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2002 AMC 10B Problem 21

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2002 AMC 10B Problem 21AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 3 of 5 · about 3 min · position P21-25

Andy's lawn has twice as much area as Beth's lawn and three times as much area as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?

  1. A)

    Andy\text{Andy}

  2. B)

    Beth\text{Beth}

  3. C)

    Carlos\text{Carlos}

  4. D)

    Andy and Carlos tie for first.\text{Andy and Carlos tie for first.}

  5. E)

    All three tie.\text{All three tie.}

Answer

B

Key insight

Pick concrete numbers: lawns of 6, 3, 2 units and mowing rates 3, 2, 1 give times 2, 1.5, 2, so Beth finishes first.

Solution

Only ratios are given, so choose convenient numbers. Andy's lawn is twice Beth's and three times Carlos', so take the areas to be
Andy 6,Beth 3,Carlos 2.\text{Andy } 6, \qquad \text{Beth } 3, \qquad \text{Carlos } 2.
Carlos' mower is half as fast as Beth's and a third as fast as Andy's, so take the rates (area per minute) to be
Andy 3,Beth 2,Carlos 1.\text{Andy } 3, \qquad \text{Beth } 2, \qquad \text{Carlos } 1.

Each finishing time is area divided by rate:
Andy: 63=2,Beth: 32=1.5,Carlos: 21=2.\text{Andy: } \tfrac63 = 2, \qquad \text{Beth: } \tfrac32 = 1.5, \qquad \text{Carlos: } \tfrac21 = 2.

Beth is done first; Andy and Carlos tie for second.

The answer is (B) Beth\boxed{\textbf{(B)}\ \text{Beth}}.

Why this works

When a problem gives only ratios, the answer cannot depend on the actual sizes, so any numbers respecting the ratios will do; picking the least common multiples (66 for areas, 33 for rates) keeps everything integral. Then the single relation time == work ÷\div rate settles it. Beth wins because her lawn is half of Andy's while her mower is two-thirds as fast, a better trade than Carlos' one-third lawn at one-third speed.

Alternative approach

Symbolically, let Andy's area be 6L6L and his rate 3r3r. Times are 6L3r=2Lr\frac{6L}{3r} = \frac{2L}{r} for Andy, 3L2r=1.5Lr\frac{3L}{2r} = \frac{1.5L}{r} for Beth, and 2Lr\frac{2L}{r} for Carlos, so Beth is fastest regardless of LL and rr.

The trap

Reading 'Carlos cuts half as fast as Beth' backwards (making Carlos the fastest), or judging by lawn size alone and picking Carlos.

Common mistakes

  • Reading 'Carlos cuts half as fast as Beth' backwards (making Carlos the fastest), or judging by lawn size alone and picking Carlos.
  • Noticing that Andy and Carlos tie and choosing (D) without computing Beth's time, which is smaller than theirs.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer