AMC 10 Step by Step

2015 AMC 10B Problem 4

P1-10~2 minPrint

Four siblings ordered an extra large pizza. Alex ate 15\frac15 , Beth 13\frac13 , and Cyril 14\frac14 of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?

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Key insight

Write every share out of 60: Alex 12, Beth 20, Cyril 15. Dan gets the rest, 60 - 47 = 13, which is more than Alex's 12.

Solution

Fractions with different bottoms are hard to compare, so put them all over the same bottom. The numbers 55, 33 and 44 all divide 6060.

  1. Change each share to sixtieths:
AlexBethCyril
15=1260\frac15 = \frac{12}{60}13=2060\frac13 = \frac{20}{60}14=1560\frac14 = \frac{15}{60}
  1. Together the first three ate 12+20+1560=4760\dfrac{12 + 20 + 15}{60} = \dfrac{47}{60} of the pizza.
  2. Dan got what was left: 60−4760=1360\dfrac{60 - 47}{60} = \dfrac{13}{60}.
  3. Now just compare the tops: Beth 2020, Cyril 1515, Dan 1313, Alex 1212.

From most to least: Beth, Cyril, Dan, Alex.

The answer is (C) Beth, Cyril, Dan, Alex\boxed{\textbf{(C)}\ \text{Beth, Cyril, Dan, Alex}}.

Why this works

Once every fraction has the same bottom number, comparing them is just comparing whole numbers. Finding Dan's share as "the whole pizza minus what the others ate" keeps his share on the same scale. Do not guess that leftovers are small; compute them.

Concise solution

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

Use the common denominator 6060:
Alex 15=1260,Beth 13=2060,Cyril 14=1560.\text{Alex } \frac15 = \frac{12}{60}, \qquad \text{Beth } \frac13 = \frac{20}{60}, \qquad \text{Cyril } \frac14 = \frac{15}{60}.
Together they ate 12+20+1560=4760\frac{12 + 20 + 15}{60} = \frac{47}{60}, so Dan's leftovers are 60−4760=1360\frac{60 - 47}{60} = \frac{13}{60}.

Ordering the numerators 20>15>13>1220 > 15 > 13 > 12 gives Beth, Cyril, Dan, Alex.

The answer is (C) Beth, Cyril, Dan, Alex\boxed{\textbf{(C)}\ \text{Beth, Cyril, Dan, Alex}}.

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.