A sum of money invested at compound interest amount to Rs. 4624 in 2 y

1.	A sum of money invested at compound interest amount to Rs. 4624 in 2 years and to Rs. 4913 in 3 years. The sum of money is:1. Rs. 40962. Rs. 42603. Rs. 43354. Rs. 4360
Answer» Correct Answer - Option 1 : Rs. 4096 Given: Amount in second year = 4624 Rs. Amount in Third year = 4913 Rs. Formula used \(Amount = P\;{\left( {1 + \frac{r}{{100}}} \right)^t}\) Where P is the principal, r is the rate of interest and t be the time Solution: Accordingly, \(4624 = \;{P\left( {1 + \frac{r}{{100}}} \right)^2}\) ----(i) \(4913 = P\;{\left( {1 + \frac{r}{{100}}} \right)^3}\) ----(ii) Divide eq (ii) by (i) \(\frac{{4913}}{{4624}} = {\left( {1 + \frac{r}{{100}}} \right)^1}\) ----(iii) \(r = \frac{{\left( {4913 - 4624} \right) × 100}}{{4624}} \) r = 6.25% 6.25% = 1/16 from Eq (i) 4624 = P× (17/16)× (17/16) ⇒ P = 4096 ∴ The sum of money is Rs. 4096

A sum of money invested at compound interest amount to Rs. 4624 in 2 years and to Rs. 4913 in 3 years. The sum of money is:1. Rs. 40962. Rs. 42603. Rs. 43354. Rs. 4360

Answer» Correct Answer - Option 1 : Rs. 4096

Given:

Amount in second year = 4624 Rs.

Amount in Third year = 4913 Rs.

Formula used

\(Amount = P\;{\left( {1 + \frac{r}{{100}}} \right)^t}\)

Where P is the principal, r is the rate of interest and t be the time

Solution:

Accordingly,

\(4624 = \;{P\left( {1 + \frac{r}{{100}}} \right)^2}\) ----(i)

\(4913 = P\;{\left( {1 + \frac{r}{{100}}} \right)^3}\) ----(ii)

Divide eq (ii) by (i)

\(\frac{{4913}}{{4624}} = {\left( {1 + \frac{r}{{100}}} \right)^1}\) ----(iii)

\(r = \frac{{\left( {4913 - 4624} \right) × 100}}{{4624}} \)

r = 6.25%

6.25% = 1/16

from Eq (i)

4624 = P× (17/16)× (17/16)

⇒ P = 4096

∴ The sum of money is Rs. 4096

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