
The present age of a man is three times that of his son. Six years ago he was four times the age than his son. Find the ratio of their ages six years later.
Answer
597.9k+ views
Hint: Assume that the present age of the son be x and that of the father be 3x. Hence form a linear equation in x using the statement of the question. Solve the equation for x and hence find the ratio of the ages 6 years later.
Complete step-by-step answer:
Let the present age of the son be x and the age of the father be 3x.
Age of father six years ago = 3x-6
Age of the son six years ago = x-6
According to the father’s age six years ago was four times that of his son.
Hence we have
$\dfrac{3x-6}{x-6}=4$
Multiplying both sides by x-6, we get
$3x-6=4\left( x-6 \right)$
Using distributive law of multiplication over subtraction, we get
$3x-6=4x-24$
Adding 6 on both sides, we get
$3x=4x-18$
Subtracting 4x from both sides, we get
$-x=-18$
Multiplying by -1 on both sides, we get
$x=18$
Hence the present age of the son is 18 and that of his father = 3x = 54.
Now the age of the father six years later = 54+6 = 60
Age of the son six years later = 18+6 = 24
Hence the ratio of the father's age to that of sons age six years later $=\dfrac{60}{24}=\dfrac{5}{2}=5:2$
Hence the ratio of age of father to that of his son six years later is 5:2
Hence option [d] is correct.
Note: Alternative solution:
We have $\dfrac{3x-6}{x-6}=\dfrac{4}{1}$
We know that if $\dfrac{a}{b}=\dfrac{c}{d}$, then for all ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ we have $\dfrac{{{x}_{1}}a+{{x}_{2}}b}{{{x}_{3}}a+{{x}_{4}}b}=\dfrac{{{x}_{1}}c+{{x}_{2}}d}{{{x}_{3}}c+{{x}_{4}}d}$
Now since $\dfrac{3x-6}{x-6}=\dfrac{4}{1}$, we have
$\begin{align}
& \dfrac{2\left( 3x-6 \right)-3\left( x-6 \right)}{-2\left( x-6 \right)+3x-6}=\dfrac{2\left( 4 \right)-3\left( 1 \right)}{-2\left( 1 \right)+4} \\
& \Rightarrow \dfrac{3x+6}{x+6}=\dfrac{5}{2} \\
\end{align}$
Hence the ratio of the age of the father to that of his son six years later is 5:2, which is the same as obtained above.
Hence option [d] is correct.
Complete step-by-step answer:
Let the present age of the son be x and the age of the father be 3x.
Age of father six years ago = 3x-6
Age of the son six years ago = x-6
According to the father’s age six years ago was four times that of his son.
Hence we have
$\dfrac{3x-6}{x-6}=4$
Multiplying both sides by x-6, we get
$3x-6=4\left( x-6 \right)$
Using distributive law of multiplication over subtraction, we get
$3x-6=4x-24$
Adding 6 on both sides, we get
$3x=4x-18$
Subtracting 4x from both sides, we get
$-x=-18$
Multiplying by -1 on both sides, we get
$x=18$
Hence the present age of the son is 18 and that of his father = 3x = 54.
Now the age of the father six years later = 54+6 = 60
Age of the son six years later = 18+6 = 24
Hence the ratio of the father's age to that of sons age six years later $=\dfrac{60}{24}=\dfrac{5}{2}=5:2$
Hence the ratio of age of father to that of his son six years later is 5:2
Hence option [d] is correct.
Note: Alternative solution:
We have $\dfrac{3x-6}{x-6}=\dfrac{4}{1}$
We know that if $\dfrac{a}{b}=\dfrac{c}{d}$, then for all ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ we have $\dfrac{{{x}_{1}}a+{{x}_{2}}b}{{{x}_{3}}a+{{x}_{4}}b}=\dfrac{{{x}_{1}}c+{{x}_{2}}d}{{{x}_{3}}c+{{x}_{4}}d}$
Now since $\dfrac{3x-6}{x-6}=\dfrac{4}{1}$, we have
$\begin{align}
& \dfrac{2\left( 3x-6 \right)-3\left( x-6 \right)}{-2\left( x-6 \right)+3x-6}=\dfrac{2\left( 4 \right)-3\left( 1 \right)}{-2\left( 1 \right)+4} \\
& \Rightarrow \dfrac{3x+6}{x+6}=\dfrac{5}{2} \\
\end{align}$
Hence the ratio of the age of the father to that of his son six years later is 5:2, which is the same as obtained above.
Hence option [d] is correct.
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