AMC 8 · 2008 · #7
Easy mode Grade 4Problem
The three fractions 53, 45M, and N60 are all equal.
Find the number M that goes on top of 45. Find the number N that goes on the bottom of 60.
What is M+N?
Pick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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.
💡 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.
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 $9$.
💡 Asking ``$5$ times what is $45$?'' is Grade 3 unknown-factor work. The same multiplier scales the top.
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 $20$.
💡 Same idea, with the unknown on the bottom this time. The multiplier is whatever turns $3$ into $60$.
4.NBT.B.4 Step 4 Add $M$ and $N$ to finish.
💡 Standard Grade 4 multi-digit addition.
4.NF.A.1 Split the chain into two separate equations. The middle and right pieces only sh 4.NF.A.1 Solve for $M$. The denominator went from $5$ to $45$, so it was multiplied by $9 4.NF.A.1 Solve for $N$. The numerator went from $3$ to $60$, so it was multiplied by $20$ 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.1Explain 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.4Fluently 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.
⭐ When two fractions are equal, the top and bottom always scale by the same number — find that scale factor and the unknown falls out.
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