Data Interpretation Challenge for SBI Clerk Mains - 2 Sets

Today I am going to share 2 Data Interpretation sets.

Set 1

Study the following table carefully and answer the questions given below:

Number of boys of standard xi participating in different games

Note:

• Every student (boy or girl) of each class of standard XI participates in a game.
• In each class, the number of girls participating in each game is 25% of the number of boys participating in each game.
• Each student (boy or girl) participates in one and only one game.

Ques 1.

All the boys of class XI D passed the annual examination but a few girls failed. If all the boys to girls as 5:1, what would be the number of girls who failed in class XI D?

Solution: Total number of boys in XI D = 40

Number of girls in XI D = 25% of 40 = 10

Since all the boys of XI D passed, so the number of boys in XII D = 40

Ratio of boys & girls in XII D is 5: 1

Number of girls in XII D = 1/5  ×40=8 

∴number of girls who failed = (10 – 8) = 2 

Ques 2.

Girls playing which of the following games need to be combined to yield a ratio of boys to girls of 4 : 1, if all the boys playing Chess and Badminton are combined?

Solution:

Total number of boys playing Chess & Badminton = (32 + 52) = 84

Number of girls playing Hockey & Football = 25% of 84 

                                                                       = 1/4  ×84=21 

Since 84: 21 is 4: 1, so the girls playing hockey and football are combined to yield a ratio of boys to girls as 4: 1.

So, Hockey and Football is the correct answer.

Ques 3.

What should be the total number of students in the school if all the boys of class XI A together with all the girls of Class XI B and Class XI C were equal to 25% of the total number of students?

Solution: Number of boys in XI A = 44

Number of girls in XI B = 25% of 48 = 12

Number of girls in XI C = 25% of 48 = 12

(44 + 12 + 12) = 68

Let x be the total number of students.

Then 25% of x = 68

Or, x = (68  × 100)/25  =272

Total number of students in the school = 272.

Ques 4

Boys of which of the following classes need to be combined to equal four times the number of girls in class XI B and class XI C were to be equal to 25% of the total number of students?

Solution:      

4 times the number of girls in XI B & XI C = 4 (12 + 12) = 96.

Ques 5.

If boys of class XI E participating in chess together with girls of class XI B and class XI C participating in Table Tennis & Hockey respectively are selected for a course at the college of sports, what percent of the students will get this advantage approximately?

Solution: Number of boys in XI E = 4

Number of girls in XI B playing Table tennis = 25% of 16 = 4

Number of girls in XI C playing Hockey  = 25% of 8 = 2

(4 + 4 + 2) = 10

Total number of students

(228 + 25% of 228) = 285

Let x% of 285 = 10

Or, x = (10  × 100)/285  =3.51

Total number of students getting advantage approximately is 3.51.

Ques 6.

If for social work every boy of class XI D and class XI C is paired with a girl of the same class, what percentage of the boys of these two classes cannot participate in social work?

Solution:       

Since the number of girls = 25% of the number of boys, so only 25% of the boys can participate in social work

• Study the table carefully and answer the questions given below:

Set 2

Financial Statement of A Company Over The Years (Rupees in Lakhs)

Ques 1. 

During which year did the ‘Net Profit’ exceed Rs. 1 crore for the first time?

Solution: 1984-85      

only a look is needed (can be studied in the table).

Ques 2.

During which year was the ‘Cross Turnover’ closest to the thrice the ‘Profit before Interest and depreciation’?

Ques 3.

During which of the given years did the ‘Net Profit’ form the highest proportion of the ‘Profit before Interest and Depreciation’?

Solution: We look at the ‘Net profit’ and ‘Profits before Interest and Depreciation’. We need to find the year in which ‘profits before……..’ is the smallest multiple of ‘Net Profits’. Use approximations, 38 ÷ 1, 40 ÷ 2, 52 ÷ 5, 60 ÷ 6.5, 80÷20, 92 ÷ 22 and make quick mental calculation. Obviously any one of the last two is the answer. We have 80 ÷20=4,92 ÷22>4,and hence 80 ÷20 is the minimum. 

Hence, 1984 – 85 is the answer.

Ques 4.

Which of the following registered the lowest increase in terms of rupees from the year 1984-85 to the year 1985 – 86?

Solution: Mental calculation with approximation is sufficient. Among 2700 - 2500, 900 – 800, 600 – 500, 99 – 92 and 220 – 212, the fourth is a single digit figure and it is the least.

Ques 5.

The ‘Gross Turnover’ for 1982 – 83 is about what per cent of the ‘Gross Turnover’ for 1984 – 85?