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Problem 491

AMC 10/12, early questions

Combinatorics Difficulty 3.0 Prove it CEMC Galois · Canada · 2017

On Monday, Daniel had 90 cups, each of which was either purple or yellow. He distributed
the cups among three boxes as follows:

Box D: 9 purple and 23 yellow cups for a total of 32 cups

Box E: 6 purple and 24 yellow cups for a total of 30 cups

Box F: 28 cups in total

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

In Box E, 6 of the 30 cups were purple.

The percentage of purple cups in Box E was $\dfrac{6}{30}\times100\%=\dfrac{2}{10}\times 100\%=20\%$ .
On Monday, 30% of Daniel’s 90 cups or $30\%\times90=\dfrac{30}{100}\times90=27$ cups were purple.

Daniel had 9 purple cups in Box D and 6 purple cups in Box E.

Therefore, the number of purple cups in Box F was $27-9-6=12$ .
Daniel had 27 purple cups and 90 cups in total.

On Tuesday, Avril added 9 more purple cups to Daniel’s cups, bringing the number of purple cups to $27+9=36$ , and the total number of cups to $90+9=99$ .

Barry brought some yellow cups and included them with the 99 cups.

Let the number of yellow cups that Barry brought be $y$ .

The total number of cups was then $99+y$ , while the number of purple cups was still 36 (since Barry brought yellow cups only).

Since the percentage of cups that were purple was again 30% or $\dfrac{30}{100}$ , then $\dfrac{30}{100}$ of $99+y$ must equal 36.

Solving, we get $\dfrac{30}{100}\times(99+y)=36$ or $30(99+y)=3600$ or $99+y=120$ , and so $y=21$ .

Therefore, Barry brought 21 cups.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.