AMC 10 · 2010 · #11
Grade 7 algebraThe length of the interval of solutions of the inequality a≤2x+3≤b is 10. What is b−a?
Pick an answer.
AMC 10 2010 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: For the compound inequality $a \le 2x + 3 \le b$, the set of $x$ that make it true is an interval of length $10$. Find $b - a$.
Givens: The inequality is $a \le 2x + 3 \le b$; The interval of $x$-values that satisfy it has length $10$; Answer choices: (A) $6$, (B) $10$, (C) $15$, (D) $20$, (E) $30$
Unknowns: The value of $b - a$
Understand
Restated: For the compound inequality $a \le 2x + 3 \le b$, the set of $x$ that make it true is an interval of length $10$. Find $b - a$.
Givens: The inequality is $a \le 2x + 3 \le b$; The interval of $x$-values that satisfy it has length $10$; Answer choices: (A) $6$, (B) $10$, (C) $15$, (D) $20$, (E) $30$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #11 Work Backwards
Tool #4 (Introduce a Variable): the inequality traps $x$ between two bounds, so isolate $x$ by doing the same steps to all three parts — subtract $3$, then divide by $2$ — turning the given bounds into plain bounds on $x$. Tool #7 (Subproblems): split the job into (i) find the two endpoints of the $x$-interval, then (ii) subtract them to get the length. Tool #11 (Work Backwards): the length $10$ is the finished result; write the length as an expression in $b - a$ and reverse the division-by-$2$ to recover $b - a$.
Execute — Answer: D
7.EE.B.4 Step 1 Isolate x in all three parts
- Whatever you do to the middle of a compound inequality you must do to both ends.
- Subtract $3$ from every part, then divide every part by $2$.
- Dividing by the positive number $2$ keeps the inequality directions the same.
💡 Peeling the $+3$ and the $\times 2$ off the middle leaves $x$ alone between two clean bounds.
7.NS.A.1 Step 2 Measure the interval length
- The length of the $x$-interval is the larger endpoint minus the smaller one.
- Subtract the lower bound from the upper bound; the two $-3$ terms cancel and the common denominator $2$ stays.
💡 Length is just the distance between the two ends, and both ends were shrunk by the same $\div 2$, so the original gap $b-a$ is halved.
6.EE.B.7 Step 3 Set the length to 10 and reverse it
- The problem says this length equals $10$, so $\dfrac{b-a}{2} = 10$.
- This is the form $p \cdot (\text{unknown}) = q$; undo the division by $2$ by multiplying both sides by $2$ to recover $b-a$.
💡 Since dividing $b-a$ by $2$ gave $10$, the original gap must have been twice as big — $20$, choice (D).
7.EE.B.4 Whatever you do to the middle of a compound inequality you must do to both ends. 7.NS.A.1 The length of the $x$-interval is the larger endpoint minus the smaller one. Sub 6.EE.B.7 The problem says this length equals $10$, so $\dfrac{b-a}{2} = 10$. This is the Review
Reasonableness: Check with concrete numbers: pick the $x$-interval $0 \le x \le 10$ (length $10$). Then $2x+3$ runs from $2(0)+3 = 3$ up to $2(10)+3 = 23$, so $a = 3$, $b = 23$, and $b - a = 20$. This matches, and it shows the choice of interval does not matter — the slope $2$ always doubles the gap. Choice (B) $10$ would be the answer only if the middle were plain $x$; the $\times 2$ is exactly why the gap grows to $20$, so (D) is right.
Alternative: Think of $2x+3$ as a machine that stretches every length by its slope $2$. As $x$ sweeps across an interval of length $10$, the value $2x+3$ sweeps across an interval $2$ times as long, namely $20$. Those swept values run from $a$ to $b$, so $b - a = 20$ with no algebra needed.
CCSS standards used (min grade 7)
7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Solving the compound inequality for $x$ by subtracting $3$ and dividing by $2$ across all three parts.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Subtracting the two endpoints to get the interval length $\frac{b-a}{2}$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $\frac{b-a}{2} = 10$ for $b-a$ by multiplying both sides by $2$.)
⭐ Solving for $x$ divides the gap by the number in front of $x$, so a slope of $2$ makes the bound-gap $b-a$ twice the length of the $x$-interval.
⭐ Solving for $x$ divides the gap by the number in front of $x$, so a slope of $2$ makes the bound-gap $b-a$ twice the length of the $x$-interval.
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