Reasoning Aptitude > Data Interpretation
BAR GRAPH MCQs
Bar Charts
Total Questions : 165
| Page 13 of 17 pages
Answer: Option B. -> B
$$\eqalign{
& \text{Number of students in:} \cr
& \text{A → } \frac{12}{100}\times6000 = 720 \cr
& \text{B → } \frac{9}{100}\times6000 = 540 \cr
& \text{C → } \frac{26}{100}\times6000 = 1560 \cr
& \text{D → } \frac{18}{100}\times6000 = 1080 \cr
& \text{E → } \frac{29}{100}\times6000 = 1740 \cr
& \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$
Number of boys in:
A → 500
B → 400
C → 900
D → 600
E → 1200
F → 100
Number of girls in:
A → 720 - 500 = 220
B → 540 - 400 = 140
C → 1560 - 900 = 660
D → 1080 - 600 = 480
E → 1740 - 1200 = 540
F → 360 - 100 = 260
Number of girls in School E = 540
= Total number of students in B
$$\eqalign{
& \text{Number of students in:} \cr
& \text{A → } \frac{12}{100}\times6000 = 720 \cr
& \text{B → } \frac{9}{100}\times6000 = 540 \cr
& \text{C → } \frac{26}{100}\times6000 = 1560 \cr
& \text{D → } \frac{18}{100}\times6000 = 1080 \cr
& \text{E → } \frac{29}{100}\times6000 = 1740 \cr
& \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$
Number of boys in:
A → 500
B → 400
C → 900
D → 600
E → 1200
F → 100
Number of girls in:
A → 720 - 500 = 220
B → 540 - 400 = 140
C → 1560 - 900 = 660
D → 1080 - 600 = 480
E → 1740 - 1200 = 540
F → 360 - 100 = 260
Number of girls in School E = 540
= Total number of students in B
Answer: Option B. -> 50%
$$\eqalign{
& \text{Number of students in:} \cr
& \text{A → } \frac{12}{100}\times6000 = 720 \cr
& \text{B → } \frac{9}{100}\times6000 = 540 \cr
& \text{C → } \frac{26}{100}\times6000 = 1560 \cr
& \text{D → } \frac{18}{100}\times6000 = 1080 \cr
& \text{E → } \frac{29}{100}\times6000 = 1740 \cr
& \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$
Number of boys in:
A → 500
B → 400
C → 900
D → 600
E → 1200
F → 100
Number of girls in:
A → 720 - 500 = 220
B → 540 - 400 = 140
C → 1560 - 900 = 660
D → 1080 - 600 = 480
E → 1740 - 1200 = 540
F → 360 - 100 = 260
Let (number of girls in A) be $$x$$% of number of students in B.
Then,
$$\eqalign{
& 220 = \frac{x}{100}\times540 \cr
& \Rightarrow x = \frac{220\times100}{540} \cr
& \Rightarrow x = \frac{1100}{27} \cr
& \Rightarrow x = 40.7 \cr
& \Rightarrow x \approx 40\% \cr} $$
$$\eqalign{
& \text{Number of students in:} \cr
& \text{A → } \frac{12}{100}\times6000 = 720 \cr
& \text{B → } \frac{9}{100}\times6000 = 540 \cr
& \text{C → } \frac{26}{100}\times6000 = 1560 \cr
& \text{D → } \frac{18}{100}\times6000 = 1080 \cr
& \text{E → } \frac{29}{100}\times6000 = 1740 \cr
& \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$
Number of boys in:
A → 500
B → 400
C → 900
D → 600
E → 1200
F → 100
Number of girls in:
A → 720 - 500 = 220
B → 540 - 400 = 140
C → 1560 - 900 = 660
D → 1080 - 600 = 480
E → 1740 - 1200 = 540
F → 360 - 100 = 260
Let (number of girls in A) be $$x$$% of number of students in B.
