2017 AMC 10A Problem 2

P1-10~2 minPrint

Linear Equations & Word Problems Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

Pablo buys popsicles for his friends. The store sells single popsicles for $\$1$ each, $3$ -popsicle boxes for $\$2$ each, and $5$ -popsicle boxes for $\$3$ . What is the greatest number of popsicles that Pablo can buy with $\$8$ ?

Show solution without answering

Key insight

The 5-popsicle box is the cheapest per popsicle. Buy two of them for $6, then spend the last $2 on a 3-popsicle box.

Solution

This asks for the greatest number, so there are two halves: find a way to buy $13$ , then show $14$ is out of reach.

Part 1: $13$ popsicles is possible.

First compare how much one popsicle costs in each option:

Option	Price	Popsicles	Cost per popsicle
Single	$\$1$	$1$	$\$1.00$
3-box	$\$2$	$3$	about $\$0.67$
5-box	$\$3$	$5$	$\$0.60$

The 5-box is the best deal, so buy as many as the money allows. Three of them cost $\$9$ , which is too much. Two of them cost $\$6$ and give $10$ popsicles.
That leaves $\$2$ . The best use of $\$2$ is one 3-box (two singles would give only $2$ popsicles).
Total: $10 + 3 = 13$ popsicles for exactly $\$8$ .

Part 2: $14$ popsicles is impossible.

No matter how Pablo buys, each popsicle costs at least $\$0.60$ , because that is the cheapest rate in the table.
So $14$ popsicles would cost at least $14 \times 0.60 = \$8.40$ , which is more than $\$8$ .

The answer is $\boxed{\textbf{(D)}\ 13}$ .

Why this works

With a fixed amount of money, buy the cheapest-per-item option first and then make the best use of what is left. The "cheapest possible price per popsicle" gives a ceiling that no plan can beat, which proves the answer is the largest.

Alternative approach

Just try every number of 5-boxes.

Two 5-boxes ( $\$6$ ), $\$2$ left: one 3-box. Total $10 + 3 = 13$ .
One 5-box ( $\$3$ ), $\$5$ left: two 3-boxes and one single. Total $5 + 6 + 1 = 12$ .
No 5-boxes, $\$8$ : four 3-boxes. Total $12$ .

The best is $13$ .

Concise solution

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

Compare the price per popsicle: singles cost $\$1$ each, a $3$ -box costs $\tfrac{2}{3}$ dollar per popsicle, and a $5$ -box costs $\tfrac{3}{5} = 0.60$ dollar per popsicle. The $5$ -boxes are the best value, so start there.

Three $5$ -boxes would cost $\$9$ , too much. Two $5$ -boxes cost $\$6$ and give $10$ popsicles, leaving $\$2$ . That $\$2$ buys a $3$ -box, for a total of $13$ popsicles for exactly $\$8$ .

Can $14$ be reached? Every popsicle costs at least $\$0.60$ under any purchase plan, so $14$ popsicles cost at least $\$8.40 > \$8$ . Hence $13$ is the maximum.

The answer is $\boxed{\textbf{(D)}\ 13}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 7: Proportion (§7.4)