AMC 8 · 2008 · #1

Easy mode Grade 3
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Problem

Susan went to the carnival with 50 dollars.

She spent 12 dollars on food. She spent twice that much on rides.

How many dollars did she have left?

Pick an answer.

(A)
12
(B)
14
(C)
26
(D)
38
(E)
50
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Toolkit + CCSS Solution

Understand

Restated: Susan brings $\$50$ to the carnival. She spends $\$12$ on food and twice that on rides. How much money does she have left?

Givens: Susan starts with $\$50$; Food cost: $\$12$; Rides cost: twice the food cost; Answer choices: (A) $12$, (B) $14$, (C) $26$, (D) $38$, (E) $50$

Unknowns: The dollars Susan has left after both purchases

Understand

Restated: Susan brings $\$50$ to the carnival. She spends $\$12$ on food and twice that on rides. How much money does she have left?

Givens: Susan starts with $\$50$; Food cost: $\$12$; Rides cost: twice the food cost; Answer choices: (A) $12$, (B) $14$, (C) $26$, (D) $38$, (E) $50$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Write an Equation

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).

Execute — Answer: B

#3 Write an Equation 3.OA.A.1 Step 1
  • Find the rides cost.
  • "Twice as much" as the $\$12$ food cost means double it.
$$\text{rides} = 2 \times 12 = 24$$

💡 "Twice" is the Grade 3 "groups of" idea — two groups of $\$12$ is $\$24$.

#7 Identify Subproblems 2.NBT.B.5 Step 2

Add the two purchases to get total spent.

$$\text{total spent} = 12 + 24 = 36$$

💡 Both costs are now known, so adding them is one straightforward step.

#7 Identify Subproblems 3.OA.D.8 Step 3

Subtract total spent from starting money to find what's left.

$$\text{left} = 50 - 36 = 14 \;\Rightarrow\; \textbf{(B)}$$

💡 "How much is left" is always a subtraction — starting amount minus what went away.

[1] #3 3.OA.A.1 Find the rides cost. "Twice as much" as the $\$12$ food cost means double it.
[2] #7 2.NBT.B.5 Add the two purchases to get total spent.
[3] #7 3.OA.D.8 Subtract total spent from starting money to find what's left.

Review

Reasonableness: 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.

Alternative: Tool #3 (Write an Equation) in one line: $\text{left} = 50 - 12 - 2 \times 12 = 50 - 12 - 24 = 14$. Same answer, fewer named stages — useful if you trust order of operations to handle the $2 \times 12$ first.

CCSS standards used (min grade 3)

  • 2.NBT.B.5 Fluently add and subtract within 100 using strategies based on place value (Adding $12 + 24 = 36$ and subtracting $50 - 36 = 14$ are Grade 2 within-$100$ arithmetic.)
  • 3.OA.A.1 Interpret products of whole numbers as equal groups (Reading "twice as much" as $2 \times 12$ is the Grade 3 equal-groups meaning of multiplication.)
  • 3.OA.D.8 Solve two-step word problems using the four operations (Chaining multiplication (rides), addition (total), and subtraction (left over) is a multi-step word problem with the four operations.)

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

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