Data Interpretation Practice Questions – Set 88

Mentor for Bank Exams
Data Interpretation Practice Questions – Set 88
Directions (1 – 5): Study the graph carefully and answer the questions that follow:
Sale of automobiles (in thousands)

1. Decline in sales of LCVs was approximately what percent over the period?
a) 4%
b) 5%
c) 3%
d) 6%
e) 7%
2. Which of the following pairs showed growth over the period?
a) HCV’s and LCV’s
b) Cars and Three-wheelers
c) Cars and Two-Wheelers
d) LCV’s and Cars
e) None of these
3. Approximately what was the difference of percentage changes in sales of HCVs and three-wheeler sales over the given period?
a) 14.0%
b) 4.5%
c) 23.5%
d) 28%
e) 36.5%
4. What approximate per cent of growth did Cars have over the period?
a) 4.5%
b) 8.5%
c) 5.0%
d) 2.5%
e) 3.0%
5. Which of the following had the maximum decline in 1999-2000 as compared to 1998-1999?
a) Three-Wheelers
b) HCV’s
c) LCV’s
d) Two-Wheelers
e) Cars
Directions (6 – 10): Study the following bar graph carefully to answer the questions.
Marks obtained By Five Students in Physics and Chemistry
6. What is the respective ratio between the total marks obtained by Q and S together in Chemistry to the total marks obtained by P and R together in Physics?
a) 23 : 25
b) 23 : 21
c) 17 : 19
d) 17 : 23
e) None of these
7. What is the respective ratio between the total marks obtained by P in Physics and Chemistry together to the total marks obtained by T in physics and Chemistry together?
a) 3 : 2
b) 4 : 3
c) 5 : 3
d) 2 : 1
e) None of these
8. Fill in the blank space in order to make the sentence correct as per the given information. Total marks obtained by T in both the subjects together is more than the marks obtained by…….
a) Q in Chemistry
b) R in Physics
c) S in Chemistry
d) P in Physics
e) R in both the subjects together
9. If the marks obtained by T in Physics were increased by 14% of the original marks, what would be his new approximate percentage in Physics if the maximum marks in Physics were 140?
a) 57
b) 32
c) 38
d) 48
e) 41
10. Marks obtained by S in Chemistry are what per cent of the total marks obtained by all the students in Chemistry?
a) 25
b) 28.5
c) 35
d) 31.5
e) 22
Directions (11 – 15): Study the following graph carefully and answer the questions given below it.
A report on the amount of water (In Million litres) sold by six different companies:
Percentage of water sold by six different companies:
The amount of water (In million litres) sold over the years:
11. If the sale of the company A is increased by 5%, how many litres of water sold by the company A from 2010 to 2012? 
a)    3245000
b)    2646000
c)     2465000
d)    3524000
e)    None of these
12. If the company R has sold the water at Rs. 20/litre and the company B has sold the water at Rs. 18/litre in 2016, what is the difference between their turnover in that year?
a)    Rs. 824000
b)    Rs. 762000
c)     Rs. 672000
d)    Rs. 726000
e)    None of these
13. What is the difference in the sale of water between the top 3 and bottom 3 companies in 2012?
a)    198900 litres
b)    1889000 litres
c)     1898000 litres
d)    1988000 litres
e)    None of these
14. In 2014, the company with lowest sale of water sold the water at Rs. 16/litre and the company with highest sale of water sold the water at Rs. 22/litre, what would be the price of water per litre sold by company with lowest sale to equal the turn over with the company with highest sale?
a)    Rs. 63.25
b)    Rs. 68.75
c)     Rs. 54.50
d)    Rs. 62.35
e)    None of these
15. What is the average amount of water sold by company B, K, and H in 2011? 
a)    150400 litres
b)    1054000 litres
c)     1504000 litres
d)    1045000 litres
e)    None of these


Solutions:
1. A) Percentage Decline = (80800 – 77300)/80800 * 100 = 4.33% ≈ 4%

2. B) Since the growth in the sales of automobiles in 1999-2000 as compared to 1998-1999as seen in the bar graph, is for Cars and three-wheelers only
Growth in the sales of cars in 1999-2000 as compared to 1998-1999 = 276000 – 262900 = 13100
Growth in the sales of three-wheelers in 1999-2000 as compared to 1998-1999 = 130000 122100 = 7900

3. C) Number of HCVs sold in 1998-1999 = 80400
Number of HCVs sold in 1999-2000 = 56200
Percentage Change in the sales of HCVs = (80400 – 56200)/80400 * 100 = 30%
Number of three-wheeler sold in 1998-1999 = 122100
Number of three-wheeler sold in 1999-2000 = 130000
Percentage Change in the sales of three-wheeler = (130000 – 122100)/122100 * 100 = 6.47%
Difference of percentage changes in sales of HCVs and three-wheeler = 30% - 6.47% = 23.53%

4. C) Percentage Growth = (276000 – 262900)/262900 * 100 = 4.98% ≈ 5%

5. B) Since it can be seen from the bar graph that decline in the sales of automobile is either for Two-wheelers, HCVs or LCVs in 1999-2000
Decline in the sales of Two-wheelers in 1999-2000 = 1719300 1714000 = 5300
Decline in the sales of HCVs in 1999-2000 = 80400 56200 = 24200
Decline in the sales of LCVs in 1999-2000 = 80800 77300 = 3500
Maximum decline is in the sales of HCVs in 1999-2000

