AMC 10 · 2008 · #6
Grade 6 rate-ratioPoints B and C lie on AD. The length of AB is 4 times the length of BD, and the length of AC is 9 times the length of CD. The length of BC is what fraction of the length of AD?
Pick an answer.
AMC 10 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: Points $B$ and $C$ sit on segment $\overline{AD}$. The piece $\overline{AB}$ is $4$ times as long as $\overline{BD}$, and the piece $\overline{AC}$ is $9$ times as long as $\overline{CD}$. Find what fraction of $\overline{AD}$ the middle piece $\overline{BC}$ takes up.
Givens: $B$ and $C$ both lie on the segment $\overline{AD}$; $AB = 4 \cdot BD$; $AC = 9 \cdot CD$; Answer choices: (A) $\frac{1}{36}$, (B) $\frac{1}{13}$, (C) $\frac{1}{10}$, (D) $\frac{5}{36}$, (E) $\frac{1}{5}$
Unknowns: The ratio $\dfrac{BC}{AD}$, i.e. what fraction of the whole segment the piece $\overline{BC}$ is
Understand
Restated: Points $B$ and $C$ sit on segment $\overline{AD}$. The piece $\overline{AB}$ is $4$ times as long as $\overline{BD}$, and the piece $\overline{AC}$ is $9$ times as long as $\overline{CD}$. Find what fraction of $\overline{AD}$ the middle piece $\overline{BC}$ takes up.
Givens: $B$ and $C$ both lie on the segment $\overline{AD}$; $AB = 4 \cdot BD$; $AC = 9 \cdot CD$; Answer choices: (A) $\frac{1}{36}$, (B) $\frac{1}{13}$, (C) $\frac{1}{10}$, (D) $\frac{5}{36}$, (E) $\frac{1}{5}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable
Everything here is about positions along one straight segment, which is exactly what Tool #1 (Draw a Diagram) is for. If I mark where $B$ and $C$ fall on $\overline{AD}$ using fractions of the whole, the answer $\overline{BC}$ is just the gap between the two marks. Tool #4 (Introduce a Variable) supports this by letting me call the whole length $\overline{AD}$ one unit, so each ratio condition turns directly into a fraction I can plot.
Execute — Answer: C
6.EE.B.6 Step 1 Call the whole segment 1
- Because both clues are ratios, the true length of $\overline{AD}$ is irrelevant — pick the most convenient value.
- Let $AD = 1$.
- Then every other length becomes the fraction of $\overline{AD}$ that it represents, which is exactly what the question asks for.
💡 Naming the whole "1" makes every piece automatically read out as a fraction of the whole.
6.RP.A.3 Step 2 Locate B
- $AB = 4 \cdot BD$ means the two pieces $\overline{AB}$ and $\overline{BD}$ are in the ratio $4 : 1$, so together they form $4 + 1 = 5$ equal parts of $\overline{AD}$.
- Thus $BD = \tfrac{1}{5}$ and $AB = \tfrac{4}{5}$, so $B$ sits $\tfrac{4}{5}$ of the way from $A$.
💡 A $4:1$ split means $5$ equal parts, and $B$ lands at the $4$-part mark.
6.RP.A.3 Step 3 Locate C
- $AC = 9 \cdot CD$ means $\overline{AC}$ and $\overline{CD}$ are in the ratio $9 : 1$, giving $9 + 1 = 10$ equal parts of $\overline{AD}$.
- Thus $CD = \tfrac{1}{10}$ and $AC = \tfrac{9}{10}$, so $C$ sits $\tfrac{9}{10}$ of the way from $A$.
💡 A $9:1$ split means $10$ equal parts, and $C$ lands at the $9$-part mark.
5.NF.A.1 Step 4 Measure the gap BC
- Plot both marks on the segment.
- Since $\tfrac{4}{5} = \tfrac{8}{10}$ is less than $\tfrac{9}{10}$, the order along $\overline{AD}$ is $A, B, C, D$.
- The piece $\overline{BC}$ is the gap between them, so subtract: $BC = AC - AB = \tfrac{9}{10} - \tfrac{4}{5} = \tfrac{9}{10} - \tfrac{8}{10} = \tfrac{1}{10}$.
- Since $AD = 1$, this fraction is the answer, choice $\textbf{(C)}$.
💡 On the diagram $\overline{BC}$ is simply the distance from $B$'s mark to $C$'s mark.
6.EE.B.6 Because both clues are ratios, the true length of $\overline{AD}$ is irrelevant 6.RP.A.3 $AB = 4 \cdot BD$ means the two pieces $\overline{AB}$ and $\overline{BD}$ are i 6.RP.A.3 $AC = 9 \cdot CD$ means $\overline{AC}$ and $\overline{CD}$ are in the ratio $9 5.NF.A.1 Plot both marks on the segment. Since $\tfrac{4}{5} = \tfrac{8}{10}$ is less tha Review
Reasonableness: $B$ is at $0.8$ and $C$ is at $0.9$ along a segment of length $1$, so $\overline{BC}$ should be a small sliver — about a tenth. The value $\tfrac{1}{10}$ matches, and it is safely between the tiny $\tfrac{1}{36}$ and the larger $\tfrac{1}{5}$. Both points also correctly land strictly inside the segment ($0.8$ and $0.9$ are between $0$ and $1$), confirming the picture is valid.
Alternative: Use a common length instead of $1$. The least common multiple of $5$ and $10$ is $10$, so let $AD = 10$. Then $AB = \tfrac{4}{5}\cdot 10 = 8$ and $AC = \tfrac{9}{10}\cdot 10 = 9$, giving whole-number positions. So $BC = 9 - 8 = 1$, and $\tfrac{BC}{AD} = \tfrac{1}{10}$ — the same answer with no fraction subtraction needed.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Setting the whole segment $AD = 1$ (a unit) so that every sub-length reads out directly as a fraction of $\overline{AD}$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Turning $AB = 4\cdot BD$ into the $4:1$ split (5 parts, $B$ at $\tfrac{4}{5}$) and $AC = 9\cdot CD$ into the $9:1$ split (10 parts, $C$ at $\tfrac{9}{10}$).)5.NF.A.1Add and subtract fractions with unlike denominators (Computing the gap $BC = \tfrac{9}{10} - \tfrac{4}{5} = \tfrac{1}{10}$ by giving the fractions a common denominator.)
⭐ Turn each "$4$ times" or "$9$ times" clue into equal parts, mark both points on one line, and the middle piece is just the gap between the marks.
⭐ Turn each "$4$ times" or "$9$ times" clue into equal parts, mark both points on one line, and the middle piece is just the gap between the marks.
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