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

Correct option is D

Given:

Numbers: 23, 40, 57, and 108

Solution:

Let the number to be subtracted from each be x.
So,

​$$\frac{23 - x}{40 - x} = \frac{57 - x}{108 - x}$$​​

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

​$$223 \times 108 - 23x - 108x + x^2 = 240 \times 57 - 40x - 57x + x^2$$​​

​$$2484 - 131x + x^2 = 2280 - 97x + x^2$$​​

2484 - 131x = 2280 - 97x

2484 - 2280 = 131x - 97x

204 = 34x

x = $$\frac{204}{34} = 6$$​​

Alternate Solution:

We can solve this by directly putting the value from the option;

​$$\frac{23 - x}{40 - x} = \frac{57 - x}{108 - x}$$​​

putting x = 6,

​$$\frac{17}{34} = \frac{51}{102}$$​​

​$$\frac12 =\frac12$$​​