AMC 8 · 2020 · #5

Grade 4 rate-ratio
fraction-arithmeticpercentagefraction-decimal-conversion identify-subproblems ↑ Prerequisites: fraction-arithmeticpercentage
📏 Short solution 💡 2 insights
📘 View easy version →

Problem

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of 55 cups. What percent of the total capacity of the pitcher did each cup receive?

(A) 5(B) 10(C) 15(D) 20(E) 25\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25

Pick an answer.

(A)
5
(B)
10
(C)
15
(D)
20
(E)
25
View mode:

Toolkit + CCSS Solution

Understand

Restated: A pitcher is $\tfrac{3}{4}$ full of pineapple juice. All of that juice is shared equally among $5$ cups. What percent of the pitcher's total capacity does the juice in each cup represent?

Givens: The pitcher is filled to $\tfrac{3}{4}$ of its total capacity with juice; All the juice is poured into $5$ cups, each receiving an equal amount; Answer choices: (A) $5$, (B) $10$, (C) $15$, (D) $20$, (E) $25$ (all in percent)

Unknowns: The amount of juice in each cup, expressed as a percent of the pitcher's total capacity

Understand

Restated: A pitcher is $\tfrac{3}{4}$ full of pineapple juice. All of that juice is shared equally among $5$ cups. What percent of the pitcher's total capacity does the juice in each cup represent?

Givens: The pitcher is filled to $\tfrac{3}{4}$ of its total capacity with juice; All the juice is poured into $5$ cups, each receiving an equal amount; Answer choices: (A) $5$, (B) $10$, (C) $15$, (D) $20$, (E) $25$ (all in percent)

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

The question hides two clean subproblems: (1) what percent of the pitcher is actually juice, and (2) once that juice is split into $5$ equal parts, how big is one part? Tool #7 (Identify Subproblems) names these two sub-questions and solves them one at a time. Because this is multiple choice, Tool #3 (Eliminate Possibilities) gives us a fast cross-check: $5$ times the correct answer must equal the total juice percent ($75\%$), so only one choice survives.

Execute — Answer: C

#7 Identify Subproblems 4.NF.A.1 Step 1
  • Subproblem 1: express the juice in the pitcher as a percent of the pitcher's total capacity.
  • Rewrite $\tfrac{3}{4}$ with denominator $100$ by multiplying top and bottom by $25$.
$$\tfrac{3}{4} = \tfrac{3 \times 25}{4 \times 25} = \tfrac{75}{100}$$

💡 Rewriting a fraction with an equivalent denominator is the Grade 4 fraction-equivalence move ($\tfrac{3}{4} = \tfrac{75}{100}$).

#7 Identify Subproblems 4.NF.C.6 Step 2
  • Read the equivalent fraction $\tfrac{75}{100}$ as a percent.
  • "Percent" literally means "per one hundred", so $\tfrac{75}{100} = 75\%$.
  • The pitcher holds juice equal to $75\%$ of its capacity.
$$\tfrac{75}{100} = 75\%$$

💡 Reading a fraction with denominator $100$ as a percent is the Grade 4 decimal/percent notation skill.

#7 Identify Subproblems 3.OA.C.7 Step 3
  • Subproblem 2: share the $75\%$ of juice equally among $5$ cups.
  • That is a plain whole-number division: $75 \div 5$.
$$75\% \div 5 = 15\%$$

💡 Dividing $75$ by $5$ within $100$ is a Grade 3 fluency fact ($5 \times 15 = 75$).

#3 Eliminate Possibilities 3.OA.C.7 Step 4
  • Verify by elimination against the answer choices.
  • Each cup's percent, multiplied by $5$, must equal the total juice percent $75\%$.
  • Test the choices: (A) $5 \times 5 = 25$; (B) $10 \times 5 = 50$; (C) $15 \times 5 = 75$ ✓; (D) $20 \times 5 = 100$; (E) $25 \times 5 = 125$.
  • Only choice (C) hits $75\%$, confirming the answer.
$$15 \times 5 = 75 \;\Rightarrow\; \textbf{(C)}$$

💡 Testing each choice against the constraint "$5$ cups must add back to $75\%$" is a Grade 3 multiplication fluency check.

[1] #7 4.NF.A.1 Subproblem 1: express the juice in the pitcher as a percent of the pitcher's tot
[2] #7 4.NF.C.6 Read the equivalent fraction $\tfrac{75}{100}$ as a percent. "Percent" literally
[3] #7 3.OA.C.7 Subproblem 2: share the $75\%$ of juice equally among $5$ cups. That is a plain
[4] #3 3.OA.C.7 Verify by elimination against the answer choices. Each cup's percent, multiplied

Review

Reasonableness: If the pitcher were $100\%$ full, splitting it among $5$ cups would give each cup $20\%$. Our pitcher is only $75\%$ full, so each cup should get a bit less than $20\%$ — and $15\%$ fits that picture exactly. The cups also account for the juice: $5 \times 15\% = 75\%$, matching the $\tfrac{3}{4}$ we started with.

Alternative: Tool #9 (Solve an Easier Related Problem): pretend the pitcher holds exactly $\$75$ instead of $75\%$ of its capacity. Splitting $\$75$ among $5$ cups gives $\$15$ per cup, so each cup gets $15$ of the same units the whole pitcher was measured in — i.e. $15\%$ of the pitcher. The dollars-and-cents version exposes the structure without any percent machinery.

CCSS standards used (min grade 4)

  • 4.NF.A.1 Explain why a fraction is equivalent to another fraction (Rewriting $\tfrac{3}{4}$ as the equivalent fraction $\tfrac{75}{100}$ by multiplying numerator and denominator by $25$.)
  • 4.NF.C.6 Use decimal notation for fractions with denominators 10 or 100 (Reading $\tfrac{75}{100}$ as $75\%$ ("per one hundred") so the juice can be expressed directly as a percent of pitcher capacity.)
  • 3.OA.C.7 Fluently multiply and divide within 100 (Dividing $75 \div 5 = 15$ to share the juice equally among the $5$ cups, and verifying with $15 \times 5 = 75$.)

⭐ This AMC 8 problem only needs Grade 4 fraction-to-percent thinking ($\tfrac{3}{4} = \tfrac{75}{100} = 75\%$) plus a Grade 3 division fact you already know!

⭐ This AMC 8 problem only needs Grade 4 fraction-to-percent thinking ($\tfrac{3}{4} = \tfrac{75}{100} = 75\%$) plus a Grade 3 division fact you already know!