AMC 8 · 2003 · #7
Grade 6 arithmeticProblem
Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?
Pick an answer.
AMC 8 2003 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: Blake and Jenny each take four $100$-point tests. Blake averages $78$. Jenny scores $+10$, $-10$, $+20$, $+20$ relative to Blake on tests $1, 2, 3, 4$. Find Jenny's average minus Blake's average.
Givens: Each student takes $4$ tests; Blake's average over the $4$ tests is $78$; Jenny's score $-$ Blake's score is $+10$ on test $1$, $-10$ on test $2$, $+20$ on test $3$, $+20$ on test $4$; Answer choices: (A) $10$, (B) $15$, (C) $20$, (D) $25$, (E) $40$
Unknowns: The difference between Jenny's average and Blake's average over the $4$ tests
Understand
Restated: Blake and Jenny each take four $100$-point tests. Blake averages $78$. Jenny scores $+10$, $-10$, $+20$, $+20$ relative to Blake on tests $1, 2, 3, 4$. Find Jenny's average minus Blake's average.
Givens: Each student takes $4$ tests; Blake's average over the $4$ tests is $78$; Jenny's score $-$ Blake's score is $+10$ on test $1$, $-10$ on test $2$, $+20$ on test $3$, $+20$ on test $4$; Answer choices: (A) $10$, (B) $15$, (C) $20$, (D) $25$, (E) $40$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #11 Work Backwards from the Mean Formula
The question only asks for the difference of two averages, not either average by itself, which is why Blake's $78$ never has to be used. Tool #7 (Identify Subproblems) splits the work into two clean pieces: (1) add the four per-test score differences to get the total difference, then (2) divide that total by the number of tests to get the difference of the averages. Tool #11 (Work Backwards) here just means reading the mean formula in reverse — average $=$ total $\div$ count, so difference of averages $=$ difference of totals $\div$ count.
Execute — Answer: A
6.NS.C.5 Step 1 - Subproblem 1: add the per-test differences.
- Let $D_i = (\text{Jenny's score on test } i) - (\text{Blake's score on test } i)$.
- The problem hands you all four values directly.
💡 The Grade 6 "signed quantities for opposite directions" move: $+$ means Jenny scored higher, $-$ means lower.
6.NS.C.6 Step 2 - Add the four signed differences.
- The $+10$ on test $1$ and the $-10$ on test $2$ cancel, leaving the two $+20$s.
💡 Adding integers with opposite signs is Grade 6 — $+10$ and $-10$ cancel to $0$, then $20 + 20 = 40$.
6.SP.B.5 Step 3 - Subproblem 2: turn the total difference into a difference of averages.
- Jenny's total over $4$ tests minus Blake's total over $4$ tests equals $40$.
- Dividing each total by $4$ gives each average, so dividing the total difference by $4$ gives the average difference.
💡 Grade 6 mean formula run in reverse — if you take the same number of tests, the gap between averages is just the gap between totals shrunk by that count.
6.SP.B.5 Step 4 Divide to finish.
💡 Notice Blake's $78$ was never used: the difference of averages only depends on the per-test gaps.
6.NS.C.5 Subproblem 1: add the per-test differences. Let $D_i = (\text{Jenny's score on t 6.NS.C.6 Add the four signed differences. The $+10$ on test $1$ and the $-10$ on test $2$ 6.SP.B.5 Subproblem 2: turn the total difference into a difference of averages. Jenny's t 6.SP.B.5 Divide to finish. Review
Reasonableness: Sanity-check with concrete scores. Pretend Blake scored $78$ on every test (total $312$, average $78$). Then Jenny scored $88, 68, 98, 98$, summing to $352$ with average $88$. Difference of averages: $88 - 78 = 10$, matching (A). Trap choices map to common mistakes: (E) $40$ stops at the total difference and forgets to divide by $4$, (C) $20$ averages the four differences but mistakenly drops the negative sign on $D_2$, and (B) $15$ / (D) $25$ are off-pattern distractors that test whether students just guess.
Alternative: Tool #9 (Solve an Easier Related Problem): forget the actual scores entirely and just average the four signed gaps. The mean of $\{+10, -10, +20, +20\}$ is $\dfrac{10 - 10 + 20 + 20}{4} = \dfrac{40}{4} = 10$. Because subtracting the same Blake score from each of Jenny's scores doesn't change the mean of the differences, this average-of-gaps equals Jenny's average minus Blake's average — answer (A).
CCSS standards used (min grade 6)
6.NS.C.5Understand positive and negative numbers as describing quantities with opposite directions (Writing each test's score gap as a signed integer ($+10$, $-10$, $+20$, $+20$) so that "higher" and "lower" can be combined in one sum.)6.NS.C.6Understand a rational number as a point on the number line; extend prior understandings of numbers to the system of rational numbers (Adding the four signed differences $10 + (-10) + 20 + 20 = 40$, using the cancellation of $+10$ and $-10$.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Using the mean formula in reverse: difference of averages $=$ (difference of totals) $\div$ (number of tests) $= 40 \div 4 = 10$.)
⭐ When a question asks for the difference of two averages over the same number of tests, you don't need either average. Add the per-test gaps and divide by the count — here, $40 \div 4 = 10$.
⭐ When a question asks for the difference of two averages over the same number of tests, you don't need either average. Add the per-test gaps and divide by the count — here, $40 \div 4 = 10$.
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