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# Quantitative Aptitude for RRB JE

By BYJU'S Exam Prep

Updated on: October 17th, 2023

**Quantitative Aptitude for the RRB JE** exam is important from a selection point of view. The Railway Recruitment Board almost every year conducts this exam to recruit suitable candidates. The RRB JE exam is conducted at the national level & a lot of aspirants apply for it making the competition level quite high. Therefore, every mark can make a big difference.

Read on to know the importance of Quantitative Aptitude for RRB JE exam. Some quantitative aptitude questions are also provided.

Table of content

## Quantitative Aptitude for RRB JE exam

The Junior Engineer post in Railway is a dream job for many aspirants. But to attain the Junior Engineer post candidates have to thoroughly master every subject listed in the official RRB JE syllabus. In the RRB JE syllabus, quantitative aptitude is one of the subjects. From quantitative aptitude, a total of 25 questions could be seen in the RRB JE exam. These 25 questions of 25 marks could make a good impact on RRB JE selection of candidates for stage 2 of RRB JE.

The Quantitative Aptitude topics are easy to learn and with little practice, aspirants can attempt almost every question of Quantitative Aptitude in the RRB JE exam.

## Best Quantitative Aptitude Books for RRB JE Exam

A good book helps in acquiring an in-depth understanding of concepts. Good knowledge of concepts helps in easily tackling difficult problems. Therefore, some of the best books for quantitative aptitude for the RRB JE exam have been listed in the table below:

Book Name |
Authors / Publications |

Quantitative Aptitude | Arun Sharma |

Fast Track Objective Arithmetic | Rajesh Verma |

Quantitative Aptitude | R.S. Aggarwal |

## Tricks to Solve Quantitative Aptitude Questions in RRB JE

Generally in competitive exams like RRB JE where speed and accuracy is the prime concern. But this can be achieved by incorporating shortcut tricks while practicing questions of quantitative aptitude. Here are some techniques by which aspirants can solve RRB JE quantitative aptitude questions in quick time.

- Strengthen arithmetic operations such as addition, subtraction, multiplication, and division.
- Practice numbers: squares, cubes, & square roots.
- Try to solve more & more unitary problems.
- Make your own techniques such as 10% of any number can be solved as 1/10 of the number.
- Focus first on the weaker sections to get a good grip on them.

## Quantitative Aptitude Questions for RRB JE Exam

To help candidates in their RRB JE preparation, we have provided some of the important quantitative aptitude questions that have been illustrated:

1. What value should come in place of the question mark (?) in the following question?

(45.8 × 6 × 5) ÷ 2 – 344 = (?)3

- (7)3
- (7)1/2
- 49
- 7
- None of these

Answer: D

Solution: (45.8 × 6 × 5) ÷ 2 – 344 = (?)3

⇒ (45.8*6*5)/2 – 344 = (?)3

⇒ 687 – 344 = 343 = (?)3

⇒ ? = 7

Hence option D is the right answer.

2. The radius of a cylindrical tank is 3 m less than the radius of a conical tank. Total time taken to fill water in cylindrical tank and in conical tank at 54 m3 per second and 66 m3 are 297 seconds and 144 seconds, respectively. If the height of the cylindrical tank is same as the height of the conical tank then find the height of the each tank

- 31.5 m
- 63 m
- 52.5
- 42 m
- None of these

Answer: B

Solution: Let, the height of each tank = ‘h’ m

And, let the radius of the cylindrical tank = ‘r’ m

The, the radius of the conical tank = (r + 3) m

Volume of the cylindrical tank = 54 x 297 = 16038 m3

So, 22/7*r2*h = 16038

So, h = (16038 * 7)/22*r2

Volume of the conical tank = 66 x 144 = 9504 m3

So, h = ( 9504*21 ) / 22* ((r+3)3)

So, (16038 * 7)/22*r2 = ( 9504*21 ) / 22* ((r+3)3)

So 9*[(r+3)2] = 16r2

= 9*(r2 + 6r + 9) = 16r2

= 9r2+54r+81 = 16r2

=7r2 – 54r – 81 = 0

=7r2 – 63r + 9r – 81 = 0

=7r (r-9) + 9(r-9) = 0

=(7r + 9)(r-9) = 0

=r=9

So, height of the tank =(16038 * 7)/22*r2 = 63 m

So option (b) is the correct answer.

3. Team A and B play a cricket match which is of 50 overs a side. Team A bats first and its run rate in the first 30 overs was 4.5 runs/over and it was 5.5 in the last 20 overs. Team B batted for all 50 overs, but lost the match by 15 runs. What was the run rate for team B?

- 4.6 runs/over
- 4.5 runs/over
- 4.2 runs/over
- 5 runs/over
- 5.1 runs/over

Answer: A

Solution: Runs scored by Team A in first 30 overs = 30*4.5 = 135

Runs scored by Team A in last 20 overs = 20 ×5.5=110

So, Total runs scored by Team A = 135+ 110 = 245 runs

Therefore, Runs scored by Team B = 245-15 = 230 runs

So run rate of Team B = 230/50 = 4.6 runs/over

4. Three-sevenths of a number k equal to 120 what is three-fifth of the number?

- 148
- 154
- 168
- 164
- None of these

Answer: C

Solution: Let K be the number.

