
A man invested one-third of his capital at 7%; one fourth at 8% and the remainder at 10%. If his annual income is Rs.5610, the capital is
\[\begin{align}
& A)Rs.54000 \\
& B)Rs.60000 \\
& C)Rs.66000 \\
& D)Rs.72000 \\
\end{align}\]
Answer
485.7k+ views
Hint: We know that if a man invested \[{{\left( \dfrac{1}{{{n}_{1}}} \right)}^{th}}\]of his capital at \[{{x}_{1}}\%\]; \[{{\left( \dfrac{1}{{{n}_{2}}} \right)}^{th}}\] at \[{{x}_{2}}\%\],…….. \[{{\left( \dfrac{1}{{{n}_{k}}} \right)}^{th}}\] of his capital at \[{{x}_{k}}\%\], the capital is equal to y and the annual income is equal to A, then \[A=\sum\limits_{i=1}^{k}{\left( \dfrac{y}{{{n}_{i}}} \right){{x}_{i}}}\]. By using this formula, we can find the value of capital.
Complete step-by-step solution:
Before solving the problem, we should know that if a man invested \[{{\left( \dfrac{1}{{{n}_{1}}} \right)}^{th}}\]of his capital at \[{{x}_{1}}\%\]; \[{{\left( \dfrac{1}{{{n}_{2}}} \right)}^{th}}\] at \[{{x}_{2}}\%\],…….. \[{{\left( \dfrac{1}{{{n}_{k}}} \right)}^{th}}\] of his capital at \[{{x}_{k}}\%\], the capital is equal to y and the annual income is equal to A, then \[A=\sum\limits_{i=1}^{k}{\left( \dfrac{y}{{{n}_{i}}} \right){{x}_{i}}}\]. By using this formula, we can find the value of capital.
Let us assume the capital is equal to x. From the question, it is clear that a man invested one-third of his capital at 7%; one fourth at 8%, and the remainder at 10%. We are also given that the annual income is equal to 5610.
\[\begin{align}
& \Rightarrow \dfrac{7}{100}\times \dfrac{x}{3}+\dfrac{8}{100}\times \dfrac{x}{4}+\left( x-\dfrac{x}{3}-\dfrac{x}{4} \right)\left( \dfrac{10}{100} \right)=5610 \\
& \Rightarrow \dfrac{7}{100}\times \dfrac{x}{3}+\dfrac{8}{100}\times \dfrac{x}{4}+\dfrac{5x}{12}\times \dfrac{10}{100}=5610 \\
& \Rightarrow \dfrac{7x}{300}+\dfrac{2x}{100}+\dfrac{5x}{120}=5610 \\
& \Rightarrow \dfrac{14x}{600}+\dfrac{12x}{600}+\dfrac{25x}{600}=5610 \\
& \Rightarrow \dfrac{51x}{600}=5610 \\
& \Rightarrow 51x=5610\times 600 \\
& \Rightarrow x=\dfrac{5610\times 600}{51} \\
& \Rightarrow x=66000.....(1) \\
\end{align}\]
From equation (1), it is clear that the capital amount is equal to 66000.
Hence, option C is correct.
Note: Students may do the solution as follows:
Let us assume the capital is equal to x. From the question, it is clear that a man invested one-third of his capital at 7%; one fourth at 8%, and the remainder at 10%. We are also have given that the annual income is equal to 5610.
\[\begin{align}
& \Rightarrow \dfrac{7}{100}\times \dfrac{5610}{3}+\dfrac{8}{100}\times \dfrac{5610}{4}+\left( 5610-\dfrac{5610}{3}-\dfrac{5610}{4} \right)\left( \dfrac{10}{100} \right)=x \\
& \Rightarrow \dfrac{7}{100}\times \dfrac{5610}{3}+\dfrac{8}{100}\times \dfrac{5610}{4}+\dfrac{5\left( 5610 \right)}{12}\times \dfrac{10}{100}=x \\
& \Rightarrow \dfrac{7(5610)}{300}+\dfrac{2(5610)}{100}+\dfrac{5(5610)}{120}=x \\
& \Rightarrow \dfrac{14(5610)}{600}+\dfrac{12(5610)}{600}+\dfrac{25(5610)}{600}=x \\
& \Rightarrow \dfrac{51(5610)}{600}=x \\
& \Rightarrow 51(5610)=x\times 600 \\
& \Rightarrow x=\dfrac{5610\times 51}{600} \\
& \Rightarrow x=476.5....(1) \\
\end{align}\]
From equation (1), it is clear that the capital amount is equal to 476.5. But we know the capital amount is equal to 66000. So, this misconception should be avoided by a student to get the wrong answer.
