AMC 10 · 2012 · #6

Grade 6 algebra
systems-of-equations convert-to-algebra ↑ Prerequisites: linear-equations-two-var
📏 Short solution 💡 2 insights
Problem
Three numbers are added two at a time, giving three pair sums. Find the middle number.

Pick an answer.

(A)
4
(B)
5
(C)
6
(D)
7
(E)
8
How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable) turns the words into three clean pair-sum equations by naming the numbers a ≤ b ≤ c. Tool #15 (Organize Information in More Ways) is what makes it easy: order the pair sums so each sum is matched to the pair it came from. Then Tool #7 (Identify Subproblems) splits the work into two small steps — first find the total of all three numbers, then peel off the middle one — instead of solving a full three-way system.

1STEP 1

Name the numbers, match the sums

Three names match the three pair sums.

a+b=12, a+c=17, b+c=19
2STEP 2

Add all three pair sums

Adding all three counts everything twice.

(a+b)+(a+c)+(b+c)=2(a+b+c)=12+17+19=48 → a+b+c=24
3STEP 3

Peel off the middle number

Subtracting one pair sum leaves 7, choice (D).

b = 24 - 17 = 7 → (D)
Answer
7
The recovered numbers 5, 7, 12 are whole numbers and their pair sums are 5+7=12, 5+12=17, 7+12=19 — exactly the three sums given, so nothing was invented. The middle value is 7, and it sits between 5 and 12 as a middle number should. A tempting trap is 5 (choice B), which is the smallest number, not the middle one; the question asks for the middle, so 7 is correct.
💡Key takeaway

Add all the pair sums to get double the total, then subtract the pair that skips the middle number to find it.

  • Name the numbers, match the sums
  • Add all three pair sums
  • Peel off the middle number