If $R s .3, 750$ amount to $R s .4, 620$ in $3$ years at simple interest, Find:
The amount of $R s .7, 500$ in $5 \frac{1}{2}$ years at the same rate of interest.

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Solution

Let $R$ be the rate of interest.
Interest $= \frac{P N R}{100}$
$\Rightarrow (4620 - 3750) = \frac{3750 \times R \times 3}{100}$
$\Rightarrow 87000 = 3750 \times 3 \times R$

$\Rightarrow R = \frac{87000}{3750 \times 3}$

Now, for principle $= 7500$ and time $= 5 \frac{1}{2} = \frac{11}{2}$ interest is,
$I = \frac{7500 \times 87000}{100 \times 3750 \times 3} \times \frac{11}{2} = 3190$
$∴$ $A = 7500 + 3190 = 10690$

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If Rs.3,750 amount to Rs.4,620 in 3 years at simple interest, Find:The amount of Rs.7,500 in 512 years at the same rate of interest.

Let R be the rate of interest.Interest =PNR100⇒(4620−3750)=3750×R×3100⇒87000=3750×3×R ⇒R=870003750×3Now, for principle =7500 and time =512=112 interest is,I=7500×87000100×3750×3×112=3190∴ A=7500+3190=10690

If $R s .3, 750$ amount to $R s .4, 620$ in $3$ years at simple interest, Find:
The amount of $R s .7, 500$ in $5 \frac{1}{2}$ years at the same rate of interest.