AMC 10 · 2009 · #14

Grade 6 arithmetic

Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share

sequences-geometricfraction-arithmeticpattern-recognition pattern-recognition ↑ Prerequisites: sequences-geometric

📏 Long solution 💡 3 insights

Problem

On Monday, Millie puts a quart of seeds, $25\%$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $25\%$ of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?

Pick an answer.

(A)

Tuesday

(B)

Wednesday

(C)

Thursday

(D)

Friday

(E)

Saturday

AMC 10 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Each day, starting Monday, Millie adds one quart of a seed mix that is $\tfrac14$ millet and $\tfrac34$ other seeds. Before each new day's seeds go in, the birds eat $25\%$ of the millet already there and $100\%$ of the other seeds. Find the first day, measured right after Millie adds the seeds, when more than half of everything in the feeder is millet.

Givens: Each added quart is $25\%$ millet ($\tfrac14$ quart) and $75\%$ other ($\tfrac34$ quart); One quart is added every day, starting Monday; Each day birds leave $75\%$ of the millet and eat all other seeds; We look at the feeder right after new seeds are added

Unknowns: The first day on which millet is more than half of all seeds

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #4 Introduce a Variable, #6 Guess and Check

The other seeds are the easy half: birds eat all of them every day, so right after adding, there is always exactly $\tfrac34$ quart of other seeds. That turns the whole question into one moving quantity — the millet. Tool #4 (Introduce a Variable) names the millet on day $n$ as $m_n$ and captures the daily rule $m_{n+1}=\tfrac34 m_n+\tfrac14$. Tool #6 (Guess and Check) runs the rule day by day, and Tool #5 (Look for a Pattern) spots the clean closed form $m_n = 1-\left(\tfrac34\right)^n$, which makes the winning condition transparent.

Execute — Answer: D

#4 Introduce a Variable 6.RP.A.3 Step 1

Fix the other seeds at 3/4

Each quart is $\tfrac14$ millet and $\tfrac34$ other seeds.
The birds eat every non-millet seed each day, so whatever other seeds were there yesterday are gone.
Right after Millie adds a fresh quart, the only other seeds present are the $\tfrac34$ quart she just poured in.
So the other-seed amount is always $\tfrac34$.
"More than half is millet" therefore means millet must beat that fixed $\tfrac34$ quart.

$$\text{other seeds} = \tfrac34 \text{ (always)} \quad\Rightarrow\quad \text{need millet} > \tfrac34$$

💡 If the other seeds are wiped out and refilled to the same amount every day, they become a fixed finish line the millet has to cross.

#4 Introduce a Variable 5.NF.B.4 Step 2

Write the daily millet rule

Let $m_n$ be the quarts of millet right after adding on day $n$.
On day 1 (Monday) that is just the fresh $\tfrac14$ quart, so $m_1=\tfrac14$.
To get the next day: the birds leave $75\%$ of the millet, which is $\tfrac34 m_n$, and then Millie adds another $\tfrac14$ quart of millet.
So $m_{n+1}=\tfrac34 m_n+\tfrac14$.

$$m_1=\tfrac14,\qquad m_{n+1}=\tfrac34 m_n+\tfrac14$$

💡 Keep three-quarters of what survives, then add one more quarter — the same two moves repeat every single day.

#6 Guess and Check 6.NS.A.1 Step 3

Run it day by day

Apply the rule and compare each result to the $\tfrac34$ finish line.
$m_1=\tfrac14$.
$m_2=\tfrac34\cdot\tfrac14+\tfrac14=\tfrac{7}{16}$.
$m_3=\tfrac34\cdot\tfrac{7}{16}+\tfrac14=\tfrac{37}{64}\approx0.578$.
$m_4=\tfrac34\cdot\tfrac{37}{64}+\tfrac14=\tfrac{175}{256}\approx0.684$, still below $0.75$.
$m_5=\tfrac34\cdot\tfrac{175}{256}+\tfrac14=\tfrac{781}{1024}\approx0.763$, which is above $0.75$.

$$m_1=\tfrac14,\; m_2=\tfrac{7}{16},\; m_3=\tfrac{37}{64},\; m_4=\tfrac{175}{256}<\tfrac34,\; m_5=\tfrac{781}{1024}>\tfrac34$$

💡 The millet climbs a little every day, so just march forward until it first passes the fixed three-quarters mark.

