Data Interpretation Questions for Competitive Exams

Data Interpretation: Data Interpretation Questions are usually asked in  all competitive exams like SSC, SBI, IBPS, RBI, LIC, RRB, AAI, DRDO, ISRO, NTR, FSSAI, CWC, LIC, SSC CGL, Railways and other state government exams. 
Data Interpretation questions require clear observation of charts like pie chart, bar chart, radar chart, line chart, table chart. On this page, we are providing all varieties of Data Interpretation Questions along with previous year Data Interpretation Questions.
Data Interpretation Questions are usually based on basic knowledge of arithmetic & fast calculation which includes good command on multiplication, division, addition, subtraction along with Vedic Math. 
Data Interpretation Questions is a very important section of Quantitative Aptitude. Most of the questions come from Data Interpretation in Quantitative Aptitude section.

Set-49
Study the following information to give answer of the following questions.
Shivam is proposed to start a business of Motor-cycles store for that he wants to buy some equipment – Fan, billing counter, banners. He also wants to buy a few material related to cycle which will include Tyre, Tube, Screw-devices and Seat (Consider only these are the part of a motor-cycle). The cost of each Fan is 13.5 times the cost of a Seat and the cost of a Billing-counter is 3/5 of the cost of a Fan. The cost of Banners is 5 times the cost of a Billing-counter. The cost of a Tyre is 20% more than the cost of a Billing Counter, a Tube costs Rs.1500 more than a Fan and the Screw-devices costs 4 times a Seat. The cost of a Seat is Rs.1000.
1. Find the cost of the Motor-cycle, if the cost of a Screw-devices and a Seat increases by 20% and 15% respectively,(considering the same cost of Tube and Tyre)?

Ans:1
Cost of a seat = Rs.1000
Cost of fan = 13.5 * Cost of a seat = 13.5 * 1000 = 13500
Cost of billing counter = 3/5 * fan = 3/5 * 13500 = 8100
Cost of banner = 5 * Cost of Billing counter = 5 * 8100 = 40500
Cost of tyre = 120/100 * Cost of billing counter = 6/5 * 8100
= 9720
Cost of tube = 1500 + fan = 13500 + 1500 = 15000
Cost of screw device = 4 * cost of seat = 4 * 1000 = 4000
20% of Screw-devices = 4000*(120/100) = Rs.4800
15% of Seat= 1000*(115/100) = Rs.1150
Tube = Rs.15000
Tyre = Rs.9720
Addition we get the cost of Motor-cycle = (4800+1150+15000+9720)= Rs.30670

 

2. It was initially decided that 4 Banners will be made but later only 3 Banners and a Sticker were made. If the cost of a Sticker is 1/10 of the cost of a banners, what was the total cost incurred?

Ans:2
Cost of a seat = Rs.1000
Cost of fan = 13.5 * Cost of a seat = 13.5 * 1000 = 13500
Cost of billing counter = 3/5 * fan = 3/5 * 13500 = 8100
Cost of banner = 5 * Cost of Billing counter = 5 * 8100 = 40500
Cost of tyre = 120/100 * Cost of billing counter = 6/5 * 8100
= 9720
Cost of tube = 1500 + fan = 13500 + 1500 = 15000
Cost of screw device = 4 * cost of seat = 4 * 1000 = 4000
Cost of a sticker = 40500/10 = Rs.4050
Now, cost of three banners and one sticker = 3*40500 + 4050
= 121500 + 4050
= Rs.125550

 

3. Shivam bought 1 Fan, 1 Billing-counter and 1 Motor-cycle for himself. What was the total cost incurred to him?

Ans:1
Cost of a seat = Rs.1000
Cost of fan = 13.5 * Cost of a seat = 13.5 * 1000 = 13500
Cost of billing counter = 3/5 * fan = 3/5 * 13500 = 8100
Cost of banner = 5 * Cost of Billing counter = 5 * 8100 = 40500
Cost of tyre = 120/100 * Cost of billing counter = 6/5 * 8100
= 9720
Cost of tube = 1500 + fan = 13500 + 1500 = 15000
Cost of screw device = 4 * cost of seat = 4 * 1000 = 4000
Total cost = 13500 + 8100 + 1000 + 4000 + 9720 + 15000
= Rs.51320

 

4. What will be the total cost of 1 Fan and 2 Billing-counter together?

Ans:4
Cost of a seat = Rs.1000
Cost of fan = 13.5 * Cost of a seat = 13.5 * 1000 = 13500
Cost of billing counter = 3/5 * fan = 3/5 * 13500 = 8100
Cost of banner = 5 * Cost of Billing counter = 5 * 8100 = 40500
Cost of tyre = 120/100 * Cost of billing counter = 6/5 * 8100
= 9720
Cost of tube = 1500 + fan = 13500 + 1500 = 15000
Cost of screw device = 4 * cost of seat = 4 * 1000 = 4000
Total Cost = 13500 + (2 * 8100) = Rs.29700

 

5. Find the ratio of the cost of a Tube to the total cost of a Seat and a Screw-device together?

Ans:4
Cost of a seat = Rs.1000
Cost of fan = 13.5 * Cost of a seat = 13.5 * 1000 = 13500
Cost of billing counter = 3/5 * fan = 3/5 * 13500 = 8100
Cost of banner = 5 * Cost of Billing counter = 5 * 8100 = 40500
Cost of tyre = 120/100 * Cost of billing counter = 6/5 * 8100
= 9720
Cost of tube = 1500 + fan = 13500 + 1500 = 15000
Cost of screw device = 4 * cost of seat = 4 * 1000 = 4000
Cost of Tube = Rs.15000
Cost of Seat and Screw-devices = Rs.5000
Required ratio = 15000:5000 = 3:1

Set-50
Study the following information carefully and answer the questions given below:

6. Sum of total number of boys, girls and teachers in school A is what percent of the sum of total number of boys, girls and teachers in school D?

Ans:2
Sum of total number of boys, girls and teachers in school A = 250 + 125 + 15
= 390
Sum of total number of boys, girls and teachers in school D = 275 + 100 + 14
= 389
Required percentage = 390/389 x 100 = 100.25%

 

7. Find the respective ratio between total number of boys, girls and teachers in school B and that in school C.

Ans:4
Total number of boys, girls and teachers in school B = 175 + 200 + 20 = 395
Total number of boys, girls and teachers in school C = 225 + 75 + 25 = 325
Required ratio = 395:325 = 79:65

 

8. Find the difference between total number of boys and total number of girls from all the schools together.

Ans:2
Total number of boys from all the schools together
= 250 + 175 + 225 + 275 + 125 + 200 = 1250
Total number of girls from all the schools together
= 125 + 200 + 75 + 100 + 225 + 175 = 900
Required difference = 1250 – 900 = 350

 

9. Total number of boys, girls and teachers of school E is what percent more/less than the total number of boys, girls and teachers of school F?

Ans:1
Total number of boys, girls and teachers of school E = 125 + 225 + 26 = 376
Total number of boys, girls and teachers of school F = 200 + 175 + 30 = 405
Required percentage = (405 – 376)/405 x 100 = 7.16% less

 

10. Find the sum of total number of teachers from all the schools together and total number of students of schools A and C.

Ans:4
Total number of teachers from all the schools together
= 15 + 20 + 25 + 14 + 26 + 30 = 130
Total number of students of school A = 250 + 125 = 375
Total number of students of school C = 225 + 75 = 300
Required sum = 130 + 375 + 300 = 805

 

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