AMC 8 · 2003 · #7

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition

mean-median-mode-rangemulti-digit-arithmeticset-partition identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmeticfraction-arithmetic

📏 Short solution 💡 2 insights

Problem

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?

$\mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#7 Identify Subproblems 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.

$$D_1 = +10,\quad D_2 = -10,\quad D_3 = +20,\quad D_4 = +20$$

💡 The Grade 6 "signed quantities for opposite directions" move: $+$ means Jenny scored higher, $-$ means lower.

#7 Identify Subproblems 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.

$$D_1 + D_2 + D_3 + D_4 = 10 + (-10) + 20 + 20 = 40$$

💡 Adding integers with opposite signs is Grade 6 — $+10$ and $-10$ cancel to $0$, then $20 + 20 = 40$.

#11 Work Backwards from the Mean Formula 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.

$$\text{Jenny's avg} - \text{Blake's avg} = \dfrac{\text{Jenny's total} - \text{Blake's total}}{4} = \dfrac{40}{4}$$

💡 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.

#7 Identify Subproblems 6.SP.B.5 Step 4

Divide to finish.

$$\dfrac{40}{4} = 10 \;\Rightarrow\; \textbf{(A)}$$

💡 Notice Blake's $78$ was never used: the difference of averages only depends on the per-test gaps.

[1] #7 6.NS.C.5 Subproblem 1: add the per-test differences. Let $D_i = (\text{Jenny's score on t

[2] #7 6.NS.C.6 Add the four signed differences. The $+10$ on test $1$ and the $-10$ on test $2$

[3] #11 6.SP.B.5 Subproblem 2: turn the total difference into a difference of averages. Jenny's t

[4] #7 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.5 Understand 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.6 Understand 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.5 Summarize 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, $40 \div 4 = 10$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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