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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
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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.