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Comparing Quantities — MATHS STD 8 — Question

Gujarat BoardEnglish MediumSTD 8MATHSComparing Quantities2 Marks
Question
Using the formula $\text A=\text P\left(1+\frac{\text R}{100}\right)^\text n,$ calculate the amount and the compound interest for the following
(i) $\text{P}=$ ₹ $500, \text R=12 \%$ and $\text n=2 \text{ yr}.$
(ii) $\text P=$ ₹ $1000, \text R=10 \%$ and $\text{n}=2\text{ yr}.$
✓

Answer

(i) $\because A=P\left(1+\frac{R}{100}\right)^n$
$\therefore A=500\left(1+\frac{12}{100}\right)^2$
$=500\left(\frac{112}{100}\right)^2$
$=500 \times \frac{12544}{10000}=$ ₹ $627.2$
$\therefore C I=A-P=$ ₹ $(627.2-500)=$ ₹ $127.2$
(ii) $\because A=P\left(1+\frac{R}{100}\right)^n$
$\therefore A=1000\left(1+\frac{10}{100}\right)^2$
$=1000 \times\left(\frac{110}{100}\right)^2$
$=1000 \times \frac{121}{100}=1210$
$\therefore CI =A-P=$ ₹ $(1210-1000)=$ ₹ $210$

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