Question

The sum of present ages of a man and his son is 45 years. 5 years ago product of their ages was 4 times that of age of man. Find their present ages. ​

Answer 1

Let the man's age be x

Let the son's age be y

Therefore,

Sum of their present ages=

$x + y = 45$

Product of their ages 5 yrs ago is equal to 4 times the father's age 5 yrs ago..... therefore,

$(x - 5)(y - 5) = 4(x - 5) \\ \\ \frac{(x - 5)(y - 5)}{(x - 5)} = 4 \\ \\ (y - 5) = 4 \\ \: \: y = 4 + 5 \\ \: \: y = 9$

Hence, son present age is 9 and father's present age is

$x + y = 45 \\ x + 9 = 45 \\ x = 45 - 9 \\ x = 36$

Man's age is 36

Answer 2

let the age of the son be x and the father be 45 - x

So,

x+ (45-x)= 45

Five years ago product of their ages was 4 times the son's age

(x-5)(45-x-5) = 4(x-5)  {Transferring (x- 5) to RHS}

40-x = 4 (x-5) / (x-5)   {Now both x-5 will be cut off in RHS}

40-x = 4

-x = 4 - 40

-x = -36   { both minus will be cut off}

x = 36

That's it!

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