A candidate with **quantitative aptitude** knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an **important** measure for a prospective business executive’s **abilities**.

The article **IBPS RRB PO Quantitative Aptitude Quiz 2** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. **IBPS RRB** has released **IBPS RRB Officer 2019 **Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article **IBPS RRB PO Quantitative Aptitude Quiz 2** will assist the students to know the expected questions from **Quantitative Aptitude**.

Click Here for Official IBPS Website

**Answer**: Option A

**Explanation** :

Assume working days

A = x, B = 2x, C = 2y, D = 3y

\(\frac{1}{x}\) + \(\frac{1}{2x}\) = \(\frac{1}{2y}\) + \(\frac{1}{3y}\)

And 2x – 2y = 16

Solving we get x = 18 days.

**Q2. A can do a piece of work in 40 days B can do the same piece of work in 60 days. A and B started the work together in the first 15 days A worked with 50% of his efficiency, in the next 15 days B worked with 50% of his efficiency. Now in how many days does the remaining work will be completed if both of them work with their full efficiencies?**

**Answer**: Option B

**Explanation** :

15*(\(\frac{1}{80}\) + \(\frac{1}{60}\)) + 15*(\(\frac{1}{120}\)+\(\frac{1}{40}\)) + x*(\(\frac{1}{40}\) + \(\frac{1}{60}\)) = 1

x = \(\frac{3}{2}\) = 1.5

Total percentage of 3 subjects = 3 × 70 = 210

% in Social = 210 – (60 + 80) = 210 – 140 = 70

**Q3. A can do a piece of work in 60 days working 14 hours. B has the same efficiency as of A. A and B started working together. A works 5,6,7 and 8 hours respectively on first four days and repeats the cycle again. Then B has to work how many hours daily if they together completed the work in 80 days? **

**Answer**: Option D

**Explanation** :

20*\(\frac{(5 + 6 + 7 +8 + 4x)}{840}\)= 1

x = 4 hours

**Q4. Sruthi, Swetha and Swati together can cut 216 Apples of the same size in 3 hours. Number of Apples cut by Sruthi in 9 hours is same as the number of Apples cut by Swati in 7 hours. In one hour, Swati can cut as many Apples more than Swetha as Swetha can cut more than Sruthi. Then the number of Apples cut by Swetha in one hour?**

**Answer**: Option B

**Explanation** :

U + v + W = 72

9U = 7W

W – V = V – U

V = 24

**Q5. If A and B work together can complete a work in 8/5 days. A started the work alone and completed. 50% of the work and left the work then B started the work alone and finished the rest of work. They took total 5 days to complete the work. Then in how many days B can complete the work if A is more efficient than A?**

**Answer**: Option D

**Explanation** :

\(\frac{1}{A}\) + \(\frac{1}{B}\) = \(\frac{5}{8}\)

\(\frac{x}{A}\) + \(\frac{y}{B}\) = 1

x + y = 5

y = 4

**Answer**: Option A

**Explanation** :

Let Abhay’s speed be x km/hr.

Then, \(\frac{30}{x}\) – \(\frac{30}{2x}\) = 3

⇒ 6x = 30

⇒ x = 5 km/hr.

**Q2. What is the number if 60% of it added to 60 gives the number itself ?**

**Answer**: Option A

**Explanation** :

40% of numb = 60

60% of 150 = 90

90 + 60 = 150

**Q3. The sum of the number and its square is 1406. What is the number ?**

**Answer**: Option D

**Explanation** :

See the last digit 06…..go for 7

Last digit ..37+(37*37) = 7 + 49 = 16

**Q4. In a bag there are coins of 25 paise and 10 paise in the ratio of 5:16. If the bag contains Rs.16 then the number of 10 paise coin is**

**Answer**: Option D

**Explanation** :

25*5+10*16 = 285

Rs 16 = 1600

No of 10p = 1600*\(\frac{16}{285}\) = 89.82 = 90

**Q5. Sum of the three consecutive number is 1956. What is 23% of the highest number ?**

**Answer**: Option A

**Explanation** :

X + x + 1 + x + 2 = 1956

3x = 1956 – 3 = 1953

X = \(\frac{1953}{3}\) = 651

X+3 = 653

23% of largest num = 23*\(\frac{653}{100}\) = 150.19

**Answer**: Option C

**Explanation**: Ratio of number of days = 9:10:15

4*4*4*4 = 256

**2. 12 students participated in the competition and each get different score. In how many ways can three different prizes given ?**

**Answer**: Option A

**Explanation**:

12*11*10 = 1320

**3. How many arrangement can be made from the word COMMERCE, such that all the vowels do not come together ?**

**Answer**: Option D

8 letters =\(\frac{8!}{2! 2!}\) = 40320/4 = 10080

6 letters = \(\frac{6!}{2!}\)= 360

Vowels = \(\frac{3!}{2! }\) = 3

No of ways vowels together = 360*3 = 1080

No of ways vowels not together = 10080 – 1080 = 9000

**4. 5 men and 3 women are to be seated such that no 2 women sit together and 2 men sit together. Find the no of ways in which this can be arranged ?**

**Answer**: Option B

**Explanation**:

5!*3! = 120*6 = 720

**5. A group consists of 3 couples in which each of the 3 men have one wife each. In how many ways could they arranged in a straight line so that the men and women occupy alternate position ?**

**Answer**: Option D

**Explanation**:

3!*3! + 3!*3! = 36 + 36 = 72