Question

In a frequency distribution, class mark of a class interval is 18 and class length is also 18. The lower boundary of the class is ___.

• A. 27
• B. 9
• C. 36
• D. 54

Answer 1

The correct option is B 9

Let,
upper boundary = x
lower boundary = y
$\text{Class mark}=\frac{\text{Upper boundary+ Lower boundary}}{2}$.
18 = $\frac{\text{x+y}}{2}$

x+y = 36 ......................(1).

Class length = Upper boundary - Lower boundary.
18 = x - y .......................(2).
x - y = 18 .......................(2).

Adding equation (1) and (2), we get
2x = 36 + 18 = 54.
So, x = 27.
By using equation (1) , y = 36 - 27 = 9.