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2015 AMC 10B Problem 4

2015 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Fractions & Decimals · difficulty 1 of 5 · about 2 min · position P1-10

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

A)
$\text{Alex, Beth, Cyril, Dan}$
B)
$\text{Beth, Cyril, Alex, Dan}$
C)
$\text{Beth, Cyril, Dan, Alex}$
D)
$\text{Beth, Dan, Cyril, Alex}$
E)
$\text{Dan, Beth, Cyril, Alex}$

Answer

Key insight

Put everything over 60: Alex 12, Beth 20, Cyril 15, so Dan gets 60 - 47 = 13, which slots between Cyril's 15 and Alex's 12.

Solution

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

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

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

Why this works

Comparing fractions with different denominators is easiest after scaling to a common denominator (the lcm $60$ here). Computing the leftover as $1$ minus the sum keeps Dan on the same scale as everyone else, so the ordering is just a comparison of integers.

The trap

Assuming the leftover must be the smallest share and putting Dan last (choice (B)); Dan's 13/60 actually beats Alex's 12/60.

Common mistakes

Assuming the leftover must be the smallest share and putting Dan last (choice (B)); Dan's 13/60 actually beats Alex's 12/60.
Ordering by denominator size backwards (a larger denominator means a smaller unit fraction), which swaps Alex and Beth.

Techniques

Set up the equation/formula and compute; no special trick needed