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Question 1. A father is four times as old as his child. In five years time, he will be 3 times as old as his child. What are their present ages.
Solution :
Present age  Five year hence 

Let the child's present age be `= x` years 
Child's age `= (x + 5)` years Father's age `= (4x + 5)` years 
In five years time, father will be 3 times as old as his child. `:. 4x  3x = 15  5` `x = 10` 

Question 2. Mrs Sharma is now twice as old as her neighbour. Twelve years ago she was three times as old as her neighbour then. How old are they now?
Solution :
Mrs sharma 
Neighbour 
Let Mrs sharma's age be `2x` Mrs sharma's age `12` years ago `= 2x  12` 
Let neighbour's age be `x` Neighbour's age `12` years ago `= 3(x  12)` 
NOTE :  As Mrs Sharma's age was three time that of her neighbour twelve years ago, therefore,
`2x  12 = 3(x  12)`
`2x  12 = 3x  36`
`2x  3x = 36 + 12`
`1x = 24`
`x = 24`
`:.` Neighbour's age `= 24` years and Mrs Sharma's age `= 2x` `=2(24)` `= 48` years 
Question 3. In 4 years time, a father will be 3 times the age of his son; 4 years ago he was 5 times the age of his son. What are their present age?

Present Age 
Age 4 years hence 
Age 4 years ago 
Father Son 
`x` `y` 
`x + 4` `y + 4` 
`x  4` `y  4` 
4 years hence, father will be 3 times the son's age. That is, `x + 4 = 3(y + 4)` `:. x + 4 = 3y + 12` `:. x  3y = 12  4` `:. x  3y = 8` ...equation number (1)

4 years ago, father was 5 times the son's age. That is, `x  4 = 5(y  4)` `:. x  4 = 5y  20` `:. x  5y = 20 + 4` `:. x  5y = 16` ...equation number (2) 
Solve equation (1) and (2)
`x  3y = 8` ...equation number (1) `x  5y = 16` ...equation number (2) Subtracting equation (2) from (1), we get, `+2y = 24` `:. y = 24/2` `:. y = 12`

Substituting y = 12 in equation (1) `x  3y = 8` `:. x  3(12) = 8` `:. x  36 = 8` `:. x = 8 + 36` `:. x = 44` 
Ans : Father's present age is 44 years and son's present age is 12 years.