Question

A man's age is four times the sum of the ages of his three children. In 8 years he will be
twice the sum of their ages. What is man's age ?​

Answer 1

Answer:

80 years

Step-by-step explanation:

Let consider sum of three children's ages = a

man's age = 4a

After 8 years :-

the sum of three children's ages = a+24 years

man's age = 2(a+24)

$= > 4a + 8 = 2(a + 24) \\ = > 4a + 8 = 2a + 48 \\ = > 4a - 2a = 48 - 8 \\ = > 2a = 40 \\ = > a = 40 \div 2 \\ = > a = 20 \: years$

The age of man's =

$= > 4a = 4 \times 20 \\ = > 80 \: years$

Answer 2

Answer:

Man's present age is 80 years.

Step-by-step explanation:

Given :

Man's age = 4 times the sum of his 3 children

After 8 years = Man will be twice the sum is his children's ages.

To find :

The present age of the man

Solution :

✯ $\textsf{\underline{\underline{Let the present ages be - }}}$

• Sum of his children = y
• The man = 4y

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{Ages after 8 years -}}}$

• Children = y + (8 + 8 + 8)
• The man = 4y + 8

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{According to the question -}}}$

After 8 years = Man will be twice the sum is his children's ages.

$\sf{\implies} \:4y + 8 = 2(y + 24) \\ \\ \sf{\implies} \: 4y + 8 = 2y + 48 \\ \\ \sf{\implies} \: 4y - 2y = 48 - 8 \\ \\ \sf{\implies} \: 2y = 40 \\ \\ \sf{\implies} \: y = \dfrac{40}{2} \\ \\ \sf{\implies} \: y = 20$

Sum of the ages of his children = 20 years

$\rule{300}{1.5}$

✯ $\textsf{\underline{\underline{Man's age - }}}$

$\sf{\implies} \: 4(20) \\ \\ \sf{\implies} \: 80$

Man's age = 80 years

$\therefore$ Man's present age is 80 years.