AMC 8 · 2012 · #7

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

mean-median-mode-rangelinear-equations-one-varbound-inequality-then-enumerate bound-inequality-then-enumerateconvert-to-algebra ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic

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Problem

Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?

Pick an answer.

(A)

$hspace{.05in}90$

(B)

$hspace{.05in}92$

(C)

$hspace{.05in}95$

(D)

$hspace{.05in}96$

(E)

$hspace{.05in}97$

AMC 8 2012 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: Isabella will take four $100$-point tests and wants an average of $95$ across all four. Her first two scores are $97$ and $91$. After seeing her third score, she sees the average-$95$ goal is still possible. What is the smallest score she could have made on the third test?

Givens: Four tests, each scored out of $100$; Target average over the four tests $= 95$; Test 1 score $= 97$; Test 2 score $= 91$; Reaching the goal is still possible after test 3; Answer choices: (A) $90$, (B) $92$, (C) $95$, (D) $96$, (E) $97$

Unknowns: The smallest possible value of the third test score

Understand

Plan

Primary tool: #11 Work Backwards

Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems

The goal (average $= 95$) is the *end state*, and we want to reverse-engineer the third score from it — that is Tool #11 (Work Backwards). Turn the average into a required total ($380$), peel away the two known scores, and we get the required sum $T_3 + T_4$. Then comes the trick: "smallest $T_3$" is hard to chase directly, but its mirror "largest $T_4$" is easy — that focus flip is Tool #16 (Change Focus). Tool #7 (Identify Subproblems) keeps the work clean: (a) what total do four tests need? (b) what's left after tests 1 and 2? (c) how small can $T_3$ be while $T_4$ stays legal?

Execute — Answer: B

#11 Work Backwards 6.SP.B.5 Step 1

Work backwards from the average.
An average of $95$ on four tests is the same as a total of $95 \times 4 = 380$ points across the four tests.
So the four scores must add to at least $380$.

$$\text{required total} = 95 \times 4 = 380$$

💡 An average is just "total divided by count", so multiplying back gives the total — a Grade 6 statistics move.

#7 Identify Subproblems 4.NBT.B.4 Step 2

Subtract the two known scores to see how many points still need to come from tests $3$ and $4$ combined.
That is the subproblem: "sum of the remaining two tests".

$$T_3 + T_4 \ge 380 - 97 - 91 = 380 - 188 = 192$$

💡 Splitting four scores into "known" and "unknown" pieces is the subproblems move; the leftover subtraction is Grade 4 multi-digit arithmetic.

#16 Change Focus / Count the Complement 6.EE.B.8 Step 3

Change the focus.
We want the smallest $T_3$, which is awkward to push down on its own.
But $T_3 + T_4 \ge 192$, so making $T_4$ as large as possible lets $T_3$ shrink the most.
The maximum legal $T_4$ is $100$.

$$T_4 \le 100, \text{ so put } T_4 = 100$$

💡 Instead of minimizing $T_3$ directly, minimize it by maximizing the *other* score — flipping the focus is Tool #16. The score-cap $T_4 \le 100$ is a simple inequality constraint.

#11 Work Backwards 6.EE.B.8 Step 4

Plug $T_4 = 100$ into $T_3 + T_4 \ge 192$ and solve for $T_3$.
This gives the lowest score the third test could have had while still leaving the average-$95$ goal reachable.

$$T_3 + 100 \ge 192 \;\Rightarrow\; T_3 \ge 92 \;\Rightarrow\; \textbf{(B)}$$

💡 Subtracting $100$ from both sides finishes the "work backwards" chain and lands on the minimum allowed third-test score.

[1] #11 6.SP.B.5 Work backwards from the average. An average of $95$ on four tests is the same as

[2] #7 4.NBT.B.4 Subtract the two known scores to see how many points still need to come from tes

[3] #16 6.EE.B.8 Change the focus. We want the smallest $T_3$, which is awkward to push down on i

[4] #11 6.EE.B.8 Plug $T_4 = 100$ into $T_3 + T_4 \ge 192$ and solve for $T_3$. This gives the lo

Review

Reasonableness: Check: if $T_3 = 92$ and $T_4 = 100$, the four scores are $97, 91, 92, 100$, summing to $380$ — exactly the total needed for an average of $95$. If $T_3$ were $91$ instead, even a perfect $100$ on test $4$ gives $97 + 91 + 91 + 100 = 379 < 380$, so the goal would be impossible. So $92$ really is the smallest workable third-test score. Answer (B) checks out.

Alternative: Tool #6 (Guess and Check) on the choices, smallest first. (A) $90$: best case sum is $97 + 91 + 90 + 100 = 378 < 380$ — fails. (B) $92$: best case sum is $97 + 91 + 92 + 100 = 380$ — works. So the smallest choice that still allows reaching the goal is (B) $92$.

CCSS standards used (min grade 6)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing $380 - 97 - 91 = 192$ to find how many points tests $3$ and $4$ must contribute together.)
6.SP.B.5 Summarize numerical data sets in relation to their context, including measures of center (Turning the target average of $95$ into a required total of $95 \times 4 = 380$ for the four tests.)
6.EE.B.8 Write an inequality of the form $x > c$ or $x < c$ to represent a constraint or condition (Writing $T_3 + T_4 \ge 192$ and $T_4 \le 100$, then solving for the smallest $T_3$ that keeps the goal possible.)

⭐ This AMC 8 problem just needs Grade 6 "average $=$ total $\div$ count" and a simple inequality — push the other score to its max to find the smallest possible one!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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