Question

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

• A. 20%
• B. 10%
• C. 30%
• D. 8%

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Work with the growth factor instead of the rate.
Call the yearly multiplier $k$, so production after one year is $40000k$ and after two years it is $40000k^{2}$.
This avoids fractions until the very last step.

Step 2: Set up and simplify.
\[ 40000k^{2} = 48400 \]
\[ k^{2} = \frac{48400}{40000} = 1.21 \]

Step 3: Take the root.
$1.21$ is a well known square, since $1.1 \times 1.1 = 1.21$.
\[ k = 1.1 \]
A multiplier of 1.1 means a 10% rise each year.

Step 4: Cross check by testing options.
Instead of solving, you could square each candidate multiplier and multiply by 40000:
20% gives $1.2^{2} = 1.44$, so $57600$.
10% gives $1.1^{2} = 1.21$, so $48400$. This matches.
30% gives $1.3^{2} = 1.69$, so $67600$.
8% gives $1.08^{2} = 1.1664$, so $46656$.
Only the 10% row hits the target exactly, so option (B) stands alone.

Step 5: Note on the two year jump.
Total rise is $\dfrac{8400}{40000} \times 100 = 21\%$ across two years. Under compounding, two years at $r\%$ give a total rise of $2r + \dfrac{r^{2}}{100}$ percent. Putting $r = 10$ gives $20 + 1 = 21$, matching perfectly.

Final Answer:
Production grew at 10% each year. \[ \boxed{10\%} \]