**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 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 $. …*” What that means then is that the price of the 2nd balloon will be . 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 $. …*” The price of the second balloon would then be $. 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 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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