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Question: The difference between CI and SI for \(3\) years is \(992\) . if the rate of interest is \(10\%\) fi...

The difference between CI and SI for 33 years is 992992 . if the rate of interest is 10%10\% find the principal?

Explanation

Solution

From the question given that we have to find the principal, given that the difference between CI and SI for 33 years is 992992, and the rate of interest is 10%10\%. As we know that the formula of simple interest is S.I=principal×rate×time100S.I=\dfrac{principal\times rate\times time}{100} and the formula for compound interest is C.I=principal×(1+rate)time−principalC.I=principal\times {{\left( 1+rate \right)}^{time}}-principal, from these we will get the principal.

Complete step by step solution:
From the question given the difference between CI and SI is
⇒C.I−S.I=992\Rightarrow C.I-S.I=992
For a time period is,
⇒T=3\Rightarrow T=3
And also, the rate of interest is,
⇒R=10%\Rightarrow R=10\%
Let the principal is P
As we know that the formula for the simple interest is
⇒S.I=principal×rate×time100\Rightarrow S.I=\dfrac{principal\times rate\times time}{100}
now by substituting the values in their respective positions we will get,
⇒S.I=P×10×3100=3P10\Rightarrow S.I=\dfrac{P\times 10\times 3}{100}=\dfrac{3P}{10}
As we know that the formula for the compound interest is
⇒C.I=principal×(1+rate)time−principal=principal((1+rate)time−1)\Rightarrow C.I=principal\times {{\left( 1+rate \right)}^{time}}-principal=principal\left( {{\left( 1+rate \right)}^{time}}-1 \right)
now by substituting the values in their respective positions we will get,
⇒C.I=P((1+110)3−1)\Rightarrow C.I=P\left( {{\left( 1+\dfrac{1}{10} \right)}^{3}}-1 \right)
As we know that the difference between the simple interest and compound interest is
⇒C.I−S.I=992\Rightarrow C.I-S.I=992
now by substituting the values in their respective positions we will get,
⇒C.I−S.I=P((1+110)3−1)−3P10=992\Rightarrow C.I-S.I=P\left( {{\left( 1+\dfrac{1}{10} \right)}^{3}}-1 \right)-\dfrac{3P}{10}=992
Now we will take P common from the all the terms, then we will get,
⇒P((1+110)3−1−310)=992\Rightarrow P\left( {{\left( 1+\dfrac{1}{10} \right)}^{3}}-1-\dfrac{3}{10} \right)=992
Now by further simplification we will get,
⇒P((1110)3−1−310)=992\Rightarrow P\left( {{\left( \dfrac{11}{10} \right)}^{3}}-1-\dfrac{3}{10} \right)=992
Now by further simplification we will get,
⇒P(1331−1000−3001000)=992\Rightarrow P\left( \dfrac{1331-1000-300}{1000} \right)=992
Now by further simplification we will get,
⇒P(311000)=992\Rightarrow P\left( \dfrac{31}{1000} \right)=992
Now by further simplification we will get,
⇒P=32000\Rightarrow P=32000
Therefore, the principal is 3200032000

Note: Students should know the formulas of simple interest and compound interest, students should not write rate in simple interest as R=110R=\dfrac{1}{10} because already the percentage is included in the formula, so students should write rate as R=10R=10.