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2009 AMC 10B Problem 3

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2009 AMC 10B Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

Paula the painter had just enough paint for 3030 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 2525 rooms. How many cans of paint did she use for the 2525 rooms?

  1. A)

    1010

  2. B)

    1212

  3. C)

    1515

  4. D)

    1818

  5. E)

    2525

Answer

C

Key insight

Losing three cans cost her five rooms, so each can paints 5/3 rooms and 25 rooms need 25 divided by 5/3 = 15 cans.

Solution

The three lost cans were exactly the paint for 30−25=530 - 25 = 5 rooms. So 33 cans cover 55 rooms, and one can covers 53\frac{5}{3} rooms.

To cover 2525 rooms she needs 25÷53=25⋅35=1525 \div \frac{5}{3} = 25 \cdot \frac{3}{5} = 15 cans.

(Originally she had 1818 cans; losing 33 leaves 1515, consistent.)

The answer is (C) 15\boxed{\textbf{(C)}\ 15}.

Why this works

A difference in outcomes caused by a known difference in inputs gives the unit rate directly: 33 cans per 55 rooms. Once the rate is known, any quantity converts by proportion. Read the final question carefully; it asks for cans used, not cans originally owned.

The trap

Answering 18, the original number of cans, instead of the number actually used after the loss.

Common mistakes

  • Answering 18, the original number of cans, instead of the number actually used after the loss.
  • Assuming one can per room and answering 25.

Techniques

Set up the equation/formula and compute; no special trick needed