Question
18 years ago, a man was three times as old as his son. Now, the man is twice as old as his son. The sum of the present ages of the man and his son is = ?
Answer: Option D
Let the son's age 18 years ago be x years, Then man's age 18 years ago = 3x years
$$\eqalign{
& \left( {3x + 18} \right) = 2\left( {x + 18} \right) \cr
& \Rightarrow 3x + 18 = 2x + 36 \cr
& \Rightarrow x = 18 \cr} $$
Sum of their present ages
$$\eqalign{
& \Rightarrow \left( {3x + 18 + x + 18} \right){\text{years}} \cr
& \Rightarrow \left( {4x + 36} \right){\text{years}} \cr
& \Rightarrow \left( {4 \times 18 + 36} \right){\text{years}} \cr
& \Rightarrow {\text{ 108 years}} \cr} $$
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Let the son's age 18 years ago be x years, Then man's age 18 years ago = 3x years
$$\eqalign{
& \left( {3x + 18} \right) = 2\left( {x + 18} \right) \cr
& \Rightarrow 3x + 18 = 2x + 36 \cr
& \Rightarrow x = 18 \cr} $$
Sum of their present ages
$$\eqalign{
& \Rightarrow \left( {3x + 18 + x + 18} \right){\text{years}} \cr
& \Rightarrow \left( {4x + 36} \right){\text{years}} \cr
& \Rightarrow \left( {4 \times 18 + 36} \right){\text{years}} \cr
& \Rightarrow {\text{ 108 years}} \cr} $$
Was this answer helpful ?
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