Math, asked by nandanisingh3832, 6 months ago

Two year ago father was three time as old as his son and two year hence twice his ago will be equal to five time that of his son .find their present ages.

Answers

Answered by NaziaFarees
8

Answer:

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Step-by-step explanation:

Let us consider

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

The present age of Father = 3X – 4

Two years hence father’s age = 3X – 4 + 2

Two years hence father’s age = 3X – 2

Two years hence son’s age = X + 2

Given:

5 × (X + 2) = 2 × (3X – 2)

5X + 10 = 6X – 4

X = 14

Son’s present age X = 14 years.

Father’s present age = (3 × 14) – 4

Father’s present age = 38 years.

Answered by innotectbusiness
1

Answer:

SON'S PRESENT AGE : 14 years.

FATHER'S PTESENT AGE: 38 years.

Step-by-step explanation:

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

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