What number should be subtracted from each of the numbers 54, 71, 75 and 99 so that the remainders are in proportion?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
Four numbers \(a, b, c, d\) are in proportion when \(a:b=c:d\), that is, when the product of the extremes equals the product of the means: \(a\times d=b\times c\).
Let \(x\) be subtracted: \((54-x)(99-x)=(71-x)(75-x)\)
\(5346-153x+x^{2}=5325-146x+x^{2}\)
\(5346-5325=153x-146x\)
\(21=7x\), so \(x=3\)
Check: the remainders 51, 68, 72, 96 give \(\displaystyle \frac{51}{68}=\frac{3}{4}=\frac{72}{96}\)
The number to be subtracted is 3.