AMC 10 Step by Step

2014 AMC 10B Problem 6

P1-10~2 minPrint

Orvin went to the store with just enough money to buy 3030 balloons. When he arrived, he discovered that the store had a special sale on balloons: buy 11 balloon at the regular price and get a second at 13\frac{1}{3} off the regular price. What is the greatest number of balloons Orvin could buy?

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Key insight

Measure money in regular balloon prices. On sale, a pair costs 1 + 2/3 = 5/3 prices, so 30 prices buy 18 pairs, which is 36 balloons.

Solution

1. Pick an easy unit. Say one balloon normally costs $1\$1. Orvin can buy exactly 3030 balloons, so he has $30\$30.
2. Price of a sale pair. The first balloon costs $1\$1. The second is 13\frac13 off, so it costs 1−13=231 - \frac13 = \frac23 of a dollar. One pair costs
1+23=53 dollars.1 + \frac23 = \frac53 \text{ dollars}.
3. How many pairs? Three pairs cost 3×53=$53 \times \frac53 = \$5. So every $5\$5 buys 33 pairs, which is 66 balloons.
4. Orvin has $30=6×$5\$30 = 6 \times \$5, so he buys 6×3=186 \times 3 = 18 pairs with nothing left over.
5. That is 18×2=3618 \times 2 = 36 balloons.

The answer is (C) 36\boxed{\textbf{(C)}\ 36}.

Why this works

Using the regular price as the unit turns the budget into a plain number (3030). A "buy one, get one at a discount" deal repeats in pairs, so work out the cost of one pair and see how many pairs fit. It is also worth checking whether money is left over, since a leftover single balloon would cost full price; here the money runs out exactly.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let one balloon regularly cost 11 unit, so Orvin has 3030 units.

Under the sale, a pair of balloons costs 1+(1−13)=1+23=531 + \left(1 - \frac13\right) = 1 + \frac23 = \frac53 units. The number of pairs he can afford is
305/3=30⋅35=18,\frac{30}{5/3} = 30 \cdot \frac35 = 18,
which is 2⋅18=362 \cdot 18 = 36 balloons, using all his money exactly.

The answer is (C) 36\boxed{\textbf{(C)}\ 36}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.