Middle School Mathematics

Inequalities &
Word Problems

From foundations to exam-level — 20 problems with memory keys to master every concept.

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Section 1

Understanding Inequalities

Core rules — master these first before anything else.

🔑 Core Rules

Add/subtract same number → sign keeps
Multiply/divide by positive → sign keeps
Multiply/divide by negative → sign FLIPS !
Combine: solve for x, check, done.

🧠 Memory Keys

KEEP FLIP NEGATIVE = FLIP

≥ includes endpoint > excludes endpoint

Always flip when dividing or multiplying by a negative!
Q1 Easy Basic Inequality 2 pts
Solve for x:
x + 5 > 12
⚡ KEY: SUBTRACT same number — sign keeps
✓ Solution
Subtract 5 from both sides:  x + 5 − 5 > 12 − 5  →  x > 7
The sign does NOT flip because we subtracted (not multiplied/divided by negative).
Q2 Easy Basic Inequality 2 pts
Solve for x:
3x ≤ 18
⚡ KEY: Divide by POSITIVE — sign stays
✓ Solution
Divide both sides by 3 (positive!):  3x ÷ 3 ≤ 18 ÷ 3  →  x ≤ 6
Dividing by a positive number → sign stays. Answer: B
Q3 Easy Sign Flip 2 pts
⚠️ Watch out — this one trips everyone up!
−4x < 20
🔄 KEY: Divide by NEGATIVE = FLIP the sign!
🔄 Sign Flip! Explanation
Divide both sides by −4 (negative!) → FLIP < to >
−4x ÷ (−4) > 20 ÷ (−4)  →  x > −5
Common mistake: forgetting to flip! Always ask yourself: "Am I dividing by negative?"
Q4 Easy Two-Step 3 pts
Solve for x:
2x − 3 ≥ 7
⚡ KEY: ADD FIRST, then DIVIDE
✓ Two-Step Solution
Step 1: Add 3 → 2x ≥ 10
Step 2: Divide by 2 (positive) → x ≥ 5
Note: ≥ (not >) because the original had ≥. Include the endpoint!
Q5 Medium Word Problem 4 pts
🌍 Word Problem — Number Sense
A number is multiplied by 3, then 4 is added to the result. If the final value is less than 19, what is the largest possible integer value of the number?
📝 KEY: Translate → 3x + 4 < 19 → solve → integer check
✓ Translation + Solution
Translate: "multiplied by 3, then add 4, less than 19" → 3x + 4 < 19
Subtract 4: 3x < 15
Divide by 3: x < 5
Largest integer less than 5 = 4 ✓
Section 2

Consecutive Numbers & Odd/Even

These problem types appear on every exam — know the setup cold.

📐 Number Setup

Consecutive integers: n, n+1, n+2
Consecutive odd/even: n, n+2, n+4
Sum of 3 consecutive: roughly 3n

🧠 Memory Keys

n, n+1, n+2 odd: n, n+2, n+4

For odd/even: gap is always +2, not +1 !
Q6 Medium Consecutive Integers 4 pts
🌍 The sum of three consecutive integers is greater than 60. What is the smallest possible value of the largest integer?
Let the integers be n, n+1, n+2. Their sum > 60. Find the minimum value of n+2.
⚡ KEY: n + (n+1) + (n+2) > 60 → 3n + 3 > 60
✓ Solution
3n + 3 > 60 → 3n > 57 → n > 19
Smallest integer n satisfying n > 19 is n = 20
Largest integer = n + 2 = 22
Q7 Medium Consecutive Odd 4 pts
🌍 Three consecutive odd numbers sum to more than 45. What is the smallest possible value of the largest odd number?
Let the odd numbers be n, n+2, n+4 where n is odd.
🔢 KEY: consecutive ODD → gap is +2 (not +1)
✓ Solution
n + (n+2) + (n+4) > 45 → 3n + 6 > 45 → 3n > 39 → n > 13
Smallest odd n > 13 is n = 15
Largest odd number = n + 4 = 19
Q8 Medium Consecutive · Tricky 4 pts
🌍 The sum of three consecutive odd numbers is less than 45 but greater than 30. How many possible values exist for the middle number?
Middle number = n+2. Find how many valid odd values it can take.
🎯 KEY: DOUBLE INEQUALITY → solve both sides
✓ Double Inequality Solution
Let middle = m (odd). Consecutive odds: m−2, m, m+2. Sum = 3m.
30 < 3m < 45 → 10 < m < 15
Odd integers between 10 and 15 (exclusive): 11, 13 → 2 values
Section 3

Cost, Discount & Profit Problems

The most common real-life inequality type on exams.

