
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
581.1k+ 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.
Recently Updated Pages
Questions & Answers - Ask your doubts

A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Two Planoconcave lenses 1 and 2 of glass of refractive class 12 physics CBSE

The compound 2 methyl 2 butene on reaction with NaIO4 class 12 chemistry CBSE

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Name the states through which the Tropic of Cancer class 8 social science CBSE

Full form of STD, ISD and PCO

Right to vote is a AFundamental Right BFundamental class 8 social science CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

