AMC 10 · 2022 · #2

Grade 6 algebra
systems-of-equationslinear-equations-two-varabsolute-value convert-to-algebrawork-backwards ↑ Prerequisites: linear-equations-one-var
📏 Short solution 💡 1 insight
Problem
Three numbers add up to 96. The first is six times the third, and the third is forty less than the second. Find the absolute difference between the first and the second.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Guess and Check

All three numbers can be described from the third: first = 6z, second = z + 40. That makes the third number a single dial we can turn. Tool #6 (Guess and Check) lets us test a small whole number for z and see if the total is 96 — much friendlier than naming three variables. Tool #11 (Work Backwards) and #7 (Subproblems) take over once z is found: plug it back to recover the first and second numbers, then compute their absolute difference.

1STEP 1

Write all three with one letter

Write all three with one letter.

first = 6z, second = z + 40, third = z
2STEP 2

Use the sum

The sum gives one equation.

z=5: 30+45+5 = 80, z=10: 60+50+10 = 120, z=7: 42+47+7 = 96 ✓
3STEP 3

Find the two numbers

Recover both numbers from the letter.

first = 42, second = 47
4STEP 4

Take the difference

The absolute difference is 5.

|42 - 47| = 5 → (E)
Answer
5
Verify the trio: 42 + 47 + 7 = 96 ✓. First is six times third: 42 = 6 · 7 ✓. Third is 40 less than second: 7 = 47 - 40 ✓. All clues hold, so |42 - 47| = 5 is solid, matching choice (E).
💡Key takeaway

This AMC 12 problem only needs Grade 6 "name the unknown and try small numbers" — once the third number turns out to be 7, the gap between 42 and 47 is just 5.