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Question

Question: If the class mark is \(25\) and class width is \(10\), then find the class....

If the class mark is 2525 and class width is 1010, then find the class.

Explanation

Solution

For solving this question we will use the formula which is CM=UL+LL2CM = \dfrac{{UL + LL}}{2}, so we will assume the lower limit be xx and the upper limit be yy. 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=UL+LL2CM = \dfrac{{UL + LL}}{2}
Here,
CMCM, will be the class mark
ULUL, will be the upper limit
LLLL, will be the lower limit

Complete step by step solution:
So let us assume the lower limit be xx and the upper limit be yy
According to the question, on framing the equation where the class width is 1010 we get
y=x+10\Rightarrow y = x + 10
By using the formula of the class mark and substituting the values, we get
25=x+y2\Rightarrow 25 = \dfrac{{x + y}}{2}
Now substituting the values, we had just obtained, we get
25=(x+(x+10))2\Rightarrow 25 = \dfrac{{(x + (x + 10))}}{2}
So on solving it we get the equation as
50=2x+10\Rightarrow 50 = 2x + 10
And on taking the constant term to one side and subtracting it, we get
2x=40\Rightarrow 2x = 40
Since 22 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
x=20\Rightarrow x = 20
Therefore from this, we have a lower limit 2020 and the upper limit y=20+10=30y = 20 + 10 = 30.

Hence, the interval of the class will be 203020 - 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.