AMC 10 · 2014 · #9

Grade 6 algebra
mean-median-mode-rangesequences-arithmetic convert-to-algebrapattern-recognition ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 1 insight
Problem
A run of consecutive numbers has an average, and that average starts a new run. Find the second run's average.

Pick an answer.

(A)
a+3
(B)
a+4
(C)
a+5
(D)
a+6
(E)
a+7
How to solve
Strategy Introduce a Variable

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.

1STEP 1

Name the five integers

The start names all five numbers.

a, a+1, a+2, a+3, a+4
2STEP 2

Average them to get b

Their average is the middle one.

b=(5a+10)/5=a+2
3STEP 3

Middle term is the average

The same rule applies to the second run.

b, b+1, b+2, b+3, b+4 → average=b+2
4STEP 4

Substitute b = a + 2

Substituting gives a plus 4, choice (B).

b+2=(a+2)+2=a+4 → (B)
Answer
a+4
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.
💡Key takeaway

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.

  • Name the five integers
  • Average them to get b
  • Middle term is the average
  • Substitute b = a + 2