Question

The cost of a cellphone is Rs.8000 and the down payment is Rs.1000. The balance amount is to be paid in 8 equal installments of Rs.1000 each. Find the rate of interest.

Answer 1

Hint: In the above question, first of all we will find the excess amount paid for the cellphone and then, we will use the formula, to find the rate of interest which is as below,
\[R=\dfrac{2400E}{n\left( \left( n+1 \right)I-2E \right)}\]
Where, E is the excess amount, I is the installment amount and 'n' is the number of installment.

Complete step-by-step answer:
We have been given, the cost of a cellphone is Rs.8000 and the down payment is Rs.1000. Also, the balance amount is to be paid in 8 equal installments of Rs.1000 each. So, we have to find the rate of interest.
Now, the balance amount (P) \[\Rightarrow 8000-1000=Rs.7000\]
Number of installment (n) \[\Rightarrow 8\]
Installment amount (I) \[\Rightarrow Rs.1000\]
We know that excess amount paid is equal to the difference of total installment amount and the balance amount.
\[\begin{align}
  & \Rightarrow \text{Excess amount(E)=nI-P} \\
 & \Rightarrow 8\times 1000-7000 \\
 & \Rightarrow 8000-7000 \\
 & \Rightarrow Rs.\text{ }1000 \\
\end{align}\]
We know, the rate of interest is given by,
\[R=\dfrac{2400E}{n\left( \left( n+1 \right)I-2E \right)}\]
Where, E is the excess amount, I is the installment amount and 'n' is the number of installment.
\[\begin{align}
  & \Rightarrow R=\dfrac{2400\times 1000}{8\left( \left( 8+1 \right)\times 1000-2\times 1000 \right)} \\
 & \Rightarrow R=\dfrac{2400\times 1000}{8\left( 9000-2000 \right)} \\
 & \Rightarrow R=\dfrac{2400\times 1000}{8\times 7000} \\
 & \Rightarrow R=\dfrac{300}{7} \\
 & \Rightarrow R=42.9\%\left( \text{approximately} \right) \\
\end{align}\]
Therefore, the rate of interest is equal to 42.9% approximately.

Note: Remember that, the formula to find the rate of interest is already multiplied by 100. So, do not multiply after the calculation to get the answer in percentage.
Also, while using the formula of rate of interest, sometimes we just substitute E=100 instead of 1000 by mistake, which gives us a wrong answer.