AMC 8 · 2022 · #19
Grade 6 arithmeticProblem
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.
AMC 8 2022 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: 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
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) $6$
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
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.
💡 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.
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 $85$.
💡 The Grade 6 definition of median — the middle value — tells us exactly which two list positions matter.
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) $85$.
💡 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.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.
💡 Guess-and-check across the four possible source scores quickly shows only $80$s clear the bar.
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 $10$.
💡 Always test the candidate minimum — Tool #6 catches the off-by-one that a quick count misses.
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 $\textbf{(C)}$.
💡 Once positions $10$ and $11$ both read $85$, the median definition forces the answer to be $85$.
6.SP.B.4 Lay out all $20$ scores in increasing order using the dot-plot counts. This make 6.SP.A.3 Confirm the original median. With $20$ scores, the median is the mean of the $10 6.SP.A.3 Reduce to a simpler question (Tool #9). The cleanest way to get a median of $85$ 6.SP.A.3 Check which scores can actually be lifted to $\ge 85$ by adding only $5$. Adding 6.SP.A.3 Putting it together: we need $3$ more scores at $\ge 85$, and the only way is to 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.4Display 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.3Recognize 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!
⭐ This AMC 8 problem only needs Grade 6 dot-plot and median skills you already know!
More like this
Same archetype — closest grade level first.