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: $100(\frac{P-Q}{Px-Qy})$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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 $\frac{(Q-P)100}{Px-Qy}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : For what sum will the simple interest at $R$% per annum for 2 years be $R$?
Question : If $x=(0.25)^\frac{1}{2}$, $y=(0.4)^{2}$, and $z=(0.216)^{\frac{1}{3}}$, then:
Question : A person borrowed some money at simple interest. After 4 years, he returned $\frac{9}{5}$th of the money to the lender. What was the rate of interest?
Question : The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:
Question : A man borrowed some money from a private organisation at 5% simple interest per annum. He lent this money to another person at 10% compound interest per annum and made a profit of Rs. 26,410 in 4 years, the man borrowed:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile