AMC 10 · 2015 · #4
Easy mode Grade 4Four siblings share one pizza. Alex eats 51 of it. Beth eats 31. Cyril eats 41. Dan eats whatever is left. List the siblings in order, from the one who ate the most to the one who ate the least.
Pick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Four siblings share one pizza. Alex eats $\frac15$ of it, Beth eats $\frac13$, and Cyril eats $\frac14$. Dan eats whatever is left over. List the four siblings from the one who ate the most down to the one who ate the least.
Givens: Alex ate $\frac15$ of the pizza; Beth ate $\frac13$ of the pizza; Cyril ate $\frac14$ of the pizza; Dan ate the leftover part — whatever the other three did not eat; Each answer choice is one ordering of the four names from most eaten to least eaten
Unknowns: The four siblings ranked in decreasing order of how much pizza each one ate
Understand
Restated: Four siblings share one pizza. Alex eats $\frac15$ of it, Beth eats $\frac13$, and Cyril eats $\frac14$. Dan eats whatever is left over. List the four siblings from the one who ate the most down to the one who ate the least.
Givens: Alex ate $\frac15$ of the pizza; Beth ate $\frac13$ of the pizza; Cyril ate $\frac14$ of the pizza; Dan ate the leftover part — whatever the other three did not eat; Each answer choice is one ordering of the four names from most eaten to least eaten
Plan
Primary tool: #4 Introduce a Variable
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
Comparing $\frac15$, $\frac13$, $\frac14$ in their raw fraction form is awkward. Tool #4 (Introduce a Variable) fixes that by letting the pizza be a concrete number of equal slices — pick $60$, a common multiple of $5$, $3$, and $4$, so every share becomes a whole number of slices and the sizes are easy to read off. Tool #16 (Count the Complement) handles Dan: instead of computing his share directly, take the whole pizza and subtract the three known shares. Tool #3 (Eliminate Possibilities) finishes the job — once the four slice counts are known, only one of the five orderings matches, so the rest are crossed off.
Execute — Answer: C
4.OA.B.4 Step 1 Cut the pizza into 60 slices
- Pick a pizza size that turns every fraction into a whole number.
- The denominators are $5$, $3$, and $4$, and $60$ is a multiple of all three, so imagine the pizza cut into $60$ equal slices.
💡 A number that all three denominators divide evenly makes each share land on a whole slice count instead of a messy fraction.
4.NF.B.4 Step 2 Count each known share
- Take each sibling's fraction of the $60$ slices.
- Alex eats $\frac15$ of $60 = 12$ slices, Beth eats $\frac13$ of $60 = 20$ slices, and Cyril eats $\frac14$ of $60 = 15$ slices.
💡 Finding a fraction of a whole pizza is just multiplying that fraction by the slice count.
4.NBT.B.4 Step 3 Dan gets the leftovers
Dan's share is whatever the other three did not eat, so subtract their slices from the whole pizza: $12 + 20 + 15 = 47$ slices are gone, leaving $60 - 47 = 13$ slices for Dan.
💡 The leftovers are easiest to find by taking the whole and removing the parts you already know.
4.NBT.A.2 Step 4 Rank from most to least
- Now compare the slice counts: Beth $20$, Cyril $15$, Dan $13$, Alex $12$.
- In decreasing order that is Beth, Cyril, Dan, Alex, which matches choice (C); the other four orderings do not fit, so the answer is $\textbf{(C)}$.
💡 Whole-slice counts line up in plain size order, so the ranking reads straight off the numbers.
4.OA.B.4 Pick a pizza size that turns every fraction into a whole number. The denominator 4.NF.B.4 Take each sibling's fraction of the $60$ slices. Alex eats $\frac15$ of $60 = 12 4.NBT.B.4 Dan's share is whatever the other three did not eat, so subtract their slices fr 4.NBT.A.2 Now compare the slice counts: Beth $20$, Cyril $15$, Dan $13$, Alex $12$. In dec Review
Reasonableness: The four slice counts must rebuild the whole pizza: $12 + 20 + 15 + 13 = 60$ slices — exactly the pizza we started with, so nothing was lost or double-counted. The ranking also passes a quick gut check: Beth's $\frac13$ is the biggest single fraction given, Alex's $\frac15$ is the smallest, and Dan's $13$ slices sit between Cyril and Alex — matching the order Beth, Cyril, Dan, Alex in (C).
Alternative: Skip the slices and compare fractions over the common denominator $60$ directly: $\frac15 = \frac{12}{60}$, $\frac13 = \frac{20}{60}$, $\frac14 = \frac{15}{60}$, and Dan $= 1 - \frac{47}{60} = \frac{13}{60}$. The numerators $20, 15, 13, 12$ give the same ranking Beth, Cyril, Dan, Alex, confirming (C).
CCSS standards used (min grade 4)
4.OA.B.4Find factor pairs and multiples of whole numbers in the range 1-100 (Choosing $60$ as a common multiple of $5$, $3$, and $4$ so the pizza splits into a whole number of slices.)4.NF.B.4Multiply a fraction by a whole number (Turning each sibling's fraction into a slice count, e.g. $\frac15$ of $60 = 12$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Finding Dan's leftover slices as $60 - (12 + 20 + 15) = 13$.)4.NBT.A.2Compare two multi-digit numbers using >, =, and < symbols (Ordering the slice counts $20 > 15 > 13 > 12$ to rank the four siblings.)
⭐ Turn the fractions into a pizza of $60$ slices and the whole AMC 10 problem becomes Grade 4 counting and comparing.
⭐ Turn the fractions into a pizza of $60$ slices and the whole AMC 10 problem becomes Grade 4 counting and comparing.
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