17. Amy, Beth and Claire each has some sweets. Amy gives one third of her sweets to Beth. Beth gives one third of all the sweets she now has to Claire. Then Claire gives one third of all the sweets she now has to Amy. All the girls end up having the same number of sweets.
Claire begins with 40 sweets.
How many sweets does Beth have originally?
A 20
B 30
C 40
D 50
E 60
Solution
D

We work backwards from the final situation when Amy, Beth and Claire end up with the same number of sweets. We let $s$ be the number of sweets they all end up with.

Claire ends up with $s$ sweets. Therefore, before she gave one third of her sweets to Amy, Claire had $t$ sweets, where $\frac{2}{3} t=s$. This gives $t=\frac{3}{2} s$. We deduce that, before giving one-third of her sweets to Amy, Claire has $\frac{3}{2} s$ sweets and gives one third of these, namely $\frac{1}{2} s$ sweets, to Amy. Hence, before she received these sweets, Amy had $\frac{1}{2} s$ sweets.
Similarly, Beth ends up with $s$ sweets after she has given one third of her sweets to Claire. Hence, before this she had $\frac{3}{2} s$ sweets. As Claire has $\frac{3}{2} s$ sweets after receiving $\frac{1}{2} s$ from Beth, she had $s$ sweets before this.
It follows that after Amy has given one third of her sweets to Beth, Amy has $\frac{1}{2} s$ sweets, Beth has $\frac{3}{2} s$ sweets, and Claire has $s$ sweets.
Therefore, before Amy gives one third of her sweets to Beth, Amy has $\frac{3}{4} s$ sweets. She gives $\frac{1}{4} s$ sweets to Beth. Hence, before receiving these, Beth had $\frac{3}{2} s-\frac{1}{4} s=\frac{5}{4} s$ sweets.
We can sum this up by the following table.
We are told that Claire begins with 40 sweets. Therefore $s=40$. So Beth begins with $\frac{5}{4}(40)=50$ sweets.