Question

A man invested Rs.X at 15% p.a. at SI for 4 years and Rs.(1.35X) at 18% p.a. at SI for 3 years. If total interest received by man is Rs.15948, then find value of Rs. (3.12X). (SI: Simple interest)​

Answer 1

$\sf\small\underline\blue{Given:-}$

$\sf{\longrightarrow Total\: interest=Rs.15948}$

$\sf\small\underline\blue{To\: Find:-}$

$\sf{\longrightarrow The\: value\:of\:Rs.3.12x=?}$

$\sf\small\underline\blue{Solution:-}$

We will solve this question by applying formula of simple interest. Then calculate the value of x as per the given clue in the question.

$\tt\small\underline\blue{Calculation\:for\:case\:-(i):-}$

$\tt\small\underline{P=Rs.x\:\:R=15\%\:\:T=4\: years:-}$

$\tt{\longrightarrow SI=\dfrac{P\times\:T\times\:R}{100}}$

$\tt{\longrightarrow SI=\dfrac{x\times\:4\times\:15}{100}}$

$\tt{\longrightarrow SI=\dfrac{60x}{100}---(i)}$

$\tt\small\underline\blue{Calculation\:for\:case\:-(ii):-}$

$\tt\small\underline{P=Rs.1.35x\:\:R=18\%\:\:T=3\: years:-}$

$\tt{\longrightarrow SI=\dfrac{P\times\:T\times\:R}{100}}$

$\tt{\longrightarrow SI=\dfrac{1.35x\times\:3\times\:18}{100}}$

$\tt{\longrightarrow SI=\dfrac{72.9x}{100}---(ii)}$

• Now calculate principal for the man :-

$\tt{\longrightarrow Case-(i) + Case-(ii)=total\: interest}$

$\tt{\longrightarrow \dfrac{60x}{100}+\dfrac{72.9x}{100}=15948}$

$\tt{\longrightarrow \dfrac{60x+72.9x}{100}=15948}$

$\tt{\longrightarrow \dfrac{132.9x}{100}=15948}$

$\tt{\longrightarrow 1.329x=15948}$

$\tt{\longrightarrow x=Rs.12000}$

• Now calculate the value of 3.12x :-

$\tt{\longrightarrow The\: value\:of\:3.12x=12000(3.12)}$

$\tt{\longrightarrow The\: value\:of\:3.12x=37440}$

$\sf\small\underline{Hence,\:\:the\: value\:of\:3.12x=Rs.37440:-}$

Answer 2

★ Step by step explanation :

• Here, we will use the formula of simple interest and then calculate the value of x.

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★ Solving for first case :

$\qquad:\implies\:\sf SI = \dfrac{P\:\times\:T\:\times\:R}{100}$

★ Values that we have :

• P = Rs. x
• T = 4 years
• R = 15%

★ Putting all values :

$\qquad:\implies\:\sf SI = \dfrac{x\:\times\:4\:\times\:15}{100}$

$\qquad:\implies\:\bf {SI = \red{ \dfrac{60x}{100}}}\qquad\qquad \lgroup 1 \rgroup$

★ Solving for second case :

$\qquad:\implies\:\sf SI = \dfrac{P\:\times\:T\:\times\:R}{100}$

★ Values that we have :

• P = 1.35x
• T = 3 years
• R = 18%

★ Putting all values :

$\qquad:\implies\:\sf SI = \dfrac{1.35x\:\times\:3\:\times\:18}{100}$

$\qquad:\implies\:\bf{SI = \red{\dfrac{72.9x}{100}}}\qquad\qquad \lgroup 2 \rgroup$

★ Principal for the man :

$\qquad:\implies\:\sf \lgroup 1 \rgroup + \lgroup 2 \rgroup = Total\:income$

★ Values that we have :

• [1] = 60x/100
• [2] = 72.9x/100
• Total income = Rs. 15,948

★ Putting all values :

$\qquad:\implies\:\sf \dfrac{60x}{100} + \dfrac{72.9x}{100} = 15948$

$\qquad:\implies\:\sf \dfrac{60x + 72.9x}{100} = 15948$

$\qquad:\implies\:\sf \dfrac{132.9x}{100} = 15948$

$\qquad:\implies\:\sf 132.9x = 15948 \:\times\:100$

$\qquad:\implies\:\sf 132.9x = 1594800$

$\qquad:\implies\:\sf x = \dfrac{1594800}{132.9}$

$\qquad:\implies\:\sf x = \dfrac{\cancel{1594800}}{\cancel{132.9}}$

$\qquad:\implies\:\bf{x = \red{Rs.\:12,000}}$

★ Value of 3.12x :

$\:\leadsto\:\sf Value\:of\:3.12x = 3.12(12000)$

$\:\leadsto\:\bf{Value\:of\:3.12x = \red{Rs.\:37,440}}$

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$\therefore\:{\underline{\frak{{Value\:of\:3.12x = Rs.\:37,440}}}}$