Question

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

Answer 1

let the upper limit and lower limit of the class be x , y
then (x+y)/2 = 42
       or x+y = 84
           x-y = 10 
solving equations x = 47 , y = 37 

Answer 2

Answer:

Step-by-step explanation:

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