Question 244884
We'll start be writing expressions for the three girls' ages.
Let x = youngest girl's age
Let y = middle girl's age
Since the oldest girl is twice the age of the middle girl:
2y = age of the oldest girl<br>
Now we can write an equation:
{{{x*y*2y = 200}}}
or
{{{2xy^2 = 200}}}<br>
Normally we need two equations to solve a problem with two variables. However, since ages are usually expressed as Natural Numbers (1, 2, 3, ...) we can use some logic and trial and error to find an answer.<br>
We'll start by dividing both sides by 2x:
{{{y^2 = 200/2x = 100/x}}}
In this form we can see that 100/x (or 100 divided by the youngest girl's age) must be a perfect square. So we just have to try dividing 100 by some Natural Numbers to see if we can find one (or more) that result in a perfect square.<br>
The first Natural Number is 1 and 100/1 is a perfect square! So if x = 1 then {{{y^2 = 100/1 = 100}}}. This makes y = 10 and 2y = 20 and the sum 1 + 10 + 20 = 31. And we have already found the answer: E.<br>
BTW, there is another possible set of ages. x = 2 and x = 3 do not work because 100/2 and 100/3 are not perfect squares. But x = 4 4 works because 100/4 = 25 which is a perfect square. This makes y = 5 and 2y = 10 and the sum 4 + 5 + 10 = 19. This sum is not one of the multiple choice answers but the ages do fit the problem.<br>
P.S. Any other set of ages will not fit because the youngest child ends up older than the middle child.