Question

Find T=?, If P= ₹600, A= ₹850, R=3%.​

Answer 1

Given :

• Principal = Rs.600
• Amount = Rs.850
• Rate = 3%

To Find :

• Time = ?

Solution :

Analysis :

Here we are given the principal, amount and rate. We don't have interest with us. From amount and principal we can find the interest. And by using the required formula to find the time using interest, principal and rate we will get the required time.

Required Formula :

$\boxed{\bf Time=\dfrac{SI\times100}{P\times R}}$

where,

• SI = Simple Interest
• P = Principal
• R = Rate
• T = Time

Explanation :

First we have to find the Interest.

We know that if we are given the amount and principal and is asked to find the interest then our required formula is,

Interest = Amount - Principal

where,

• Amount = Rs.850
• Principal = Rs.600

Using the required formula and substituting the required formula,

Interest = Amount - Principal

Interest = Rs.(850 - 600)

Interest = Rs.250.

• Simple Interest = Rs.250.

Now the time,

We know that if we are given the principal, simple interest and rate and is asked to find the time then our required formula is,

$\bf Time=\dfrac{SI\times100}{P\times R}$

where,

• SI = Rs.250
• P = Rs.600
• R = 3%

Using the required formula and substituting the required formula,

$\\ :\implies\sf Time=\dfrac{250\times100}{600\times3}$

$\\ :\implies\sf Time=\dfrac{250\times1\!\!\!\!\not{0}\!\!\!\!\not{0}}{6\!\!\!\!\not{0}\!\!\!\!\not{0}\times3}$

$\\ :\implies\sf Time=\dfrac{250\times1}{6\times3}$

$\\ :\implies\sf Time=\dfrac{\cancel{250}\ \ ^{125}}{\not{6}\ \ ^3\times3}$

$\\ :\implies\sf Time=\dfrac{125}{3\times3}$

$\\ :\implies\sf Time=\dfrac{125}{9}$

$\\ :\implies\sf Time=13\dfrac{8}{9}$

$\\ \therefore\boxed{\bf Time=13\dfrac{8}{9}\ years.}$

$\\ \\ \therefore\underline{\sf The\ required\ time\ is\ \bf 13\dfrac{8}{9}\ years.}$

Verification :

LHS :

$\\ \therefore\boxed{\bf250.}$

RHS :

• Taking time as 125/9 years,

$\\ :\implies\sf\dfrac{P\times R\times T}{100}$

$\\ :\implies\sf\dfrac{600\times3\times125}{100\times9}$

$\\ :\implies\sf\dfrac{6\!\!\!\!\not{0}\!\!\!\!\not{0}\times3\times125}{1\!\!\!\not{0}\!\!\!\!\not{0}\times9}$

$\\ :\implies\sf\dfrac{6\times3\times125}{1\times9}$

$\\ :\implies\sf\dfrac{\not{6}\ \ ^2\times3\times125}{\not{9}\ \ ^9}$

$\\ :\implies\sf\dfrac{2\times3\times125}{3}$

$\\ :\implies\dfrac{2\times\not{3}\times125}{\not{3}}$

$\\ :\implies\sf2\times\times125$

$\\ :\implies\sf250$

$\\ \therefore\boxed{\bf250.}$

LHS = RHS.

• Hence verified.