AMC 10 · 2014 · #10
Grade 6 arithmeticFive positive consecutive integers starting with a have average b. What is the average of 5 consecutive integers that start with b?
Pick an answer.
AMC 10 2014 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: Five consecutive integers begin at $a$, and their average is $b$. Find the average of five consecutive integers that begin at $b$, written in terms of $a$.
Givens: The first list is five consecutive integers starting at $a$: $a,\;a+1,\;a+2,\;a+3,\;a+4$; The average of that first list is called $b$; A second list is five consecutive integers starting at $b$; Answer choices: (A) $a+3$, (B) $a+4$, (C) $a+5$, (D) $a+6$, (E) $a+7$
Unknowns: The average of the second list, expressed in terms of $a$
Understand
Restated: Five consecutive integers begin at $a$, and their average is $b$. Find the average of five consecutive integers that begin at $b$, written in terms of $a$.
Givens: The first list is five consecutive integers starting at $a$: $a,\;a+1,\;a+2,\;a+3,\;a+4$; The average of that first list is called $b$; A second list is five consecutive integers starting at $b$; Answer choices: (A) $a+3$, (B) $a+4$, (C) $a+5$, (D) $a+6$, (E) $a+7$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #5 Look for a Pattern
Everything hangs on one letter, $a$, so Tool #4 (Introduce a Variable) lets me write all five integers as $a, a+1, a+2, a+3, a+4$ and turn the words into an exact expression for the average. Tool #5 (Look for a Pattern) supplies the shortcut that does the heavy lifting: for five consecutive integers the average is always the middle one, so I never have to add long lists twice.
Execute — Answer: B
6.EE.A.2 Step 1 Name the five integers
- Start at $a$ and step up by $1$ each time.
- The five consecutive integers are $a,\;a+1,\;a+2,\;a+3,\;a+4$.
- Writing them this way keeps everything in terms of the single unknown $a$.
💡 Consecutive integers just march up by one, so one letter plus $0,1,2,3,4$ names them all.
6.SP.B.5 Step 2 Average them to get b
- Add the five numbers and divide by $5$.
- The sum is $a+(a+1)+(a+2)+(a+3)+(a+4)=5a+10$.
- Dividing by $5$ gives $b=\dfrac{5a+10}{5}=a+2$.
- So the average $b$ is just the middle integer of the list.
💡 The mean is the sum shared equally, and for a balanced run of numbers that share lands on the middle one.
6.SP.A.3 Step 3 Middle term is the average
- The second list is five consecutive integers starting at $b$: $b,\;b+1,\;b+2,\;b+3,\;b+4$.
- By the same pattern, the average of five consecutive integers is the middle one, so its average is $b+2$ — no need to add and divide again.
💡 Five in a row always balance on their center value, so the middle term is the average every time.
6.EE.A.3 Step 4 Substitute b = a + 2
- The second average is $b+2$.
- Replace $b$ with $a+2$: $b+2=(a+2)+2=a+4$.
- That matches choice (B).
- The trap answers come from stopping too early — $a+3$ if you forget one of the two "add $2$" shifts, or $a+5$ and up if you double-count them.
💡 Each averaging step nudges the start up by $2$, and doing it twice moves $a$ up by $4$.
6.EE.A.2 Start at $a$ and step up by $1$ each time. The five consecutive integers are $a, 6.SP.B.5 Add the five numbers and divide by $5$. The sum is $a+(a+1)+(a+2)+(a+3)+(a+4)=5a 6.SP.A.3 The second list is five consecutive integers starting at $b$: $b,\;b+1,\;b+2,\;b 6.EE.A.3 The second average is $b+2$. Replace $b$ with $a+2$: $b+2=(a+2)+2=a+4$. That mat Review
Reasonableness: Test with real numbers. Let $a=1$: the first list is $1,2,3,4,5$ with average $b=3$. The second list starts at $3$: $3,4,5,6,7$ with average $5$. And $a+4=1+4=5$, which agrees. Each averaging pass moved the starting point up by exactly $2$, and two passes give $+4$, so (B) is consistent and the other choices are ruled out.
Alternative: Skip the algebra and pick a number: with $a=10$ the first list $10,11,12,13,14$ averages to $b=12$; the list starting at $12$ is $12,13,14,15,16$, averaging $14$. Since $14=10+4=a+4$, the answer is (B). Plugging in one value confirms the general result quickly.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Representing the five consecutive integers as $a, a+1, a+2, a+3, a+4$ and the second list starting from $b$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Computing the mean of the first list, $b=\dfrac{5a+10}{5}=a+2$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Using the middle term as the average of five consecutive integers to get $b+2$ without re-summing.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Substituting $b=a+2$ into $b+2$ to simplify to $a+4$.)
⭐ The average of five numbers in a row is just the middle one, and averaging twice bumps the starting number up by $2$ each time — so $a$ climbs to $a+4$.
⭐ The average of five numbers in a row is just the middle one, and averaging twice bumps the starting number up by $2$ each time — so $a$ climbs to $a+4$.
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