Dimensions Math 7B · Ch 11.1–11.4
Linear Inequalities
Practice Quiz
20 challenge-level questions. Choose the best answer. Instant feedback with full explanation.
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🔁
Flip the Sign
Multiply or divide by negative → reverse < / >
🗣️
Key Words
"at most" → ≤ · "at least" → ≥ · "no more" → ≤ · "exceed" → >
🔵
Number Line
Open circle ○ = strict (< >) · Closed ● = equal (≤ ≥)
⚖️
Balance Rule
Do same operation BOTH sides — inequality stays valid
⚠️
Trap Alert
"6 less than x" = x − 6, NOT 6 − x
📐
Variable Side
x < 3 same as 3 > x. But moving x to right flips direction!
Q 1 TRAP ⚠️
Which inequality correctly represents: "8 less than a number n is greater than 20"?
🧠 "less than" flips order → n − 8, not 8 − n
A \(n - 8 > 20\)
B \(8 - n > 20\)
C \(n + 8 > 20\)
D \(n - 8 \geq 20\)
📖 Explanation
"8 less than
n" means start with
n and subtract 8 → \(n - 8\). "is greater than 20" → \(> 20\). Combined: \(n - 8 > 20\). Option B reverses the subtraction (a very common trap). Option D uses ≥ instead of strict >.
Q 2 HARD
Translate: "The product of 5 and a number p, decreased by 3, is at most 12."
🧠 "at most" = ≤ (no more than)
A \(5p - 3 < 12\)
B \(5p - 3 \leq 12\)
C \(5(p - 3) \leq 12\)
D \(5p + 3 \leq 12\)
📖 Explanation
"Product of 5 and p" = \(5p\). "Decreased by 3" = subtract 3 → \(5p - 3\). "At most 12" = \(\leq 12\). Answer: \(5p - 3 \leq 12\). C is wrong because "decreased by 3" applies after the product, not inside the parenthesis.
Q 3 TRAP ⚠️
Which phrase matches \(\dfrac{x}{4} \geq 9\)?
🧠 "quotient of x and 4" = x÷4
A 4 divided by x is at most 9
B The quotient of x and 4 is greater than 9
C The quotient of x and 4 is at least 9
D x divided by 4 is less than or equal to 9
📖 Explanation
\(\frac{x}{4}\) = "quotient of x and 4" (x is numerator, 4 is denominator). \(\geq 9\) = "at least 9" or "greater than or equal to 9". A swaps the order of division. B uses strict > not ≥. D uses ≤.
Q 4 HARD
A roller coaster requires riders to be at least 48 inches tall. If h = height in inches, which inequality models this?
🧠 "at least" → the minimum → ≥
A \(h > 48\)
B \(h \leq 48\)
C \(48 > h\)
D \(h \geq 48\)
📖 Explanation
"At least 48" means 48 is included as acceptable → use ≥. A excludes exactly 48. C is equivalent to h < 48 (wrong direction). B means at most 48, the opposite.
Q 5 MEDIUM
A taxi charges a flat fee of $3.50 plus $2 per mile. You have no more than $15. Let m = miles driven. Which inequality models the situation?
🧠 "no more than" = ≤ (≤ is the ceiling)
A \(2m + 3.50 < 15\)
B \(2m + 3.50 \leq 15\)
C \(3.50m + 2 \leq 15\)
D \(2m - 3.50 \leq 15\)
📖 Explanation
Total cost = flat fee + (rate × miles) = \(3.50 + 2m\). "No more than $15" = \(\leq 15\). So \(2m + 3.50 \leq 15\). A uses strict < (you could spend exactly $15). C swaps the values. D subtracts the flat fee instead of adding.
Q 6 TRAP ⚠️
Which of the following is NOT equivalent to "x is no less than 7"?
🧠 "no less" = cannot go below = ≥
A \(x \geq 7\)
B \(7 \leq x\)
C \(x > 7\)
D x is at least 7
📖 Explanation
"No less than 7" means x ≥ 7 (7 is included). A, B, and D are all equivalent to x ≥ 7. C says x > 7 which excludes exactly 7 — so it is NOT equivalent.
Q 7 HARD
A store sells notebooks for $2.25 each. Sam wants to spend less than $20 but must buy at least 3. Which pair of inequalities correctly models both conditions? Let n = number of notebooks.
