AMC 10 Step by Step

2010 AMC 10A Problem 11

P11-15~2 minPrint

The length of the interval of solutions of the inequality a≤2x+3≤ba \le 2x + 3 \le b is 1010 . What is b−ab - a ?

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Key insight

Subtracting 3 slides the interval without changing its length; dividing by 2 halves it. So the x-interval is half of b - a, and b - a = 20.

Solution

Get xx by itself, then measure the interval.

1. Start with a≤2x+3≤ba \le 2x + 3 \le b. Subtract 33 from all three parts:
a−3≤2x≤b−3.a - 3 \le 2x \le b - 3.
2. Divide all three parts by 22:
a−32≤x≤b−32.\frac{a - 3}{2} \le x \le \frac{b - 3}{2}.
3. The length of an interval is the right end minus the left end:
b−32−a−32=(b−3)−(a−3)2=b−a2.\frac{b - 3}{2} - \frac{a - 3}{2} = \frac{(b - 3) - (a - 3)}{2} = \frac{b - a}{2}.
4. This length is 1010, so b−a2=10\dfrac{b - a}{2} = 10 and b−a=20b - a = 20.

The answer is (D) 20\boxed{\textbf{(D)}\ 20}.

Why this works

Adding or subtracting the same number from every part of an inequality just slides the interval along; its length stays the same. Dividing every part by 22 shrinks the interval to half its length. So the interval for xx is half as long as the interval from aa to bb.

Alternative approach

Try an example. Pick an easy interval for xx of length 1010, like 0≤x≤100 \le x \le 10. Then 2x+32x + 3 runs from 2(0)+3=32(0) + 3 = 3 to 2(10)+3=232(10) + 3 = 23. So a=3a = 3, b=23b = 23, and b−a=20b - a = 20.

Concise solution

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

Isolate xx in the compound inequality. Subtracting 33 throughout and then dividing by 22 gives
a−32≤x≤b−32.\frac{a - 3}{2} \le x \le \frac{b - 3}{2}.

The solution interval runs from a−32\dfrac{a-3}{2} to b−32\dfrac{b-3}{2}, so its length is
b−32−a−32=b−a2.\frac{b - 3}{2} - \frac{a - 3}{2} = \frac{b - a}{2}.

Setting this equal to 1010 gives b−a=20b - a = 20.

The answer is (D) 20\boxed{\textbf{(D)}\ 20}.

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.