Question

Four years ago a father was six times a s old as his son . Ten years later, the father will be two and a half times as old as his son determine

Answer 1

Correct Question :

Four years ago a father was six times as old as his son. Ten years later, the father will be two and a half times as old as his son. Determine their present ages.

Answer :

The present age of father = 60 years

The present age of son = 18 years

Step-by-step explanation :

Given :

• Four years ago a father was six times as old as his son
• Ten years later, the father will be two and a half times as old as his son.

To find :

• the present ages of father and son

Solution :

Let the present age of father be x years

and the present age of son be y years

Four years ago,

Father's age = (x - 4) years

Son's age = (y - 4) years

Father's age = 6(son's age)

 x - 4 = 4(y - 4)

 x - 4 = 4y - 16

 4y - x = 16 - 4

 4y - x = 12

  x = 4y - 12

Ten years later,

Father's age = (x + 10) years

Son's age = (y + 10) years

the father will be two and a half times as old as his son

      $\sf x+10=2\dfrac{1}{2}(y+10) \\\\ x+10=\dfrac{5}{2}(y+10) \\\\ 2(x+10)=5(y+10) \\\\ 2x+20=5y+50 \\\\ 2x-5y=50-20 \\\\ 2x-5y =30$

Substitute x = 4y - 12

    2(4y - 12) - 5y = 30

     8y - 24 - 5y = 30

     3y = 30 + 24

     3y = 54

      y = 54/3

      y = 18

• x = 4y - 12

        x = 4(18) - 12

        x = 72 - 12

        x = 60

The present age of father = 60 years

The present age of son = 18 years

Verification :

• Four years ago,

 Father's age = 60 - 4 = 56 years

 Son's age = 18 - 4 = 14 years

Condition : father was six times as old as his son

     56 = 4(14)

      56 = 56

     LHS = RHS

• Ten years later,

 Father's age = 60 + 10 = 70 years

 Son's age = 18 + 10 = 28 years

Condition : father will be two and a half times as old as his son

       $\sf 70=2\dfrac{1}{2}(28) \\\\ \sf 70=\dfrac{5}{2} \times 28 \\\\ \sf 70=5 \times 14 \\\\ \sf 70=70$

      LHS = RHS

Hence verified!