Data Interpretation MCQ Questions and Answers - Bar Chart

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$

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

Required ratio
= 900 : 140 : 1740
= 90 : 14 : 174
= 45 : 7 : 87

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$

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

Required sum
= 660 + 540 + 600
= 1800

$$\eqalign{ & \text{Number of students in:} \cr & \text{A → } \frac{12}{100}\times6000 = 720 \cr & \text{B → } \frac{9}{100}\times6000 = 540 \cr & \text{C → } \frac{26}{100}\times6000 = 1560 \cr & \text{D → } \frac{18}{100}\times6000 = 1080 \cr & \text{E → } \frac{29}{100}\times6000 = 1740 \cr & \text{F → } \frac{6}{100}\times6000 = 360 \cr} $$

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 girls in School E = 540
= Total number of students in B