Question:

In a factory, the production of scooters rose to 48400 from 40000 in 2 years. The rate of growth per annum is

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Growth compounds year on year, so use A = P(1 + r/100)^2. The ratio 48400/40000 = 1.21, and 1.21 is the square of 1.1.
Updated On: Jul 17, 2026
  • 20%
  • 10%
  • 30%
  • 8%
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The Correct Option is B

Solution and Explanation

Step 1: Recognise the type of growth.
When production grows at the same rate every year, each year's growth applies to the previous year's output, not to the original one. That is compound growth, so the compound interest formula applies.
\[ A = P\left(1 + \frac{r}{100}\right)^{n} \]
Here \( P = 40000 \) is the starting production, \( A = 48400 \) is the production after \( n = 2 \) years, and \( r \) is the rate we want.

Step 2: Substitute the values.
\[ 48400 = 40000\left(1 + \frac{r}{100}\right)^{2} \]

Step 3: Isolate the bracket.
\[ \left(1 + \frac{r}{100}\right)^{2} = \frac{48400}{40000} = \frac{484}{400} \]
Both 484 and 400 are perfect squares, which is a strong hint that the numbers were built for this.
\[ \frac{484}{400} = \frac{22^{2}}{20^{2}} = \left(\frac{22}{20}\right)^{2} \]

Step 4: Take the square root.
Production is positive, so we keep only the positive root:
\[ 1 + \frac{r}{100} = \frac{22}{20} = 1.1 \]
\[ \frac{r}{100} = 0.1 \]
\[ r = 10 \]

Step 5: Check year by year.
Year 1: \( 40000 \times 1.1 = 44000 \).
Year 2: \( 44000 \times 1.1 = 48400 \). This is exactly the given figure.

Step 6: Why the other options are wrong.
At 20%, production would reach \( 40000 \times 1.44 = 57600 \), far too high.
At 30%, it would reach \( 40000 \times 1.69 = 67600 \), higher still.
At 8%, it would reach \( 40000 \times 1.1664 = 46656 \), short of 48400.
A tempting shortcut is total growth of 8400 on 40000, which is 21% over two years, then halving it to 10.5%. That simple average is wrong because growth compounds, but it does land close to 10 and confirms the answer.

Final Answer:
The rate of growth is 10% per annum. \[ \boxed{10\%} \]
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