Data Interpretation for IBPS Clerk Mains 2017
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Data Interpretation for IBPS Clerk Mains 2017
Directions:(1-5) Refer to the bar graphs and answer the given questions. A test cricket match is played between team X and team Y. It is assumed that team have only eight players. The following bar graph shows the number of runs scored by different players of both the teams in Inning 1 and 2. Team X won the toss and elected to bat first.
1.What is the margin in Team X’s win?
a. 49 runs
b. 50 runs
c. 51 runs
d. 52 runs
e. 55 runs
Show Correct Answers
C. 51 runs
|
Team X |
Team Y |
|
1st Innings |
2nd Innings |
Total |
1st Innings |
2nd Innings |
Total |
| Player 1 |
60 |
32 |
92 |
22 |
24 |
46 |
| Player 2 |
35 |
53 |
88 |
29 |
41 |
70 |
| Player 3 |
94 |
58 |
152 |
17 |
26 |
43 |
| Player 4 |
29 |
15 |
44 |
129 |
7 |
136 |
| Player 5 |
8 |
26 |
34 |
82 |
32 |
114 |
| Player 6 |
78 |
2 |
80 |
26 |
28 |
54 |
| Player 7 |
99 |
7 |
106 |
36 |
76 |
112 |
| Player 8 |
22 |
44 |
66 |
12 |
24 |
36 |
| Total |
425 |
237 |
662 |
353 |
258 |
611 |
Winning run margin = 662 – 611=51
2.If the man of the match award is given based on the highest average runs scored in an inning, then who won the man of the match award?
a. Player 5 from Team Y
b. Player 7 from Team Y
c. Player 7 from Team X
d. Player 4 from Team Y
e. None of these
Show Correct Answers
E. None of these
|
Team X |
Team Y |
|
1st Innings |
2nd Innings |
Total |
1st Innings |
2nd Innings |
Total |
| Player 1 |
60 |
32 |
92 |
22 |
24 |
46 |
| Player 2 |
35 |
53 |
88 |
29 |
41 |
70 |
| Player 3 |
94 |
58 |
152 |
17 |
26 |
43 |
| Player 4 |
29 |
15 |
44 |
129 |
7 |
136 |
| Player 5 |
8 |
26 |
34 |
82 |
32 |
114 |
| Player 6 |
78 |
2 |
80 |
26 |
28 |
54 |
| Player 7 |
99 |
7 |
106 |
36 |
76 |
112 |
| Player 8 |
22 |
44 |
66 |
12 |
24 |
36 |
| Total |
425 |
237 |
662 |
353 |
258 |
611 |
The person whose score is the highest in the match would have the highest average, which is Player 3.
3.The total runs scored by lowest run scorer of team X is by what percentage greater/lower than the total runs scored by lowest run scorer of team Y?
a. 3.35%
b. 3.80%
c. 4.45%
d. 5.55%
e. 5.90%
Show Correct Answers
D. 5.55%
|
Team X |
Team Y |
|
1st Innings |
2nd Innings |
Total |
1st Innings |
2nd Innings |
Total |
| Player 1 |
60 |
32 |
92 |
22 |
24 |
46 |
| Player 2 |
35 |
53 |
88 |
29 |
41 |
70 |
| Player 3 |
94 |
58 |
152 |
17 |
26 |
43 |
| Player 4 |
29 |
15 |
44 |
129 |
7 |
136 |
| Player 5 |
8 |
26 |
34 |
82 |
32 |
114 |
| Player 6 |
78 |
2 |
80 |
26 |
28 |
54 |
| Player 7 |
99 |
7 |
106 |
36 |
76 |
112 |
| Player 8 |
22 |
44 |
66 |
12 |
24 |
36 |
| Total |
425 |
237 |
662 |
353 |
258 |
611 |
Total runs scored by lowest run scorer of team X = 34 (player 5)
Total runs scored by lowest run scorer of team Y = 36 (player 8)
Required percentage = (2/36)*100 = 5.55%
4.Run rate is calculated by run scored per over. Team X played 132 and 68 overs in 1st and 2nd inning respectively and team Y played 98 and 78 overs in 1st and 2nd inning respectively, then which of the following has maximum run rate?
