Question

sum of the present ages of two friends are 23 years,five years ago product of their ages was 42.find their ages 5 years hence​

Answer 1

Answer:

the age of the friends 5 yrs ago is 6 and 7 yrs.

Step-by-step explanation:

the product of ages 5 yrs ago= factor of 42

the sum of present ages is 23{11+12}

so their ages 5 yrs ago was 6,7 yrs

Answer 2

$\huge{\underline{\underline{Answer\::-}}}$

$\underline{\underline{Given\::-}}$

Sum of ages of both Friends = 23

Product of ages of both Friends 5 years ago = 42

$\underline{\underline{Solution\::-}}$

• Let the age of one friend be x

• Then age of another friend = (23 - x)

• Then age of both 5 years ago

=( x - 5 ) and ((23 - x) - 5)

= (x - 5 ) and ( 18 - x)

• Then the product of (x - 5 ) and ( 18 - x) = 42

=> (x - 5 ) × ( 18 - x ) = 42

=> x(18 - x ) - 5(18 - x) = 42

=> 18x - $x^2$ - 90 + 5x = 42

=> 23x - $x^2$ - 90 = 42

=> $x^2$ - 23x + 132 = 0

♦Factoring further

=> $x^2$ - 12x - 11x + 132 = 0

= x(x - 12 ) -11(x - 12 )

= (x - 11) (x - 12 ).

♦ Now x = 11 or 12

>> If x = 11

Then

Age of one friend = 11 (x)

Age of another friend = 12 (23 - x)

>> If x = 12

Age of one friend = 12 (x)

Age of another friend = 11 (23 - x)

OR POSSIBLE AGE'S OF BOTH

x = 11 or 12

(23 - x ) = 12 or 11