Example A father is twice as old as his son; 20 years back he was twelve times as old as the son; what are their present
ages ?

Solution :

Let 'x years' be the present age of the son

His father is twice as old as he i.e. 2 x years

20 years back,

the son was (x - 20) years old and

the father was (2 x - 20) years old.

Now we set up the equation as:

( 2 x - 20 ) = 12 ´ ( x - 20)

2 x - 20 = 12 x - 240

- 20 + 240 = 12 x - 2 x . . . (transposing)

220 = 10 x

x = 22

Therefore, the son’s present age is 22 years and the father’s present age is 44 years

Check: 22 ´ 2 = 44 and

12 ( 22 - 20 ) = (44 - 20)

24 = 24

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Index

9.1 - Definition and Solving Techniques
9.2 - Use of Simultaneous Linear Equations

Chapter 1