6. B) Total marks obtained by Q and S together in Chemistry = (110 + 120)
= 230
Total marks obtained by P and R together in Physics = (130 + 80)
= 210
Ratio between the total marks obtained by Q and S together in Chemistry to the total marks obtained by P and R together in Physics = 230 : 210
= 23 : 21

7. D) Total marks obtained by P in Physics and Chemistry together = (130 + 90)
= 220
Total marks obtained by T in physics and Chemistry together = (50 + 60)
= 110
Ratio between the total marks obtained by P in Physics and Chemistry together to the total marks obtained by T in physics and Chemistry together = 220 : 110
= 2 : 1

8. B) Total marks obtained by T in both the subjects together = (50 + 60)
= 110
Marks obtained by Q in Chemistry = 110
Marks obtained by R in Physics = 80
Marks obtained by S in Chemistry = 120
Marks obtained by P in Physics = 130
Total marks obtained by R in both the subjects together = (80 + 100)
= 180
Therefore,
Total marks obtained by T in both the subjects together is more than the marks obtained by marks obtained by R in Physics

9. E) Marks obtained by T in Physics = 50
If the marks obtained by T in Physics were increased by 14% of the original marks
Then,
New marks of T in physics = 50 + 14% of 50
= 50 + (14/100) × 50
= 50 + 7
= 57
Maximum marks in Physics = 140
Percentage of marks obtained by T in physics = (57/140) × 100
= 40.71% ≈ 41%

10. A) Marks obtained by S in Chemistry = 120
The total marks obtained by all the students in Chemistry = (90 + 110 + 100 + 120 + 60)
= 480
Percentage of marks obtained by S in Chemistry to the total marks obtained by all the students in Chemistry = (120/480) × 100
= 25%

11. B) From the pie - chart and line graph,
Percentage of water sold by company A = 9
The sale of the company A is increased by 5%.
So, the amount of water sold by company A in 2010 = 5600000 × (14/100) = 784000 litres.
The amount of water sold by company A in 2011 = 6200000 × (14/100) = 868000 litres.
And, the amount of water sold by company A in 2012 = 7100000 × (14/100) = 994000 litres.
Hence, total amount of water sold by company A from 2010 to 2012
= (784000 + 868000 + 994000) litres = 2646000 litres.

12. B) From the pie - chart and line graph,
The total amount of water sold in 2016 = 12.7 million litres.
Percentage of water sold by company R = 21
So, the amount of water sold by company R in 2016 = 12.7 × (21/100) = 2.667 million litres.
The company R has sold the water at Rs. 20/litre. So, the turnover of the company R in 2016
= Rs. 2667000 × 20 = Rs. 53340000 = Rs. 5.334 Crores.
And,
Percentage of water sold by company B = 23
So, the amount of water sold by company B in 2016 = 12.7 × (23/100) = 2.921 million litres.
The company R has sold the water at Rs. 18/litre. So, the turnover of the company B in 2016
= Rs. 2921000 × 18 = Rs. 52578000 = Rs. 5.2578 Crores.
Hence, the difference between their turnover in that year
= Rs. (5.334 - 5.2578) Crores = Rs. 0.0762 Crores = Rs. 762000.

13. D) From the pie - chart and line graph,
The total amount of water sold in 2012 = 7.1 million litres.
We can find the top 3 and bottom 3 companies in sale of water from the percentage of water sold by different companies.
Top 3 companies = B, K, R
Bottom 3 companies = A, H, O
The amount of water sold by top 3 companies in 2012 = 7.1 × (23 + 21 + 20)% = 7.1 × (64/100) = 4.544 million litres.
And, the amount of water sold by bottom 3 companies in 2012
= 7.1 × (19 + 9 + 8) = 7.1 × (36/100) = 2.556 million litres.
The difference in the sale of water between the top 3 and bottom 3 companies in 2012
= (4.544 - 2.556) million litres = 1.988 million litres = 1988000 litres.

14. A) From the pie - chart and line graph,
The total amount of water sold in 2014 = 9.5 million litres.
We can find the company with lowest and highest sale of water from the pie - chart.
Company with lowest sale of water = H (8%)
Company with highest sale of water = B (23%)
The amount of water sold by company H in 2014 = 9.5 × (8/100) = 0.76 million litres.
The amount of water sold by company B in 2014 = 9.5 × (23/100) = 2.185 million litres.
According to the question, company B has sold water at Rs. 22/litre. So, the turnover of company B
= Rs. (2.185 × 22) millions = Rs. 48.07 millions.
The required price of water per litre = Rs. (48.07/0.76) = Rs. 63.25.

15. B) From the pie - chart and line graph,
The total amount of water sold in 2011 = 6.2 million litres.
Percentage of water sold by company B = 23
So, the amount of water sold by company B in 2011 = 6.2 × (23/100) = 1.426 million litres
Percentage of water sold by company K = 20
So, the amount of water sold by company K in 2011 = 6.2 × (20/100) = 1.24 million litres
Percentage of water sold by company H = 8
So, the amount of water sold by company H in 2011 = 6.2 × (8/100) = 0.496 million litres
Then, the total amount of water sold by these three companies in 2011
= (1.426 + 1.24 + 0.496) million litres = 3.162 million litres.
Hence, the average amount of water sold by company B, K, and H in 2011
= 3.162/3 = 1.054 million litres = 1054000 litres.