Question

A father has 3 children with a gap of 2 years in every two consecutive children. The sum of present ages of children is half the present age of the father. 4 years before, the sum of ages of children was 1 year more than the one fourth of age of the father. Find the present ages of children and the father

Answer 1

if the 3 children present ages are x , x-2 , x+2 ( because gap between two children is 2)
according to question
x+x-2+x+2 = father age/2
3x = 6x
so father's present age is = 6x
_
4 year before children year 4 decrease
again according to question
3x -12 (3 × 4 year before) ={ (6x- 4)/4 } +1
3x-12 = 6x-4+4/4
3x-12 = 6x/4
4(3x-12)=6x
12x- 48= 6x
12x-6x= 48
6x = 48
x = 48/6=8

present age of children are x=8 , x-2=6, x+2=10
and father's present age is = 6x = 6×8=48

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Answer 2

Answer:

Step-by-step explanation:

if the 3 children present ages are x , x-2 , x+2 ( because gap between two children is 2)

according to question

x+x-2+x+2 = father age/2

3x = 6x

so father's present age is = 6x

_

4 year before children year 4 decrease

again according to question

3x -12 (3 × 4 year before) ={ (6x- 4)/4 } +1

3x-12 = 6x-4+4/4

3x-12 = 6x/4

4(3x-12)=6x

12x- 48= 6x

12x-6x= 48

6x = 48

x = 48/6=8

present age of children are x=8 , x-2=6, x+2=10

and father's present age is = 6x = 6×8=48

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