# Reverse plugging in answers to questions

How do I solve such questions without reverse plugging in the answers into the question itself?

First, I will exclude all answers which are not multiple of fives, based on the statement "five times the age..."

Second, I will take the oldest child age + 2 and check if it is twice of the middle child age.

How do I represent these word problems in algebraic equations?

Mary has three children. The oldest child is five times the age of her youngest, and in two years will be twice as old as the middle child.

Which of the following ages could the oldest child be?

1. 10
2. 12
3. 15
4. 18
• What exactly do you mean? Do you mean by solving it directly, as if it weren't multiple-choice, as opposed to just plugging in an answer and seeing if it works? – Eevee Trainer May 10 at 5:21
• It's either 10 or 15 by your reasoning, which then has to be an even number when you add 2 (Twice the age of the middle child) . This makes it 10. – Phil H May 10 at 5:27
• How do I solve it algebraically? Let's say I let X be the age of the youngest child and 5x as the age of the oldest child. But, how do I represent the age of the middle child? – ilovetolearn May 10 at 5:31
• Let $x,y,z$ be the ages of oldest, middle, youngest, respectively. Then, now: $x=5z$, in $2$ years $x+2=2(y+2)$. However, it should state the ages are positive integers, otherwise any choice could be true: $\color{red}{10},4,2;\color{red}{12},5,2.4;\color{red}{15},5,3;\color{red}{18},8,3.6$. – farruhota May 10 at 6:33

Your multiples of $$5$$ are $$10$$ and $$15$$. In two years you get $$12$$ and $$17$$.
Well, $$12$$ is twice $$6$$ which is the middle child age but $$17$$ is an odd number and it is not twice anybody's age.
Thus the answer is $$10$$.
• Let the age of the middle child be $x$. We have $2x-2=10\implies 2x=12\implies x=6$ – Mohammad Riazi-Kermani May 10 at 10:39