AMC 8 · 2000 · #23
Grade 6 arithmeticProblem
There is a list of seven numbers. The average of the first four numbers is , and the average of the last four numbers is . If the average of all seven numbers is , then the number common to both sets of four numbers is
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Seven numbers sit in a row. The average of the first four is $5$, and the average of the last four is $8$. The average of all seven is $6\frac{4}{7}$. The first-four group and the last-four group overlap on one number (the same number is in both). Find that overlapping number.
Givens: There are $7$ numbers in total; Average of the first $4$ numbers $= 5$; Average of the last $4$ numbers $= 8$; Average of all $7$ numbers $= 6\tfrac{4}{7}$; Answer choices: (A) $5\tfrac{3}{7}$, (B) $6$, (C) $6\tfrac{4}{7}$, (D) $7$, (E) $7\tfrac{3}{7}$
Unknowns: The single number that belongs to both the first-four group and the last-four group
Understand
Restated: Seven numbers sit in a row. The average of the first four is $5$, and the average of the last four is $8$. The average of all seven is $6\frac{4}{7}$. The first-four group and the last-four group overlap on one number (the same number is in both). Find that overlapping number.
Givens: There are $7$ numbers in total; Average of the first $4$ numbers $= 5$; Average of the last $4$ numbers $= 8$; Average of all $7$ numbers $= 6\tfrac{4}{7}$; Answer choices: (A) $5\tfrac{3}{7}$, (B) $6$, (C) $6\tfrac{4}{7}$, (D) $7$, (E) $7\tfrac{3}{7}$
Plan
Primary tool: #11 Find an Invariant
Secondary: #9 Solve an Easier Problem
We never need the seven individual numbers — only their sums. Tool #11 (Find an Invariant) says: the grand total of all seven numbers is the same no matter how you slice them. Tool #9 (Solve an Easier Problem) is the matching move — instead of seven unknowns, turn each average into a sum and reason about three totals: first-four sum, last-four sum, and all-seven sum. Adding the first-four sum and the last-four sum double-counts exactly the overlapping number, so (first-four sum) $+$ (last-four sum) $-$ (all-seven sum) $=$ overlap. No variables, no algebra.
Execute — Answer: B
6.SP.B.5 Step 1 - Turn each average into a sum.
- The mean formula says sum $=$ average $\times$ count.
- Apply it three times: to the first four, to the last four, and to all seven.
💡 Sums are easier to combine than averages. Convert once and the rest is plain arithmetic.
6.EE.A.3 Step 2 - Count what gets added when you combine the two group sums.
- The first-4 sum covers numbers $1, 2, 3, 4$.
- The last-4 sum covers numbers $4, 5, 6, 7$.
- Adding them touches numbers $1, 2, 3, 5, 6, 7$ once each and the overlapping number $4$ twice.
💡 Two groups of $4$ over $7$ slots forces exactly one number into both groups. That is the invariant: combined sum $=$ grand total $+$ one extra copy of the overlap.
6.EE.B.7 Step 3 - Solve for the overlap.
- The combined sum $52$ equals the grand total $46$ plus one extra copy of the overlap, so subtract.
💡 The double-counted amount is exactly the gap between the two ways of measuring the same numbers.
6.SP.B.5 Turn each average into a sum. The mean formula says sum $=$ average $\times$ cou 6.EE.A.3 Count what gets added when you combine the two group sums. The first-4 sum cover 6.EE.B.7 Solve for the overlap. The combined sum $52$ equals the grand total $46$ plus on Review
Reasonableness: The overlap $6$ should be a plausible value for a number that belongs to both groups. The first-four average is $5$, so numbers there hover near $5$; the last-four average is $8$, so numbers there hover near $8$. A shared number sitting between $5$ and $8$ fits — and $6$ lies in that range. Also, $7 \times 6\tfrac{4}{7} = 46$ matches: $20 + 32 - 6 = 46$, exactly the all-seven sum. Choices (A) $5\tfrac{3}{7}$ and (E) $7\tfrac{3}{7}$ would also balance the arithmetic only if the totals were different, so we can rule them out by the calculation above.
Alternative: Tool #4 (Introduce a Variable) gives the algebraic route. Let the seven numbers be $a_1, a_2, \dots, a_7$ and let $d = a_4$ be the overlap. Then $a_1 + a_2 + a_3 + a_4 = 20$ and $a_4 + a_5 + a_6 + a_7 = 32$. Add these to get $a_1 + a_2 + a_3 + 2a_4 + a_5 + a_6 + a_7 = 52$. Subtract the all-seven sum $a_1 + \cdots + a_7 = 46$ to leave $a_4 = 6$. Same answer (B), same idea, more notation.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Using the definition of mean (sum $\div$ count) to convert the three given averages into three sums: $20$, $32$, and $46$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Recognizing that (first-4 sum) $+$ (last-4 sum) counts every number once except the overlap, which is counted twice — so the combined expression equals (all-7 sum) $+$ (overlap).)6.EE.B.7Solve real-world and mathematical problems by writing and solving one-variable equations of the form $x + p = q$ (Solving $46 + \text{overlap} = 52$ for the overlap, giving $6$.)
⭐ Convert the averages into sums: $20$, $32$, $46$. When you add $20$ and $32$, the shared number gets counted twice — so $52 - 46 = 6$ is exactly that shared number. Answer (B).
⭐ Convert the averages into sums: $20$, $32$, $46$. When you add $20$ and $32$, the shared number gets counted twice — so $52 - 46 = 6$ is exactly that shared number. Answer (B).
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