Data Interpretation Practice Questions – Set 87
Directions
(1 – 5): Below is given a pair of pie chart and bar graph representing
number of students in different classes in a school in 2015 and growth in 2016.
Based on this information, answer the questions that follow.
1. The
students in 2016 in different classes are represented in a pie chart. What will
be the difference between angles subtended by sectors representing 7thand
9th classes?(in°)
(a) 30
(b) 36.24
(c) 39.93
(d) 45
(e) 48.2
2. If the
increment in all the classes has only been in the form of girl students, then
what will be the ratio of total number of boys and girls in 2016?
(a) 43 : 41
(b) 23 : 27
(c) 8 : 9
(d) 2 : 3
(e) Cannot be
determined
3. Instead
of 20%, what should be the minimum growth in number of students for class 7th,
so that it does not have minimum number of students in 2016?
(a) 41.12
(b) 43.89
(c) 46.67
(d) 48
(e) 49.52
4. What
will be the average number of students in classes 8th and 9th taken
together in 2015 and 2016?
(a) 164
(b) 171
(c) 177
(d) 183
(e) 189
5. What
should be growth in number of students for class 8th in 2017 so
that their number becomes same as number of students in class 9th in
2016?
(a) 10%
(b) 12%
(c) 15%
(d) 8%
(e) 7.5%
Directions
(6 – 10): Study the table carefully and answer the following question.
The given table
represents 10th and 12th class students reading
different subject in a school.
6. What
reading the difference between the number of 10th class
students reading English subject and the number of 12th class
students reading Math?
(a) 532
(b) 680
(c) 482
(d) 702
(e) None of
these
7. Which
subject has the maximum number of 10th class students reading?
(a) Hindi
(b) Physics
(c) Chemistry
(d) Computer
(e) Economics
8. The 10th class
students reading Hindi subject are how much percent more than 12th class
students reading Hindi subject?
(a) 35%
(b) 48%
(c) 75%
(d) 55%
(e) None of
these
9. What is
the approximate average number of 12th class students reading
subjects in English language?
(a) 572
(b) 463
(c) 372
(d) 562
(e) 509
10. What is
the difference between the number of 12th students reading
computer and the number of 10th students reading economics?
(a) 432
(b) 558
(c) 468
(d) 238
(e) None of
these
Directions
(11 – 15): Study the following graphs carefully and
answer the questions that follow.
Percentage profit
earned by two companies over the given years
%
Profit = (Income – Expenditure)/Expenditure × 100
11. If the
total income of Company A in all the years together was equal to the total
expenditure of Company B in all the years together, which was Rs. 265 lakhs,
what was the total percentage profit earned by Company A for all the years
together?
(a) 45
(b) 37
(c) 52
(d) Cannot be
determined
(e) None of
these
12. If the income
of Company A in 1999 was equal to the expenditure of Company B in 2001, then
what was the ratio of expenditure of Company A in 1999 to the income of Company
B in 2001?
(a) 25: 56
(b) 66: 25
(c) 10: 13
(d) 13: 10
(e) None of
these
13. If the
total expenditure of the two Companies in 2000 was Rs. 18 lakhs and expenditure
of Companies A & B in that year were in the ratio of 4: 5 respectively,
then what was the income of Company B in that year (in lakh)?
(a) 8
(b) 10
(c) 10.4
(d) Cannot be
determined
(e) None of
these
14. If the
income of Company B in 1999 was Rs. 18.6 lakhs and ratio of incomes of
Companies A % B in 1999 was 4: 3, What was the expenditure of company A in 1999
(in Rs. lakh)?
(a) 12.5
(b) 14.5
(c) 15.5
(d) 16.5
(e) None of
these
15. If the
income of Company A in 1998 was equal to its expenditure in 2000, what was the
ratio between Company’s expenditure in the years 1998 and 2000 respectively?
(a) 3: 2
(b) 2: 3
(c) 1: 2
(d) Cannot be
determined
(e) None of
these
Solutions:
(1 – 5): Explanation:
Number
of students in class 6th in 2016 = 60 × (1 + 10/100) = 66
Number
of students in class 7th in 2016 = 45 × (1 + 20/100) = 54
Number
of students in class 8th in 2016 = 75 × (1 + 20/100) = 90
Number
of students in class 9th in 2016 = 90 × (1 + 10/100) = 99
Number
of students in class 10th in 2016 = 120 × (1 + 15/100) = 138
1. B) ∴ difference between angles subtended by sectors
representing 7th and 9thclasses = [(99 - 54)/(66 +
54 + 90 + 99 + 138)] × 360° = 36.24°
2. E) We do not
have the original distribution of boys and girls in 2015. So, actual ratio
cannot be determined neither for 2015 and nor for 2016 after increment.
