 ## Data Interpretation for IBPS Clerk Mains 2017

Data Interpretation for IBPS Clerk Mains. Welcome to the www.letsstudytogether.co online Quant section. If you are preparing for IBPS PO/Clerk 2017 exam, you will come across a section on Data Interpretation Section. Here we are providing you with Data Interpretation for IBPS Clerk Mains 2017 based on the latest pattern of your daily practice.

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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

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

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%

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

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

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

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

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

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

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

E. Cannot be determined

No bifurcation of students among boys and girls is given.

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