What number must be subtracted from each term of the ratio 17: 33 so t

1.	What number must be subtracted from each term of the ratio 17: 33 so that the ratio becomes 7: 15?
Answer» Let the number be x, Then the two term becomes, = (17 – x): (33 – x) = 7: 15 = (17 – x) / (33 – x) = (7/15) By cross multiplication, = 15 × (17 – x) = 7 × (33 – x) = 255 – 15x = 231 – 7x Transposing -15x to RHS and 231 to LHS = 255 – 231 = 15x -7x = 8x = 24 = x = (24/8) = x = 3 Hence, the number be subtracted from each term of the ratio 17:33 to make the ratio 7: 15 is 3

What number must be subtracted from each term of the ratio 17: 33 so that the ratio becomes 7: 15?

Answer»

Let the number be x,

Then the two term becomes,

= (17 – x): (33 – x) = 7: 15

= (17 – x) / (33 – x) = (7/15)

By cross multiplication,

= 15 × (17 – x) = 7 × (33 – x)

= 255 – 15x = 231 – 7x

Transposing -15x to RHS and 231 to LHS

= 255 – 231 = 15x -7x

= 8x = 24

= x = (24/8)

= x = 3

Hence, the number be subtracted from each term of the ratio 17:33 to make the ratio 7: 15 is 3