AMC 8 · 2022 · #19

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangegraph-readingsystematic-enumeration caseworkbound-inequality-then-enumerate ↑ Prerequisites: mean-median-mode-rangegraph-reading

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores.

Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$ . What is the minimum number of students who received extra points?

(Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.)

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A class of $20$ students took a test whose scores are shown on a dot plot. After regrading, the teacher added $5$ extra points to some students, and the new median jumped to $85$. We want the smallest possible number of students who received the $5$-point bonus.

Givens: Score counts from the dot plot: $65$ has $2$, $70$ has $2$, $75$ has $4$, $80$ has $5$, $85$ has $2$, $90$ has $3$, $95$ has $1$, $100$ has $1$ (total $20$); Each adjusted student's score increases by exactly $5$ points; The new median of the $20$ adjusted scores is $85$; With $20$ scores, the median is the average of the $10$th and $11$th values when sorted; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$

Unknowns: The minimum number of students who must receive the $5$-point bonus to push the median up to $85$

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #9 Solve an Easier Related Problem, #6 Guess and Check

The dot plot is small enough to lay out every score as an ordered list (Tool #2), which lets us see exactly which positions are the $10$th and $11$th. Tool #9 (Easier Problem) is the key insight: instead of tracking many students, ask the simpler question "how many of the $20$ scores need to be at least $85$?". Because the median position is fixed (positions $10$ and $11$), we only need both of those slots to read $85$. Tool #6 (Guess and Check) verifies that $4$ works and $3$ does not — a clean minimum-existence argument.

Execute — Answer: C

#2 Make a Systematic List 6.SP.B.4 Step 1

Lay out all $20$ scores in increasing order using the dot-plot counts.
This makes the $10$th and $11$th slots — the ones that determine the median — easy to spot.

$$\underbrace{65, 65}_{\text{1-2}},\, \underbrace{70, 70}_{\text{3-4}},\, \underbrace{75, 75, 75, 75}_{\text{5-8}},\, \underbrace{80, 80, 80, 80, 80}_{\text{9-13}},\, \underbrace{85, 85}_{\text{14-15}},\, \underbrace{90, 90, 90}_{\text{16-18}},\, 95,\, 100$$

💡 Writing the dot plot as one ordered list is the Grade 6 way of organizing dot-plot data so we can read off particular positions.

#2 Make a Systematic List 6.SP.A.3 Step 2

Confirm the original median.
With $20$ scores, the median is the mean of the $10$th and $11$th scores.
From the list both are $80$, so the original median is $80$ — too low.
We need the $10$th and $11$th scores to average $85$.

$$\text{old median} = \dfrac{80 + 80}{2} = 80$$

💡 The Grade 6 definition of median — the middle value — tells us exactly which two list positions matter.

#9 Solve an Easier Related Problem 6.SP.A.3 Step 3

Reduce to a simpler question (Tool #9).
The cleanest way to get a median of $85$ is to make BOTH the $10$th and $11$th scores equal to $85$.
That means we need at least $10$ scores in the final list to be at least $85$ (so that $85$ lands at positions $10$ and $11$).
Currently only $2 + 3 + 1 + 1 = 7$ scores are already $\ge 85$, so we need to lift $10 - 7 = 3$ more scores up to (or above) $85$.

$$\text{scores already} \ge 85: \; 2 + 3 + 1 + 1 = 7. \quad \text{need: } 10 - 7 = 3 \text{ more.}$$

💡 Asking "how many slots at or above $85$ do I need?" is much easier than tracking each student — the same Grade 6 median idea, viewed from the other side.

#6 Guess and Check 6.SP.A.3 Step 4

Check which scores can actually be lifted to $\ge 85$ by adding only $5$.
Adding $5$ to a $65$, $70$, or $75$ score gives $70$, $75$, or $80$ — still below $85$.
Only a score of $80$ becomes $85$ when boosted by $5$.
So every "useful" bonus must go to an $80$ student.

$$80 + 5 = 85 \;\checkmark, \quad 75 + 5 = 80 < 85, \quad 70 + 5 = 75 < 85, \quad 65 + 5 = 70 < 85$$

💡 Guess-and-check across the four possible source scores quickly shows only $80$s clear the bar.

