Question

The sum of the age of a man and his son is 45 years.five years ago,the product of their age was four times the man's age at that time.find their present age

Answer 1

Answer:

if this helps u then mark me as brainliest and follow me

Step-by-step explanation:

Ans : 36 years, 9 years

Hint :

Let the present age of the man be x years. Then, present age of his son is (45 - x) years. Five years ago, man's age = (x - 5) years. Son's age = (45 - x - 5) years.

∴ (x - 5) (45 - x - 5) = 4(x - 5)

Answer 2

Solution :

Let the present age of a man be r years & let the present age of son be m years respectively;

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

$\underbrace{\bf{1^{st} \:Case\::}}$

$\mapsto\tt{r+m=45}$

$\mapsto\tt{m=45-r...................(1)}$

$\underbrace{\bf{2^{nd} \:Case\::}}$

$\underbrace{\bf{5\:years\:ago\::}}$

The age of man's was = (r - 5) years .

The age of son's was = (m - 5) years .

Now;

$\mapsto\tt{(r-5) (m-5) = 4(r-5)}$

$\mapsto\tt{(r-5) (45-r-5) = 4(r-5)\:\:\:[from(1)]}$

$\mapsto\tt{(r-5) (40-r) = 4(r-5)}$

$\mapsto\tt{r(40-r) -5(40-r) = 4r - 20}$

$\mapsto\tt{40r-r^{2} -200 +5r = 4r - 20}$

$\mapsto\tt{45r-r^{2} -200 +20 -4r =0}$

$\mapsto\tt{41r -r^{2} -180 = 0}$

$\mapsto\tt{r^{2} -41r +180 = 0}$

$\mapsto\tt{r^{2} -36r - 5r +180 = 0}$

$\mapsto\tt{r(r-36) -5(r-36) = 0}$

$\mapsto\tt{(r-36) (r-5) = 0}$

$\mapsto\tt{r-36 =0\:\:\:Or\:\:\:r-5=0}$

$\mapsto\bf{r=36\:\:\:Or\:\:\:r\neq 5}$

As we know that never may 5 years the age of man, so not acceptable .

∴ Putting the value of r = 36 in equation (1),we get;

$\mapsto\tt{m=45-36}$

$\mapsto\bf{m=9\:years}$

Thus;

The present age of man's = r = 36 years .

The present age of son's = m = 9 years .