a. Team X in 1st inning
b. Team Y in 1st inning
c. Team Y in 2nd inning
d. Team X in 2nd inning
e. Cannot determined
Show Correct Answers
B.Team Y in 1st inning
|
Team X |
Team Y |
|
1st Innings |
2nd Innings |
Total |
1st Innings |
2nd Innings |
Total |
| Player 1 |
60 |
32 |
92 |
22 |
24 |
46 |
| Player 2 |
35 |
53 |
88 |
29 |
41 |
70 |
| Player 3 |
94 |
58 |
152 |
17 |
26 |
43 |
| Player 4 |
29 |
15 |
44 |
129 |
7 |
136 |
| Player 5 |
8 |
26 |
34 |
82 |
32 |
114 |
| Player 6 |
78 |
2 |
80 |
26 |
28 |
54 |
| Player 7 |
99 |
7 |
106 |
36 |
76 |
112 |
| Player 8 |
22 |
44 |
66 |
12 |
24 |
36 |
| Total |
425 |
237 |
662 |
353 |
258 |
611 |
Run rate: Team X (1st inning) = 425/132 = 3.22
Team X (2nd inning) = 237/68 = 3.48
Team Y (1st inning) = 353/98 = 3.60 [maximum]
Team Y (2nd inning) = 258/78 = 3.30
5.What is the ratio of runs scored by first 5 players of team X & team Y together in 1st inning to 2nd inning respectively?
a. 505:314
b. 515:314
c. 505:318
d. 515:315
e. 505:315
Show Correct Answers
A. 505:314
|
Team X |
Team Y |
|
1st Innings |
2nd Innings |
Total |
1st Innings |
2nd Innings |
Total |
| Player 1 |
60 |
32 |
92 |
22 |
24 |
46 |
| Player 2 |
35 |
53 |
88 |
29 |
41 |
70 |
| Player 3 |
94 |
58 |
152 |
17 |
26 |
43 |
| Player 4 |
29 |
15 |
44 |
129 |
7 |
136 |
| Player 5 |
8 |
26 |
34 |
82 |
32 |
114 |
| Player 6 |
78 |
2 |
80 |
26 |
28 |
54 |
| Player 7 |
99 |
7 |
106 |
36 |
76 |
112 |
| Player 8 |
22 |
44 |
66 |
12 |
24 |
36 |
| Total |
425 |
237 |
662 |
353 |
258 |
611 |
Runs scored by first 5 players of team X & team Y together in 1st inning = 505
Runs scored by first 5 players of team X & team Y together in 2nd inning = 314
Required ratio = 505:314
Directions:(6-10) Study the graph carefully and answer the questions.
In St. Mary University, the first year is a foundation program for all the students after which in second year the students can opt for their major subject. In 2013, only 70% of the students passed the first year examination and all of those who passed joined in the second year. In 2013, only 20% of students of first year who passed the first year examination joined Physical Sciences, 30% joined the Arts stream, while remaining joined Commerce stream. Apart from these some students from other Universities who passed their first year also joined St. Mary University. Such students were 20% of the number of first year students from St. Mary University who passed 2013 examination. The break-up of the number of students from other universities across Physical Sciences, Arts and Commerce streams at St. Mary University was in ratio 2:3:5 respectively. In 2013, the number of students in Arts stream St. Mary University in Second year was 504.
6.How many first year students of St. Mary University appeared for the first year examination in 2013?
a. 2000
b. 2500
c. 3000
d. 2600
e. None of these
Show Correct Answers
A. 2000
Let us assume the total number of first year students in St. Mary University be p
Then, as per data, number of students who passed in first year = .7p
Students in science ,commerce and arts stream becomes : .14p , .35p , .21p respectively
Total number of students from other universities = 20% of .7p = .14p
Ratio of students from other universities in different streams is Science =.028p , Arts = .042p , Commerce = .07p
Given that number of students in arts stream = 504
Therefore .042p + .21p = 504
.252p = 504, therefore, p= 2000 .
