AMC 8 · 2019 · #7
Grade 6 arithmeticProblem
Shauna takes five tests, each worth a maximum of 100 points. Her scores on the first three tests are 76 , 94 , and 87 . In order to average 81 for all five tests, what is the lowest score she could earn on one of the other two tests?
Pick an answer.
AMC 8 2019 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: Shauna takes $5$ tests, each scored from $0$ to $100$. Her first three scores are $76$, $94$, and $87$. She wants the average of all $5$ tests to be exactly $81$. Of the two scores still to come, what is the smallest one of them could possibly be?
Givens: Five tests, each capped at a maximum of $100$ points; First three scores: $76$, $94$, $87$; Target average across all $5$ tests: $81$; Answer choices: (A) $48$, (B) $52$, (C) $66$, (D) $70$, (E) $74$
Unknowns: The lowest possible value of one of the two remaining test scores
Understand
Restated: Shauna takes $5$ tests, each scored from $0$ to $100$. Her first three scores are $76$, $94$, and $87$. She wants the average of all $5$ tests to be exactly $81$. Of the two scores still to come, what is the smallest one of them could possibly be?
Givens: Five tests, each capped at a maximum of $100$ points; First three scores: $76$, $94$, $87$; Target average across all $5$ tests: $81$; Answer choices: (A) $48$, (B) $52$, (C) $66$, (D) $70$, (E) $74$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #11 Work Backwards, #3 Eliminate Possibilities
The question hides three little subproblems inside one prompt, so Tool #7 (Identify Subproblems) is the cleanest entry point: (a) what total do all five tests need, (b) what total has she already earned, and (c) given the leftover total for two tests, how small can one of them be? Tool #11 (Work Backwards) is the move that turns 'I want an average of $81$' into 'I need a total of $405$' — we start from the desired ending and undo the averaging. Tool #3 (Eliminate Possibilities) is then a free sanity check against the multiple-choice list.
Execute — Answer: A
6.SP.A.3 Step 1 - Work backwards from the average.
- An average of $81$ across $5$ tests means the five scores must add to $5 \times 81 = 405$.
- (Reverse of 'sum divided by $5$ equals $81$' is 'sum equals $81$ times $5$'.)
💡 The mean is a single number that stands in for the whole list, so multiplying the mean by the count recovers the total — that is the Grade 6 definition of the mean.
4.NBT.B.4 Step 2 Add up what she has already earned on tests $1$, $2$, $3$ to see how much of the $405$-point budget is already in the bank.
💡 Fluently adding multi-digit whole numbers is a Grade 4 skill — no fractions, no algebra needed.
4.NBT.B.4 Step 3 - Subtract to find the total the last two tests must contribute together.
- The two remaining scores must add to $405 - 257 = 148$.
💡 Splitting the $405$-point target into 'already earned' plus 'still needed' is the Tool #7 subproblems move, finished with one Grade 4 subtraction.
4.OA.A.3 Step 4 - Make one of the last two scores as small as possible by pushing the other one as high as the rules allow.
- Since every test maxes out at $100$ points, the best case is $100$ on one of the remaining tests, which forces the other to be $148 - 100 = 48$.
- That score is between $0$ and $100$, so it is allowed.
💡 Reasoning 'biggest possible $+$ smallest possible $= $ fixed sum' is a Grade 4 multi-step word-problem skill — no variables required.
4.OA.A.3 Step 5 - Check against the choices.
- $48$ matches choice (A); the other options ($52, 66, 70, 74$) would each force the partner test to be less than $100$, wasting room that could have been spent lowering this test further.
💡 Plugging back into the constraint confirms $48 + 100 = 148$ — the only choice that uses the full $100$-point cap.
6.SP.A.3 Work backwards from the average. An average of $81$ across $5$ tests means the f 4.NBT.B.4 Add up what she has already earned on tests $1$, $2$, $3$ to see how much of the 4.NBT.B.4 Subtract to find the total the last two tests must contribute together. The two 4.OA.A.3 Make one of the last two scores as small as possible by pushing the other one as 4.OA.A.3 Check against the choices. $48$ matches choice (A); the other options ($52, 66, Review
Reasonableness: Sanity check the totals: $76 + 94 + 87 + 100 + 48 = 405$, and $405 \div 5 = 81$, exactly the target average. The $48$ feels low compared to her first three scores ($76, 87, 94$), but the problem asks for the lowest *possible* score, not the most likely — and the only way to be that lopsided is to ace the other remaining test ($100$), which is allowed. Any answer larger than $48$ would mean she is leaving 'cap room' on the other test unused, so $48$ is the true minimum.
Alternative: Tool #3 (Eliminate Possibilities) on its own also nails it: for each choice $X$, the partner test would have to score $148 - X$. Choices (B)-(E) give partners $96, 82, 78, 74$ — all under $100$, meaning the partner could have been bigger and this test could have been smaller. Only choice (A) $48$ pairs with the maximum allowed $100$, so (A) is the smallest *achievable* score.
CCSS standards used (min grade 6)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the first three scores ($76 + 94 + 87 = 257$) and subtracting to find the leftover two-test total ($405 - 257 = 148$).)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Reasoning that maximizing one of the two remaining scores at the $100$-point cap forces the other down to $148 - 100 = 48$, and confirming this against the choices.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Translating the target average of $81$ into the required five-test total of $5 \times 81 = 405$ — the definition of the arithmetic mean.)
⭐ This AMC 8 problem only needs Grade 6 understanding of the average (total $\div$ count) you already know — plus a bit of Grade 4 add-and-subtract!
⭐ This AMC 8 problem only needs Grade 6 understanding of the average (total $\div$ count) you already know — plus a bit of Grade 4 add-and-subtract!
More like this
Same archetype — closest grade level first.