Math, asked by risuuu, 9 months ago

Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their presents ages.

Answers

Answered by amishafilomeena1003
1

Answer:

Let son's age = x

Two years ago son's age= x-2.

His Father's age at that time = 3(x-2).

Present age of Father = 3x-6+2=3x-4

Two years hence father's age=3x-4+2=3x-2.

Two years hence son's age = x+2.

Given :-5*(x+2)=2*(3x-2)

5x + 10= 6x - 4

10+4=6x-5x

14=x

SON'S PRESENT AGE : 14 years.

FATHER'S PTESENT AGE: 38 years.

Step-by-step explanation:

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Answered by goutam9573
1

Let son's age = x

Two years ago son's age= x-2.

His Father's age at that time = 3(x-2).

Present age of Father = 3x-6+2=3x-4

Two years hence father's age=3x-4+2=3x-2.

Two years hence son's age = x+2.

Given :-5*(x+2)=2*(3x-2)

5x + 10= 6x - 4

10+4=6x-5x

14=x

SON'S PRESENT AGE : 14 years.

FATHER'S PRESENT AGE: 38 years.

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