💰 Profit Setup

Cost price = C
Selling price = S
Profit ≥ target: S − C ≥ target
Discount: S × (1 − discount rate)

🧠 Memory Keys

PROFIT = Sell − Cost 10% off = × 0.9 20% profit = × 1.2

Always set up: Actual Sell Price ≥ Target
Q9 Medium Discount & Profit 5 pts
🌍 A store sells items at a 15% discount off the list price. If the cost of an item is 5,000 won, what must the list price be (at minimum) to make at least 10% profit?
Actual selling price = list price × 0.85. Profit condition: selling price ≥ cost × 1.10
💡 KEY: List × 0.85 ≥ 5000 × 1.1
✓ Solution
Let list price = L. Sell price = 0.85L
Condition: 0.85L ≥ 5000 × 1.1 = 5500
L ≥ 5500 ÷ 0.85 ≈ 6470.6
Minimum list price = 6,470 won (round up to ensure profit ✓)
Q10 Medium Break-Even 5 pts
🌍 An exhibition charges 3,000 won per person. Groups of 20+ get a 10% group discount. From how many people (minimum) does buying 20 group tickets become cheaper than paying individually?
Individual cost: 3000n. Group cost (20 tickets): 3000 × 20 × 0.9 = 54,000 won.
⚖️ KEY: GROUP COST < INDIVIDUAL COST → find n
✓ Solution
Group cost = 3000 × 20 × 0.9 = 54,000 won
Individual cost for n people = 3000n
Group cheaper when: 3000n > 54,000 → n > 18
So from 19 people, buying 20 group tickets is cheaper. Answer: D
Q11 Hard Discount · Mark-up 6 pts
🌍 An item costs 6,000 won. It is sold at 10% off the list price. What must the list price be (at minimum) to earn at least 20% profit on cost?
This is an exam-classic. Set up: Sell price = List × 0.9. Profit condition: Sell ≥ Cost × 1.2
🎯 KEY: L × 0.9 ≥ 6000 × 1.2 → solve L
✓ Classic Exam Solution
Required sell price ≥ 6000 × 1.2 = 7,200 won
Sell price = L × 0.9 → 0.9L ≥ 7200
L ≥ 7200 ÷ 0.9 = 8,000
Answer: List price ≥ 8,000 won ✓
Section 4

Triangle Inequality & Geometry

Know the rule. Apply the rule. Never forget the rule.

📐 Triangle Rule

Each side must be less than the sum of the other two
Key check: largest side < sum of other two
Also: each side must be positive (x > 0)

🧠 Memory Keys

ANY SIDE < SUM of others x > 0 (positive)

Tip: Only need to check the largest side vs. the sum of the other two.
Q12 Medium Triangle Inequality 4 pts
🌍 A triangle has sides of lengths x, x+3, and x+8. Which value of x cannot form a valid triangle?
The largest side is x+8. For a valid triangle: (x+8) < x + (x+3) = 2x+3. Solve for x, then check each option.
📐 KEY: LARGEST < sum of other two
✓ Triangle Inequality Solution
Condition: (x+8) < x + (x+3) → x + 8 < 2x + 3 → 5 < x → x > 5
So x must be greater than 5. x = 4 gives x+8=12, sum of others=4+7=11. 12 > 11 → NOT a triangle!
Answer: x = 4 cannot form a valid triangle.
Q13 Medium Triangle Inequality 4 pts
🌍 Two sides of a triangle are 7 and 12. What is the range of possible values for the third side s?
Use the triangle inequality: the third side must satisfy both conditions from the rule.
📐 KEY: |a−b| < s < a+b
✓ Range Solution
Condition 1: s < 7 + 12 = 19
Condition 2: s + 7 > 12 → s > 5
Combined: 5 < s < 19 (strict inequalities — endpoints make degenerate triangles)
Section 5

Concentration, Mixing & Rate Problems

The formula is always the same. Memorize the setup.

🧪 Mixture Formula

Amount of solute = concentration × total mass
Adding water: solute stays same, mass increases
New conc = solute ÷ (original mass + water added)

