Math, asked by cujhb, 1 year ago

a man is three times as old as his son and 6 years ago the product of their ages was 288 find their present ages

Answers

Answered by misa3
14
let the son's age be x

then the man's age = 3x

A/Q.
(x-6)(3x-6) = 288

==> 3x^2 -6x - 18x + 36 = 288

==> 3x^2 - 24x + 36 - 288 = 0

==> 3x^2 - 24x - 252 = 0

==> x^2 - 8x - 84 = 0

==> x^2 - 14x + 6x - 84 = 0

==> x (x - 14) + 6 (x - 14) = 0

==> (x -14)(x+ 6) = 0

either of them = 0

if (x-14 )= 0
then x= 14

if (x+6) = 0
then x = -6

since, age cannot be negative
therefore, x=14

hence, son's age = 14 years
and man's age = 3×14years = 42 years

with regards....

#misa
Answered by yusufkhanstar29
0

Answer:

Man= 42 years

Son= 14 years

Step-by-step explanation:

Concept= Time age problems.

Given = Age relation and product of six years ago age.

To find= The present ages

Explanation=

We have been given the question as a man is three times as old as his son and 6 years ago the product of their ages was 288 find their present ages.

So we know that in present time the age of man is three times that of his son.

Let the age of Son be x and then age of Man is 3x.

Son= x

Man = 3x

The information is given that 6 years ago their age product is 288.

Before 6 years:

Son= x-6

Man= 3x-6

Product of ages = 288.

so,

(3x-6)(x-6)=288

=> 3x² -18x -6x +36 = 288

=> 3x² - 24x + 36 = 288

=> 3x² - 24x + 36 -288 = 0

=> 3x² - 24x -252 = 0

(Dividing the whole equation by 3 and making coefficient of x² as 1)

=> x² -8x - 84 =0

=> x² - 14x + 6x - 84 = 0

=> x(x - 14) + 6(x - 14) = 0

=> (x - 14)( x + 6) = 0

Now we see that value of x is 14 and -6.

Age cannot be negative so -6 is neglected hence the value of x is 14.

Therefore present age of Son = x = 14 years and Man= 3x= 3*14=42 years.

So , Man = 42 years and Son= 14 years.

#SPJ2

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