Question

The age of a man is three times the sum of the ages of his two sons.Five years hence,his age will be double of the sum of the ages of his sons.The father's present age is:

Answer 1

Son's age = x + y

Man's age = 3(x+y)

After 5 years Sons' age = x + 5 + y + 5  = x + y + 10

After 5 years age father's age = 3(x+y) + 5 or 2(x+y+10)

3x + 3y + 5 = 2 x + 2y + 20

x + y = 15

Father's present age = 3(x+y) = 3 x 15 = 45

Answer 2

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Present age of man is 45 years.

Given :-

The age of a man is three times the sum of the ages of his two sons.

5 years hence, his age will be double the sum of their ages.

To find :-

Present age of man.

Solution :-

Let ,

Man's present age = x years

Present age of 1st son = y years

Present age of 2nd son = z years

$\red{\underline{\sf{According\:to\:the\:1st\: condition:-}}}$

The age of a man is three times the sum of the ages of his two sons.

➪ x = 3(y+z)

➪ y+z = x/3 ..................(i)

$\red{\underline{\sf{According\:to\:the\:2nd\: condition:-}}}$

5 years hence, his age will be double the sum of their ages

5 years hence,

Man's age = (x+5) years

Age of 1st son = (y+5) years

Age of 2nd son = (z+5) years

➪ x+5 = 2[(y+5)+(z+5)]

➪ x+5 = 2(y+z+10)

[ put y+z = x/3 from eq (i)]

➪ x+5 = 2(x/3 + 10)

➪ x+5 = 2x/3 + 20

➪ x - 2x/3 = 20-5

➪ x/3 = 15

➪ x = 45

† Present age of man is 45 years.

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