Three seventh of K is 120 ⇒ 3K/7 = 120 ⇒ K = 120 × 7/3 = 40 × 7 = 280

Now three-fifth of K = 3 × 280/5 = 3 × 56 = 168

5. The number of votes not cast for a particular candidate increased by 12.5% in the National General Election over those not casted for him in the previous Assembly Polls, and the that candidate lost by a majority twice as large as that by which he had won the Assembly Polls. If a total 1,30,000 people voted each time, how many voted for the candidate in the previous Assembly Polls?

- 82,400
- 67,600
- 62,400
- 54,300
- None of these

Answer: B

Solution: Total Votes = 1,30,000

Let x voters voted against the Candidate in the Assembly Poll.

Then votes in favour = 130000 – x

Therefore, majority of votes by which candidate won in previous poll = 130000 – x – x = 130000 – 2x

Next year votes against the Candidate increases by 12.5%

So, votes against the candidate in general election = 1.125 x

And votes polled in favor of the candidate = total votes – votes against = 130000 – 1.125x

Therefore, majority of votes by which Candidate lost in general election = 1.125x – (130000 – 1.125x) = 2.25x – 130000

Now, it is given that the Candidate lost by a majority twice as large as that by which it had won the Assembly Polls, Therefore

2.25x – 130000 = 2(130000 – 2 x)

⇒ x = 62,400

Therefore, votes polled by the voters for the party in Assembly Polls for previous year

= (1,30,000 – x) = (1,30,000 – 62400) = 67,600

6. Monika bought a piece of cotton cloth for Rs. 750 and spent Rs. 250 on designing it. At what price should she sell it to make 25% profit?

- 1250
- 950
- 1155
- 1125
- None of these

Answer: A

Solution: Actual cost price of cloth = Rs. (750 + 250) = Rs. 1000

To gain 25% SP = Rs. 125/100 × 1000

= Rs. 1250

7. Ratio of present age of Akhil and Bhavna is 3 : 5. After 8 years, the ratio of ages of Akhil and Bhavna will be in the ratio 11 : 13. Find the age of Akhil after 4 years.

- 10
- 12
- 5
- 6
- 7

Answer: E

Solution: Let present age of Akhil and Bhavna be 3x and 5x respectively

ATQ,

(3x + 8) / (5x + 8) = 11/13

39x + 104 = 55x + 88

16x = 16

x =1

Therefore, age of Akhil 4 year hence = 3 + 4 = 7 years

8. A principal of approx. Rs. 9998.69 becomes approx. Rs. 14398.1136 in 2 years when compounded annually at some rate of interest. How much will be the approx amount if the same principal was compounded half-yearly?

- Rs. 14639.08
- Rs. 16105.10
- Rs. 13176.90
- Rs. 13294.028
- Rs. 15225.80

Answer: A

Solution: P ≈ 10000, A ≈ 14400, t = 2 years

14400 = 10000 (1 + (r/100))2

(1.44)1/2 = 1+ r/100 => r = 20%

Now the amount when the same principal is compounded half-yearly for the same time period can be given as

A liters0(1 + (10/100))4 = 14641 equivalent to 14639.08

9. In a mixture, the ratio of the alcohol and water is 9 : 11. When 40 litre mixture are replaced by water, the ratio becomes 9 : 16. What is the initial quantity of alcohol?

- 108 litres
- 80 litres
- 70 litres
- 90 litres
- None of these.

Answer: D

Solution: Alcohol in 40 litres =(9/20)*40 = 18

Water in 40 litres = (11/20)*40 = 22

Let initial quantity of alcohol and water be 9k and 11k respectively.

(9k – 18)/11k – 22 + 40 = 9/16

(k-2) / 11k + 18 = 1/16

16k – 32 = 11k + 18

5k = 50

k = 10

The initial quantity of alcohol = 9k = 90 litres.

10. A contractor undertakes to make a mall in 60 days and he employs 30 men. After 30 days it is found that only one- third of the work is completed. How many extra men should he employ so that the work is completed on time?

- 20 men
- 25 men
- 30 men
- 40 men
- None of these

Answer: C

Solution: Let total work is w and it is given that one-third of the work is completed after 30 days. Means

M*D = 30*30 = w/3, so total work = 30*30*3

2700 = 30*30 + (30+p)*30, so we get P = 30 (p = additional men)

## Importance of Quantitative Aptitude for RRB JE Exam

The importance of quantitative aptitude for the RRB JE exam can be understood from the mentioned points:

- 25 questions from Quantitative Aptitude come in RRB JE.
- QA carries a total of 25 marks in the RRB JE exam.
- The quantitative aptitude topic is easy to understand & with some practice QA questions can be solved in quick time.

So, If an aspirant can score 25 out of 25 marks in one section of the RRB JE exam it will help them to achieve the RRB JE cut-off marks with a lot of ease.

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