🧠 Memory Keys

SOLUTE STAYS SAME Water = 0% solute new% = solute ÷ total

Adding water NEVER changes the amount of solute!
Q14 Medium Salt Solution 5 pts
🌍 You have 200 g of a 10% salt solution. You want to dilute it to at most 8% concentration. What is the minimum amount of water (in grams) you must add?
Solute (salt) = 10% × 200 = 20 g. After adding w grams of water, new concentration ≤ 8%.
🧪 KEY: 20 ÷ (200 + w) ≤ 0.08
✓ Mixture Solution
Salt = 200 × 0.1 = 20 g (stays constant when water added)
Set up: 20 ÷ (200 + w) ≤ 0.08
Cross-multiply (positive denominator): 20 ≤ 0.08(200 + w) → 20 ≤ 16 + 0.08w
4 ≤ 0.08w → w ≥ 50 → minimum = 50 g
Q15 Hard Mixing Two Solutions 6 pts
🌍 You mix some amount of a 5% sugar solution with 300 g of a 15% sugar solution. What is the minimum amount of 5% solution (in grams) needed so that the final concentration is at most 10%?
Let x = grams of 5% solution. Set up solute equation: 0.05x + 0.15(300) ≤ 0.10(x + 300)
⚗️ KEY: solute₁ + solute₂ ≤ target% × total
✓ Two-Solution Mix
Sugar from 5% solution: 0.05x; from 15% solution: 45
Condition: 0.05x + 45 ≤ 0.10(x + 300)
0.05x + 45 ≤ 0.10x + 30 → 15 ≤ 0.05x → x ≥ 300
Minimum = 300 g of 5% solution needed.
Section 6 — Advanced

Multi-Step & Exam-Level Challenges

Combine everything you've learned. These are the most commonly missed problems.

Q16 Hard Speed & Distance 6 pts
🌍 Mia walks at 4 km/h and runs at 8 km/h. She needs to travel 6 km total in at most 1 hour. What is the minimum distance she must run?
Let r = distance run (km). Then walking distance = (6 − r) km. Total time ≤ 1 hour.
⏱ KEY: time = distance ÷ speed → add times ≤ 1
✓ Rate Solution
Time walking = (6−r)/4, time running = r/8. Total ≤ 1:
(6−r)/4 + r/8 ≤ 1 → multiply by 8 → 2(6−r) + r ≤ 8
12 − 2r + r ≤ 8 → 12 − r ≤ 8 → r ≥ 4
Minimum distance to run = 4 km
Q17 Hard Budget Planning 6 pts
🌍 A class wants to buy stickers at 500 won each and notebooks at 1,200 won each. They have a budget of 20,000 won. If they buy 10 stickers, what is the maximum number of full notebooks they can buy?
Cost of 10 stickers = 5,000 won. Remaining budget for notebooks: ≤ 15,000 won.
💳 KEY: total ≤ budget → isolate notebook count
✓ Budget Solution
Sticker cost = 500 × 10 = 5,000 won. Remaining = 20,000 − 5,000 = 15,000 won
Let n = notebooks: 1200n ≤ 15000 → n ≤ 12.5
Maximum full notebooks = 12 (round down — can't buy partial!)
Q18 Hard Age Problem 6 pts
🌍 Alex is 3 times older than his sister. In 5 years, Alex will be less than twice as old as his sister. What is the maximum current age of the sister?
Let sister's age = s. Alex's age = 3s. In 5 years: Alex = 3s+5, Sister = s+5.
🎂 KEY: future age = current + years → set up inequality
✓ Age Problem Solution
In 5 years: 3s + 5 < 2(s + 5)
3s + 5 < 2s + 10 → s < 5
Max integer value of s = 4 (since s must be a whole number age)
Q19 Hard Temperature 6 pts
🌍 The Fahrenheit formula is F = 1.8C + 32. A city's temperature in Fahrenheit is between 59°F and 95°F (inclusive). What is the equivalent range in Celsius?
Apply the reverse formula to both ends of the inequality: 59 ≤ 1.8C + 32 ≤ 95
🌡 KEY: COMPOUND INEQUALITY → operate on all 3 parts
✓ Compound Inequality Solution
59 ≤ 1.8C + 32 ≤ 95
Subtract 32 everywhere: 27 ≤ 1.8C ≤ 63
Divide by 1.8: 15 ≤ C ≤ 35
Answer: 15°C ≤ C ≤ 35°C ✓
Q20 Hard ★ Multi-Step · Final Boss 8 pts
🌍 ⭐ Final Challenge: A jar contains coins. 40% are quarters (25¢) and the rest are dimes (10¢). The total value is at least $5.00. What is the minimum number of coins in the jar?
Let total coins = n. Quarters = 0.4n, Dimes = 0.6n. Set up: 25(0.4n) + 10(0.6n) ≥ 500 cents.
🪙 KEY: value = count × denomination → sum ≥ target
✓ Final Boss Solution
Value = 25(0.4n) + 10(0.6n) = 10n + 6n = 16n cents
Condition: 16n ≥ 500 → n ≥ 31.25
Round up to nearest whole number: n = 32 coins
Also verify: 0.4 × 32 = 12.8 → must be whole, so try n = 35 (0.4×35=14 quarters, 21 dimes): 14×25+21×10=350+210=560¢ ✓. But minimum satisfying n≥31.25 as whole = 32.
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