Mathematics exam question.

Question

I am having some trouble working out this question. Wanted to know if someone could help me out with it.

Amy, Ben, Colleen and Dave share some money.

Amy has $\dfrac16$ of the money.

Ben has $\dfrac15$ of the money.

The difference between Amy's share and Ben's share is added to Colleen's share. The answer is equal to half the money.

Show that Amy and Dave each have the same amount of money.

Thanks.

Answer 1

Set $A:=$ Amy, $B:=$ Ben, $C:=$ Colleen, and $D:=$ David. From what we are told, we may first solve for $C$ \begin{aligned}C+\frac{1}{5}-\frac{1}{6}=\frac{1}{2} \\ \implies C+\frac{6}{30}-\frac{5}{30}=\frac{15}{30} \\ \implies C+\frac{1}{30}=\frac{15}{30} \\ \implies C=\frac{14}{30} \end{aligned}

Now notice that $D=1-(A+B+C)$.

Answer 2

Note that in this answer I assume that my interpretation (as illustrated in the comments) is correct.

Best thing to do with starting these kinds of word questions is to turn the words into math (equations and the like). We want to assign variables for the unknown parts of the problem, and we want to create equations relating the variables based on the given information. We have: $$\begin{align}&(1)\qquad\text{Amy, Ben, Colleen and Dave share some money$~\to$}\\&\qquad\qquad\text{let $a,b,c,d$ represent the amount of money from the pile that the respective} \\&\qquad\qquad\text{person has ($a$ for Amy, etc.)}\\&(2)\qquad\text{Amy has $\frac{1}{6}$-th of the money}\to a=\frac{1}{6}(a+b+c+d)\\&(3)\qquad\text{Ben has $\frac{1}{5}$-th of the money}\to b=\frac{1}{5}(a+b+c+d)\\&(4)\qquad\text{The difference between Amy's share and Ben's share is added to Colleen's share.}\\&\qquad\qquad\text{ The answer is equal to half the money.}\to b-a+c=\frac{1}{2}(a+b+c+d)\end{align}$$ If we let $t:=a+b+c+d$ then we have the system: $$\begin{cases}a=\frac{1}{6}t\\b=\frac{1}{5}t\\b-a+c=\frac{1}{2}t\end{cases}$$ and we want to show that $a=d$. Like Lulu says in the comments, we can assume that $t=1$ for simplification purposes. I will then leave the system for you to solve, as the toughest and most important part is setting up the words into equations, which I have shown how to do above.