🧠 Two separate constraints → two separate inequalities
A \(2.25n < 20\) and \(n \geq 3\)
B \(2.25n \leq 20\) and \(n > 3\)
C \(2.25n < 20\) and \(n > 3\)
D \(2.25 + n < 20\) and \(n \geq 3\)
📖 Explanation
Spending constraint: each notebook costs $2.25, so total = \(2.25n\). "Less than $20" → strict <. Quantity constraint: "at least 3" → \(n \geq 3\) (3 is allowed). B changes both symbols incorrectly. C uses n > 3 (excludes exactly 3). D adds instead of multiplies.
Q 8 TRAP ⚠️
Solve: \(-3x + 6 > 15\)
🔁 Dividing by NEGATIVE → FLIP the sign!
A \(x > -3\)
B \(x < -3\)
C \(x > 3\)
D \(x < 3\)
📖 Explanation
Step 1: Subtract 6 → \(-3x > 9\). Step 2: Divide both sides by \(-3\). Because we divide by a negative, FLIP the sign: \(x < -3\). Many students forget to flip and get A or D.
Q 9 HARD
Solve: \(\dfrac{2x - 1}{3} \leq 5\)
🧠 Multiply both sides by 3 FIRST (positive → no flip)
A \(x \leq 7\)
B \(x \leq 8\)
C \(x \leq 8\)
D \(x < 8\)
📖 Explanation
Multiply both sides by 3: \(2x - 1 \leq 15\). Add 1: \(2x \leq 16\). Divide by 2: \(x \leq 8\). Note: multiplying by positive 3 does NOT flip the sign. Answer is C (same symbol ≤ preserved).
Q 10 TRAP ⚠️
Solve: \(4 - 2(x + 3) \geq -6\)
🧠 Distribute FIRST before moving terms
A \(x \leq 5\)
B \(x \geq 5\)
C \(x \geq 2\)
D \(x \leq 2\)
📖 Explanation
Distribute: \(4 - 2x - 6 \geq -6\) → \(-2x - 2 \geq -6\). Add 2: \(-2x \geq -4\). Divide by \(-2\) (FLIP!): \(x \leq 2\). Many students forget to distribute the negative sign over both terms in the parenthesis.
Q 11 HARD
Which value of x is NOT a solution to \(3x - 5 \leq 4\)?
🧠 Solve for x first, then test which value falls outside
A \(x = 0\)
B \(x = 3\)
C \(x = -2\)
D \(x = 4\)
📖 Explanation
Solve: \(3x \leq 9 \Rightarrow x \leq 3\). Any x ≤ 3 is valid. Test x = 4: \(3(4)-5 = 7 > 4\) ✗ Not a solution. x = 3 gives exactly 4 ≤ 4 ✓ (included because of ≤).
Q 12 MEDIUM
Solve and identify the correct number line: \(5 - x < 2\)
🔁 Subtracting x from both sides or isolating by adding x
A \(x > 3\) — open circle at 3, arrow right
B \(x < 3\) — open circle at 3, arrow left
C \(x > -3\) — open circle at −3, arrow right
D \(x \geq 3\) — closed circle at 3, arrow right
📖 Explanation
\(5 - x < 2\) → subtract 5: \(-x < -3\) → divide by \(-1\) and FLIP: \(x > 3\). This is an open circle at 3 (not included) with arrow pointing right (values larger than 3).
Q 13 HARD
Solve: \(\dfrac{x}{-5} + 2 > 4\)
🔁 Multiplying by −5 → FLIP the sign
A \(x > -10\)
B \(x > 10\)
C \(x < -10\)
D \(x < 10\)
📖 Explanation
Subtract 2: \(\frac{x}{-5} > 2\). Multiply both sides by \(-5\) (negative → FLIP): \(x < 2 \times (-5) = -10\). So \(x < -10\). A common error is forgetting to flip when multiplying by \(-5\).
Q 14 TRAP ⚠️
A student solved \(2(3 - x) \geq 10\) and got \(x \geq 2\). What mistake did the student make?
🔁 Check: did they flip after dividing by negative?
A They forgot to distribute 2 into the parenthesis
B They did not flip the inequality sign when dividing by a negative
C They added instead of subtracted 6
D There is no mistake; the answer is correct
📖 Explanation
Correct work: Distribute → \(6 - 2x \geq 10\). Subtract 6 → \(-2x \geq 4\). Divide by \(-2\) — must FLIP → \(x \leq -2\). The student got \(x \geq 2\), meaning they failed to flip the sign AND made an arithmetic error. The key error is not flipping.
