Linear Inequalities Pattern - Solve

This pattern asks you to set up and solve a linear inequality to find a minimum, maximum, or greatest/least integer that satisfies a constraint.

 

One-Step: "At Least" with Addition

The simplest version: you have some amount, need at least a total, and must find the minimum additional amount.

A student must read at least 250 pages. She has already read 195 pages. What is the minimum number of additional pages she must read?

A) 55
B) 195
C) 445
D) 0

Let $p$ = additional pages. The total is $195 + p$, which must be at least 250:

$195 + p \geq 250$ $p \geq 55$

The minimum is 55. The answer is A. Option C adds instead of subtracts ($195 + 250 = 445$).

A charity needs at least $500. They've raised $385. What minimum additional amount is needed?

A) 115
B) 125
C) 385
D) 885

$385 + x \geq 500$, so $x \geq 115$. The answer is A.

 

One-Step: "At Least" with Multiplication

When each unit contributes a fixed amount, divide the total requirement by the per-unit rate.

A school club sells cupcakes for $3 each and needs at least $150. What is the minimum number of cupcakes to sell?

A) 3
B) 50
C) 147
D) 150

Let $c$ = cupcakes. Revenue: $3c$. Need $3c \geq 150$, so $c \geq 50$. The answer is B.

130 students need buses that hold 28 each. What is the minimum number of buses?

A) 4
B) 5
C) 6
D) 28

$28b \geq 130$, so $b \geq \dfrac{130}{28} = 4.64...$. Since you can't use a fraction of a bus, round up to 5. The answer is B.

This is a critical point: when the variable must be a whole number and the division doesn't come out even, always round up for "at least" problems (you need more than the fractional amount, not less).

A charity wants at least $750 by selling cookie boxes at $8 profit each. Minimum boxes to sell?

$8c \geq 750$, so $c \geq 93.75$. Round up: $c = 94$. The answer is 94.

 

Two-Step: Fixed Cost Plus Variable Cost

When there's a fixed amount plus a per-unit cost, set up a two-step inequality.

An elevator has a 2,000-lb capacity. Cargo weighing 450 lbs is loaded. If each passenger weighs 140 lbs, what is the greatest number of passengers?

A) 12
B) 14
C) 11
D) 15

Total weight: $450 + 140p \leq 2000$. Subtract: $140p \leq 1550$. Divide: $p \leq 11.07...$

Since $p$ must be a whole number, the greatest is 11. The answer is C. Round down for "at most" / "maximum" problems.

A data plan has 20 GB/month. Background apps use 1.25 GB. Streaming uses 0.35 GB per episode. Maximum whole episodes?

A) 53
B) 54
C) 57
D) 58

$1.25 + 0.35e \leq 20$, so $0.35e \leq 18.75$, giving $e \leq 53.57...$. Round down to 53. The answer is A.

A gift card has $250. A required textbook costs $89.50. Notebooks cost $3.25 each. Maximum notebooks?

A) 50
B) 76
C) 77
D) 49

$89.50 + 3.25n \leq 250$, so $3.25n \leq 160.50$, giving $n \leq 49.38...$. Round down to 49. The answer is D.

A banquet budget is at most $8,000. Venue rental is $1,500, and catering costs $45 per guest. Maximum guests?

A) 145
B) 177
C) 144
D) 178

$1500 + 45g \leq 8000$, so $45g \leq 6500$, giving $g \leq 144.44...$. Round down: 144. The answer is C.

 

Converting Units Before Solving

Sometimes the fixed cost and the variable cost are in different units. Convert everything to the same unit first.

A machine runs at most 40 hours/week. Setup takes 90 minutes. Each widget takes 8 minutes. Greatest number of widgets?

A) 289
B) 288
C) 300
D) 311

Convert 40 hours to $40 \times 60 = 2400$ minutes.

$90 + 8w \leq 2400$, so $8w \leq 2310$, giving $w \leq 288.75$. Round down: 288. The answer is B.

 

Budget with Percentage Surcharge

A harder variant adds a percentage fee on top of the base cost.

A university has a $250,000 scholarship budget for 45 students. A 4.5% admin fee is applied to the total scholarship amount, and the combined total (scholarships + fee) cannot exceed the budget. What is the maximum scholarship per student?

A) $5,805.56
B) $5,555.56
C) $5,316.32
D) $5,305.56

Let $s$ = scholarship per student. Total scholarships: $45s$. With 4.5% fee: $45s(1.045) \leq 250{,}000$.

$45s \leq \dfrac{250{,}000}{1.045} = 239{,}234.45...$

$s \leq \dfrac{239{,}234.45}{45} = 5{,}316.32...$

Rounded to the nearest cent: $5,316.32. The answer is C.

 

Translate-then-Evaluate

Some questions give a verbal description of an inequality, ask you to translate it, then evaluate at a specific value.

A quantity $a$ is no more than 8 greater than 5 times quantity $b$. If $b = -6$, what is the maximum value of $a$?

"No more than 8 greater than $5b$" means $a \leq 5b + 8$.

Substitute $b = -6$: $a \leq 5(-6) + 8 = -30 + 8 = -22$.

The maximum value of $a$ is $-22$. The answer is $-22$.

Temperature $C$ is no less than 10 more than half of $T$. If $T = -50$, what is the least possible $C$?

"No less than" means $\geq$. So $C \geq \dfrac{1}{2}T + 10$.

Substitute $T = -50$: $C \geq \dfrac{1}{2}(-50) + 10 = -25 + 10 = -15$.

The least value of $C$ is $-15$. The answer is $-15$.

 

Two-Variable Optimization

The hardest questions give two constraints (a budget/capacity and a minimum total) with two types of items, and ask for the maximum of the more expensive item.

A cargo plane holds at most 45,000 kg. It must carry at least 150 containers. Standard containers weigh 120 kg; heavy-duty containers weigh 350 kg. What is the maximum number of heavy-duty containers?

Let $s$ = standard, $h$ = heavy-duty.

Weight constraint: $120s + 350h \leq 45{,}000$ Minimum count: $s + h \geq 150$, so $s \geq 150 - h$

To maximize $h$, minimize $s$. Set $s = 150 - h$ (the smallest allowed):

$120(150 - h) + 350h \leq 45{,}000$ $18{,}000 - 120h + 350h \leq 45{,}000$ $230h \leq 27{,}000$ $h \leq 117.39...$

Round down: $h = 117$. The answer is 117.

An organization has a $40,000 budget and must assemble at least 500 kits. Food kits cost $55; medical kits cost $95. Maximum medical kits?

$55f + 95m \leq 40{,}000$ and $f + m \geq 500$, so $f \geq 500 - m$.

$55(500 - m) + 95m \leq 40{,}000$ $27{,}500 - 55m + 95m \leq 40{,}000$ $40m \leq 12{,}500$ $m \leq 312.5$

Round down: $m = 312$. The answer is 312.

 

Rounding Rules Summary

The direction you round depends on the inequality:

For "at least" / "minimum" ($\geq$): round up to the next whole number. For "at most" / "maximum" ($\leq$): round down to the previous whole number.

If the division comes out exactly even, no rounding is needed.