Question

The simple interest. on a certain sum for 4 years at 9% per annum is 216 less than the simple interest
on the same sum for 3 years at 18% per annum. Find the sum.​

Answer 1

$\huge{\underline{\red{\mathfrak{AnsWer:}}}}$

$\implies \bf Sum=Rs.\;1200$

$\huge{\underline{\red{\mathfrak{Step\;By\;Step\;Explanation:}}}}$

$\underline{\bf Given:}\\ \\ \implies \sf S.I._{2}-S.I._{1}=Rs.\;216\\ \\ \\ \underline{\bf To\;Find:}\\ \\ \implies \sf Find\;the\;sum\;(P).\\ \\ \\ \underline{\bf Formula\;Used:}\\ \\ \implies \sf S.I.=\dfrac{PRT}{100}\\ \\ \rule{200}{2}$

$\underline{\sf Let\;sum\;be\;'x',}$

$\underline{\bf Case-1}\\ \\ \implies \sf Rate\;(R) = 9\%\\ \\ \implies \sf Time\;(T)=4\;years\\ \\ \\ \underline{\sf We\;know\;that,}\\ \\ \implies \bf S.I._{1}=\dfrac{PRT}{100}\\ \\ \\ \implies \sf S.I._{1}=\dfrac{x\times 9\times 4}{100}\\ \\ \\ \implies \bf S.I._{1}=\dfrac{36x}{100}\;\;\;\;\;\;...........(1)\\ \\ \rule{200}{2}$

$\underline{\bf Case-2}\\ \\ \implies \sf Rate\;(R) = 18\%\\ \\ \implies \sf Time\;(T)=3\;years\\ \\ \\ \underline{\sf We\;know\;that,}\\ \\ \implies \bf S.I._{2}=\dfrac{PRT}{100}\\ \\ \\ \implies \sf S.I._{2}=\dfrac{x\times 18\times 3}{100}\\ \\ \\ \implies \bf S.I._{2}=\dfrac{54x}{100}\;\;\;\;\;\;...........(2)\\ \\ \rule{200}{2}$

$\underline{\bf Now,\;it\;is\;given\;that,}\\ \\ \implies \bf S.I._{2}-S.I._{1}=216\\ \\ \\ \implies \sf \dfrac{54x}{100}-\dfrac{36x}{100}=216\\ \\ \\ \implies \sf \dfrac{54x-36x}{100}=216\\ \\ \implies \sf \dfrac{18x}{100}=216\\ \\ \\ \implies \sf 18x=21600\\ \\ \\ \implies \sf x=\dfrac{21600}{18}\\ \\ \\ \implies x=Rs.\;1200\\ \\ \rule{200}{2}$

$\bigstar{\large{\boxed{\green{\bf Hence,\;Sum=Rs.\;1200}}}}$

Answer 2

AnswEr :

Rs.1200.

$\bf{\Large{\green{\underline{\underline{\sf{Given\::}}}}}}$

The simple Interest on a certain sum for 4 years at 9% per annum is 216 less than the simple Interest on the same sum for 3 years at 18% per annum.

$\bf{\Large{\red{\underline{\underline{\sf{To\:find\::}}}}}}$

The sum.

$\bf{\Large{\underline{\underline{\tt{\blue{Explanation\::}}}}}}$

Let the sum be R

We know that formula of the Simple Interest :

$\bf{\large{\boxed{\sf{Simple\:Interest\:(S.I.)=\frac{P*R*T}{100} }}}}}}$

We have;

• Principal = Rs. R
• Rate = 9%
• Time = 4 years

$\dashrightarrow\tt{(S.I.)_{1}=\dfrac{P*R*T}{100} }\\\\\\\\\dashrightarrow\tt{(S.I.)_{1}=\dfrac{R*9*4}{100} }\\\\\\\\\dashrightarrow\tt{\green{(S.I.)_{1}=\dfrac{36R}{100} }}$

Now, same sum for 3 years at 18% per annum.

$\dashrightarrow\tt{(S.I)_{2}=\dfrac{P*R*T}{100} }\\\\\\\\\dashrightarrow\tt{(S.I)_{2}=\dfrac{R*18*3}{100} }\\\\\\\\\dashrightarrow\tt{\green{(S.I)_{2}=\dfrac{54R}{100} }}$

$\bf{\large{\underline{\underline{\tt{\red{A.T.Q\::}}}}}}$

$\longrightarrow\tt{(S.I)_{2}-(S.I)_{1}=216}\\\\\\\\\longrightarrow\tt{\dfrac{54R}{100} -\dfrac{36R}{100} =216}\\\\\\\\\longrightarrow\tt{\dfrac{54R-36R}{100} =216}\\\\\\\\\longrightarrow\tt{\dfrac{18P}{100} =216}\\\\\\\\\longrightarrow\tt{18R\:=\:216\times 100}\\\\\\\\\longrightarrow\tt{18R\:=\:21600}\\\\\\\\\longrightarrow\tt{R\:=\:\cancel{\dfrac{21600}{18} }}\\\\\\\\\longrightarrow\tt{\red{R\:=\:Rs.1200}}$

∴ The sum is Rs.1200 .