Question

What integer must be added to each of the four numbers 10, 18, 22, 38 so that they become a proportion?

Answer 1

Let the required integer be x.
When x is added, they are in proportion.
So, (10 + x) / (18 + x ) = (22 + x ) / (38 + x )
By cross multiplication,
(10 + x ) × ( 38 + x ) = ( 18 + x ) × (22 + x)
380 + 38x + 10x + x^2 = 396 + 18x + 22x + x^2
380 + 48x + x^2 = 396 + 40x + x^2
380 + 48x + x^2 - x^2 - 396 - 40x = 0
8x - 16 = 0
8x = 16
x = 16 ÷ 8
x = 2

So, 2 must be added to each so that they become a proportion.
You can check answer by yourself.

Hope it helps !! ;-)

Answer 2

Let the required integer be = x
So, the numbers will be:
(10+x) , (18+x) , (22+x) & (38+x)
Since the numbers are in proportion,
a/b = c/d

BTP,
(10+x)÷(18+x) = (22+x)÷(38+x)

380 + 38x + 10x +k^2 = 396 + 22k + 18k + k^2

48k - 40k = 396 - 380

8k = 16

k = 2

Therefore, the required integer is 2