2014 AMC 10B Problem 6 / AMC 12B Problem 2

2014 AMC 10B 6 / AMC 12B 2. Orvin went to the store with just enough money to buy 30 balloons. When he arrived, he discovered that the store had a special sale on balloons: buy 1 balloon at the regular price and get a second at \frac{1}{3} off the regular price. What is the greatest number of balloons Orvin could buy?

(A) 33   (B) 34   (C) 36   (D) 38   (E) 39

This is a straightforward problem but what I want to illustrate here is that in a multiple-choice contest, sometimes we can make life a little easier with some simplifying assumptions.

A naive approach to this problem would start with “Let the price of 1 balloon be $x. …” What that means then is that the price of the 2nd balloon will be \frac{2x}{3}. It’s very easy to make silly mistakes with fractions, and even if you are careful, adding fractions usually takes a longer time than adding integers.

A slightly better approach would be to start with “Let the price of 1 balloon be $3x. …” The price of the second balloon would then be $2x. This is slightly better than working with the fraction above.

The best approach would be to notice that the value of x is quite irrelevant: it is a scaling factor. Whatever x is, the cost of each balloon will be something times x, the amount of money Orvin has will be something times x. Why not just assume that x = 1?

Assume that the regular price of the balloon costs $3. Then Orvin has \$3 \times 30 = \$90 to begin with. With the discount, every second balloon costs $2, i.e. every pair of balloons costs $5. Orvin has enough money to buy $90 / $5 = 18 pairs of balloons, or 36 balloons. The answer is (C).

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