AMC 8 · 2009 · #21

Grade 6 arithmetic

Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known

mean-median-mode-rangeratio-proportionset-partition identify-subproblemspattern-recognition ↑ Prerequisites: mean-median-mode-rangeratio-proportion

📏 Medium solution 💡 3 insights

Problem

Andy and Bethany have a rectangular array of numbers with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A$ . Bethany adds the numbers in each column. The average of her $75$ sums is $B$ . What is the value of $\frac{A}{B}$ ?

Pick an answer.

(A)

$\frac{64}{225}$

(B)

$\frac{8}{15}$

(C)

(D)

$\frac{15}{8}$

(E)

$\frac{225}{64}$

AMC 8 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: A rectangular table of numbers has $40$ rows and $75$ columns. Andy sums each row and averages his $40$ row sums to get $A$. Bethany sums each column and averages her $75$ column sums to get $B$. Find $\dfrac{A}{B}$.

Givens: The array has $40$ rows and $75$ columns; $A$ = the average of the $40$ row sums; $B$ = the average of the $75$ column sums; Answer choices: (A) $\tfrac{64}{225}$, (B) $\tfrac{8}{15}$, (C) $1$, (D) $\tfrac{15}{8}$, (E) $\tfrac{225}{64}$

Unknowns: The value of the ratio $\dfrac{A}{B}$

Understand

Plan

Primary tool: #11 Find an Invariant

Secondary: #4 Introduce a Variable

The actual numbers in the array are never given, which is the tell that Tool #11 (Find an Invariant) applies: there must be a quantity that does not depend on the specific entries. That invariant is $S$, the grand total of all numbers in the array — counted by rows or by columns, it is the same value. Tool #4 (Introduce a Variable) lets us name $S$ so we can write $A$ and $B$ in terms of it. Once both averages share $S$, the ratio $A/B$ falls out as a ratio of two counts.

Execute — Answer: D

#11 Find an Invariant 6.EE.A.2 Step 1

Name the grand total.
Let $S$ be the sum of every number in the array.
Whether you add row by row or column by column, you visit each cell exactly once, so both methods give $S$.

$$S = \text{sum of all entries in the } 40 \times 75 \text{ array}$$

💡 Naming the unchanging total with a single letter is the Grade 6 "write expressions with variables" move.

#4 Introduce a Variable 6.SP.B.5 Step 2

Write $A$ in terms of $S$.
Andy's $40$ row sums add up to $S$, and $A$ is their average, so $A = \dfrac{S}{40}$.
Rearrange to get $S$ alone.

$$A = \dfrac{S}{40} \;\Rightarrow\; S = 40A$$

💡 The Grade 6 mean formula says average $\times$ count $=$ total. Here count $= 40$, so total $= 40A$.

#4 Introduce a Variable 6.SP.B.5 Step 3

Write $B$ in terms of $S$ the same way.
Bethany's $75$ column sums also add up to $S$, and $B$ is their average.

$$B = \dfrac{S}{75} \;\Rightarrow\; S = 75B$$

💡 Same mean formula, this time with count $= 75$, so the same $S$ also equals $75B$.

#11 Find an Invariant 6.RP.A.1 Step 4

Equate the two expressions for $S$ and solve for the ratio.
Both $40A$ and $75B$ equal the same $S$, so they equal each other.

$$40A = 75B \;\Rightarrow\; \dfrac{A}{B} = \dfrac{75}{40} = \dfrac{15}{8} \;\Rightarrow\; \textbf{(D)}$$

💡 Dividing both sides by $40B$ turns the equation into a Grade 6 ratio. Simplifying $\tfrac{75}{40}$ by the common factor $5$ gives $\tfrac{15}{8}$.

[1] #11 6.EE.A.2 Name the grand total. Let $S$ be the sum of every number in the array. Whether y

[2] #4 6.SP.B.5 Write $A$ in terms of $S$. Andy's $40$ row sums add up to $S$, and $A$ is their

[3] #4 6.SP.B.5 Write $B$ in terms of $S$ the same way. Bethany's $75$ column sums also add up t

[4] #11 6.RP.A.1 Equate the two expressions for $S$ and solve for the ratio. Both $40A$ and $75B$

Review

Reasonableness: Sanity check with an all-$1$s array: every entry is $1$, so each row sum is $75$ and each column sum is $40$. Then $A = 75$, $B = 40$, and $\dfrac{A}{B} = \dfrac{75}{40} = \dfrac{15}{8}$. The ratio matches answer (D), independent of the specific entries — exactly the invariant we expected. Also, $A > B$ makes sense because each row has more cells ($75$) than each column has ($40$), so row sums are typically larger than column sums.

Alternative: Tool #1 (Try a Simpler Case): replace the array with one where every entry is the same number, say $1$. Row sums are all $75$ (so $A = 75$) and column sums are all $40$ (so $B = 40$), giving $\dfrac{A}{B} = \dfrac{75}{40} = \dfrac{15}{8}$. Because the problem's answer must be the same for every array, this simpler case pins down (D).

CCSS standards used (min grade 6)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the grand total of the array with the variable $S$ so that both averages can be written as algebraic expressions.)
6.SP.B.5 Summarize numerical data sets, including reporting the number of observations and measures of center (Using the definition of mean (sum $\div$ count) to write $A = S/40$ and $B = S/75$, which gives $S = 40A = 75B$.)
6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship (Reading $40A = 75B$ as the ratio $\dfrac{A}{B} = \dfrac{75}{40} = \dfrac{15}{8}$.)

⭐ Row sums and column sums add up to the same grand total — once you spot that invariant, this AMC 8 problem becomes a Grade 6 mean-and-ratio exercise!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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