AMC 10 · 2020 · #1
Easy mode Grade 5Problem
Picture an equation with a missing number x.
On the left side, you have x with 43 taken away from it. On the right side, you have 125 with 31 taken away from it. Both sides are equal:
x−43=125−31
What number does x have to be?
Pick an answer.
AMC 10 2020 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: Find the number $x$ that makes $x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3}$ true.
Givens: The left side is $x - \frac{3}{4}$; The right side is $\frac{5}{12} - \frac{1}{3}$; Answer choices: (A) $-\frac{2}{3}$, (B) $\frac{7}{36}$, (C) $\frac{7}{12}$, (D) $\frac{2}{3}$, (E) $\frac{5}{6}$
Unknowns: The single number $x$
Understand
Restated: Find the number $x$ that makes $x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3}$ true.
Givens: The left side is $x - \frac{3}{4}$; The right side is $\frac{5}{12} - \frac{1}{3}$; Answer choices: (A) $-\frac{2}{3}$, (B) $\frac{7}{36}$, (C) $\frac{7}{12}$, (D) $\frac{2}{3}$, (E) $\frac{5}{6}$
Plan
Primary tool: #11 Work Backwards
Secondary: #3 Eliminate Possibilities
Tool #11 (Work Backwards) fits perfectly: $x$ went through one operation (subtract $\frac{3}{4}$) to reach the right-hand value, so we undo it by adding $\frac{3}{4}$. The arithmetic on the right is a small fraction sum on the common denominator $12$. Tool #3 (Eliminate) is a quick sanity check — the right side is clearly small and positive after adding back $\frac{3}{4}$, so negative answer (A) and tiny $\frac{7}{36}$ (B) can be eliminated by feel before the final compute.
Execute — Answer: E
5.NF.A.1 Step 1 - First compute the right side.
- The denominators $12$ and $3$ have a common denominator of $12$, so rewrite $\frac{1}{3} = \frac{4}{12}$ and subtract.
💡 Same-denominator subtraction is straightforward — rewrite $\frac{1}{3}$ over $12$ first.
1.OA.D.8 Step 2 - Undo the $-\frac{3}{4}$ on $x$ by adding $\frac{3}{4}$ to both sides.
- The equation becomes $x = \frac{1}{12} + \frac{3}{4}$.
💡 If a number minus $\frac{3}{4}$ gives $\frac{1}{12}$, then the number is $\frac{1}{12}$ plus $\frac{3}{4}$ — Grade 1 unknown-addend logic with fractions.
5.NF.A.1 Step 3 Add $\frac{1}{12} + \frac{3}{4}$ on the common denominator $12$: $\frac{3}{4} = \frac{9}{12}$, so $x = \frac{1}{12} + \frac{9}{12} = \frac{10}{12} = \frac{5}{6}$.
💡 Rewrite over the common denominator $12$ and simplify by dividing top and bottom by $2$.
4.NF.A.2 Step 4 - Quick elimination check.
- The right side $\frac{1}{12} + \frac{3}{4}$ is clearly close to $\frac{3}{4} \approx 0.83$, so $x$ must be near $\frac{5}{6} \approx 0.83$.
- Choices (A) negative, (B) $\approx 0.19$, (C) $\approx 0.58$, (D) $\approx 0.67$ all miss; only (E) $\frac{5}{6}$ matches.
💡 Comparing fractions to $\frac{3}{4}$ rules out four of the five choices instantly.
5.NF.A.1 First compute the right side. The denominators $12$ and $3$ have a common denomi 1.OA.D.8 Undo the $-\frac{3}{4}$ on $x$ by adding $\frac{3}{4}$ to both sides. The equati 5.NF.A.1 Add $\frac{1}{12} + \frac{3}{4}$ on the common denominator $12$: $\frac{3}{4} = 4.NF.A.2 Quick elimination check. The right side $\frac{1}{12} + \frac{3}{4}$ is clearly Review
Reasonableness: Plug $x = \frac{5}{6}$ back in: $\frac{5}{6} - \frac{3}{4} = \frac{10}{12} - \frac{9}{12} = \frac{1}{12}$, and the right side was $\frac{5}{12} - \frac{4}{12} = \frac{1}{12}$. Both sides match, so (E) is correct.
Alternative: Tool #3 (Eliminate) alone — plug each choice into the left side and stop when it equals $\frac{1}{12}$. For (E): $\frac{5}{6} - \frac{3}{4} = \frac{10 - 9}{12} = \frac{1}{12}\ \checkmark$. This is the AMC multiple-choice instinct: when the work is short, just verify the answer choices directly.
CCSS standards used (min grade 5)
1.OA.D.8Determine the unknown whole number in an addition or subtraction equation (Recognizing that $x - \frac{3}{4} = \frac{1}{12}$ means $x = \frac{1}{12} + \frac{3}{4}$ — the unknown-addend idea applied to fractions.)4.NF.A.2Compare two fractions with different numerators and different denominators (Eliminating answer choices by comparing their sizes to $\frac{3}{4}$.)5.NF.A.1Add and subtract fractions with unlike denominators (Computing $\frac{5}{12} - \frac{1}{3}$ and $\frac{1}{12} + \frac{3}{4}$ by rewriting over the common denominator $12$.)
⭐ This AMC 10 problem only needs Grade 5 "add and subtract fractions with unlike denominators" you already know — write everything over $12$, undo the $-\frac{3}{4}$ by adding it back, and the answer falls out as $\frac{5}{6}$.
⭐ This AMC 10 problem only needs Grade 5 "add and subtract fractions with unlike denominators" you already know — write everything over $12$, undo the $-\frac{3}{4}$ by adding it back, and the answer falls out as $\frac{5}{6}$.
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