AMC 8 · 2019 · #7

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

mean-median-mode-rangelinear-equations-one-var identify-subproblemsconvert-to-algebra ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Work backwards: an average of 81 over 5 tests means the five scores must total 5 × 81 = 405.

required total = 5 × 81 = 405

2STEP 2

Add the three tests she already took to see how much of the 405 budget is banked: 76 + 94 + 87 = 257.

76 + 94 + 87 = 257

3STEP 3

Subtract to get what the last two tests must total together: 405 - 257 = 148.

405 - 257 = 148

4STEP 4

To make one score smallest, push its partner to the 100 cap; that forces the other to 148 - 100 = 48, which sits between 0 and 100.

min score = 148 - 100 = 48

5STEP 5

Check the choices: only 48 spends the full 100 cap, so it is the true minimum — 52, 66, 70, 74 all leave room unused.

48 = (A)

Answer

Sanity check the totals: 76 + 94 + 87 + 100 + 48 = 405, and 405 ÷ 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.

💡Key takeaway

This AMC 8 problem only needs Grade 6 understanding of the average (total ÷ count) you already know — plus a bit of Grade 4 add-and-subtract!

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