Question

THE product of two number is 336.Their sum exceed their difference by 32 . what are the numbers?​

Answer 1

Let us say a,b(a>b) be those two numbers. No possibility of a=b because 336 is not a perfect square

Given (a+b) = (a-b) + 32. (Since sum exceeds it's difference by 32.

Therefore, on solving 2b = 32 => b= 16.

Also a*b = 336 => a= 336/16

Therefore , a = 21.

So the numbers are 21 & 16

Answer 2

Let, one of the number be x and the other be y.

Given, their sum exceed their difference by 32

By condition,

x+y = x-y+32

or, x+y-x+y = 32

or,2y=32

or, y = 32/2

or, y=16

Therefore, one number is 16

As we know that x*y=336

By condition,

x*y=336

or,x*16 =336

or,x=336/16

or,x=21

Hence , the numbers are 16 and 21.