Competition · AMC preparation · step 4 of 4

AMC 8 · 2008 · #1

Grade 3 arithmetic
multi-digit-arithmeticmental-arithmetic identify-subproblems ↑ Prerequisites: multi-digit-arithmeticorder-of-operations
📏 Short solution 💡 2 insights
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Problem
Susan brings $50 to the carnival. She spends $12 on food and twice that on rides. How much money does she have left?

Pick an answer.

(A)
12
(B)
14
(C)
26
(D)
38
(E)
50

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Three things happen in order — food cost, rides cost, money left — and only the food cost is given directly. Tool #7 (Identify Subproblems) breaks the story into three single-step computations so each one stays simple. Tool #3 (Write an Equation) gives us the one short formula we need for each step (multiply for rides, add for total spent, subtract for what's left).

1STEP 1

Find the rides cost

Rides cost twice the $12 food, so doubling gives $24.

rides = 2 × 12 = 24
2STEP 2

Add the two purchases

Add both purchases: $12 plus $24 gives $36 total spent.

total spent = 12 + 24 = 36
3STEP 3

Subtract from the starting money

Subtract the $36 spent from the $50 start to get $14 left.

left = 50 - 36 = 14 → (B)
Answer
14
Quick sanity pass: $12 on food plus $24 on rides is $36, and $36 + $14 = $50 matches the starting amount, so the answer balances. Also $14 is less than the $50 Susan started with and less than the $36 she spent, which is exactly what "left over" should look like. Choice (E) $50 would mean she spent nothing, and (D) $38 ignores the rides — neither fits the story.
💡Key takeaway

Three small steps in the right order — double, add, subtract — turn this AMC 8 problem into easy Grade 3 arithmetic.

  • Find the rides cost
  • Add the two purchases
  • Subtract from the starting money

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