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.​

Answer 1

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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.

Hope this answer helps you.......

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Answer 2

S O L U T I O N :

Let the present age of son's be r years & let the present age of father's be m years respectively.

A/q

$\underbrace{\bf{2\:years\:ago\::}}$

The son age was = (r - 2) years.

The father age was = (m - 2) years.

$\mapsto\tt{(m-2) = 3(r-2)}$

$\mapsto\tt{m-2 = 3r-6}$

$\mapsto\tt{m-3r = -6+2}$

$\mapsto\tt{m-3r = -4}$

$\mapsto\tt{m=-4+3r..............(1)}$

$\underbrace{\bf{After\:2\:years\::}}$

The son age will be = (r + 2) years.

The father age will be = (m + 2) years.

$\mapsto\tt{2(m+2) = 5(r+2)}$

$\mapsto\tt{2m +4 = 5r+ 10}$

$\mapsto\tt{2m -5r = 10-4}$

$\mapsto\tt{2m -5r = 6}$

$\mapsto\tt{2(-4 + 3r) -5r = 6\:\:\: [from(1)]}$

$\mapsto\tt{-8 + 6r - 5r = 6}$

$\mapsto\tt{r= 6 + 8}$

$\mapsto\bf{r= 14\:years}$

∴ Putting the value of r in equation (1),we get;

$\mapsto\tt{m = -4 + 3(14)}$

$\mapsto\tt{m = -4 + 42}$

$\mapsto\bf{m = 38\:years}$

Thus,

The present age of son = r = 14 years .

The present age of father = m = 38 years .