The compound interest accrued on an amount of Rs. 8000 at the end of two years is Rs. 1680. What would be the simple interest accrued on the same amount at twice the rate in the half the period?
Solution
Correct Answer: Option D
C.I = Amount – Principal (P)
Amount \(= \;P\;{\left( {1 + \frac{r}{{100}}} \right)^n}\) where n = time period; r = rate % p. a
\(\begin{array}{l} \therefore \;1680\; = \;8000\;{\left( {1\; + \frac{r}{{100}}} \right)^2}\;-\;8000\\ \Rightarrow \frac{{9680}}{{8000}}\; = \;{\left( {1\; + \frac{r}{{100}}} \right)^2}\\ \Rightarrow \frac{{121}}{{100}}\; = \;{\left( {1\; + \frac{r}{{100}}} \right)^2}\\ \Rightarrow \frac{{11}}{{10}}\; = \;1\; + \frac{r}{{100}} \end{array}\)
⇒ r = 10% p. a ---(1)
Now, S.I = Pnr/100
Where, P = principal amount, n = time period; r = rate % p. a
For this case, r = 2 × 10 = 20% p.a.
And n = 2/2 = 1 year
⇒ S.I = (8,000 × 20 × 1)/100 (from 1)
Thus, the simple interest is Rs. 1600
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