Question

A car manufacturer increased his productions of cars from 80,000 to 92,610 in 3 years. Find the rate of growth of production.
A) 3%
B) 4%
C) 2%
D) 5%

Answer 1

Hint: The growth of production varies each year and the increased growth gets added to previous years base, so to calculate the rate of growth in 3 years we can use the formula for the amount of compound interest where the principal will be the initial and amount will be the final value.
 Formula to be used:
 $ A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t} $ where,
A = Amount
P = Principal
R = Rate
t = Time

Complete step-by-step answer:
We can calculate the rate of growth of production using the formula for amount used in case of compound interest because the increase in growth varies each year and gets added to the previous base.
Here,
Amount (A) = 92,610
Principal (P) = 80,000
Time (t) = 3 years
Rate (R) = ?
Applying the formula of amount and substituting the values:
 $\Rightarrow A = P{\left( {1 + \dfrac{R}{{100}}} \right)^t} $
\[
\Rightarrow 92610 = 80000{\left( {1 + \dfrac{R}{{100}}} \right)^3} \\
\Rightarrow \dfrac{{92610}}{{80000}} = {\left( {1 + \dfrac{R}{{100}}} \right)^3} \\
\Rightarrow {\left( {1 + \dfrac{R}{{100}}} \right)^3} = \dfrac{{21 \times 21 \times 21}}{{20 \times 20 \times 20}} \\
\Rightarrow {\left( {1 + \dfrac{R}{{100}}} \right)^3} = {\left( {\dfrac{{21}}{{20}}} \right)^3} \\
 \]
Taking cube root on both the sides, we get:
 $
\Rightarrow 1 + \dfrac{R}{{100}} = \dfrac{{21}}{{20}} \\
\Rightarrow \dfrac{R}{{100}} = \dfrac{{21}}{{20}} - 1 \\
\Rightarrow \dfrac{R}{{100}} = \dfrac{1}{{20}} \\
\Rightarrow R = \dfrac{{100}}{{20}} \\
\Rightarrow R = 5%$

So, the correct answer is “Option D”.

Note: We have used the formula for the amount of compound interest because the value that increases each year gets added in the principal for the next subsequent years and goes on.
Compound interest is generally used in finance sectors, but other areas of its application are:
Increase / decrease in population, the growth of bacteria, rise / depreciation of quantities.