Question

10 years ago, father was 12 times as old as son and 8 Years hence will be twive as old as is sum of their present age​

Answer 1

Correct Question :-

10 years ago, father was 12 times as old as son and 10 Years hence will be twice as old as

To Find :-

Sum

Solution :-

Let

Age of son = x

Age of father = y

Ten years ago

Age of son = (x - 10)

Age of father = (y - 10)

Case 1

(y - 10) = 12(x - 10)

y - 10 = 12x - 120

y - 12x = -120 + 10

y - 12x = -110

y = 12x - 110(i)

Case 2

After 8 years

Age of son = (x + 10)

Age of father = (y + 10)

2(x + 10) = y + 10

2x + 20 = y + 10

2x - y = 10 - 20

2x - y = -10

2x - 12x + 110 = -10

-10x = -10 - 110

-10x = -120

x = -120/-10

x = 12

Using 1

y = 12x - 110

y = 12(12) - 110

y = 144 - 110

y = 34 years

Answer 2

Appropriate Question:

• Ten years ago father was 12 times as old as his son at that time and 10 years hence he will be twice as old as his son. Find their present ages.

Let us assume:

Their present ages be,

• Father's age = x
• Son's age = y

10 years ago:

• Father's age = x - 10
• Son's age = y - 10

First equation,

• x - 10 = 12(y - 10)

10 years hence:

• Father's age = x + 10
• Son's age = y + 10

Second equation,

• x + 10 = 2(y + 10)

Subtracting eqⁿ (i) from eqⁿ (ii):

↠ x + 10 - (x - 10) = 2(y + 10) - 12(y - 10)

↠ x + 10 - x + 10 = 2y + 20 - 12y + 120

↠ 20 = - 10y + 140

↠ 10y = 140 - 20

↠ 10y = 120

↠ y = 120/10

↠ y = 12

In first equation:

↠ x - 10 = 12(y - 10)

Putting the value of y.

↠ x - 10 = 12(12 - 10)

↠ x - 10 = 12(2)

↠ x - 10 = 24

↠ x = 24 + 10

↠ x = 34

Hence,

Their present ages are:

• Father's age = 34 years
• Son's age = 12 years