A borrowed Rs \(P\) at \(x\%\) per annum and B borrowed Rs \(Q\) \((Q>P)\) at \(y\%\) per annum, both at simple interest and at the same time. After how much time will the amounts of their debts be equal?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle 100\left(\frac{Q-P}{Px-Qy}\right)\) years
Each debt is principal + simple interest. Set the two amounts equal after \(t\) years and solve for \(t\).
Debt of A \(\displaystyle =P+\frac{Pxt}{100}\); debt of B \(\displaystyle =Q+\frac{Qyt}{100}\)
\(\displaystyle P+\frac{Pxt}{100}=Q+\frac{Qyt}{100}\)
\(\displaystyle \frac{t\,(Px-Qy)}{100}=Q-P\)
\(\displaystyle t=100\left(\frac{Q-P}{Px-Qy}\right)\)
The debts become equal after \(\displaystyle 100\left(\frac{Q-P}{Px-Qy}\right)\) years. A's smaller debt can catch up only if it grows faster, that is \(Px>Qy\), which keeps this time positive.
Why the other options are incorrect
Option (2) \(\displaystyle 100\left(\frac{Px+Qy}{Q-P}\right)\) years — Incorrect. It adds the yearly interests and turns the fraction upside down.
Option (3) \(\displaystyle 100\left(\frac{Px-Qy}{P-Q}\right)\) years — Incorrect. The fraction is upside down and its sign is negative when \(Px>Qy\).
Option (4) \(\displaystyle 100\left(\frac{P-Q}{Px-Qy}\right)\) years — Incorrect. The top has the wrong sign, so the time comes out negative.