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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. A person borrowed Rs. 10000 on compound interest at the rate of 40 percent per annum. If the interest is compounded half yearly, then what will be the amount to be paid after 1.5 years?

  2. The difference between the compound interest (compounding annually) and the simple interest on a sum of money at the rate of 40 per cent per annum for 2 years is Rs. 2400. What is the amount?

  3. In how many years will a sum of Rs.1875 amount to Rs.2187 at 8 percent p.a. compound interest?

  4. A sum of money has increased by 45% in 9 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

  5. At a certain rate of compound interest a certain sum amounts to Rs. 64800 in 4 years and Rs. 93312 in 6 years. What is the compound interest earned in fifth year?

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