Question

A man is 37 years old and his two sons are 8 years and 3 years old. After how many
years will he be twice as old as the sum of the ages of his sons?

Answer 1

❍ Let the number of years in which his age will be twice as old as the sum of his sons be x years respectively.

Present age of Man is 37 years old and Present age of his first son is 8 years old and second one is 3 years old.

Therefore,

— In x years their ages;

Age of Man = (37 + x)

Age of First son = (8 + x)

Age of younger son = (3 + x)

$\rule{250px}{.3ex}$

⠀⠀$\qquad\quad{\underline{ \underline{ \pink{\bigstar\:\frak{According \: to \: the \: Given \: Question :}}}}}$

⠀⠀

The age of father will be twice as old as the sum of ages of his sons.

⠀⠀

Therefore,

⠀

★ Father's age = 2(Sum of ages of son) ★

⠀⠀

$:\implies\sf (37 + x) = 2\Big[(8 + x) + (3 + x)\Big]\\\\\\:\implies\sf 37 + x = 2 (11 + 2x) \\\\\\:\implies\sf 37 + x = 22 + 4x\\\\\\:\implies\sf 3x = 15\\\\\\:\implies\sf x = \cancel\dfrac{15}{3}\\\\\\:\implies{\underline{\boxed{\frak{\pink{x = 5}}}}}\;\bigstar$

⠀⠀

$\therefore{\underline{\textsf{After\;\textbf{5 years}\;his\;age\;will\;be\;twice\;as\;old\;sum\;of\;ages\;his\;sons.}}}$

Answer 2

Given man’s age 37 and his elder son’s age is 8, So age difference is ED =37–8 = 29 between the man and his elder son. And his younger son’s age is 3, age difference is YD= 37–3 = 34 between the man and his younger son.

Let x be the required man’s which is such that x equals twice as old as the sum of the age of his sons and note that differences remain same always

x = 2 * [(x - ED) + (x - YD)] = 2 * [ 2x - ED - YD ]

x = 2 * [ 2x -29 -34] = 2 * [ 2x - 63]

x = 4x - 126

3x = 126 => x = 42

Thus after (42–37 = ) 5 years he becomes twice as old as the sum of the age of his sons.