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Question

At what rate of interest per annum will a sum of ₹8,000 amount to ₹15,625 in 1 year and 6 months, if the interest is compounded half-yearly?

The correct answer is
50%

This problem requires us to find the annual rate of interest ('R') given the principal amount ('P'), the final amount ('A'), the time period ('t'), and the compounding frequency (half-yearly).

Compound Interest Problem Setup

We are given the following information:

  • Principal Amount (P) = ₹8,000
  • Final Amount (A) = ₹15,625
  • Time Period (t) = 1 year and 6 months = 1.5 years
  • Interest is compounded half-yearly.

We need to find the Rate of Interest (R) per annum.

Calculating the Interest Rate

Since the interest is compounded half-yearly, we need to adjust the time period and the rate of interest accordingly.

  • Number of compounding periods (n): The time period is 1.5 years. Since compounding occurs twice a year (half-yearly), the total number of periods is $n = \text{Time in years} \times 2$. $n = 1.5 \times 2 = 3$ periods.
  • Rate of interest per period (r): The annual rate is R. For half-yearly compounding, the rate per period is half of the annual rate, so $r = \frac{R}{2}$. However, in the compound interest formula, we usually use the rate per period directly as a decimal. So, the rate per period is $\frac{R/2}{100} = \frac{R}{200}$.

The formula for the final amount (A) with compound interest is:

$$ A = P \left(1 + \frac{r}{100}\right)^n $$

Substituting the rate per period (R/2) into the formula, we get:

$$ A = P \left(1 + \frac{R}{2 \times 100}\right)^n $$

$$ A = P \left(1 + \frac{R}{200}\right)^n $$

Now, let's plug in the given values:

$$ 15625 = 8000 \left(1 + \frac{R}{200}\right)^3 $$

To solve for R, we first isolate the term containing R:

$$ \frac{15625}{8000} = \left(1 + \frac{R}{200}\right)^3 $$

Let's simplify the fraction $\frac{15625}{8000}$:

  • Divide both numerator and denominator by 5: $\frac{3125}{1600}$
  • Divide by 5 again: $\frac{625}{320}$
  • Divide by 5 again: $\frac{125}{64}$

So, the equation becomes:

$$ \frac{125}{64} = \left(1 + \frac{R}{200}\right)^3 $$

We need to find the cube root of both sides. We know that $5^3 = 125$ and $4^3 = 64$. Therefore:

$$ \left(\frac{5}{4}\right)^3 = \left(1 + \frac{R}{200}\right)^3 $$

Now, we can equate the bases:

$$ \frac{5}{4} = 1 + \frac{R}{200} $$

Determining the Annual Rate

Let's solve for R:

  • Subtract 1 from both sides: $$ \frac{5}{4} - 1 = \frac{R}{200} $$
  • Simplify the left side: $$ \frac{5 - 4}{4} = \frac{R}{200} $$ $$ \frac{1}{4} = \frac{R}{200} $$
  • Multiply both sides by 200 to find R: $$ R = \frac{1}{4} \times 200 $$ $$ R = 50 $$

The rate of interest is 50% per annum.

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Important Questions from Compound Interest

  1. The certain sum amounts to Rs. 9,982.50 in \(2\frac{1}{2}\)  years at 12% p.a., interest compounded 10-monthly. The sum (in Rs.) is:

  2. The difference between the simple interest and the compound interest compounded annually on a certain sum of money for 2 years at a rate of 8% per annum is Rs. 16.80. Find the principle amount. 

  3. If a sum of ₹ 2000 is lent at 10% p.a. compound interest, what is the interest for the second year?

  4. A sum becomes 5 times of itself in 3 years. at compound interest (interest is compounded annually). In how many years. will the sum becomes 125 times of itself?

  5. If the compound interest on a certain sum of money for two years at 9% p.a. is Rs. 3,762, then the sum is:

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