Complete step-by-step solution:
Before solving the problem, we should know that if a man invested \[{{\left( \dfrac{1}{{{n}_{1}}} \right)}^{th}}\]of his capital at \[{{x}_{1}}\%\]; \[{{\left( \dfrac{1}{{{n}_{2}}} \right)}^{th}}\] at \[{{x}_{2}}\%\],…….. \[{{\left( \dfrac{1}{{{n}_{k}}} \right)}^{th}}\] of his capital at \[{{x}_{k}}\%\], the capital is equal to y and the annual income is equal to A, then \[A=\sum\limits_{i=1}^{k}{\left( \dfrac{y}{{{n}_{i}}} \right){{x}_{i}}}\]. By using this formula, we can find the value of capital.
Let us assume the capital is equal to x. From the question, it is clear that a man invested one-third of his capital at 7%; one fourth at 8%, and the remainder at 10%. We are also given that the annual income is equal to 5610.
\[\begin{align}
& \Rightarrow \dfrac{7}{100}\times \dfrac{x}{3}+\dfrac{8}{100}\times \dfrac{x}{4}+\left( x-\dfrac{x}{3}-\dfrac{x}{4} \right)\left( \dfrac{10}{100} \right)=5610 \\
& \Rightarrow \dfrac{7}{100}\times \dfrac{x}{3}+\dfrac{8}{100}\times \dfrac{x}{4}+\dfrac{5x}{12}\times \dfrac{10}{100}=5610 \\
& \Rightarrow \dfrac{7x}{300}+\dfrac{2x}{100}+\dfrac{5x}{120}=5610 \\
& \Rightarrow \dfrac{14x}{600}+\dfrac{12x}{600}+\dfrac{25x}{600}=5610 \\
& \Rightarrow \dfrac{51x}{600}=5610 \\
& \Rightarrow 51x=5610\times 600 \\
& \Rightarrow x=\dfrac{5610\times 600}{51} \\
& \Rightarrow x=66000.....(1) \\
\end{align}\]
From equation (1), it is clear that the capital amount is equal to 66000.
Hence, option C is correct.
Note: Students may do the solution as follows:
Let us assume the capital is equal to x. From the question, it is clear that a man invested one-third of his capital at 7%; one fourth at 8%, and the remainder at 10%. We are also have given that the annual income is equal to 5610.
\[\begin{align}
& \Rightarrow \dfrac{7}{100}\times \dfrac{5610}{3}+\dfrac{8}{100}\times \dfrac{5610}{4}+\left( 5610-\dfrac{5610}{3}-\dfrac{5610}{4} \right)\left( \dfrac{10}{100} \right)=x \\
& \Rightarrow \dfrac{7}{100}\times \dfrac{5610}{3}+\dfrac{8}{100}\times \dfrac{5610}{4}+\dfrac{5\left( 5610 \right)}{12}\times \dfrac{10}{100}=x \\
& \Rightarrow \dfrac{7(5610)}{300}+\dfrac{2(5610)}{100}+\dfrac{5(5610)}{120}=x \\
& \Rightarrow \dfrac{14(5610)}{600}+\dfrac{12(5610)}{600}+\dfrac{25(5610)}{600}=x \\
& \Rightarrow \dfrac{51(5610)}{600}=x \\
& \Rightarrow 51(5610)=x\times 600 \\
& \Rightarrow x=\dfrac{5610\times 51}{600} \\
& \Rightarrow x=476.5....(1) \\
\end{align}\]
From equation (1), it is clear that the capital amount is equal to 476.5. But we know the capital amount is equal to 66000. So, this misconception should be avoided by a student to get the wrong answer.
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