AMC 8 · 1999 · #18
Grade 6 rate-ratioProblem
At Central Middle School the students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of cookies, lists this items: cups flour, eggs, tablespoons butter, cups sugar, and package of chocolate drops. They will make only full recipes, not partial recipes.
They learn that a big concert is scheduled for the same night and attendance will be down . How many recipes of cookies should they make for their smaller party?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Of the $108$ AMC8 students, attendance will be down $25\%$ because of a concert, so only the remaining $75\%$ will show up. Each student eats an average of $2$ cookies. A recipe makes a pan of $15$ cookies, and only full recipes can be baked. How many recipes must Walter and Gretel make?
Givens: Normal attendance: $108$ students; Attendance is down $25\%$, so $75\%$ of $108$ will attend; Average cookies per attending student: $2$; One recipe yields a pan of $15$ cookies; Only whole recipes can be baked (no half-pans); Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Unknowns: The smallest whole number of recipes that produces at least enough cookies
Understand
Restated: Of the $108$ AMC8 students, attendance will be down $25\%$ because of a concert, so only the remaining $75\%$ will show up. Each student eats an average of $2$ cookies. A recipe makes a pan of $15$ cookies, and only full recipes can be baked. How many recipes must Walter and Gretel make?
Givens: Normal attendance: $108$ students; Attendance is down $25\%$, so $75\%$ of $108$ will attend; Average cookies per attending student: $2$; One recipe yields a pan of $15$ cookies; Only whole recipes can be baked (no half-pans); Answer choices: (A) $6$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Plan
Primary tool: #7 Identify Subproblems
The wording stacks four small questions on top of each other — attendance, total cookies, raw recipe count, then round up. Tool #7 (Identify Subproblems) is the cleanest fit: peel the problem apart into a chain of one-step calculations, finish each, then feed it into the next. No algebra needed; each sub-question is a single elementary operation (percent of a number, multiplication, division, ceiling). The only trap is the final "full recipes only" line, which is just a real-world constraint reminding us to round up, not down.
Execute — Answer: E
6.RP.A.3 Step 1 - Sub-question 1: how many students actually show up?
- Attendance is down $25\%$, so $75\%$ of the usual $108$ students come.
- Compute $75\%$ of $108$.
💡 Grade 6 percent-of-a-number: $25\%$ off means $75\%$ left, and $75\%$ of $108$ is three-quarters of $108$, namely $81$.
6.RP.A.3 Step 2 - Sub-question 2: how many cookies does the party need?
- Each of the $81$ attendees eats $2$ on average, so multiply.
💡 A Grade 6 unit-rate calculation: $2$ cookies per student times $81$ students gives $162$ cookies total.
6.RP.A.3 Step 3 - Sub-question 3: how many recipes would exactly cover $162$ cookies?
- One recipe is $15$ cookies, so divide.
💡 Another unit-rate step, this time the other direction: total cookies divided by cookies-per-recipe gives recipes. The $.8$ tells us $10$ recipes is not enough.
4.OA.A.3 Step 4 - Sub-question 4: round to a whole number of recipes.
- Since only full recipes are allowed and $10$ pans give only $10 \times 15 = 150$ cookies (short of $162$), we must bake one more.
- Eleven pans give $11 \times 15 = 165$ cookies, comfortably covering the $162$ needed.
💡 Grade 4 "interpret remainders": when leftover cookies are still needed, the answer rounds up, not down. Ten pans leave the party twelve cookies short, so eleven it is.
6.RP.A.3 Sub-question 1: how many students actually show up? Attendance is down $25\%$, s 6.RP.A.3 Sub-question 2: how many cookies does the party need? Each of the $81$ attendees 6.RP.A.3 Sub-question 3: how many recipes would exactly cover $162$ cookies? One recipe i 4.OA.A.3 Sub-question 4: round to a whole number of recipes. Since only full recipes are Review
Reasonableness: Quick sanity pass. Attendance dropped by a quarter, so $108 - 27 = 81$ — agrees with $0.75 \times 108 = 81$. Total cookies $81 \times 2 = 162$ is between $150$ ($10$ recipes) and $165$ ($11$ recipes), so the answer must be $11$. Choice (D) $10$ would leave $162 - 150 = 12$ students short a cookie; choice (E) $11$ leaves $165 - 162 = 3$ cookies over — exactly the small surplus you'd expect when forced to round up. None of the smaller choices ($6$, $8$, $9$) come anywhere near $162$ cookies, so they are clearly too few.
Alternative: Tool #3 (Eliminate Possibilities). Start from the answer choices. Six recipes give $6 \times 15 = 90$ cookies; eight give $120$; nine give $135$; ten give $150$ — all below $162$. Only eleven recipes ($165$ cookies) clear the $162$-cookie bar, so (E) is the only choice that works. This skips the percent calculation if you first compute $162$, but more importantly it doubles as a check that no smaller choice fits.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems, including percent problems (Computing $75\%$ of $108$ to find $81$ attendees, multiplying by the $2$-cookies-per-student rate to get $162$ cookies, and dividing by the $15$-cookies-per-recipe rate to get $10.8$ recipes.)4.OA.A.3Solve multistep word problems posed with whole numbers, including problems in which remainders must be interpreted (Rounding $10.8$ recipes up to $11$ because $10$ full pans ($150$ cookies) fall short of the $162$ needed — the textbook "interpret the remainder" move.)
⭐ Chain the questions: $75\% \times 108 = 81$ attendees, $81 \times 2 = 162$ cookies, $162 \div 15 = 10.8$ recipes — and since you cannot bake $0.8$ of a pan, round up to $11$, answer (E).
⭐ Chain the questions: $75\% \times 108 = 81$ attendees, $81 \times 2 = 162$ cookies, $162 \div 15 = 10.8$ recipes — and since you cannot bake $0.8$ of a pan, round up to $11$, answer (E).