AMC 10 Step by Step

2009 AMC 10B Problem 2

P1-10~1 minPrint

Which of the following is equal to 13−1412−13\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}} ?

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Key insight

Work out the top and the bottom separately first: the top is 1/12 and the bottom is 1/6, and 1/12 is half of 1/6.

Solution

A fraction built out of fractions is easiest if you simplify the top and the bottom on their own before dividing.

  1. Top: 13−14\frac13 - \frac14. The common denominator is 1212, so this is 412−312=112\frac{4}{12} - \frac{3}{12} = \frac{1}{12}.
  2. Bottom: 12−13\frac12 - \frac13. The common denominator is 66, so this is 36−26=16\frac{3}{6} - \frac{2}{6} = \frac{1}{6}.
  3. Now divide: 112÷16\frac{1}{12} \div \frac{1}{6}. Dividing by 16\frac16 is the same as multiplying by 66, so this is 612=12\frac{6}{12} = \frac12.
  4. Quick sense check: a twelfth of a pizza is half of a sixth of a pizza, so the answer 12\frac12 makes sense.

The answer is (C) 12\boxed{\textbf{(C)}\ \frac12}.

Why this works

Stacked fractions look scary, but each layer is an ordinary subtraction. Simplify each layer to a single fraction, then divide by flipping the bottom fraction and multiplying. Keeping the numbers small at each step makes mistakes less likely.

Alternative approach

Clear all the small fractions at once. Multiply the top and the bottom by 1212, which does not change the value:

  • Top: 12×13−12×14=4−3=112 \times \frac13 - 12 \times \frac14 = 4 - 3 = 1.
  • Bottom: 12×12−12×13=6−4=212 \times \frac12 - 12 \times \frac13 = 6 - 4 = 2.

So the whole thing equals 12\frac12.

Concise solution

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

Simplify numerator and denominator separately:
13−14=4−312=112,12−13=3−26=16.\frac13 - \frac14 = \frac{4 - 3}{12} = \frac{1}{12}, \qquad \frac12 - \frac13 = \frac{3 - 2}{6} = \frac16.
Dividing, 1/121/6=112⋅6=12\dfrac{1/12}{1/6} = \dfrac{1}{12} \cdot 6 = \dfrac{1}{2}.

The answer is (C) 12\boxed{\textbf{(C)}\ \frac12}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.