Question

If the class mark is $25$ and class width is $10$, then find the class.

Answer 1

Hint:
For solving this question we will use the formula which is $CM = \dfrac{{UL + LL}}{2}$, so we will assume the lower limit be $x$ and the upper limit be $y$. And with the conditions given, we will end up with an interval of class which is said to be as a class.

Formula used:
Class mark,
$CM = \dfrac{{UL + LL}}{2}$
Here,
$CM$, will be the class mark
$UL$, will be the upper limit
$LL$, will be the lower limit

Complete step by step solution:
So let us assume the lower limit be $x$ and the upper limit be $y$
According to the question, on framing the equation where the class width is $10$ we get
$ \Rightarrow y = x + 10$
By using the formula of the class mark and substituting the values, we get
$ \Rightarrow 25 = \dfrac{{x + y}}{2}$
Now substituting the values, we had just obtained, we get
$ \Rightarrow 25 = \dfrac{{(x + (x + 10))}}{2}$
So on solving it we get the equation as
$ \Rightarrow 50 = 2x + 10$
And on taking the constant term to one side and subtracting it, we get
$ \Rightarrow 2x = 40$
Since $2$ is in multiplication so taking it to the right side of the equation then it will come in divide, So on dividing the number we get
$ \Rightarrow x = 20$
Therefore from this, we have a lower limit $20$ and the upper limit $y = 20 + 10 = 30$.

Hence, the interval of the class will be $20 - 30$.

Note:
For solving this type of question, we just need the formulas and then we can easily solve it. It should be noted that the upper limit will always be higher as compared to the lower limit. So the lower limit has a lower value. So in this way, we can answer such a type of question, where the interval plays a major role.