#5 Look for a Pattern 6.EE.A.1 Step 4

See the pattern behind the numbers

The fractions $\tfrac14,\tfrac{7}{16},\tfrac{37}{64},\tfrac{175}{256}$ look messy, but writing each as "$1$ minus something" reveals it: $\tfrac14=1-\tfrac{3}{4}$, $\tfrac{7}{16}=1-\tfrac{9}{16}$, $\tfrac{37}{64}=1-\tfrac{27}{64}$.
The subtracted piece is a power of $\tfrac34$, so $m_n=1-\left(\tfrac34\right)^n$.
Then "millet $>\tfrac34$" becomes $1-\left(\tfrac34\right)^n>\tfrac34$, i.e.
$\left(\tfrac34\right)^n<\tfrac14$.

$$m_n=1-\left(\tfrac34\right)^n \quad\Rightarrow\quad m_n>\tfrac34 \iff \left(\tfrac34\right)^n<\tfrac14$$

💡 The leftover "not-yet-millet" share shrinks by three-quarters each day, so the gap to a full quart is exactly a power of $\tfrac34$.

#6 Guess and Check 6.NS.C.7 Step 5

Find the first winning day

Check the powers of $\tfrac34$: $\left(\tfrac34\right)^4=\tfrac{81}{256}\approx0.316$, still bigger than $\tfrac14=0.25$; $\left(\tfrac34\right)^5=\tfrac{243}{1024}\approx0.237$, which is finally less than $\tfrac14$.
So the condition first holds at $n=5$.
Counting from Monday as day 1, day 5 is Friday.
The answer is $\textbf{(D)}$.

$$\left(\tfrac34\right)^4\approx0.316>\tfrac14,\quad \left(\tfrac34\right)^5\approx0.237<\tfrac14 \;\Rightarrow\; n=5=\text{Friday} \Rightarrow \textbf{(D)}$$

💡 The tipping point is the first exponent that drags $\left(\tfrac34\right)^n$ under one-quarter — pick that day and count off from Monday.

[1] #4 6.RP.A.3 Each quart is $\tfrac14$ millet and $\tfrac34$ other seeds. The birds eat every

[2] #4 5.NF.B.4 Let $m_n$ be the quarts of millet right after adding on day $n$. On day 1 (Monda

[3] #6 6.NS.A.1 Apply the rule and compare each result to the $\tfrac34$ finish line. $m_1=\tfra

[4] #5 6.EE.A.1 The fractions $\tfrac14,\tfrac{7}{16},\tfrac{37}{64},\tfrac{175}{256}$ look mess

[5] #6 6.NS.C.7 Check the powers of $\tfrac34$: $\left(\tfrac34\right)^4=\tfrac{81}{256}\approx0

Review

Reasonableness: On day 4 the millet is $\tfrac{175}{256}\approx0.684$ quart against $0.75$ quart of other seeds — still losing, so Thursday is too early, which rules out (C) and everything before it. On day 5 the millet is $\tfrac{781}{1024}\approx0.763$ against $0.75$, a slim but real win, so Friday is exactly the first day it crosses over. The margin being small and positive is a good sign we did not overshoot to Saturday (E). Everything is consistent with (D).

Alternative: Skip fractions and think about the shortfall directly. Right after each day the feeder holds $1$ quart of millet minus whatever has not accumulated yet, and that missing piece is multiplied by $\tfrac34$ daily starting from $\tfrac34$. Millet beats the fixed $\tfrac34$ of other seeds once that missing piece drops below $\tfrac14$: $\tfrac34,\tfrac{9}{16},\tfrac{27}{64},\tfrac{81}{256},\tfrac{243}{1024}$ — the fifth term is the first under $\tfrac14$, again giving Friday.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Reading $25\%$ millet as $\tfrac14$ quart and recognizing that "more than half is millet" means millet must exceed the fixed $\tfrac34$ quart of other seeds.)
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Taking $\tfrac34$ of the surviving millet each day to build the recurrence $m_{n+1}=\tfrac34 m_n+\tfrac14$.)
6.NS.A.1 Interpret and compute quotients of fractions and solve word problems (Computing the day-by-day millet amounts $\tfrac14,\tfrac{7}{16},\tfrac{37}{64},\tfrac{175}{256},\tfrac{781}{1024}$ with fraction arithmetic.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Spotting and evaluating the closed form $m_n = 1-\left(\tfrac34\right)^n$ and the powers $\left(\tfrac34\right)^n$.)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Comparing each $\left(\tfrac34\right)^n$ to $\tfrac14$ to find the first day the millet share crosses one half.)

⭐ When one part gets wiped out and refilled to the same size each day, it becomes a fixed target the other part just has to grow past.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Fix the other seeds at 3/4

Write the daily millet rule

Run it day by day

See the pattern behind the numbers

Find the first winning day

Review

CCSS standards used (min grade 6)

More like this