Question

need this answer. answer is (required class is 7-13)
I need method for it.
The best will be marked as BRAINLIST!

Answer 1

Answer:

$\frac{x + y}{2} = 10 \\ \: \: \: \: and \: \\ \: \: \: y - x = 6$

Now we have

$x + y = 20 \\ y - x = 6$

So, y =13 and x=7

And class is( 7-13)

Hope it helps you.

Pls mark it as brainliest.

Answer 2

Answer:

Class : 7 - 13

Before we get into solving the question, we'll define the terms given in the question.

Class mark: It refers to the sum of the lower limit and the upper limit of a particular class, divided by 2.

$\boxed{\sf \ Class \ Mark = \dfrac{Lower \ Limit + Upper \ Limit}{2} \ }$

Class width: It refers to the difference between the upper limit and the lower limit of a particular class.

$\boxed{\sf Class \ Width = Upper \ limit - Lower \ limit}}$

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It's given that we have to find the class, which comprises of the lower and upper limits.

Let us assume the lower limit of the class to be 'x', and the upper limit to be 'y'.

ATQ,

$\Longrightarrow \ \sf Class \ mark = 10 \\ \\ \\ \Longrightarrow \sf \ \dfrac{Lower \ Limit + Upper \ Limit}{2} = 10\\ \\ \\ \Longrightarrow \sf \ \dfrac{x+y}{2} = 10\\ \\ \\ \Longrightarrow \sf \ x+y= 20\\ \\ \\ \Longrightarrow \sf \ y+x= 20 \ \ ..Eq(1)$

Also, The Class width is 6.

$\Longrightarrow \sf \ Class \ Width = 6\\ \\ \\ \Longrightarrow \sf \ Upper \ limit - Lower \ limit = 6\\ \\ \\ \Longrightarrow \sf \ y-x=6 \ \ ..Eq(2)$

Adding equations 1 and 2 we get,

     $\sf y \ + \ \bcancel{x} \ = \ 20\\ \\\sf y \ - \ \bcancel{x} \ = \ 6\\ \rule{70}{1}$

             $\sf 2y = \ 26\\ \\ \sf y \ = \ \dfrac{26}{2}\\ \\ \sf y \ = \ 13$

Substitute the value of 'y' in Eq(1)

⇒  y + x = 20

⇒  13 + x = 20

⇒  x = 20 - 13

⇒  x = 7

Lower Limit 'x' = 7

Upper Limit 'y' = 13

Class = Lower Limit - Upper Limit

         = 7 - 13

$\large\boxed{\large\boxed{\sf Class = 7 - 13}}$