
⇒ x = –1/2, –3/4
II. 10y² + 38y + 24 = 0
⇒ 5y² + 19y + 12 = 0
⇒ 5y² + 15y + 4y + 12 = 0
⇒ (y + 3) (5y + 4) = 0
y = –3, –4/5
x > y

Q8. I. 8x² + 6x = 5
II. 12y² – 22y + 8 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 8x² + 6x – 5 = 0
⇒ 8x² + 10x – 4x – 5 = 0
⇒ (4x + 5) (2x – 1) = 0
⇒ x = ½, –5/4
II. 12y² – 22y + 8 = 0
⇒ 6y² – 11y + 4 = 0
⇒ 6y² – 3y – 8y + 4 = 0
⇒ (2y – 1) (3y – 4) = 0
⇒ y = 1/2, 4/3
y ≥ x

Q9. I. 17x² + 48x = 9
II. 13y² = 32y – 12

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 17x² + 48x – 9 = 0
⇒ 17x² + 51x – 3x – 9 = 0
⇒ (x + 3) (17x – 3) = 0
⇒ x = 3/17, – 3
II. 13y² – 32y + 12 = 0
⇒ 13y² – 26y – 6y + 12 = 0
⇒ (y – 2) (13y – 6) = 0
⇒ y = 2, 6/13
y > x

Q10. I. 8x² + 26x + 15 = 0
II. 4y² + 24y + 35 = 0

if x < y

if x ≤ y

relationship between x and y cannot be determined

if x ≥ y

if x > y

Solution:

I. 8x² + 26x + 15 = 0
⇒ 8x² + 20x + 6x + 15 = 0
⇒ 4x (2x + 5) + 3(2x + 5) = 0
⇒ (2x + 5) (4x + 3) = 0
⇒ x = – 5/2, –3/4
II. 4y² + 24y + 35 = 0
⇒ 4y² + 10y + 14y + 35 = 0
⇒ 2y (2y + 5) + 7 (2y + 5) = 0
⇒ (2y + 5) (2y + 7) = 0
⇒ y = –5/2, –7/2
x ≥ y

Directions (11-15): Study the following graph and answer the questions accordingly:

The following pie-chart shows the production of slippers in India during the month of March 2018