# Quantitative Aptitude: Data Interpretation Questions – Set 78

Questions (1-5):  Following table shows the total number of students appeared from five different institutes, ratio of boys and girls among those appeared students, percentage of passed students and number of passed girls among them. 1. What is the average number of Girls appeared in the examination from all six cities?
A) 1855
B) 1960
C) 1720
D) 1880
E) None
Option B
Solution:

Appeared Girls A
9 == 4500
4 ? = 2000
Appeared Girls B
5 == 5000
2 ? = 2000
Appeared Girls C
12 == 3600
5 ? = 1500
Appeared Girls D
11 == 4400
7 ? = 2800
Appeared Girls E
13 == 3900
5 ?  = 1500
So, Average=(2000+2000+1500+2800+1500)/5
=1960.
2. The total number of girls passed from Institute C is what percentage of the total number of girls appeared from City C?
A) 88%
B) 75%
C) 82%
D) 72%
E) None
Option A
Solution:

From Ques 1 we know that
Appeared Girls C
12 == 3600
5 ? = 1500
Then 1500 = 100
1320 ? ==> 88%.
3. What is the total number of failed students in the examination from all six cities together?
A) 7250
B) 6800
C) 7570
D) 7360
E) None
Option C
Solution:
A:  (100-55)% of 4500= 2025
B:  (100-60)% of 5000=2000
C:  (100-70)% of 3600=1080
D: (100-75)% of 4400=1100
E: (100-65)% of 3900=1365
Total =2025+2000+1080+1100+1365=7570.
4. The total number of girls passed in the examination is approximately what percentage of the total number of girls appeared in the examination, taking all cities together?
A) 75.7%
B) 78%
C) 69.5%
D) 65%
E) none
Option A
Solution:

From Ques 1 we know that
Total girls appeared=2000+2000+1500+2800+1500=9800.
Total girls passed=1250+1300+1320+2100+1450=7420.
9800 == 100
7420 ? ==>75.7%.
5. The total number of boys passed from City B is what percentage more than the total number of girls passed from that city?
A) 1750
B) 1600
C) 1560
D) 1700
E) None
Option D
Solution:

Total no of passed students in city B 60% of 5000=3000.
No of girls passed in city B =1300
Then No of boys passed in city B=3000-1300=1700.

Questions (6-10) : Study the following table carefully to answer the following questions given below: 1. The average no of B.Ed students and B.Sc students in T was 520. What was the total no of students in T ?
A) 1500
B) 1450
C) 1600
D) 1560
E) None
Option C
Solution:

B.Ed students in T = 100-75 = 25%
65% = 2*520 = 1040
100% = 1040*100/65 = 1600
2. If the ratio of no of B.Sc students to that of B.Ed students in P was 9:6, what was the no of B.Ed students in P ?
A) 620
B) 588
C) 565
D) 482
E) None
Option B
Solution:

P = 2100
B.Sc students + B.Ed students in P = 100-30 = 70%
B.Ed students = 70*6/15 = 28% = 28% of 2100 = 588.
3. If the difference between the no of B.Ed students in S and that in R was 300. What was the total no of students in R? (Assume no of students in R and S are same)
A) 1110
B) 1200
C) 1000
D) 1250
E) None
Option C
Solution:

% of B.Ed students in R = 100-50 = 50%
Difference between the no of B.Ed students in S and that in R = 50-20 = 30%
X *30/100 = 200
X = 300*100/30 = 1000
4. The total no of students in Q is increased by 20% and then decreased by 20%, what is the total number of B.Com students in Q ?
A) 160
B) 175
C) 180
D) 195
E) None
Option D
Solution:

Q = 700*120/100 = 840*80/100 = 672
B.Com = 672 * 29/100 = 195.
5. If the total no of students in S is equal to twice the number of students in R .What is the ratio of number of B.Sc students in Q to the no of B.Ed students in S(Total no of students in R is 1000) ?
A) 217:400
B) 215:206
C) 145:132
D) 120:133
E) None
Option A
Solution:

No of B.Sc students in Q = 700*31/100 = 217
No of B.Ed students in S = 2000* 20/100 = 400
Ratio 217:400.

## 3 Thoughts to “Quantitative Aptitude: Data Interpretation Questions – Set 78”

1. karthik

the 8th question was worthy..

1. Jeevan

Usme 200 ki jagah 300 hai. Type galt huwa shayad jaldi me

2. lokesh

i think 5th ans is wrong