The Quantitative Aptitude section for IBPS RRB Assistant prelims examination consists of 40 questions each carrying 1 mark. This section is to test the candidate’s mental awareness, problem solving ability and accuracy in mathematical calculations. The important miscellaneous topics for IBPS RRB Assistant prelims examination are being discussed below.

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Topics Covered

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**Miscellaneous Quant topics for IBPS RRB Assistant Prelims**

1) Speed, Time and Distance

2) Interest

3) Profit and Loss

4) Age Problems

5) Mixtures and Alligations

6) Percentages

7) Time and Work

8) Partnership

9) Ratios and Averages

10) Probability

**Latest Update | IBPS RRB** **2021**

**IBPS RRB Clerk (PET) admit card** is expected soon. The admit card for IBPS RRB pre-exam training was expected on July 9 at the official website that is ibps.in but is delayed for now. IBPS in its official notice said, “the pre-exam training will only be held if it is safe and possible to conduct it due to the COVID-19 situation”. “Pre-Examination Training may be arranged by the Regional Rural Banks to a limited number of candidates belonging to Scheduled Caste/ Scheduled Tribes/ Minority Communities/ Ex-Servicemen/ Persons With Benchmark Disabilities for the Post of Office Assistant (Multipurpose) and Scheduled Caste/ Scheduled Tribes/ Minority Communities for the Post of Officer Scale-I at some centers,” the notification added.

Some of the exam centers will be Warangal, Anantapur, Naharlagun (Papumpare), Guwahati, Ajmer, Raibareilly, Guntur, Raipur, Gandhinagar, Srinagar, Lucknow, Mandi, Jammu, Ranchi, Varanasi, Patna, Imphal, Jodhpur, Shillong, Aizawl, Kohima, Indore, Bhubaneshwar, Salem, Howrah, Moradabad, Puducherry, Ludhiana, Gorakhpur, Rohtak, Rajkot, Hyderabad, Agartala, Muzaffarpur, Dehradun, and Nagpur.

**IBPS RRB PET: Important Dates**

- Training is likely to begin on- 19 July 2021
- Training likely to end on- 25 July 2021

**Practice Questions for IBPS RRB Assistant Prelims**

Let us look at one question of each of the topics mentioned above. Similar type of questions are expected to come in the IBPS RRB Assistant prelims examination 2020.

1) A train of length 130 m passes a bridge in 30 seconds. If the speed of the train is 45km/hr, then what is the length of the bridge?

(a) 245 m

(b) 185 m

(c) 225 m

(d) 195 m

(e) 230 m

Answer Key: a

Solution:

Distance covered in 1 minute = 45/60 km = 0.75 km

Distance covered in 30s = 0.75/2 = 0.375 km = 375m

Distance covered = Length of train + length of bridge

So, length of bridge = 375 – 130 = 245m

2) The difference between Compound Interest and Simple Interest on a sum of Rs. 3125 for 2 years at 4% per annum is

(a) Rs.10

(b) Rs.8

(c) Rs.4

(d) Rs. 5

(e) Rs. 7

Answer key: d

Solution:

Difference between SI and CI for 2 years = Sum * (Rate/100)2

Required Difference = 3125 * (4/100)2 = 3125 * 16/10000 = Rs.5

3) A shopkeeper bought 100 kg of rice for Rs. 1800 and sold it to make profit which was equal to the selling price of 20 kg of rice. What is the selling price of rice per kg?

(a) Rs. 25

(b) Rs. 22.5

(c) Rs. 24

(d) Rs. 20

(e) Rs. 26

Answer key: b

Solution:

SP of 100 kg – CP of 100 kg = SP of 20 kg.

SP of 80 kg = CP of 100 kg

SP of 80 kg = Rs. 1800

SP of 1 kg = Rs. 1800/80 = Rs 22.5

4) In 10 years time, Ram will be 2 times older than his son. 5 years ago he was 9 times as old as his son. What will be the sum of the ages of Ram and his son 3 years from now?

(a) 46

(b) 66

(c) 86

(d) 76

(e) None of these

Answer key: b

Solution:

Let current age of Ram and his son be R,S

(R+10) = 3(S+10) [2 times older = 3 times the age]
R = 3S + 20

(R – 5) = 9(S-5)

3S + 20 – 5 = 9S – 45

3S + 15 = 9S – 45

or S = 10, R = 50

Sum of ages 3 years from now = 10 + 3 + 50 + 3 = 66.

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5) In what ratio should two varieties of baking powder at Rs.80/kg and Rs. 120/ kg respectively, be mixed to make the mixture at Rs. 100/kg?

(a) 1:2

(b) 2:1

(c) 1:1

(d) 2:3

(e) None of these

Answer key: c

Solution:

Let the required ratio be k:1

80k/(k+1) + 120/(k+1) =100

80k +120=100k +100

20=20k

So, k=1 ; hence ratio =1:1.

6) A man bought some apples of which 13% of them were rotten. He sold 75% of the balance and was left with 261 apples. How many apples did he have originally?

(a) 1800

(b) 1500

(c) 1200

(d) 1000

(e) 800

Answer key: c

Let the total apples be ‘n’

13% are rotten, 25% of the rest = 261

So, 261 = (1-0.13) x 0.25n

Or, 0.2175n = 261

n = 1200

7) A certain number of men complete a piece of work in 60 days. If there were 8 more men, the work could be finished in 10 days less. How many men were originally there?

(a) 30

(b) 50

(c) 60

(d) 80

(e) 40

Answer key:e

Solution:

Let the number of men be ‘x’

60 * x = 50 * (x+8)

60x = 50x + 400

10x = 400

x=40

Hence, initially 40 men were there

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8) What is the sum of the value of profit shares of A, B, and C if A, B, C and D invested their money in the ratio 1:2:2:1 for a whole year and the profit at the end of the year is Rs 54,000?

(a) Rs. 40,000

(b) Rs. 36,000

(c) Rs. 45,000

(d) Rs. 30.000

(e) Rs. 42,000

Answer key: c

Solution:

Sum of the value of shares of A, B and C = Total profit – share of D

Value of share of D = (1/6)*54,000 = Rs 9,000

Sum of the value of shares of A, B and C = 54,000 – 9000 = Rs 45,000

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9) There are 9 sections of grade 10 in a school. The average number of students in 1st five sections is 46 and the average number of students in last five sections is 49 and the average of all sections is 48, then the number of students in 5th section is:

(a) 43

(b) 44

(c) 45

(d) 46

(e) 47

Answer key:1

Solution:

Number of students in 1st 5 sections = 46 × 5 = 230

Number of students in last 5 sections = 49 × 5 = 245

Number of students in 9 sections = 48 × 9 = 432

Number of students in 5th section = 230 + 245 – 432 = 43

10) A box contains 300 cards on which number from 0 to 299 is written. If a card is picked at random, then what is the probability that the number on card is divisible by 3 but not divisible by 6?

(a) 1/2

(b) 1/4

(c) 1/8

(d) 1/6

(e) 1/3

Answer key:d

Solution:

All the numbers divisible by 6 are divisible by 3

Numbers divisible by 6 from 0 to 299 are 6, 12…..294

Total numbers = (294/6) = 49

Numbers divisible by 3 from 0 to 299 are 3, 6,……297

Total numbers = (297/3) = 99

Numbers which are divisible by 3 but not divisible by 6 = 99-49 = 50

Required probability

= (Number of cards containing number divisible by 3 but not divisible by 6)/(Total number of cards) = 50/300 =1/6.

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