# Quantitative Aptitude: Data Interpretation Questions for IBPS PO Prelims Set 68

Directions (1-5): A company has 4 offices – P, Q, R, and S in different cities. Each office has 5 departments – A, B, C, D, and E. The company has decided to keep same % of employees in each of the 5 departments in each office. (In each office number of total number of employees vary)

1. In office P, there are total 6300 employees, what is the total number of employees in its A and D departments?A) 2691
B) 2987
C) 2365
D) 2898
E) 2765
Â Â Option D
Solution:
Total number of employees in its A and D departments = (26+20)/100 * 6300 = 2898
2. In office P, there are total 6500 employees. Office Q has 20% more employees than in office P. What is the difference between the employees in E department of office Q and the employees in B departments of office P?A) 672
B) 724
C) 546
D) 598
E) 627
Â Â Option Â C
Solution:
P = 6500, Q = 120/100 * 6500 = 7800
in E in Q = 22/100 * 7800 = 1716
in B in P = 18/100 * 6500 = 1170
Required difference = 1716 – 1170 = 546
3. Total number of employees in offices P and R is 13,600. The difference between the employees in departments E and D in offices R and P respectively is 304. Find the total number of employees in departments D and E in offices P and R respectively.
A) 3265
B) 2864
C) 2756
D) 2891
E) 3167
Â Â Option B
Solution:
22/100 * x – 20/100 * (13600 – x) = 304
Solve, x = 7200
So employees in office R = 7200, in P = 13600-7200 = 6400
So total number of employees in departments E and D in offices R and P = 22/100 * 7200 + 20/100 * 6400 = 2864
4. In office Q, total number of employees in departments A and B is 2860. Also in office S, total number of employees in departments C and E is 2664. Find the total number of employees in offices Q and S.
A) 13,900
B) 15,600
C) 14,300
D) 13,400
E) 15,200
Â Â OptionÂ  A
Solution:
In Q = 2860 * 100/(26+18) = 6500
In S = 2664 * 100/(14+22) = 7400
So required total = 6500 + 7400 = 13,900
5. Total number of employees in departments E, B, and C in offices P, R and S respectively is 2924. Also total number of employees in offices P and S is 10600. If total number of employees in departments P, R and S is 16200, find the number of employees in department R.
A) 5800
B) 5200
C) 4900
D) 4500
E) 5600
Â Â OptionÂ  E
Solution:
Employees in P = x, in S = y
So x + y = 10600
22/100 * x + 18/100 (16200 – x – y) + 14/100 * y = 2924
4x/100 – 4y/100 + 2916 = 2924
Solve, x – y = 200
x + y = 10600
Solve, x = 5400, y = 5200
So in R = (16200 – 5400 – 5200) = 5600

Directions (6-10): Study the following bar graph and answer the questions that follow:
The following bar graph shows the percentage of students passing in various standards in a school. Maximum marks for each subject of every class is same.

1. If the number of boys and girls passing in standard 5th are same, then what is the ratio between the number of boys and number of girls in standard 5th?
A) 3 : 7
B) 3 : 4
C) 2 : 5
D) 4 : 7
E) 1 : 5
Â Â Option B
Solution:
In 5th standard, let no. of boys = x, girls = y
Number of boys passing = 0.8x
Number of girls passing = 0.6y
0.8x = 0.6y
x/y = 3/4
2. In 7th standard, 44% of the total students got passed. If total number of boys in 7th standard is 200, then what is the total number of girls in 7th standard?
A) 50
B) 78
C) 67
D) 45
E) 58
Â Â OptionÂ  A
Solution:
Let the total no. of girls in 7th standard = xTotal no. of boys passed = 200*0.4 = 80
Total no. of girls passed = x*0.6 = 0.6x
So, passed % of students = (80+0.6x)/(200+x)
Now, 44/100 = (80+0.6x)/(200+x)
Solve, x = 50
3. If the total number of boys and girls in each standard is 150 and 120 respectively, then what is overall pass percentage of the school?
A) 75%
B) 81%
C) 73%
D) 69%
E) 63%
Â Â Option D
Solution:

Total number of students = 6 * (150+120) = 1620
So overall pass % = [150(0.8+0.8+0.4+0.9+0.7+0.7) + 120(0.6+0.7+0.6+0.6+0.8+0.6)]/1620 * 100 = (645+468)/1620 * 100 = 69%
4. If the ratio between the number of boys and the number of girls in standard 9th is 4:1, then what is the ratio between number of boys passed and number of girls passed in standard 9th?
A) 8 : 3
B) 8 : 5
C) 7 : 2
D) 9 : 4
E) 7 : 1
Â Â OptionÂ  C
Solution:
Let there are 400 boys and 100 girls in 9th
Number of boys passed = 0.7*400 = 280
Number of girls passed = 0.8*100 = 80
Required ratio = 280 : 80 = 7 : 2
5. Assuming data in question 8, if the over-all average marks of boys is 50 and that of girls is 60, then what is average marks of the students in the school?
A) 49.13%
B)Â 30.66%
C) 52.17%
D)Â 42.02%
E) 54.44%
Â Â OptionÂ  E
Solution:
Required average marks = (50*150 + 60*120)/(150+120) = 14700/270 = 54.44%

## 18 Thoughts to “Quantitative Aptitude: Data Interpretation Questions for IBPS PO Prelims Set 68”

1. Ambiiii

Thankyou mam :))))

2. prediction baba

Thks mam

1. Navya

Baba wt wud b the IBPS PO Pre cutoff .? ðŸ˜›

depends on paper level….:)

1. Navya

50 to fir b man k chalr mai

3. Prabakar

Try to put cat level di..mod….or mmore tuffer than that.tq u

1. In prelims, this level will be asked

1. Wikki Waran

22/100 * x â€“ 20/100 * (13600 â€“ x) = 304
Solve, x = 7200
how to solve this equation mam?

1. 42x/100 = 304 + 20*136
42x/100 = 3024

2. Prabakar

yes mam..tq u..please start mains mock in testzone

4. DHANANJAY

thx mam
pls daily post one set of DI for pre level

5. deepali

thnkx mam ,,pls post daily fo po prre ðŸ™‚

6. aspirant

ty mam

7. Wikki Waran

Thank you

8. Wikki Waran

22/100 * x â€“ 20/100 * (13600 â€“ x) = 304
Solve, x = 7200
how to solve this equation anyone ?

9. jaga

thank u…mam

10. Jitu

ty..

11. kumkum ahuja

ty mam