Question

the mid value of a class interval in 42 if the class size is 10 then the upper and lower limit of the class

Answer 1

Let the upper and lower limits are x and y respectively.

Now, (x + y)/2 = 42

=> x + y = 42*2

=> x + y = 84 ..........1

Again given,

class size = 10

=> x - y = 10 ........2

Add equation 1 and 2, we get

2x = 94

=> x = 94/2

=> x = 47

From equation 2

47 - y = 10

=> y = 47 - 10

=> y = 37

So, lower limit = 37

and upper limit = 47

Answer 2

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