AMC 10 · 2022 · #3

Grade 6 arithmetic
systems-of-equationslinear-equations-two-varabsolute-value convert-to-algebrawork-backwards ↑ Prerequisites: linear-equations-one-var
📏 Short solution 💡 1 insight
Problem
Three numbers (call them first, second, third) add up to 96. The first is 6 times the third, and the third is 40 less than the second. Find |first - second|.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Use the third number z as a dial: first = 6z, second = z + 40, so the trio is 6z, z+40, z.

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

Guess z: z=5 gives 80 (low), z=10 gives 120 (high), z=7 gives 42+47+7 = 96 — a match, so z = 7.

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

Plug z=7 back: first = 6 · 7 = 42, second = 7 + 40 = 47.

first = 42, second = 47
4STEP 4

The second number is larger, so |first - second| = |42 - 47| = 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 10 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.