Q 15 MEDIUM
A parking garage charges $4 to enter plus $1.50 per hour. Mia wants to pay at most $16. What is the maximum number of whole hours she can park?
🧠 Set up inequality, solve, then round DOWN (whole hours)
A 7 hours
B 9 hours
C 8 hours
D 10 hours
📖 Explanation
\(4 + 1.5h \leq 16\) → \(1.5h \leq 12\) → \(h \leq 8\). Maximum whole hours = 8. Check: \(4 + 1.5(8) = 4 + 12 = 16 \leq 16\) ✓
Q 16 HARD
On a number line, which graph represents \(-2 \leq x < 5\)?
🔵 ≤ → closed ● | < → open ○
A Closed dot at −2, open dot at 5, shaded between
B Open dot at −2, closed dot at 5, shaded between
C Closed dots at both −2 and 5, shaded between
D Open dots at both −2 and 5, shaded between
📖 Explanation
\(-2 \leq x\) → −2 IS included → closed (filled) circle at −2. \(x < 5\) → 5 is NOT included → open (empty) circle at 5. The region between them is shaded. This is a compound inequality.
Q 17 HARD
A school fundraiser needs to raise more than $500. Each ticket costs $8. They already raised $180. How many more tickets t must they sell?
🧠 Already raised = starting amount → add to ticket revenue
A \(t > 40\)
B \(t > 40\) — they must sell at least 41 tickets
C \(t \geq 40\)
D \(t > 85\)
📖 Explanation
\(180 + 8t > 500\) → \(8t > 320\) → \(t > 40\). Since t must be a whole number and strictly greater than 40, they need to sell at least 41 tickets. "More than 500" is strict, so > not ≥.
Q 18 TRAP ⚠️
The speed limit sign shows 65 MPH maximum. If s = your speed, which is the correct model AND a valid speed?
🧠 Maximum = at most = ≤. You cannot exceed it.
A \(s < 65\); valid speed: 65 mph
B \(s \geq 65\); valid speed: 70 mph
C \(s \leq 65\); valid speed: 60 mph
D \(s \leq 65\); valid speed: 66 mph
📖 Explanation
"Maximum 65" means you can go up to AND including 65 → \(s \leq 65\). A uses strict < (65 itself is allowed). B means at least 65 (wrong direction). D has correct model but 66 > 65 is NOT valid. Only C has both correct model and valid example.
Q 19 HARD
If \(x\) satisfies \(2x + 1 > 9\) AND \(3x - 2 < 16\), which value of \(x\) satisfies both inequalities?
🧠 Solve each separately, find the overlap (intersection)
A \(x = 4\)
B \(x = 5\)
C \(x = 6\)
D \(x = 3\)
📖 Explanation
Inequality 1: \(2x + 1 > 9 \Rightarrow x > 4\). Inequality 2: \(3x - 2 < 16 \Rightarrow x < 6\). Together: \(4 < x < 6\). Test: x=4: not valid (need x>4). x=5: \(5>4\) ✓ and \(5<6\) ✓. x=6: not valid (need x<6). x=3: fails first. Only x=5 works.
Q 20 TRAP ⚠️ BOSS LEVEL
James earns $12/hour. He wants to save at least $200 after spending $45 on groceries. He can work at most 25 hours this week. Which system correctly models this, and does a valid solution exist?
🧠 Earnings − spending ≥ savings goal. Also check max hours.
A \(12h - 45 \geq 200\) and \(h \leq 25\); yes, \(h = 21\) works
B \(12h - 45 \geq 200\) and \(h \leq 25\); no valid solution exists
C \(12h + 45 \geq 200\) and \(h \leq 25\); yes, \(h = 13\) works
D \(12h \geq 200\) and \(h \leq 25\); yes, \(h = 17\) works
📖 Explanation
Savings = earnings − groceries = \(12h - 45\). "At least $200" → \(12h - 45 \geq 200\) → \(12h \geq 245\) → \(h \geq 20.42\) → must work at least 21 hours (whole number). Max hours: \(h \leq 25\). So \(21 \leq h \leq 25\) — valid solutions exist. Check h=21: \(12(21)-45 = 252-45 = 207 \geq 200\) ✓ and \(21 \leq 25\) ✓. Answer: A.