Question: The difference of the ages of a sister and her younger brother is 6 years. The numerical value of the product of their ages is equal 4 times the sum of their ages. Find the present ages of the sister and her brother.

Solution:

Let the present ages of the sister and her younger brother be x years and y years respectively.

According to the question,

Condition I,
The difference of their ages is 6 years
$or, x - y = 6$
$or, x = 6 +y$ - (i)

Condition II,
The product of their ages is equal 4 times the sum of their ages
$or, xy = 4(x +y)$

[Put value of x from equation (i) ]

$or, (6+y)y = 4(6 + y +y)$
$or, 6y +y² = 4(6 +2y)$
$or, 6y +y² = 24 +8y$
$or, y² -8y +6y -24 = 0$
$or, y² - 2y -24 = 0$
$or, y²- (6-4)y -24 = 0$
$or, y² -6y +4y -24 = 0$
$or, y(y -6) +4(y -6) = 0$
$or, (y +4)(y -6) = 0$

Either,
$y +4 = 0$
$\therefore y = -4$

Or,
$y -6  = 0$
$\therefore y = 6$

Since, age in years is positive, put value of y = 6 in equation (i), we get,

$or, x = 6+6$
$\therefore x = 12$

So, (x,y) = (12,6)

Therefore, the required ages of the sister and her younger brother are 12 years and 6 years, respectively.

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