Question

Simple Interest= ₹100
Time= 24 months
Amount= ₹1100
Find the Rate? In full Solution
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Answer 1

Answer:

5%

Step-by-step explanation:

amount = 1100/-

SI = 100/-

principal = 1100 - 100/-

= 1000/-

time = 24 months = 2 years

SI = prt / 100

so, r = (SI × 100) / (p × t)

= (100 × 100) / (1000 × 2)

= 5%

Answer 2

➤ Given :-

Simple Interest :- ₹100

Time :- 24 months = 2 years

Total Amount :- ₹1100

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➤ To Find :-

The rate of interest applied to the given sum...

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➤ Formula required :-

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

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★ How to do :-

Here, we are given with the simple interest, the time taken to return the money back and also the total amount that should be returned at end of the period. We are asked to find the rate of interest that should be applied in this sum. So, first we should find the principle amount by subtracting the total amount and the simple interest. The obtained answer will be the principle amount. Later, to find the rate of interest we can use the given formula.

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➤ Solution :-

Principle :-

${\tt \leadsto 1100 - 100}$

${\tt \leadsto Rs \: \: 1000}$

Here, we had found with the principle amount by subtracting the total amount and the simple interest. So, now we should substitute the values of the formula.

${\tt \leadsto \dfrac{100 \times 100}{1000 \times 2}}$

${\tt \leadsto \dfrac{100 \times \cancel{100}}{\cancel{1000} \times 2} = \dfrac{100 \times 1}{10 \times 2}}$

${\tt \leadsto \dfrac{\cancel{100} \times 1}{\cancel{10} \times 2} = \dfrac{10 \times 1}{1 \times 2}}$

${\tt \leadsto \cancel \dfrac{10}{2} = \boxed{\tt 5 \bf\%}}$

$\Huge\therefore$ The rate of interest being paid is 5%.

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$\dashrightarrow$ Some related formulas :-

$\small\boxed{\begin{array}{cc}\large\sf\dag \: {\underline{More \: Formulae}} \\ \\ \bigstar \: \sf{Simple \: Interest :- \: \dfrac{P \times R \times T}{100}} \\ \\ \bigstar \: \sf{Principle :- \: \dfrac{SI \times 100}{R \times T}} \\ \\ \bigstar \: \sf{Time :- \: \dfrac{SI \times 100}{P \times R}}\end{array}}$