AMC 8 · 2004 · #6

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

percentagefraction-arithmeticlinear-equations-one-var identify-subproblemsconvert-to-algebra ↑ Prerequisites: multi-digit-arithmeticfraction-decimal-conversion

📏 Short solution 💡 2 insights

Problem

After Sally takes $20$ shots, she has made $55\%$ of her shots. After she takes $5$ more shots, she raises her percentage to $56\%$ . How many of the last $5$ shots did she make?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2004 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Sally has made $55\%$ of her first $20$ shots. After she takes $5$ more shots, her overall success rate rises to $56\%$. How many of those last $5$ shots did she make?

Givens: First $20$ shots taken, $55\%$ made; $5$ more shots taken, raising the overall total to $25$ shots; New overall success rate is $56\%$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The number of made shots among the last $5$ attempts

Understand

Restated: Sally has made $55\%$ of her first $20$ shots. After she takes $5$ more shots, her overall success rate rises to $56\%$. How many of those last $5$ shots did she make?

Plan

Primary tool: #3 Set Up an Equation

Secondary: #4 Introduce a Variable

The problem links two percentages of the same shooter, so the natural move is Tool #3 (Set Up an Equation): write the new success rate as a fraction whose numerator and denominator both depend on what happens in the last $5$ shots. Tool #4 (Introduce a Variable) names the unknown count $x$ of made shots among those last $5$, so we can turn the percentage sentence into a single linear equation and solve. Once $x$ is isolated, the answer is one subtraction away.

Execute — Answer: C

#3 Set Up an Equation 6.RP.A.3 Step 1

Find how many of the first $20$ shots Sally made.
A $55\%$ success rate over $20$ shots means $55\%$ of $20$ are made.

$$\text{first made} = 0.55 \times 20 = 11$$

💡 Percent of a whole is the Grade 6 "percent times total" move: $55\%$ of $20$ is $11$.

#4 Introduce a Variable 6.EE.A.2 Step 2

Name the unknown.
Let $x$ be the number of made shots among the last $5$.
After all $25$ shots, total made is $11 + x$ out of $25$.

$$\text{total made} = 11 + x, \quad \text{total taken} = 20 + 5 = 25$$

💡 One letter for the unknown count keeps the rest of the work as plain arithmetic.

#3 Set Up an Equation 6.EE.B.7 Step 3

Translate "new rate is $56\%$" into an equation.
The new success rate equals total made divided by total taken.

$$\dfrac{11 + x}{25} = 0.56$$

💡 A percent equation is just a fraction set equal to a decimal — exactly what Grade 6 "solve $px = q$" handles.

#3 Set Up an Equation 6.EE.B.7 Step 4

Solve for $x$.
Multiply both sides by $25$, then subtract $11$.

$$11 + x = 0.56 \times 25 = 14 \;\Rightarrow\; x = 14 - 11 = 3 \;\Rightarrow\; \textbf{(C)}$$

💡 Clearing the denominator turns the percent equation into one-step arithmetic.

[1] #3 6.RP.A.3 Find how many of the first $20$ shots Sally made. A $55\%$ success rate over $20

[2] #4 6.EE.A.2 Name the unknown. Let $x$ be the number of made shots among the last $5$. After

[3] #3 6.EE.B.7 Translate "new rate is $56\%$" into an equation. The new success rate equals tot

[4] #3 6.EE.B.7 Solve for $x$. Multiply both sides by $25$, then subtract $11$.

Review

Reasonableness: Plug $x = 3$ back in. Total made is $11 + 3 = 14$ out of $25$, and $14 \div 25 = 0.56 = 56\%$. That matches the problem exactly. The answer is also a whole number from $0$ to $5$, which it must be since you cannot make a fractional shot. A quick sanity check on the size: the rate went up by only $1$ percentage point, so Sally must have made a bit more than $55\%$ of the last $5$ shots — and $3$ out of $5$ is $60\%$, just a touch above her old rate, which fits.

Alternative: Tool #6 (Guess and Check): Sally needs a whole number of made shots from $0$ to $5$. After $25$ shots, $56\%$ means $0.56 \times 25 = 14$ total makes. She already had $11$, so she needs exactly $14 - 11 = 3$ more makes. Same answer (C), reached by computing the required total first instead of writing an equation.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including finding a percent of a quantity (Computing $55\%$ of $20$ to find the $11$ shots Sally made in her first $20$ attempts.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Introducing the variable $x$ for the unknown number of made shots in the last $5$ attempts.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Writing $(11 + x)/25 = 0.56$ and solving step by step to get $x = 3$.)

⭐ Percent problems get easy once you turn the percent into a fraction with the right top and bottom. Here, $56\%$ of $25$ shots is exactly $14$ makes, and Sally already had $11$ — so she needed $14 - 11 = 3$ more makes in those last $5$ shots.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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