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ML Aggarwal Solutions Class 9 Mathematics Solutions for Compound Interest Exercise 2.2 in Chapter 2 - Compound Interest
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A certain sum amounts to ₹ 5292 in 2 years and to ₹ 5556.60 in 3 years at compound interest. Find the
rate and the sum.
It is given that
Amount after 2 years = ₹ 5292
Amount after 3 years = ₹ 5556.60
So the difference = 5556.60 – 5292 = ₹ 264.60
Here ₹ 264.60 is the interest on ₹ 5292 for one year
We know that
Rate % = (SI × 100)/ (P × t)
Substituting the values
= (264.60 × 100)/ (5292 × 1)
Multiply and divide by 100
= (26460 × 100)/ (100 × 5292)
= 5%
Here
\begin{aligned} &\mathrm{A}=\mathrm{P}(1+\mathrm{r} / 100)^{\mathrm{n}}\\ &\text { Substituting the values }\\ &5292=\mathrm{P}(1+5 / 100)^{2} \end{aligned}
By further calculation
\begin{aligned} &\mathrm{P}=5292 \div(1+5 / 100)^{2}\\ &\text { So we get }\\ &\mathrm{P}=5292 \div(21 / 20)^{2} \end{aligned}
P = 5292 × 21/20 × 21/20
P = ₹ 4800
Hence, the rate is 5% and the sum is ₹ 4800.
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