Question: The product of the present ages of two sisters is 150. 5 years ago, the elder sister was twice as old as her younger sister. Find their present ages.

Solution:

Let the present ages of the two sisters be x and y years respectively.

According to the question,

Condition I,
The product of their ages is 150.
or, xy = 150 - (i)

Condition II,
5 years ago, the elder sister was twice as old as her younger sister.
or, (x -5) = 2(y -5)
or, x -5 = 2y -10
or, x = 2y -10+5
or, x = 2y -5 - (ii)

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

or, (2y -5)y = 150
or, 2y² -5y = 150
or, 2y² -5y -150 = 0
or, 2y² -(20-15)y -150 = 0
or, 2y² -20y +15y -150 = 0
or, 2y(y -10) + 15(y -10) = 0
or, (2y +15)(y -10) = 0

Either,
2y +15 = 0
or, 2y = -15
\therefore y = - \frac{15}{2}

Or,
y -10 = 0
\therefore y = 10

Since, age in years can not be negative, put value of y = 10 in equation (ii), we get,

or, x = 2×10 - 5
or, x = 20 -5
\therefore x = 15

Therefore, the required present ages of the two sisters are 15 years and 10 years, respectively.

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