Question

What must be subtracted from each of the numbers 23,40,57 and 108 so that the remainders are in proportion

Answer 1

Let the no. To be subtracted be x . Then ,
(23-x):(40-x)::(57-x):(108-x)
=(23-x)/(40-x)=(57-x)/(108-x)
=(23-x)(108-x)=(40-x)(57-x)
=2484-131x+2x=2280-97+2x
=34x=204 =
x = 204/34=6

Answer 2

let no to be subtracted be x

then,

(23-x) : (40-x) = (57-x) : (108-x)

                                           After transposing

(23-x) (108-x ) = (57-x) (40-x)

2484-23x-108x-x^2 = 2280-57x-40x-x^2

              x^2 gets cancelled from both the sides

2484-131x = 2280-97x

2484-2280 = -97x + 131x

204=34x

204/34=x

     204/34 gets divided and become 6

6=x