Question

Match the two columns
$\begin{array}{cc}Column-I& Column-II\\ \text{(A) The class mark of the class interval}& \left(p\right)13\\ 145-150is& \\ \text{(B) In a frequency distribution, the mid – value of a}& \left(q\right)110\\ \text{class is 10 and width of each class is 6. The upper}& \\ \text{limit of the class is}& \\ \text{(C) The mean of eight numbers is 40. If one}& \left(r\right)22.5–27.5\\ \text{number is excluded, their mean becomes}30.& \\ \text{The excluded number is}& \\ \text{(D) The class marks of a frequency distribution are}& \left(s\right)147.5\\ 15,20,25,30,\dots \dots \text{The class corresponding to the class}& \\ mark25is& \end{array}$
Choose the correct option:

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

Answer 1

The correct option is D

A-s; B-p; C-q; D-r

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

(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) Sum of the 8 observation $=40\times 8=320$
Sum of the 7 observation $=30\times 7=210$
Excluded observation is 320 - 210 = 110

(D) Class width = 20 - 15 = 5
Class mark = 25
$\therefore$ Required class,
$(25-\frac{5}{2})-(25+\frac{5}{2})\Rightarrow 22.5-27.5$