Business Analytics Quiz: Linear Programming

10 questions · exam conditions

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Linear ProgrammingQuestion 1 of 10

A furniture manufacturer produces desks and chairs. Each desk contributes $70$ dollars and requires $3$ carpentry hours and $2$ finishing hours. Each chair contributes $110$ dollars and requires $4$ carpentry hours and $3$ finishing hours. Weekly availability is $240$ carpentry hours and $180$ finishing hours. A contract requires at least $20$ desks, while demand limits chair sales to $40$ . Let $x$ and $y$ denote the numbers of desks and chairs produced.

Which linear programming formulation correctly represents the manufacturer's weekly production decision?

\max 70x+110y

subject to

3x+4y\le240

2x+3y\le180

x\ge20

y\le40

, and

x,y\ge0

\max 110x+70y

subject to

3x+4y\le240

2x+3y\le180

x\ge20

y\le40

, and

x,y\ge0

\max 70x+110y

subject to

3x+4y\ge240

2x+3y\ge180

x\le20

y\ge40

, and

x,y\ge0

\max 70x+110y

subject to

4x+3y\le240

3x+2y\le180

x\ge20

y\le40

, and

x,y\ge0

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Business Analytics Quiz

Business Analytics Quiz: Linear Programming

Practice Linear Programming in Business Analytics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Linear Programming, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Analytics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which linear programming formulation correctly represents the manufacturer's weekly production decision?

$\max 70x+110y$ subject to $3x+4y\le240$ , $2x+3y\le180$ , $x\ge20$ , $y\le40$ , and $x,y\ge0$ . (correct answer)
$\max 110x+70y$ subject to $3x+4y\le240$ , $2x+3y\le180$ , $x\ge20$ , $y\le40$ , and $x,y\ge0$ .
$\max 70x+110y$ subject to $3x+4y\ge240$ , $2x+3y\ge180$ , $x\le20$ , $y\ge40$ , and $x,y\ge0$ .
$\max 70x+110y$ subject to $4x+3y\le240$ , $3x+2y\le180$ , $x\ge20$ , $y\le40$ , and $x,y\ge0$ .

Explanation: When building a linear programming model, your job is to match each element of the problem narrative to the right mathematical structure: an objective function that captures what you're optimizing, constraints that capture resource limits, and bounds that capture contractual or market requirements. Here, the manufacturer wants to maximize weekly contribution, and the contributions are

\$70

per desk (

x

) and

\$110

per chair (

y

). So the objective must be

\max\ 70x + 110y

. The carpentry constraint uses

3

hours per desk and

4

per chair against a 240-hour limit:

3x + 4y \le 240

. The finishing constraint uses

2

hours per desk and

3

per chair against 180 hours:

2x + 3y \le 180

. The contract floor on desks gives

x \ge 20

, and the demand ceiling on chairs gives

y \le 40

. That's exactly Answer A. Answer B swaps the contribution coefficients, assigning

\$110

to desks and

\$70

to chairs — the opposite of what the problem states. This would steer the model toward the wrong product mix. Answer C gets the objective right but flips all the inequality directions: resource constraints become

\ge

(implying you must exceed capacity, which makes no sense), and the desk/chair bounds are inverted (

x \le 20

and

y \ge 40

), reversing the contract minimum and demand maximum. Answer D also has the right objective but swaps the carpentry and finishing coefficients between the two constraints, misattributing resource usage and producing a structurally different feasible region. A reliable strategy: read the problem column by column — identify the decision variables, then map each resource row carefully. Swapped coefficients and flipped inequalities are the two most common traps in LP formulation questions.

Question 2

A retailer allocates a promotional budget between digital and traditional media. Let $d$ and $t$ be the amounts, in thousands of dollars, allocated to digital and traditional media. Management requires digital media to receive at least $40\%$ of total promotional spending. Total spending cannot exceed $120$ thousand dollars, and at least $30$ thousand dollars must be allocated to traditional media.

Which set of constraints correctly represents all three requirements?

$d\ge0.40(d+t)$ , $d+t\le120$ , $t\ge30$ , and $d,t\ge0$ . (correct answer)
$d\ge0.40t$ , $d+t\le120$ , $t\ge30$ , and $d,t\ge0$ .
$d\le0.40(d+t)$ , $d+t\le120$ , $t\ge30$ , and $d,t\ge0$ .
$d\ge0.60(d+t)$ , $d+t\le120$ , $t\ge30$ , and $d,t\ge0$ .

