AMC 10 Step by Step

2016 AMC 10B Problem 1

P1-10~1 minPrint

What is the value of 2a−1+a−12a\frac{2a^{-1}+\frac{a^{-1}}{2}}{a} when a=12a= \tfrac{1}{2} ?

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Key insight

A negative exponent means flip the number over, so with a = 1/2 the flipped value is 2. Put in numbers first and the problem becomes simple arithmetic.

Solution

The expression looks messy, but aa is a friendly number, so put numbers in right away.

1. A power of −1-1 means "one over": a−1=1aa^{-1} = \frac{1}{a}. With a=12a = \frac12, that is 11/2=2\frac{1}{1/2} = 2.
2. Top of the fraction: 2a−1=2⋅2=42a^{-1} = 2 \cdot 2 = 4, and a−12=22=1\frac{a^{-1}}{2} = \frac{2}{2} = 1. So the top is 4+1=54 + 1 = 5.
3. Bottom of the fraction: a=12a = \frac12.
4. Dividing by 12\frac12 means asking "how many halves fit into 55?" The answer is 1010, because each whole holds two halves. So
51/2=5⋅2=10.\frac{5}{1/2} = 5 \cdot 2 = 10.

The answer is (D) 10\boxed{\textbf{(D)}\ 10}.

Why this works

When a variable is given a simple value, substitute it before doing any algebra. A negative exponent just flips the number, so (12)−1\left(\frac12\right)^{-1} becomes the whole number 22. Also remember: dividing by a number smaller than 11 makes the answer bigger, not smaller.

Alternative approach

Simplify first, then substitute. The top is 2a−1+12a−1=52a−12a^{-1} + \frac12 a^{-1} = \frac52 a^{-1}. Dividing by aa gives 52a−2\frac52 a^{-2}. With a=12a = \frac12, a−2=22=4a^{-2} = 2^2 = 4, so the value is 52⋅4=10\frac52 \cdot 4 = 10.

Concise solution

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

Since a=12a = \tfrac12, its reciprocal is a−1=2a^{-1} = 2. Substitute into the numerator:
2a−1+a−12=2⋅2+22=4+1=5.2a^{-1} + \frac{a^{-1}}{2} = 2\cdot 2 + \frac{2}{2} = 4 + 1 = 5.
Now divide by a=12a = \tfrac12. Dividing by one half is the same as multiplying by 22:
51/2=5⋅2=10.\frac{5}{1/2} = 5 \cdot 2 = 10.

The answer is (D) 10\boxed{\textbf{(D)}\ 10}.

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.