AMC 10 · 2014 · #10
Grade 6 algebraPick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Name the five integers
Start at a and step up by 1, so the five integers are a, a+1, a+2, a+3, a+4 — everything in one unknown.
Consecutive integers just march up by one, so one letter plus 0,1,2,3,4 names them all.
6.EE.A.2Introduce A VariableAverage them to get b
The five add up to 5a+10, so dividing by 5 gives b=a+2 — 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.B.5Introduce A VariableMiddle term is the average
The second list runs b, b+1, b+2, b+3, b+4, so by the same rule its average is the middle term b+2.
Five in a row always balance on their center value, so the middle term is the average every time.
Five numbers in a row always balance on their centre value, so the middle is the average.
▸ Why?
Each term is the same fixed step from the next, so the terms sit symmetrically about the middle.
▸ Why?
An average is a total shared over a count, and those symmetric offsets cancel in the total.
Substitute b = a + 2
Replace b with a+2: the second average is b+2=(a+2)+2=a+4, which is choice (B).
Each averaging step nudges the start up by 2, and doing it twice moves a up by 4.
6.EE.A.3Introduce A VariableThe 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.
- Name the five integers
- Average them to get b
- Middle term is the average
- Substitute b = a + 2