Question

Adjusted Class Mark =
(a) Number of classes class interval
(b) (Adjusted Upper class limit + Adjusted Lower class limit)/2
(c) (Upper class limit+ Lower class limit)/2
(d) None

Answer 1

Hint: To do this question, firstly we will see the meaning of class limit with some examples and then we will see the meaning of Adjusted Class Limit with example and also, how we adjust class limit. Then, according to that, we will choose the right option.
Before we solve the question, let us see what class limits are and how the adjusted class limits are different from class limits.

Complete step by step answer:
Class limits are also known as class Interval or Class Mark. Class limit has two components which are Upper class limit and Lower class Limit. Lower class limit of a class is the smallest data value that can go into the class and Upper class limit of a class is the largest data value that can go into the class.
Now, in general, in corresponding classes, if the upper limit of first class is equal to the lower limit of the next class then the class limit is maintained and balanced and requires no adjustment in class limit.
For, example 10-15, 15-20, 20-25,--
But, when the class intervals are discontinuous, so to get the continuous interval we need to adjust the interval. Here, the difference between the upper limit of any class and the lower limit of the next class is calculated and then is divided by factor of 2, to calculate the value of Adjusted Class Limit.
So, $\text{Adjusted Class Limit=}\dfrac{\text{Adjusted Upper Class Limit + Adjusted Lower Class Limit}}{\text{2}}$
For example if we have class interval as 10-15, 17-22, 24-29,--
We can see that the difference between the upper limit of any class and the lower limit of the next class is 2.
So, $\text{Adjusted Class Limit=}\dfrac{\text{2 + 0}}{\text{2}}$
$\text{Adjusted Class Limit = 1}$

Hence, option (b) is correct.

Note: As this question has no numerical based concept and has no calculation, so these questions need basic knowledge of the theory and sometimes solving as many as different types of questions gives the hint to solve these types of questions. Also, one of the ways we can use it is we can discard the options by using counter examples in such questions.