Math, asked by bn661293, 2 months ago

find the simple interest in which principle is 2.200 Rs , rate is 3% and time is 6 months...
plz answer this question ​

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Answers

Answered by radhabrijraj
0

Answer:

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Step-by-step explanation:

So here it is given that

P = 2.200

R = 3%

T = 6 months

SI =PRT /100

SI = 2200*3*6/100

SI = 396

Answered by MasterDhruva
6

Given :-

Principle :- ₹2200

Rate of interest :- 3%

Time :- 6 months

\:

To Find :-

Simple Interest to be paid at end of the period.

\:

Formula required :-

{\tt \large \dashrightarrow \orange{\boxed{\tt \gray{\dfrac{P \times R \times T}{100}}}}}

\:

How to do :-

Here, we are given with a principle amount, the rate of interest and the time taken to give back money. We should find the simple interest to be given at end of the period. Here, we can notice that the time taken to give back money is given in the form of months. We know that to calculate the simple interest or the total amount, we always need the time taken in the years format. The other rule is the rate of interest should be in percentage form. So, first we should right the months format in years format by writing it as a fraction part. It can be written as the months given divided by the total number of months in a year. Later, we can find the simple interest by using the given formula. So, let's solve!!

\:

Solution :-

{\tt \leadsto \dfrac{2200 \times 3 \times 6}{100 \times 12}}

{\tt \leadsto \dfrac{2200 \times 3 \times \cancel{6}}{100 \times \cancel{12}} = \dfrac{2200 \times 3 \times 1}{100 \times 2}}

{\tt \leadsto \dfrac{\cancel{2200} \times 3 \times 1}{\cancel{100} \times 2} = \dfrac{22 \times 3}{2}}

{\tt \leadsto \cancel \dfrac{66}{2} = \boxed{\tt Rs \: \: 33}}

\Huge\therefore The simple interest to be paid at end of the period is 33.

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\dashrightarrow Some related formulas :-

{\longrightarrow \boxed{\sf Principle = \dfrac{SI \times 100}{R \times T}}}

{\longrightarrow \boxed{\sf Rate \: of \: Interest = \dfrac{SI \times 100}{P \times T}}}

{\longrightarrow \boxed{\sf Time = \dfrac{SI \times 100}{P \times R}}}

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