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, 2012

Algebra Difficulty 1.0 Junior Prove it Canada

In Carrotford, candidate A ran for mayor and received 1008 votes out of a total of 5600 votes. What percentage of all votes did candidate A receive?
In Beetland, exactly three candidates, B, C and D, ran for mayor. Candidate B won the election by receiving $\frac{3}{5}$ of all votes, while candidates C and D tied with the same number of votes. What percentage of all votes did candidate C receive?
In Cabbagetown, exactly two candidates, E and F, ran for mayor and 6000 votes were cast. At 10:00 p.m., only 90% of these votes had been counted. Candidate E received 53% of those votes. How many more votes had been counted for candidate E than for candidate F at 10:00 p.m.?
In Peaville, exactly three candidates, G, H and J, ran for mayor. When all of the votes were counted, G had received 2000 votes, H had received 40% of the votes, and J had received 35% of the votes. How many votes did candidate H receive?

Solution

Candidate A received $\frac{1008}{5600}\times100\%=0.18\times100\%$ or $18\%$ of all votes.
Solution 1

Since $\frac{3}{5}\times100\%=0.60\times100\%$ , Candidate B received $60\%$ of all votes.

Since Candidates C and D tied, they equally shared the remaining $100\%-60\%=40\%$ of the votes.

Therefore, Candidate C received $\frac{1}{2}$ of $40\%$ of the votes, or $20\%$ of all votes.

Solution 2

Since Candidate B received $\frac{3}{5}$ of all votes, then Candidates C and D shared the remaining $1-\frac{3}{5}=\frac{2}{5}$ of all votes.

Candidates C and D tied, thus they shared equally the remaining $\frac{2}{5}$ of the votes.

Therefore, Candidate C received $\frac{1}{2}$ of $\frac{2}{5}$ or $\frac{1}{2}\times\frac{2}{5}=\frac{2}{10}=\frac{1}{5}$ of the votes.

Since $\frac{1}{5}\times100\%=0.20\times100\%$ , Candidate C received $20\%$ of all votes.
Solution 1

At 10:00 p.m., $90\%$ of 6000 votes or $\frac{90}{100}\times6000=5400$ votes had been counted.

Of those 5400 votes that had been counted, Candidate E received $53\%$ .

Therefore at 10 p.m., $\frac{53}{100}\times5400=2862$ votes had been counted for Candidate E.

Since there were only 2 candidates, the remaining $5400-2862$ or 2538 votes must have been counted for Candidate F.

Thus, there were $2862-2538$ or 324 more votes counted for Candidate E than for Candidate F.

Solution 2

At 10:00 p.m., $90\%$ of 6000 votes or $\frac{90}{100}\times6000=5400$ votes had been counted.

Of those 5400 votes that had been counted, Candidate E received $53\%$ .

Since there are only 2 candidates, then Candidate F must have received the remaining $100\%-53\%$ or $47\%$ .

Thus, Candidate E received $53\%-47\%$ or $6\%$ more votes than Candidate F.

Since there were a total of 5400 votes that had been counted at 10:00p.m., then Candidate E received $6\%$ of 5400 or 324 more votes than Candidate F.
Candidate H received 40% of the votes and Candidate J received 35% of the votes.

Thus, the only other candidate, G, received the remaining $100\%-40\%-35\%=25\%$ of the votes.

Since Candidate G received 2000 votes representing 25% of all votes cast, then the total number of votes cast was $2000\times4=8000$ (since $25\%\times4=100\%$ ).

Thus, Candidate H received $40\%$ of 8000 votes or $\frac{40}{100}\times8000=3200$ votes.

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