The sum of the present ages of elder and younger brothers is 46 years and the product of their ages is 504. Find their present ages. | Quadratic Equation | SciPiPupil

Question: The sum of the present ages of elder and younger brothers is 46 years and the product of their ages is 504. Find their present ages.

Solution:

Let the present ages of two brothers be x years and y years, respectively. Let x years be the age of elder brother.

According to the question,

Condition I,

The sum of their present ages is 46 years.

$or, x + y = 46$ - (i)

Condition II,

The product of their ages is 504 years

$or, xy = 504$

$or, x = \frac{504}{y}$ - (ii)

Put value of x from equation (ii) in equation (i), we get,

$or, \frac{504}{y} + y = 46$

$or, \frac{504 + y²}{y} = 46$

$or, 504 + y² = 46y$

$or, y² - 46y +504 = 0$

$or, y² - (28+18)y +504 = 0$

$or, y² -28y -18y +504 = 0$

$or, y (y -28) -18(y -28) = 0$

$or, (y -18)(y -28) = 0$

Either,

$y -18 = 0$

$\therefore y = 18$

Or,

$y -28 = 0$

$\therefore y = 28$

Put value of y = 18 in equation (ii), we get,

$or, x = \frac{504}{18} = 28$

Put value of y = 28 in equation (ii), we get,

$or, x = \frac{504}{28} = 18$

Therefore, the required ages of the two brothers is 28 years and 18 years respectively.

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