Hence total number of students in first year = 2000
Total students in promoted to second year = 1400
Students from other universities joining St. Mary University = .14p = 280
Total students studying in St. Mary University in second year = 1400+280= 1680.
7.What was the number of students in second year Commerce stream in St. Mary University in 2013?
a. 900
b. 840
c. 880
d. 700
e. None of these
Show Correct Answers
B. 840
Let us assume the total number of first year students in St. Mary University be p
Then, as per data, number of students who passed in first year = .7p
Students in science, commerce and arts stream becomes: .14p, .35p, .21p respectively
Total number of students from other universities = 20% of .7p = .14p
Ratio of students from other universities in different streams is Science =.028p , Arts = .042p , Commerce = .07p
Given that number of students in arts stream = 504
Therefore .042p + .21p = 504
.252p = 504, therefore, p= 2000 .
Hence total number of students in first year = 2000
Total students in promoted to second year = 1400
Students from other universities joining St. Mary University = .14p = 280
Total students studying in St. Mary University in second year = 1400+280= 1680.
Total number of students in Commerce stream in 2013 = 0.35p + 0.07p = 0.42p = 0.42 × 2000 = 840.
8.In 2013, what percent of science stream students in second year at St. Mary University were first year pass outs from St. Mary University only?
a. 16.66%
b. 83.33%
c. 86%
d. 20%
e. cannot be determined
Show Correct Answers
B. 83.33%
Let us assume the total number of first year students in St. Mary University be p
Then, as per data, number of students who passed in first year = .7p
Students in science, commerce and arts stream becomes: .14p, .35p, .21p respectively
Total number of students from other universities = 20% of .7p = .14p
Ratio of students from other universities in different streams is Science =.028p , Arts = .042p , Commerce = .07p
Given that number of students in arts stream = 504
Therefore .042p + .21p = 504
.252p = 504, therefore, p= 2000 .
Hence total number of students in first year = 2000
Total students in promoted to second year = 1400
Students from other universities joining St. Mary University = .14p = 280
Total students studying in St. Mary University in second year = 1400+280= 1680.
Total science students in St. Mary university = .14p + .028p = 336 students.
First year pass-outs of St. Mary joining science stream = 0.14p = 280
Required percentage = 280/336 × 100 = 83.33%.
9.If after second year, 10% of Commerce stream students changed to Physical Science stream then what would have been the number of students in third year in Physical Science at St. Mary University in 2014? (Assume no outside student was admitted to third year and no student failed in second year examination in St. Mary University)?
a. 420
b. 380
c. 480
d. 460
e. None of these
Show Correct Answers
A. 420
Let us assume the total number of first year students in St. Mary University be p
Then, as per data, number of students who passed in first year = .7p
Students in science, commerce and arts stream becomes: .14p, .35p, .21p respectively
Total number of students from other universities = 20% of .7p = .14p
Ratio of students from other universities in different streams is Science =.028p , Arts = .042p , Commerce = .07p
Given that number of students in arts stream = 504
Therefore .042p + .21p = 504
.252p = 504, therefore, p= 2000 .
Hence total number of students in first year = 2000
Total students in promoted to second year = 1400
Students from other universities joining St. Mary University = .14p = 280
Total students studying in St. Mary University in second year = 1400+280= 1680.
Total number of commerce students in 2013 = 0.35p + 0.07p = 0.42p = 840
Total number of science students in 2013 = 0.14p + 0.028p = 0.168p = 336
10% of Commerce students =84
So, total number of science students in 2014 = 336 + 84 = 420.
10.If the number of girls in second year at St. Mary University in 2013 are distributed in the ratio 2:5:3 across Physical Sciences, Commerce and Arts streams respectively, then what is the number of the girls in second year in 2013 at St. Mary University?
a. 350
b. 255
c. 380
d. 385
e. Cannot be determined
Show Correct Answers
E. Cannot be determined
No bifurcation of students among boys and girls is given.