Then,
$$\eqalign{
& 220 = \frac{x}{100}\times540 \cr
& \Rightarrow x = \frac{220\times100}{540} \cr
& \Rightarrow x = \frac{1100}{27} \cr
& \Rightarrow x = 40.7 \cr
& \Rightarrow x \approx 40\% \cr} $$
Answer: Option B. -> 860
$$\eqalign{
& \text{Number of students in:} \cr
& \text{A → } \frac{12}{100}\times6000 = 720 \cr
& \text{B → } \frac{9}{100}\times6000 = 540 \cr
& \text{C → } \frac{26}{100}\times6000 = 1560 \cr
& \text{D → } \frac{18}{100}\times6000 = 1080 \cr
& \text{E → } \frac{29}{100}\times6000 = 1740 \cr
& \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$
Number of boys in:
A → 500
B → 400
C → 900
D → 600
E → 1200
F → 100
Number of girls in:
A → 720 - 500 = 220
B → 540 - 400 = 140
C → 1560 - 900 = 660
D → 1080 - 600 = 480
E → 1740 - 1200 = 540
F → 360 - 100 = 260
(Number of boys in E) - (Number of students in F)
= 1200 - 360
= 840
$$\eqalign{
& \text{Number of students in:} \cr
& \text{A → } \frac{12}{100}\times6000 = 720 \cr
& \text{B → } \frac{9}{100}\times6000 = 540 \cr
& \text{C → } \frac{26}{100}\times6000 = 1560 \cr
& \text{D → } \frac{18}{100}\times6000 = 1080 \cr
& \text{E → } \frac{29}{100}\times6000 = 1740 \cr
& \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$
Number of boys in:
A → 500
B → 400
C → 900
D → 600
E → 1200
F → 100
Number of girls in:
A → 720 - 500 = 220
B → 540 - 400 = 140
C → 1560 - 900 = 660
D → 1080 - 600 = 480
E → 1740 - 1200 = 540
F → 360 - 100 = 260
(Number of boys in E) - (Number of students in F)
= 1200 - 360
= 840
Answer: Option B. -> 6 : 5
(Number of students appeared from C) : (Number of students qualified from C)
$$\eqalign{
& = \frac{3250}{2250} \cr
& = \frac{13}{9} \cr
& = 13:9 \cr} $$
(Number of students appeared from C) : (Number of students qualified from C)
$$\eqalign{
& = \frac{3250}{2250} \cr
& = \frac{13}{9} \cr
& = 13:9 \cr} $$
Answer: Option D. -> 18%
Required %
$$= \left(\frac{2500}{3000+2250+3250+2500+3000}\times100\right)\%$$
$$\eqalign{
& = \left(\frac{2500}{14000}\times100\right)\% \cr
& = \frac{125}{7}\% \cr
& \approx 18\% \cr} $$
Required %
$$= \left(\frac{2500}{3000+2250+3250+2500+3000}\times100\right)\%$$
$$\eqalign{
& = \left(\frac{2500}{14000}\times100\right)\% \cr
& = \frac{125}{7}\% \cr
& \approx 18\% \cr} $$
Question 126. Directions (1 - 5): Study the following graph carefully and answer the questions given below:
What is the difference between the average number of students appeared in the scholarship exam from all the given schools and the average number of students qualified from all the school together?
What is the difference between the average number of students appeared in the scholarship exam from all the given schools and the average number of students qualified from all the school together?
Answer: Option A. -> 950
Average number of students appeared from all the schools
= $$\frac{1}{5}$$(3000 + 2250 + 3250 + 2500 + 3000)
= $$\frac{14000}{5}$$
= 2800
Average number of students qualified from all the schools
= $$\frac{1}{5}$$(1750 + 1250 + 2250 + 2000 + 2000)
= $$\frac{9250}{5}$$
= 1850
∴ Required difference
= 2800 - 1850
= 950
Average number of students appeared from all the schools
= $$\frac{1}{5}$$(3000 + 2250 + 3250 + 2500 + 3000)
= $$\frac{14000}{5}$$
= 2800
Average number of students qualified from all the schools
= $$\frac{1}{5}$$(1750 + 1250 + 2250 + 2000 + 2000)
= $$\frac{9250}{5}$$
= 1850
∴ Required difference
= 2800 - 1850
= 950
Answer: Option D. -> 7 : 5
(Number of students qualified from A) : (Number of students qualified from B)
$$\eqalign{
& = \frac{1750}{1250} \cr
& = \frac{7}{5} \cr
& = 7:5 \cr} $$
(Number of students qualified from A) : (Number of students qualified from B)
$$\eqalign{
& = \frac{1750}{1250} \cr
& = \frac{7}{5} \cr
& = 7:5 \cr} $$
Question 128. Directions (1 - 5): Study the following graph carefully and answer the questions given below:
The average number of students qualified in the examination from Schools C and D is what percent of the average number of students appeared for the examination from the same school? ( round off to 2 digits after decimal).
The average number of students qualified in the examination from Schools C and D is what percent of the average number of students appeared for the examination from the same school? ( round off to 2 digits after decimal).
Answer: Option B. -> 73.91%
Average number of students qualified from C and D
$$\eqalign{
& = \frac{1}{2}\left(2250+2000\right) \cr
& = \frac{4250}{2} \cr
& = 2125 \cr} $$
Average number of students appeared from C and D
$$\eqalign{
& = \frac{1}{2}\left(3250+2500\right) \cr
& = \frac{5750}{2} \cr
& = 2875 \cr
& \therefore \text{ Required %} \cr
& = \left(\frac{2125}{2875}\times100\right)\% \cr
& = \left(\frac{85}{115}\times100\right)\% \cr
& = \frac{1700}{23}\% \cr
& = 73.91\% \cr} $$
Average number of students qualified from C and D
$$\eqalign{
& = \frac{1}{2}\left(2250+2000\right) \cr
& = \frac{4250}{2} \cr
& = 2125 \cr} $$
Average number of students appeared from C and D
$$\eqalign{
& = \frac{1}{2}\left(3250+2500\right) \cr
& = \frac{5750}{2} \cr
& = 2875 \cr
& \therefore \text{ Required %} \cr
& = \left(\frac{2125}{2875}\times100\right)\% \cr
& = \left(\frac{85}{115}\times100\right)\% \cr
& = \frac{1700}{23}\% \cr
& = 73.91\% \cr} $$
Answer: Option B. -> 10000
No. of other vehicles registered in January 1991
= 27 - 21
= 6 thousand
No. of other vehicles registered in April 1991
= 35 - 20
= 15 thousand
Hence, increased
= 15 - 6
= 9 thousand
No. of other vehicles registered in January 1991
= 27 - 21
= 6 thousand
No. of other vehicles registered in April 1991
= 35 - 20
= 15 thousand
Hence, increased
= 15 - 6
= 9 thousand
Answer: Option B. -> 6000
Registered increase
= 28 - 21
= 7 thousand
Registered increase
= 28 - 21
= 7 thousand
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