The simple interest on a sum of money is 1-9th of the principal and the number of years is equal to the rate

The correct option is A 31/3% I=1/9P, R=T We have I=PTR100 1/9P=P× R× R100 R× R=1009 R× R=10/3×10/3 ∴ R=10/3 ...

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Standard XII

Chemistry

First Order Reaction

The simple in...

Question

The simple interest on a sum of money is $\frac{1}{9} t h$ of the principal and the number of years is equal to the rate % per annum. The rate % per annum is:

3 \frac{1}{3} %

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\frac{1}{3} %

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Solution

The correct option is A $3 \frac{1}{3} %$
$I = \frac{1}{9} P, R = T$
We have
$I = \frac{P T R}{100}$
$\frac{1}{9} P = \frac{P \times R \times R}{100}$
$R \times R = \frac{100}{9}$
$R \times R = \frac{10}{3} \times \frac{10}{3}$
$∴ R = \frac{10}{3}$ % = $3 \frac{1}{3}$ %

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The simple interest on a sum of money is 19th of the principal and the number of years is equal to the rate % per annum. The rate % per annum is:

The correct option is A 313%I=19P,R=TWe haveI=PTR10019P=P×R×R100R×R=1009R×R=103×103∴R=103 % = 313 %

The simple interest on a sum of money is $\frac{1}{9} t h$ of the principal and the number of years is equal to the rate % per annum. The rate % per annum is:

The correct option is A $3 \frac{1}{3} %$
$I = \frac{1}{9} P, R = T$
We have
$I = \frac{P T R}{100}$
$\frac{1}{9} P = \frac{P \times R \times R}{100}$
$R \times R = \frac{100}{9}$
$R \times R = \frac{10}{3} \times \frac{10}{3}$
$∴ R = \frac{10}{3}$ % = $3 \frac{1}{3}$ %