Question

Disha borrowed some money for 2 years at
simple interest. The rate of interest is 11 p.c.p.a.
If she returned 18,300 altogether at the end of
2 years then,
was borrowed buy her.​

Answer 1

Given:

• Time = 2 years
• Rate = 11%
• Amount = 18300

Solution:

Let the Principal be x.

Then SI =  $\sf{\frac{P\times R\times T}{100}}$

$\mathtt{=\frac{x\times11\times2}{100}}\\\\\sf{=\frac{11x}{50}}$

Then, Amount = Principal + SI

$\sf{=x+\frac{11x}{50}}\\\\\mathtt{=\frac{50x+11x}{50}}\\\\\sf{=\frac{61x}{50}}$

But Amount is given 18300.

$\therefore \mathtt{\frac{61x}{50}=18300}\\\\\implies \mathtt{x=18300\times50}\\\\\implies \mathtt{61x=915000}\\\\\implies \mathtt{x=\frac{915000}{61}}\\\\\implies \sf{x=15000}$

Hence, Disha borrowed 15000.

Extra Information:

• Principal is the money borrowed. It is also called sum.
• Interest is the additional money paid by the borrower.
• Amount is total money paid by the borrower (Principal + Interest).
• Formulas used in Simple Interest:

$1. P=\frac{100\times SI}{R\times T}$

$2. R=\frac{100\times SI}{P\times T}$

$3. T=\frac{100\times SI}{R\times P}$

Answer 2

Time, T= 2 years

Rate, R = 11 %

Amount = 18,300

Principal, P = ?

$= > Simple \: Interest = \frac{P \times 2 \times 11}{100}$

$= > Simple Interest = \frac{11P}{50}$

$= > Amount = Simple Interest+ P$

$= > 18300 = \frac{11P}{50} + P$

$= > 18300 = \frac{61P}{50}$

$= > P = 15000$

Hence, she borrowed 15,000.

Hope this answer helps you ^_^ !