Question

Match the two columns
$\begin{array}{ccc}& \text{Column-I \& Column-II}& \\ & \text{(A)The class mark of the class interval}& \text{(p)}13\\ & \text{145-150 is}& \\ & \text{(B)In a frequency distribution, the mid – value of a}& \text{(q)22.5–27.5}\\ & \text{class is 10 and width of each class is 6. The upper}& \\ & \text{limit of the class is}& \\ & \text{(C)The class marks of a frequency distribution are}& \text{(r)147.5}\\ & \text{15, 20, 25, 30, .....}\text{The class corresponding to the class}& \\ & \text{mark 25 is}& \end{array}$

Choose the correct option:

• A. A-p; B-r; C-q
• B. A-p; B-q; C-r
• C. A-q; B-p; C-r
• D. A-r; B-p; C-q

Answer 1

The correct option is D

A-r; B-p; C-q

$\left(A\right)\to r;\left(B\right)\to p;\left(C\right)\to q$

(A) Classmark

$=\frac{145+150}{2}=\frac{295}{2}=147.5$.

(B) Let the upper limit = x

Lower limit = y

$\therefore x-y=6$ and $\frac{x+y}{2}=10\Rightarrow x+y=20$

On solving both the equations, we get x = 13 and y = 7

(C) Class width = 20 - 15 = 5

Class mark = 25

$\therefore$ Required class,

$(25-\frac{5}{2})-(25+\frac{5}{2})\Rightarrow 22.5-27.5$