#6 Guess and Check 6.SP.A.3 Step 5

Putting it together: we need $3$ more scores at $\ge 85$, and the only way is to convert three $80$s into $85$s.
But we should double-check whether $3$ bonuses actually push the MEDIAN to $85$.
With $3$ bonuses, the $80$ count drops from $5$ to $2$ and the $85$ count rises from $2$ to $5$.
The first $9$ positions are $65, 65, 70, 70, 75, 75, 75, 75, 80$, position $10$ is $80$, position $11$ is $85$.
Median $= \tfrac{80 + 85}{2} = 82.5 \ne 85$.
So $3$ is NOT enough — one $80$ still sneaks into position $10$.

$$\text{with 3 bonuses: positions 10 and 11 are } 80, 85 \Rightarrow \text{median} = \dfrac{80 + 85}{2} = 82.5$$

💡 Always test the candidate minimum — Tool #6 catches the off-by-one that a quick count misses.

#6 Guess and Check 6.SP.A.3 Step 6

Try $4$ bonuses (the next choice).
Convert four of the five $80$s into $85$s.
Now the count below $85$ is $2 + 2 + 4 + 1 = 9$, so positions $1$ through $9$ end at score $80$ and positions $10$ and $11$ are both $85$.
Median $= \tfrac{85 + 85}{2} = 85$ exactly — it works.
Since $3$ failed and $4$ succeeds, the minimum is $4$, which is choice $\textbf{(C)}$.

$$\text{with 4 bonuses: positions 10 and 11 are both } 85 \Rightarrow \text{median} = \dfrac{85 + 85}{2} = 85 \;\Rightarrow\; \textbf{(C)}$$

💡 Once positions $10$ and $11$ both read $85$, the median definition forces the answer to be $85$.

[1] #2 6.SP.B.4 Lay out all $20$ scores in increasing order using the dot-plot counts. This make

[2] #2 6.SP.A.3 Confirm the original median. With $20$ scores, the median is the mean of the $10

[3] #9 6.SP.A.3 Reduce to a simpler question (Tool #9). The cleanest way to get a median of $85$

[4] #6 6.SP.A.3 Check which scores can actually be lifted to $\ge 85$ by adding only $5$. Adding

[5] #6 6.SP.A.3 Putting it together: we need $3$ more scores at $\ge 85$, and the only way is to

[6] #6 6.SP.A.3 Try $4$ bonuses (the next choice). Convert four of the five $80$s into $85$s. No

Review

Reasonableness: The answer $4$ sits in the middle of the choice list and is clearly the smallest value that works: $3$ bonuses give a median of $82.5$ (still short), $4$ bonuses give exactly $85$, and any larger number $5$ or $6$ would also work but is wasteful. Also note that the $5$ students at $80$ are the only viable targets, so the answer cannot exceed $5$ — and $4 \le 5$ is consistent.

Alternative: Tool #16 (Change Focus / Complement) gives the same answer faster: ask "how many of the $20$ scores can stay strictly below $85$?". For the median to be $85$, at most $9$ scores can be $< 85$. Originally $2 + 2 + 4 + 5 = 13$ scores are $< 85$, so we must lift $13 - 9 = 4$ of them above $85$. The only $5$-point lift that crosses the threshold is $80 \to 85$, so all $4$ bonuses must go to $80$-scoring students — answer $\textbf{(C)}\; 4$.

CCSS standards used (min grade 6)

6.SP.B.4 Display numerical data in plots on a number line including dot and box plots (Reading the dot plot and turning it into an ordered list of all $20$ scores so the middle positions can be located.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Using the Grade 6 definition of median for $20$ numbers (average of the $10$th and $11$th values) to test how many bonuses produce a median of $85$.)

⭐ This AMC 8 problem only needs Grade 6 dot-plot and median skills you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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