AMC 10 Step by Step

2015 AMC 10A Problem 6

P1-10~1 minPrint

The sum of two positive numbers is 55 times their difference. What is the ratio of the larger number to the smaller number?

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Key insight

The sum equals 5 times the difference. Gathering terms gives 4 times the larger equals 6 times the smaller, so the ratio is 3 to 2.

Solution

Turn the sentence into an equation, then find the ratio.

1. Call the larger number aa and the smaller number bb. "The sum is 55 times the difference" says
a+b=5(a−b).a + b = 5(a - b).
2. Multiply out the right side: a+b=5a−5ba + b = 5a - 5b.
3. Gather the aa's on one side and the bb's on the other. Subtract aa from both sides and add 5b5b to both sides:
6b=4a.6b = 4a.
4. Divide both sides by 4b4b to get the ratio the question wants:
ab=64=32.\frac{a}{b} = \frac{6}{4} = \frac{3}{2}.
5. Check with actual numbers a=3a = 3, b=2b = 2: the sum is 55, the difference is 11, and 55 is indeed 5×15 \times 1.

The answer is (B) 32\boxed{\textbf{(B)}\ \frac{3}{2}}.

Why this works

One equation with two unknowns cannot tell you both numbers, but it can tell you how they compare. Any doubled pair, like 66 and 44, fits too. So when a problem gives one condition and asks for a ratio, collect terms and divide.

Alternative approach

Test the choices with real numbers. For 32\frac{3}{2}, use 33 and 22: sum 55, difference 11, and 5=5×15 = 5 \times 1. It works. For 54\frac{5}{4}, use 55 and 44: sum 99, difference 11, and 99 is not 55. The other choices fail the same way.

Concise solution

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

Let the numbers be a>b>0a > b > 0. The condition says
a+b=5(a−b)=5a−5b.a + b = 5(a - b) = 5a - 5b.
Move everything to one side: 6b=4a6b = 4a, so
ab=64=32.\frac{a}{b} = \frac{6}{4} = \frac{3}{2}.

Quick check with a=3a = 3, b=2b = 2: the sum is 55 and the difference is 11, and indeed 5=5⋅15 = 5 \cdot 1.

The answer is (B) 32\boxed{\textbf{(B)}\ \frac{3}{2}}.

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.