Re: $ P were deposited in each of two different accounts, one gives sim
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06 Jun 2025, 04:00
Let's denote the principal amount by $P$ and the annual interest rate by $R \%$. So, the interest rate as a decimal is $\frac{R}{100}$.
Simple Interest Account:
For simple interest, the interest earned each year is the same, calculated only on the principal amount.
Interest earned in the 1st year $\(=P \times \frac{R}{100}\)$
Interest earned in the 2nd year $\(=P \times \frac{R}{100}\)$
Compound Interest Account (compounded annually):
For compound interest, the interest earned in a given year is calculated on the principal amount plus any accumulated interest from previous years.
- 1st Year:
Amount at the end of 1st year $\(=P\left(1+\frac{R}{100}\right)\)$
Interest earned in the 1st year $\(=P\left(1+\frac{R}{100}\right)-P=P \times \frac{R}{100}\)$
- 2nd Year:
The principal for the 2nd year's interest calculation is the amount at the end of the 1st year.
Interest earned in the 2nd year = (Amount at end of 1st year) $\(\times \frac{R}{100}\)$
Interest earned in the 2nd year $\(=P\left(1+\frac{R}{100}\right) \times \frac{R}{100}\)$
Quantity B: Difference between the interests earned in accounts in the 1st year
Interest earned in simple interest account in 1st year $\(=P \times \frac{R}{100}\)$
Interest earned in compound interest account in 1st year $\(=P \times \frac{R}{100}\)$
Difference in 1st year $=$ (Interest in compound account) - (Interest in simple account)
Difference in 1st year $\(=P \times \frac{R}{100}-P \times \frac{R}{100}=0\)$
Quantity A: Difference between the interests earned in accounts in the 2nd year
Interest earned in simple interest account in 2nd year $\(=P \times \frac{R}{100}\)$
Interest earned in compound interest account in 2nd year $\(=P\left(1+\frac{R}{100}\right) \times \frac{R}{100}\)$
Difference in 2nd year = (Interest in compound account) - (Interest in simple account)
Difference in 2nd year $\(=P\left(1+\frac{R}{100}\right) \times \frac{R}{100}-P \times \frac{R}{100}\)$
Difference in 2nd year $\(=P \times \frac{R}{100}\left(1+\frac{R}{100}-1\right)\)$
Difference in 2nd year $\(=P \times \frac{R}{100}\left(\frac{R}{100}\right)\)$
Difference in 2nd year $\(=P\left(\frac{R}{100}\right)^2\)$
Since $P>0$ and $R>0$, we know that $\(\left(\frac{R}{100}\right)^2\)$ will also be greater than 0 .
Therefore, the Difference in 2nd year $\(=P\left(\frac{R}{100}\right)^2>0\)$.
Comparison:
Quantity $\(\mathrm{A}=P\left(\frac{R}{100}\right)^2\)$
Quantity $\(B=0\)$
Since $\(P\left(\frac{R}{100}\right)^2>0\)$, Quantity A is greater than Quantity B.
The final answer is Quantity A is greater.