AMC 8 · 2008 · #7

Grade 4 arithmeticrate-ratio

Archetype Arithmetic Expression Decomposition

ratio-proportionfraction-arithmeticlinear-equations-one-var identify-subproblems ↑ Prerequisites: fraction-arithmeticratio-proportion

📏 Short solution 💡 2 insights

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Problem

If $\frac{3}{5}=\frac{M}{45}=\frac{60}{N}$ , what is $M+N$ ?

Pick an answer.

(A)

(B)

(C)

(D)

105

(E)

127

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Toolkit + CCSS Solution

Understand

Restated: The fractions $\dfrac{3}{5}$, $\dfrac{M}{45}$, and $\dfrac{60}{N}$ are all equal. Find $M + N$.

Givens: $\dfrac{3}{5} = \dfrac{M}{45}$; $\dfrac{3}{5} = \dfrac{60}{N}$; Answer choices: (A) $27$, (B) $29$, (C) $45$, (D) $105$, (E) $127$

Unknowns: The numerator $M$; The denominator $N$; The sum $M + N$

Understand

Restated: The fractions $\dfrac{3}{5}$, $\dfrac{M}{45}$, and $\dfrac{60}{N}$ are all equal. Find $M + N$.

Givens: $\dfrac{3}{5} = \dfrac{M}{45}$; $\dfrac{3}{5} = \dfrac{60}{N}$; Answer choices: (A) $27$, (B) $29$, (C) $45$, (D) $105$, (E) $127$

Plan

Primary tool: #3 Write an Equation

Secondary: #7 Identify Subproblems

The chain $\dfrac{3}{5} = \dfrac{M}{45} = \dfrac{60}{N}$ is really two independent statements glued together. Tool #7 (Identify Subproblems) splits it into one equation for $M$ and one equation for $N$. Each piece is a Grade 4 equivalent-fraction question: scale the top and bottom of $\dfrac{3}{5}$ by the same factor. Tool #3 (Write an Equation) lets us name that factor in each case and solve quickly.

Execute — Answer: E

#7 Identify Subproblems 4.NF.A.1 Step 1

Split the chain into two separate equations.
The middle and right pieces only share the value $\dfrac{3}{5}$, so handle them one by one.

$$\dfrac{3}{5} = \dfrac{M}{45} \quad\text{and}\quad \dfrac{3}{5} = \dfrac{60}{N}$$

💡 The Grade 4 equivalent-fraction rule says two fractions are equal when one comes from the other by multiplying top and bottom by the same number. Each sub-equation can be solved on its own.

#3 Write an Equation 4.NF.A.1 Step 2

Solve for $M$.
The denominator went from $5$ to $45$, so it was multiplied by $9$.
The numerator must be multiplied by the same $9$.

$$5 \times 9 = 45 \;\Rightarrow\; M = 3 \times 9 = 27$$

💡 Asking ``$5$ times what is $45$?'' is Grade 3 unknown-factor work. The same multiplier scales the top.

#3 Write an Equation 4.NF.A.1 Step 3

Solve for $N$.
The numerator went from $3$ to $60$, so it was multiplied by $20$.
The denominator must be multiplied by the same $20$.

$$3 \times 20 = 60 \;\Rightarrow\; N = 5 \times 20 = 100$$

💡 Same idea, with the unknown on the bottom this time. The multiplier is whatever turns $3$ into $60$.

#3 Write an Equation 4.NBT.B.4 Step 4

Add $M$ and $N$ to finish.

$$M + N = 27 + 100 = 127 \;\Rightarrow\; \textbf{(E)}$$

💡 Standard Grade 4 multi-digit addition.

[1] #7 4.NF.A.1 Split the chain into two separate equations. The middle and right pieces only sh

[2] #3 4.NF.A.1 Solve for $M$. The denominator went from $5$ to $45$, so it was multiplied by $9

[3] #3 4.NF.A.1 Solve for $N$. The numerator went from $3$ to $60$, so it was multiplied by $20$

[4] #3 4.NBT.B.4 Add $M$ and $N$ to finish.

Review

Reasonableness: Check each piece is really equal to $\dfrac{3}{5}$. $\dfrac{27}{45}$: divide top and bottom by $9$ to get $\dfrac{3}{5}$. $\dfrac{60}{100}$: divide top and bottom by $20$ to get $\dfrac{3}{5}$. Both match, so $M = 27$ and $N = 100$ are correct, and $M + N = 127$. Answer choices (A) $27$ and (D) $105$ are traps for students who stop after finding $M$ or who add only one of the two unknowns to something else.

Alternative: Tool #3 (Write an Equation) via cross-multiplication: from $\dfrac{3}{5} = \dfrac{M}{45}$, cross-multiply to get $5M = 3 \times 45 = 135$, so $M = 27$. From $\dfrac{3}{5} = \dfrac{60}{N}$, cross-multiply to get $3N = 5 \times 60 = 300$, so $N = 100$. Then $M + N = 127$, same answer.

CCSS standards used (min grade 4)

4.NF.A.1 Explain why a fraction $a/b$ is equivalent to $(n \times a)/(n \times b)$ by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size (Recognizing $\dfrac{3}{5} = \dfrac{M}{45}$ and $\dfrac{3}{5} = \dfrac{60}{N}$ as equivalent-fraction equations and scaling the numerator and denominator by the same factor ($9$ and $20$).)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the final sum $M + N = 27 + 100 = 127$.)

⭐ When two fractions are equal, the top and bottom always scale by the same number — find that scale factor and the unknown falls out.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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