Mahibur paid an interest of ₹750 for 2 years on a sum of ₹4500. If the rate of interest is $x \frac{1}{3}$ , then find the value of ' $x$ '.

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Solution

Simple Interest= $\frac{P R T}{100}$
Principle, P=4500.
Rate of Interest, R= $x \frac{1}{3}$ .
Time, T=2.

750 = $\frac{4500 \times R \times 2}{100}$

R = $\frac{25}{3}$

R = 8 $\frac{1}{3}$

comparing 8 $\frac{1}{3}$ with $x \frac{1}{3}$
$x$ = 8

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Mahibur paid an interest of ₹750 for 2 years on a sum of ₹4500. If the rate of interest is x13, then find the value of 'x'.

Simple Interest = PRT100 Principle, P = 4500. Rate of Interest, R = x13. Time, T = 2. 750 = 4500×R×2100 R = 253 R = 813 comparing 813 with x13 x = 8

Mahibur paid an interest of ₹750 for 2 years on a sum of ₹4500. If the rate of interest is $x \frac{1}{3}$ , then find the value of ' $x$ '.

Simple Interest= $\frac{P R T}{100}$
Principle, P=4500.
Rate of Interest, R= $x \frac{1}{3}$ .
Time, T=2.

750 = $\frac{4500 \times R \times 2}{100}$

R = $\frac{25}{3}$

R = 8 $\frac{1}{3}$

comparing 8 $\frac{1}{3}$ with $x \frac{1}{3}$
$x$ = 8