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Some important Formulas

June 15, 2020
Category: Some important Formulas

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☼ Some important Formulas :

$\large \mapsto$ A = P

e.g. Find the amount on Rs. 2000 for 3 yrs. at 5 % rate of interest.

Soln. Here, P = Rs. 2000, N = 3 yrs. and R = 5

$\therefore$ A = P

= 2000 $\times \left (1+\frac{5\times3}{100} \right )$ = Rs. 2300

$\large \mapsto$ Difference in I = $\frac{\left (\textup{Diffrence in P} \right ) \times R \times N}{100}$

e.g. There is a capital of Rs. 4000 for 5 years at 8 % rate of interest. Find the increment in an interest, if the principal is increased to Rs. 5000.

Soln. Here, R = 8, N = 5 and Difference of P = 5000 – 4000 = 1000

Difference of I = $\frac{\left (\textup{Diffrence in P} \right ) \times R \times N}{100}$

= $\frac{1000\times8\times 5}{100}$ = Rs. 400

$\large \mapsto$ Difference in I = $\frac{P\times\left ( \textup{Diffrence in R} \right ) \times N}{100}$

e.g. Due to a fall in the rate of interest from 8 $\frac{1}{2}$ % p.a. to 7 % p.a. , a money lender’s yearly income decreased by Rs. 300. His capital is :

Soln. Difference in R = 8 $\frac{1}{2}$ – 7 = 1 $\frac{1}{2}$ = $\frac{3}{2}$

Now,

Difference in I = $\frac{P\times\left ( \textup{Diffrence in R} \right ) \times N}{100}$

$\therefore$ 300 = $\frac{\textup{P}\times 3\times1}{100\times2}$ P = Rs. 20000

$\large \mapsto$ Difference in I = $\frac{P\times R \times \left ( \textup{Diffrence in N} \right )}{100}$

e.g. There is a capital of Rs. 4000 for 5 years at 8 % rate of interest. Find the increment in an interest, if number of years is increased to 7 yrs..

Soln. Difference of N = 7 – 5 = 2 yrs.

Now,

Difference in I = $\frac{P\times R \times \left ( \textup{Diffrence in N} \right )}{100}$

= $\frac{4000\times8\times2}{100}$ = Rs. 640