Explanation: When translating word problems into linear constraints, your job is to convert each plain-language requirement into a precise mathematical inequality — paying close attention to what each percentage or threshold is actually relative to. The first requirement says digital spending must be at least 40% of total spending, meaning

d \ge 0.40(d+t)

. Notice the base is the total

(d+t)

, not just the other category. The second constraint caps total spending:

d+t \le 120

. The third sets a floor on traditional spending:

t \ge 30

. Together with non-negativity, this is exactly what Answer A states — making it correct. Here's why the distractors fail: Answer B writes

d \ge 0.40t

, which compares digital spending to traditional spending alone, not to the total budget. This is a classic misreading — "40% of total" is not the same as "40% of the other category." Answer C flips the inequality to

d \le 0.40(d+t)

, which would mean digital receives at most 40% of total spending — the opposite of the requirement. Answer D uses

d \ge 0.60(d+t)

, which would require digital to be at least 60% of total spending. While 60% is the complement of 40% (i.e., the share that could go to traditional), the requirement is specifically about digital's minimum share, which is 40%, not 60%. A useful habit: whenever you see "at least X% of total," immediately write it as

\text{amount} \ge X\% \times (\text{sum of all categories})

. This prevents both the complement-confusion trap (D) and the wrong-base trap (B).

Question 3

A service center forecasts $1{,}800$ routine tickets requiring $8$ minutes each and $600$ complex tickets requiring $18$ minutes each. Existing staff can supply $260$ productive hours during the planning period. Each additional full-time staff equivalent supplies $32$ productive hours and costs $1{,}200$ dollars; each part-time staff equivalent supplies $14$ productive hours and costs $600$ dollars. For an initial linear programming plan, fractional staff equivalents are allowed. Let $x$ and $y$ be additional full-time and part-time staff equivalents.

Which model component correctly represents the staffing objective and coverage requirement?

$\min 1200x+600y$ subject to $32x+14y\ge160$ and $x,y\ge0$ . (correct answer)
$\min 1200x+600y$ subject to $32x+14y\ge420$ and $x,y\ge0$ .
$\min 1200x+600y$ subject to $32x+14y\ge260$ and $x,y\ge0$ .
$\min 1200x+600y$ subject to $32x+14y\le160$ and $x,y\ge0$ .

Explanation: When tackling workforce planning LP problems, your first move should always be to convert demand into the unit your constraint is measured in — here, productive hours — then compare that total against existing supply to find the gap that additional staff must cover. Start with total hours required:

1{,}800 \times \frac{8}{60} + 600 \times \frac{18}{60} = 240 + 180 = 420

hours demanded. Existing staff already supplies

260

hours, leaving a shortfall of

420 - 260 = 160

hours. The objective is to minimize the cost of covering that gap:

\min 1200x + 600y

, subject to additional staff providing at least

160

hours — giving

32x + 14y \ge 160

. That's exactly answer A, which is correct. Answer B uses

32x + 14y \ge 420

, which ignores existing staff entirely and treats the full demand as if zero coverage already exists — a classic trap when you forget to subtract current capacity from total need. Answer C uses

32x + 14y \ge 260

, which confuses existing staff supply (

260

hours) for the required additional hours — essentially setting the constraint equal to what you already have rather than what you still need. Answer D flips the inequality to

\le 160

, which would limit additional hours rather than guarantee a minimum — this would minimize cost by hiring nobody, completely defeating the purpose of a coverage constraint. As a strategy tip: in any resource-gap LP, always compute (total demand) − (existing supply) = required addition, and remember that coverage constraints use

\ge

(you must meet or exceed the gap, not cap it).

Question 4

An analyst is optimizing advertising allocations using predicted lead rates. For each thousand impressions, channel A generates $24$ leads with a contribution of $50$ dollars per lead and costs $400$ dollars. Channel B generates $18$ leads with a contribution of $70$ dollars per lead and costs $350$ dollars. Let $x$ and $y$ be thousands of impressions purchased from channels A and B. The media budget is $35{,}000$ dollars.

Which formulation correctly maximizes predicted net contribution while enforcing the media budget?

$\max 800x+1260y$ subject to $400x+350y\le35000$ and $x,y\ge0$ .
$\max 1200x+1260y$ subject to $400x+350y\le35000$ and $x,y\ge0$ .
$\max 800x+910y$ subject to $400x+350y\le35000$ and $x,y\ge0$ . (correct answer)
$\max 0.80x+0.91y$ subject to $400x+350y\le35000$ and $x,y\ge0$ .

