Question

a father's age is 3 times the sum of ages of his sons . five years later he will be twice the sum of ages of his two sons. find the present age of farther.​

Answer 1

Answer:

Step-by-step explanation:

Let the age of father be x years and sum of the ages of his children by y years.

After 5 years,

Father's age = (x + 5) years

Sum of ages of his children = (y + 10) years

From the given information, we have:

x = 3y ...(1)

and x + 5 = 2(y + 10)

x - 2y = 15 ...(2)

From (1) and (2), we have,

3y - 2y = 15  y = 15

x = 3y = 45

Father's age = 45 years

Answer 2

45 is the required answer.

CONSIDER:

• Let the father's age be x
• Let the sum of ages of his two sons be y

• Let the age of first son be a
• Let the age of second son be b

• So, sum of ages of two sons, y = a + b

GIVEN:

• Father's age is three times the sum of ages of his sons .
• x = 3 (a + b)
• x = 3y

• Five years later he will be twice the sum of ages of his two sons.
• x + 5 be the father's age after five years.
• a + 5 be the first son's age after 5 years.
• b + 5 be the second son's age after 5 years.

• By data, x + 5 = 2 (a + 5 + b + 5 )
• x +5 = 2 (a + b + 10)
• x + 5 = 2 (y + 10)

TO FIND:

The present age of father , x

SOLUTION:

At present, x = 3y

After five years, x + 5 = 2 (y + 10)

Substitute, x = 3y in x + 5 = 2 (y + 10)

3y + 5 = 2y + 20

3y - 2y = 20 - 5

y = 15

The sum of ages of his two sons is 15

Substitute y = 15 in x = 3y

x = 3 × 15

x = 45

ANSWER:

The present age of father is 45.

The father is 45 years old now.