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2008 AMC 10A Problem 4

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2008 AMC 10A Problem 4AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

Suppose that 23\tfrac{2}{3} of 1010 bananas are worth as much as 88 oranges. How many oranges are worth as much as 12\tfrac{1}{2} of 55 bananas?

  1. A)

    22

  2. B)

    52\frac{5}{2}

  3. C)

    33

  4. D)

    72\frac{7}{2}

  5. E)

    44

Answer

C

Key insight

Twenty thirds of a banana equal 8 oranges, so one banana is worth 6/5 of an orange; five halves of a banana is then 3 oranges.

Solution

Two thirds of 1010 bananas is 203\frac{20}{3} bananas, and this equals 88 oranges in value. So one banana is worth
8÷203=2420=65 oranges.8 \div \frac{20}{3} = \frac{24}{20} = \frac{6}{5}\ \text{oranges}.

Half of 55 bananas is 52\frac{5}{2} bananas, worth
52⋅65=3 oranges.\frac{5}{2} \cdot \frac{6}{5} = 3\ \text{oranges}.

The answer is (C) 3\boxed{\textbf{(C)}\ 3}.

Why this works

Any "worth as much as" statement is an exchange rate. Reduce the given relation to the value of one unit (one banana), then multiply by the new quantity. Keeping the fractions as fractions, rather than decimals, lets the 55's cancel cleanly.

Alternative approach

Scale the given relation directly: 203\frac{20}{3} bananas =8= 8 oranges. Multiplying both sides by 38\frac{3}{8} gives 52\frac{5}{2} bananas =3= 3 oranges, exactly the quantity asked about.

The trap

Reading 'two thirds of 10 bananas' as 10 bananas, or inverting the exchange rate to 5/6 orange per banana.

Common mistakes

  • Reading 'two thirds of 10 bananas' as 10 bananas, or inverting the exchange rate to 5/6 orange per banana.
  • Computing 52\frac{5}{2} bananas and reporting 52\frac{5}{2} (choice B) without converting to oranges.

Techniques

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