Question

Angela deposited 15000 rupees in a bank at a rate of 9 p.c.p.a. She got simple interest
amounting to 5400 rupees. For how many years had she deposited the amount?
​

Answer 1

$\begin{gathered}\sf Given \: that \begin{cases} & \sf{Simple \: interest \: = \bf{Rs \: 5400}} \\ & \sf{Rate \: of \: interest \: = \bf{9\:\% \: per \: annum}} \\ & \sf{Principal \: = \bf{Rs \: 15000}} \end{cases} \end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\bf{Time \: in \: years}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

Understanding the concept :-

Principal: The principal is the amount that initially borrowed from the bank or invested. The principal is denoted by P.

Rate: Rate is the rate of interest at which the principal amount is given to someone for a certain time. The rate of interest is denoted by R.

Time: Time is the duration for which the principal amount is given to someone. Time is denoted by T.

Amount: When a person takes a loan from a bank, he/she has to return the principal borrowed plus the interest amount, and this total returned is called Amount.

Amount = Principal + Simple Interest

${\bf{\ Simple \: interest \: = \: \dfrac{P \times R \times T}{100}}}$

where,

• P denotes Principal

• R denotes Rate

• T denotes Time

$\large\underline{\bold{Solution-}}$

Given that,

• Principal, P = Rs 15, 000

• Rate of interest, R = 9 % per annum

• Simple Interest = Rs 5400

Let time period, for which Principal is invested is 't' years.

On substituting all the values in formula of Simple interest,

${\sf{Simple \: interest \: = \: \dfrac{P \times R \times T}{100}}}$

$\: \sf \: 5400 = \dfrac{150 \: \: \cancel{00} \times 9 \times t}{ \cancel{100}}$

$\bf \: t = \: \sf \: \dfrac{5400}{150 \times 9} = 4 \: years$

So,

• Required time is 4 years.