Target Exam

CUET

Subject

General Aptitude Test

Chapter

Numerical Ability

Topic

Number System

Question:

The age distribution of 40 children is as follows

Age (in yr.)

5-6

6-7

7-8

8-9

9-10

10-11

No. of children

4

7

9

12

6

2

Consider the following statements in respect of the above frequency distribution.

(A) The median of the age distribution is 7 yr.

(B) 70% of the children are in the age group 6-9 yr.

(C) The modal class of the children is 8-9 yr.

Which of the above statements are true?

Choose the correct answer from the options given below:

Options:

(A) and (B) only

(B) and (C) only

(A) and (C) only

(A), (B) and (C) only

Correct Answer:

(B) and (C) only

Explanation:

The correct answer is Option (2) → (B) and (C) only

Given data:

Age (yr)

5–6

6–7

7–8

8–9

9–10

10–11

Total

Children

4

7

9

12

6

2

40

(A) Median

  • Total children = 40 → Median position = $\frac{40+1}{2} = 20.5^{th}$ child
  • Cumulative frequency:

Age

Frequency

Cumulative

5–6

4

4

6–7

7

11

7–8

9

20

8–9

12

32

9–10

6

38

10–11

2

40

  • Median class = 8–9 yr (since 20.5th child falls in 8–9 yr)

So median is NOT 7 yr
(A) is false

(B) 70% of children in age group 6–9 yr

  • Age groups 6–7, 7–8, 8–9 → 7 + 9 + 12 = 28 children
  • Percentage = $\frac{28}{40} \times 100 = 70\%$

(B) is true

(C) Modal class

  • Mode = class with highest frequency
  • Frequencies: 4, 7, 9, 12, 6, 2 → Highest = 12 → 8–9 yr

(C) is true

Correct answer: (B) and (C) only