AMC 8 · 2008 · #20
Grade 6 rate-ratioProblem
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 43 of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
Pick an answer.
AMC 8 2008 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: In Mr. Neatkin's class, $\tfrac{2}{3}$ of the boys passed a penmanship test and $\tfrac{3}{4}$ of the girls passed. The number of boys who passed equals the number of girls who passed. Find the smallest possible total number of students in the class.
Givens: $\tfrac{2}{3}$ of the boys passed; $\tfrac{3}{4}$ of the girls passed; The number of passing boys equals the number of passing girls; Answer choices: (A) $12$, (B) $17$, (C) $24$, (D) $27$, (E) $36$
Unknowns: The smallest possible total number of students (boys $+$ girls)
Understand
Restated: In Mr. Neatkin's class, $\tfrac{2}{3}$ of the boys passed a penmanship test and $\tfrac{3}{4}$ of the girls passed. The number of boys who passed equals the number of girls who passed. Find the smallest possible total number of students in the class.
Givens: $\tfrac{2}{3}$ of the boys passed; $\tfrac{3}{4}$ of the girls passed; The number of passing boys equals the number of passing girls; Answer choices: (A) $12$, (B) $17$, (C) $24$, (D) $27$, (E) $36$
Plan
Primary tool: #3 Set Up an Equation
Secondary: #4 Introduce a Variable, #14 Consider Extreme Cases
The two fractions describe two counts that the problem says are equal — that is exactly an equation waiting to be written, so Tool #3 (Set Up an Equation) drives the work. Tool #4 (Introduce a Variable) gives us letters for the number of boys and the number of girls so we can write the equation. The word "minimum" is the cue for Tool #14 (Consider Extreme Cases): once the equation forces a ratio between boys and girls, the smallest whole-number pair fitting that ratio gives the answer.
Execute — Answer: B
6.EE.A.2 Step 1 - Name the unknowns.
- Let $b$ be the number of boys and $g$ be the number of girls in the class.
- Then $\tfrac{2}{3}b$ boys passed and $\tfrac{3}{4}g$ girls passed.
💡 Giving the two unknown counts their own letters is the standard Grade 6 move before any equation.
6.EE.B.7 Step 2 - Translate "equal number of boys and girls passed" into an equation.
- The boys who passed are $\tfrac{2}{3}b$ and the girls who passed are $\tfrac{3}{4}g$, and those two amounts are equal.
💡 "Equal" in English becomes "$=$" in math — a one-step Grade 6 translation.
6.NS.B.4 Step 3 - Clear the fractions to read the ratio of boys to girls.
- Multiply both sides by $12$ (the least common multiple of $3$ and $4$) to get rid of the denominators.
💡 Multiplying by the LCM is the Grade 6 way to turn a fraction equation into a clean whole-number equation.
6.RP.A.1 Step 4 - Read $8b = 9g$ as a ratio.
- Since $8$ and $9$ share no common factor, $b$ must be a multiple of $9$ and $g$ must be a multiple of $8$ for the equation to hold in whole numbers.
- Write the ratio of boys to girls.
💡 An equation like $8b = 9g$ with coprime coefficients $8$ and $9$ pins the ratio $b:g$ at $9:8$.
6.NS.B.4 Step 5 - Pick the smallest whole-number boys/girls counts in that ratio.
- Because $\gcd(8,9) = 1$, the smallest pair satisfying $b:g = 9:8$ is $b = 9$, $g = 8$.
- Check the constraints: $\tfrac{2}{3}(9) = 6$ boys passed and $\tfrac{3}{4}(8) = 6$ girls passed — equal, as required.
💡 "Smallest counts in a fixed ratio" is the Grade 6 GCF idea — take the numerator and denominator of the already-reduced fraction.
6.RP.A.3 Step 6 Add boys and girls for the smallest possible class size.
💡 Once the smallest ratio pair is found, the total is just their sum.
6.EE.A.2 Name the unknowns. Let $b$ be the number of boys and $g$ be the number of girls 6.EE.B.7 Translate "equal number of boys and girls passed" into an equation. The boys who 6.NS.B.4 Clear the fractions to read the ratio of boys to girls. Multiply both sides by $ 6.RP.A.1 Read $8b = 9g$ as a ratio. Since $8$ and $9$ share no common factor, $b$ must be 6.NS.B.4 Pick the smallest whole-number boys/girls counts in that ratio. Because $\gcd(8, 6.RP.A.3 Add boys and girls for the smallest possible class size. Review
Reasonableness: Check the next-smallest pair: $b = 18$, $g = 16$ also satisfies $8b = 9g$, and gives $\tfrac{2}{3}(18) = 12$ boys and $\tfrac{3}{4}(16) = 12$ girls passing — equal, but the total is $34$, larger than $17$. Every valid pair is a whole-number multiple of $(9, 8)$, so $17$ is truly the minimum. The wrong choices line up with traps: (A) $12$ ignores that $b$ and $g$ must be different sizes; (C) $24$, (D) $27$, (E) $36$ are larger multiples of the ratio.
Alternative: Tool #6 (Guess and Check) on the answer choices: (A) $12$ can't split as a multiple of $3$ plus a multiple of $4$ with equal passers ($\tfrac{2}{3}b = \tfrac{3}{4}g$ forces $b > g$, but if $b + g = 12$ the only candidates are $b = 9, g = 3$ giving passers $6$ vs $\tfrac{9}{4}$, not a whole number). For (B) $17$, try $b = 9, g = 8$: passers are $6$ and $6$, equal. Answer (B).
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the unknown counts of boys and girls with the letters $b$ and $g$ so the passing counts $\tfrac{2}{3}b$ and $\tfrac{3}{4}g$ can be written.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $px = q$ (Turning "an equal number of boys and girls passed" into the equation $\tfrac{2}{3}b = \tfrac{3}{4}g$.)6.NS.B.4Find the greatest common factor and least common multiple (Multiplying by $\mathrm{lcm}(3,4) = 12$ to clear fractions, and using $\gcd(8,9) = 1$ to find the smallest whole-number pair $(b, g) = (9, 8)$.)6.RP.A.1Understand the concept of a ratio and use ratio language to describe a ratio relationship (Reading $8b = 9g$ as the ratio of boys to girls is $9 : 8$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Choosing the smallest whole-number pair in the ratio $9 : 8$ to minimize the total $b + g$.)
⭐ When two fractions of two groups give the same count, write one equation, clear the fractions, and the smallest whole numbers in the resulting ratio give the answer — here $9$ boys and $8$ girls, total $17$.
⭐ When two fractions of two groups give the same count, write one equation, clear the fractions, and the smallest whole numbers in the resulting ratio give the answer — here $9$ boys and $8$ girls, total $17$.
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