Problems based

on Age

x – 12y = –110

x + 10

y + 10

Let present age of Father be 'x'

years and that of Son be 'y' years

…….(i)

As per the first condition

Sol.

Son

Father

Present age

y

10 years hence

x + 10

y + 10

x

As per the second condition

=

( )

∴ x + 10 = 2y + 20

∴ x – 2y = 20 – 10

x – 2y = 10

…….(ii)

2

10 Years ago

x – 10

y – 10

x – 10

y – 10

( )

( )

=

12 ×

∴ x – 10 = 12y – 120

∴ x – 12y = – 120 + 10

Q. 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.

What do we need to find ?

Present ages of father & son

Was means equal to

Number it as equation (i)

12 times means 12 ×

Number it as equation (ii)

Here the coefficient of X is same in equation (i) and (ii),i.e 1

Since the coefficients are same we will subtract the two equations

Subtracting (ii) from (i)

x – 12y

= – 110

x – 2y

= 10

(–) (+) (+)

–10y

= – 100

∴ y

= 10

Substituting y=10 in(ii),

we get

∴ x – 2(10)

= 10

∴ x

= 10 + 20

∴ x

= 30

The present age of father is 30 years and the age of son is 10 years