AMC 10 · 2014 · #2

Grade 6 rate-ratio
fraction-arithmeticratio-proportion identify-subproblems ↑ Prerequisites: fraction-arithmetic
📏 Medium solution 💡 2 insights
Problem
A deal discounts every second item, and the money is fixed. Find the most that can be bought.

Pick an answer.

(A)
33
(B)
34
(C)
36
(D)
38
(E)
39
How to solve
Strategy Introduce a Variable

No dollar amount is given, so Tool #4 (Introduce a Variable) fixes a convenient regular price; because the count of balloons cannot depend on the price, any handy value works, and 3 makes the 1/3 discount a whole number. Tool #8 (Analyze the Units) tracks dollars per balloon and dollars per pair. Tool #7 (Identify Subproblems) turns the sale into one repeatable unit — a pair of balloons at a fixed cost — so the whole purchase is just 'how many pairs fit in the money.'

1STEP 1

Pick a convenient price

A convenient price makes everything whole.

money = 30 × 3 = 90 dollars
2STEP 2

Price the discounted balloon

The discounted item costs 2.

1/3 × 3 = 1, 3 - 1 = 2 dollars
3STEP 3

Cost of one pair

So one pair costs 5.

3 + 2 = 5 dollars per 2 balloons
4STEP 4

Fit the pairs into the money

The money holds 18 whole pairs.

90 ÷ 5 = 18 pairs
5STEP 5

Count the balloons

That is 36 items, choice (C).

18 × 2 = 36 → (C)
Answer
36
Without the sale, 90 dollars buys 30 balloons, so the answer must be more than 30; 36 is a believable bump. Check the saving directly: each pair costs 5 instead of 6, a 1/6 saving, so the money stretches to 30 × 6/5 = 36 balloons. Both routes give 36, and the money is used up exactly, matching choice (C).
💡Key takeaway

Buy balloons two at a time: every second one is a third off, so five dollars' worth of deal gets you six dollars' worth of balloons, stretching money for 30 into 36.

  • Pick a convenient price
  • Price the discounted balloon
  • Cost of one pair
  • Fit the pairs into the money
  • Count the balloons