Data Interpretation Practice Set - 3

Mentor for Bank Exams
Data Interpretation Practice Set - 3
Directions (1 – 5): study the following pie-chart and table and answer the given questions.
The following pie-chart shows the percentage number of students passed in matriculate examination from six different schools across the country in the year 2015. Also the bar graph shows the percentage of female students who passed their exam in 2015.
1. If in 2015, total number of female students from school F was 336, then how many male students passed the exam from school B?
(a) 1825 (b) 1342 (c) 1345 (d) 1056 (e) None of these
2. If total passed students from school D in 2015 was 2400, then what is the ratio of the numbers of female students who passed from school C to that of male students who passed from school F?
(a) 54:19 (b) 14:45 (c) 77:540 (d) 7:38 (e) None of these
3. If in the year 2016, there is an increase of 16% and 24% in the strength of passing students from school A and F respectively and the total number of passed students from school C in the year 2015 was 1650. What would be the total number os students who passed from school A and F in 2016?
(a) 7245 (b) 5862 (c) 7074 (d) 8746 (e) None of these
4. If the male students passed from school A in 2015 were 306. What was the total number of passed students from all schools together?
(a) 1875 (b) 1650 (c) 1550 (d) 1700 (e) None of these
5. If in 2015, the total number of students who passed the exam in all school E was 1750. What was the total number of male students from school E who passed the matriculation exam?
(a) 126 (b) 154 (c) 134 (d) 180 (e) None of these
Directions (6 – 10): Refer to the following table and answer the given questions:
Batsman
No. of Matches Played
Average Runs Scored
Total Balls faced
Strike Rate
M
22
56
-
-
N
18
-
-
153.6
O
-
45
900
110
P
-
36
-
84
Q
-
-
-
140
R
24
51
1368
-
NOTE:
Strike rate = (total runs scored/total balls faced) * 100.
All the given batsmen could bat in all the given matches played by them.
6. If the runs scored by R in last 3 matches of the tournament are not considered, his average runs scored in the tournament will decrease by 9, if the runs scored by R in the 21st and 22nd match are below are 140 and no two scores among these 3 scores are equal, what is the minimum possible runs scored by R in the 24th match?
a) 64 b) 85 c) 70 d) 60 e) 75
7. The respective ratio between total number of balls faced by O and that of Q in the tournament is 5:3. Total number of runs scored by Q in the tournament is what percent less than the total number of runs scored by O in the tournament?
a) 21 (3/11)% b) 25 (9/11)% c) 29 (1/11)% d) 27 (5/11)% e) 23 (7/11)%
8. Batsman M faced equal number of balls in the first 11 matches he played in the tournament and last 11 matches he played in the tournament. If his strike rate in first 11 matches and last 11 matches of the tournament are 83 and 72 respectively, what is the total number of balls faced by him in the tournament?
a) 1800 b) 1500 c) 1700 d) 1600 e) 1400
9. In the tournament, the total number of balls faced by batsman N is 402 less than the total number of runs scored by him. What is the average number of runs scored by batsman N in the tournament?
a) 32 b) 54 c) 64 d) 62 e) 71
10. What is the number of matches played by batsman O in the tournament?
a) 16 b) 28 c) 18 d) 22 e) 24
Solutions:
1. (d), let the total number of students who passed from school F be x, then 28% of x = 336 or x = 1200, now,
let the total students passed from all school be y, then, 15% of y = 1200 or y =8000. Now total students passed from school B = 22% of 8000 =1760. Therefore number of male students passed from school B = 60% of 1760 = 1056.
2. (c), we can say, 12% of x = 2400(x = total number of passed students in all schools or, x= 20000. Number of
female passed students from school C = 14% of 11% of 20000 = 308
Number of male passed students from school F = 72% of15% of 20000 = 2160.
Required ratio = 308:2160 = 77/540
3. (c), 11% of x = 1650( x = total number of passed students in all school or x = 15000. Total students passed from school F = 15% of 15000 = 2250, total students passed from school A = 24% of 15000 = 3600.
In the year 2016, total students passed from school A = 116% of 2250 = 2610, total students passed from school A = 124% of 3600 = 4464, total number of students passed in both school in 2008 = 2610+4464 = 7074
4. (a), it is given that, 68% of x = 306 (x = total number of A passed students ) or x = 450, now, 24% of y = 450(y= total number of passed students in all schools or, y = 1875
5. (b), total number of students who passed from school E = 16% of 1750 = 280, number of male students who passed from school E = 55% of 280 = 154.
6. (a), total runs scored by R = 24*51 = 1224
Total runs scored by R in 21 matches = 21*(51-9) = 21*42 = 882 runs
Therefore runs scored by R in last # matches = 1224-882 = 342
So maximum runs scored by R in 22nd and 23rd matches = 139*2 = 278
So minimum possible runs scored by R in 24th match = 342-278 = 64
7. (e), total balls faced by O = 900
Therefore total balls faced by Q = 3/5 *900 = 540
Total runs of O = strike rate*total balls/100= 140*540 / 100 = 756
Therefore required % = (990-756)/990*100 = 23 7/11%
8. (d), total runs scored by batsman M in 22 matches = 22*56 = 1232
Now, total number of balls faced by batsman M in tournament = (2 * total runs * 100)/(strike rate of first 11 matches + strike rate of last 11 matches) =(21232100)/(83+72) ≈ 1600
9. (c), let the average runs scored by N = x
Then 153.6 = [18𝑥/(18𝑥 −402)] 100 => 𝑥 = 64
10. (d), let, total matches played by batsman O = x
Now the strike rate = total runs scored*100/total faced off balls
110 = total runs scored*100/900, so total runs scored = 990
Now, total runs scored = total match played*average run
990 *45, then x = 22, total matches played = 22.

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