∴ Ratio
cannot be uniquely determined.
3. C) To not have
the minimum value, number of students for class 7th should be
more than 66.
∴ Minimum
growth = [(66 - 45)/45] × 100 = 46.67%
4. C) ∴ average
number of students in classes 8th and 9th taken
together in 2015 and 2016 ((75 + 90) + (90 + 99))/2 = 177
5. A) Suppose
growth is T% for class 8th students in 2017.
⇒ 90 × (1 + T/100) = 99
⇒ T = (99/90
- 1) × 100 = 10
∴ Growth
should be 10%.
6. B) It is given
that number of 10th class students reading English subject
=
9/13 × 1352 = 936
Number
of 12th class students reading Math’s = 1/4 × 1024 = 256
∴ Required
difference = 936 – 256 = 680
7. B) Given that,
number of 10th class students reading Physics
=
4/7 × 2401 = 1372
Number
of 10th class students reading Chemistry = 2/3 × 729 = 486
Number
of 10th class students reading Hindi = 7/11 × 1331 = 847
Number
of 10th class students reading Computer = 7/10 × 1500 = 1050
Number
of 10th class students reading language = 4/13 × 2197 = 676
Number
of 10th class students reading Math = 3/4 × 1024 = 768
Number
of 10th class students reading Economics = 7/12 × 1728 = 1008
Number
of 10th class students reading English = 9/13 × 1352 = 936
Since,
the 10th class students are reading in maximum for Physics.
Therefore this subject has the maximum percentage of 10th class
students.
8. C) It is given
that number of 10th class students reading Hindi = 7/11 × 1331 = 847
Number
of 12th class students reading Hindi = 4/11 × 1331 = 484
Therefore
required percentage = (847 – 484)/484 × 100 = 363/484 × 100 = 75%
9. A) Required
Average = (1029 + 243 + 450 + 720 + 416)/5 = 2858/5 = 571.6 ≈ 572
10. B) According
to question, number of 12th class students reading computer
=
3/10 × 1500 = 450
Number
of 10th class students reading economics = 7/12 × 1728 = 1008
∴ Required
difference = 1008 – 450 = 558
11. D) The answer
can’t be determined. Because we need more data to solve this type of question.
We
don’t have the Expenditure of all over the year for company A that is essential
to solve this question.
12. A) Suppose
Income of Company A in 1999 = X
Therefore
from the question, the expenditure of Company B in 2001 = X
Also
Expenditure of company A in 1999 = Y
And
Income of company B in 2001 = Z
⇒ 0.4 = Z/X
- 1
⇒1.4 = Z/X
⇒0.714 = X/Z
Ratio
between the Expenditure of company A in 1999 to income of company B in 2001 is
=
Y / Z
=
(Y/X) × (X/Z)
=
0.625 × 0.714 (∵
Substituting Values of Y/Z and X/Z))
=
0.446
=
25 / 56
13. E) The total
expenditure of the two Companies in 2000 was Rs. 18 lakhs
Expenditure
of Companies A & B in that year were in the ratio of 4: 5
If
ratio factor is X lakhs then
Expenditure
of Company A = 4X lakhs
Expenditure
of Company B = 5X lakhs
Total
Expenditure = 18 lakhs
4X
+ 5X = 18
⇒ 9X =18
⇒ X = 2
Expenditure
of Company A = 4 × X = 4 × 2 = 8 lakhs
Expenditure
of Company B = 5 × X = 5 × 2 = 10 lakhs
Profit
percentage of company B in 2000 = 60 %
14. C) Income of
Company B in 1999 was Rs 18.6 lakhs
Ratio
of incomes of Companies A and B in 1999 was 4: 3
Income
of company B in 1999 = 18.6 lakhs
Income
of company A in 1999/Income of company B in 1999 = 4/3
⇒ Income of
company A in 1999 = 4 × 18.6 /3
⇒ Income of
company A in 1999 = 24.8 lakhs
Profit
percentage of company A in 1999 = 60 %
⇒ 1.6 = 24.8
/ Expenditure of company A in 1999
⇒
Expenditure of company A in 1999 = 24.8 / 1.6
⇒
Expenditure of company A in 1999 = 15.5 lakhs
15. B) Percentage
of profit of company A in 1998 is 50%
Percentage
profit earned by two companies over the given years
By
the formula we get,
⇒ 1.5 = Y/X
⇒ X/Y =
1/1.5
⇒ X/Y =
0.666
⇒X/Y = 2/3
