Question : If A borrowed Rs. $P$ at $x$% and B borrowed Rs. $Q (> P)$ at $y$% per annum in simple interest at the same time, then the amount of their debts will be equal after:
Option 1: $100(\frac{Q-P}{Px-Qy})$
Option 2: $100(\frac{Px-Qy}{Q-P})$
Option 3: $100(\frac{Px-Qy}{P-Q})$
Option 4: $100(\frac{P-Q}{Px-Qy})$
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Correct Answer: $100(\frac{Q-P}{Px-Qy})$
Solution :
Given: $P$ is the principal, $x$ is the rate for 1st case, $Q$ is the principal and $y$ is the rate for the second case.
Let the amount be equal in $T$ years.
Simple interest = $\frac{P×R×T}{100}$
According to the question:
$P+\frac{P×x×T}{100}=Q+\frac{Q×y×T}{100}$
⇒ $\frac{PxT}{100}-\frac{QyT}{100}=Q–P$
⇒ $T(\frac{Px}{100}-\frac{Qy}{100})=Q–P$
⇒ $ T(\frac{Px-Qy}{100})$ = $Q–P$
⇒ $T=\frac{(Q-P)100}{Px-Qy}$
Hence, the answer is $\frac{(Q-P)100}{Px-Qy}$.
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