Math, asked by ITZROMANCE, 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 Anonymous
2

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

SON'S PRESENT AGE : 14 years.

FATHER'S PTESENT AGE: 38 years.

hope it helps u ✌

Answered by ThakurRajSingh24
16

QUESTION :-

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.

SOLUTION :-

=>Let us consider,

=>Son’s age = R

=>Two years ago son’s age = R – 2

=>His Father’s age at that time = 3(R– 2)

=>Present age of Father = 3R – 6 + 2

=>.°. The present age of Father = 3R – 4.

=>Two years hence father’s age = 3R – 4

+ 2

=>Two years hence father’s age = 3R – 2

=>.°. Two years hence son’s age = R + 2.

Given:

=>5 × (R + 2) = 2 × (3R – 2)

=>5R + 10 = 6R – 4

=> R = 14

=> Therefore, Son’s present age R = 14 years.

=>Father’s present age = ( 3R - 4 ) = (3 × 14 – 4)=(42 - 4) = 38 years.

=>Hence, Father’s present age is 38 years.


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