AMC 8 · 2012 · #7

Easy mode Grade 6

Problem

Isabella takes four math tests. Each test is worth $100$ points.

She wants her average score on the four tests to be exactly $95$ . On her first two tests she scored $97$ and $91$ . She has just taken the third test, and she sees that she can still reach an average of $95$ if she does well enough on the fourth test.

What is the lowest score she could have gotten on the third test and still have a chance to average $95$ ?

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.

View mode:

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)

More like this