
What sum of money will amount to INR 2832 at $ 9\dfrac{1}{2}\% $ per annum in 5 years?
Answer
569.7k+ views
Hint: In this question, we need to determine the principal sum of money that will amount to INR 2832 in the time period of 5 years with $ 9\dfrac{1}{2}\% $ as the rate of interest. For this, we will use the relation between the amount, principal amount, rate of interest and time period.
Complete step-by-step answer:
Here, the rate of interest has been given as in mixed fraction, so first, we will convert the mixed fraction to the proper fraction by following $ a\dfrac{b}{c} = \dfrac{{ac + b}}{c} $ .
Here, a=9, b=1 and c=2 so,
$
9\dfrac{1}{2} = \dfrac{{9 \times 2 + 1}}{2} \\
= \dfrac{{19}}{2}\% \\
$
Now, the product of the principal amount, rate of interest (in decimal equivalent) and the time period result in the interest on the amount that one will get after the specified time period. Mathematically, $ I = \dfrac{{prt}}{{100}} $ where, I is the interest gained, p is the principal amount, r is the rate of interest and t is the time period.
So, substitute $ r = \dfrac{{19}}{2}\% $ and t=5 years in the formula $ I = \dfrac{{prt}}{{100}} $ to determine the expression for the interest gained in terms of the principal amount.
$
\Rightarrow I = \dfrac{{prt}}{{100}} \\
= \dfrac{{p\left( {\dfrac{{19}}{2}} \right)5}}{{100}} \\
= \dfrac{{95p}}{{200}} \\
= \dfrac{{19p}}{{40}} \\
$
The amount is the summation of the principal amount and the interest gained on the principal amount.
According to the question, the amount $ (A = p + I) $ is INR 2832. So, substitute $ I = \dfrac{{19p}}{{40}} $ in the equation $ (A = p + I) $ to determine the principal amount.
$
\Rightarrow A = p + I \\
\Rightarrow 2832 = p + \dfrac{{19p}}{{40}} \\
\Rightarrow 2832 = \dfrac{{40p + 19p}}{{40}} \\
\Rightarrow 69p = 2832 \times 40 \\
\Rightarrow p = \dfrac{{2832 \times 40}}{{69}} \\
= 1641.7 \\
$
Hence, the INR 1641.7 sum of money amounts to INR 2832 in 5 years at $ 9\dfrac{1}{2}\% $ the rate of interest.
Note: Students should know how to convert mixed fraction to proper fraction. Moreover, whenever the question is not indicating the type of interest (simple or compound) then, we should by default always consider for the simple interest only.
Complete step-by-step answer:
Here, the rate of interest has been given as in mixed fraction, so first, we will convert the mixed fraction to the proper fraction by following $ a\dfrac{b}{c} = \dfrac{{ac + b}}{c} $ .
Here, a=9, b=1 and c=2 so,
$
9\dfrac{1}{2} = \dfrac{{9 \times 2 + 1}}{2} \\
= \dfrac{{19}}{2}\% \\
$
Now, the product of the principal amount, rate of interest (in decimal equivalent) and the time period result in the interest on the amount that one will get after the specified time period. Mathematically, $ I = \dfrac{{prt}}{{100}} $ where, I is the interest gained, p is the principal amount, r is the rate of interest and t is the time period.
So, substitute $ r = \dfrac{{19}}{2}\% $ and t=5 years in the formula $ I = \dfrac{{prt}}{{100}} $ to determine the expression for the interest gained in terms of the principal amount.
$
\Rightarrow I = \dfrac{{prt}}{{100}} \\
= \dfrac{{p\left( {\dfrac{{19}}{2}} \right)5}}{{100}} \\
= \dfrac{{95p}}{{200}} \\
= \dfrac{{19p}}{{40}} \\
$
The amount is the summation of the principal amount and the interest gained on the principal amount.
According to the question, the amount $ (A = p + I) $ is INR 2832. So, substitute $ I = \dfrac{{19p}}{{40}} $ in the equation $ (A = p + I) $ to determine the principal amount.
$
\Rightarrow A = p + I \\
\Rightarrow 2832 = p + \dfrac{{19p}}{{40}} \\
\Rightarrow 2832 = \dfrac{{40p + 19p}}{{40}} \\
\Rightarrow 69p = 2832 \times 40 \\
\Rightarrow p = \dfrac{{2832 \times 40}}{{69}} \\
= 1641.7 \\
$
Hence, the INR 1641.7 sum of money amounts to INR 2832 in 5 years at $ 9\dfrac{1}{2}\% $ the rate of interest.
Note: Students should know how to convert mixed fraction to proper fraction. Moreover, whenever the question is not indicating the type of interest (simple or compound) then, we should by default always consider for the simple interest only.
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