Explanation: When building a linear programming model for net contribution, you need to carefully distinguish between gross revenue and net contribution — that is, revenue minus cost. For each unit of decision variable (thousands of impressions), the objective coefficient must capture what's left after paying for that channel. For Channel A, one thousand impressions generates

24

leads at

\$50

each, producing

24 \times 50 = \$1{,}200

in gross contribution. Subtract the

\$400

cost, and the net contribution per thousand impressions is

\$800

. For Channel B,

18 \times 70 = \$1{,}260

gross, minus

\$350

cost, yields

\$910

net. So the correct objective is

\max\ 800x + 910y

, subject to the budget constraint

400x + 350y \le 35{,}000

, confirming C. Choice A uses

800x

correctly for Channel A but mistakenly uses

1{,}260y

for Channel B — that's the gross contribution, ignoring the

\$350

cost. Choice B compounds the error by using gross values for both channels:

1{,}200x + 1{,}260y

omits all media costs from the objective entirely. Choice D appears to compute some ratio (roughly gross/cost), but

0.80

and

0.91

have no coherent interpretation as contribution dollars — they're dimensionally wrong and would give a meaningless optimization. A reliable strategy: always build objective coefficients by asking "if I buy one unit of this variable, what do I earn minus what do I spend?" Confusing gross revenue with net contribution is the most common trap in media-mix optimization problems, so make that subtraction explicit before writing your objective function.

Question 5

A manufacturer produces products $x$ and $y$ . The proposed linear program includes the constraints $3x+2y\le120$ for material, $x+4y\le100$ for labor, $x\ge20$ and $y\ge10$ for customer commitments, and $x+y\ge45$ for total contracted output.

Which proposed production plan is feasible?

Produce $28$ units of $x$ and $17$ units of $y$ . (correct answer)
Produce $31$ units of $x$ and $14$ units of $y$ .
Produce $26$ units of $x$ and $19$ units of $y$ .
Produce $32$ units of $x$ and $12$ units of $y$ .

Explanation: When checking feasibility in a linear program, your job is simple: plug each proposed solution into every constraint and confirm all of them are satisfied simultaneously. One violation disqualifies the entire plan. The five constraints here are:

3x+2y\le120

x+4y\le100

x\ge20

y\ge10

, and

x+y\ge45

. Option A (

x=28, y=17

) passes every check:

3(28)+2(17)=118\le120

✓,

28+4(17)=96\le100

✓,

28\ge20

✓,

17\ge10

✓, and

28+17=45\ge45

✓. This is your feasible plan. Option B (

x=31, y=14

) fails the labor constraint:

31+4(14)=87\le100

✓, but

3(31)+2(14)=121>120

— the material constraint is violated by 1 unit. Option C (

x=26, y=19

) fails the minimum output commitment:

x\ge20

✓ and

y\ge10

✓, but

26+19=45\ge45

✓ — actually check material:

3(26)+2(19)=116\le120

✓, labor:

26+4(19)=102>100

✗. Labor is violated. Option D (

x=32, y=12

) also fails labor:

32+4(12)=80\le100

✓, but

3(32)+2(12)=120\le120

✓,

32+12=44<45

✗. The total contracted output minimum is not met. A practical strategy: always test the tightest or most easily overlooked constraints first — minimums like

x+y\ge45

are traps students frequently skip because they focus on the "less than" resource constraints.

Question 6

A distributor begins the month with $80$ units of inventory and must satisfy demand for $500$ units during the month. Let $x$ be units purchased during the month and $e$ be ending inventory. All demand must be met, and no lost sales or disposal are allowed.

Which linear constraint correctly represents inventory flow?

$80+e=500+x$ , with $x,e\ge0$ .
$80+x+e=500$ , with $x,e\ge0$ .
$x=80+500+e$ , with $x,e\ge0$ .
$80+x=500+e$ , with $x,e\ge0$ . (correct answer)

Explanation: Whenever you see an inventory or material-flow problem, your instinct should be to write a simple conservation equation: what you start with, plus what comes in, must equal what goes out, plus what remains. Think of it like a bathtub — beginning water plus water added equals water drained plus water left. Here, the distributor begins with

80

units and purchases

x

units, so total supply available is

80 + x

. That supply is used to satisfy demand of

500

units, with

e

units left over as ending inventory. Setting supply equal to demand plus leftovers gives you

80 + x = 500 + e

, with

x, e \ge 0

— which is exactly answer D. You can verify the logic by rearranging:

x = 420 + e

, meaning at minimum (when

e = 0

) you must purchase

420

units to cover the

420

-unit shortfall between beginning inventory and demand. A rearranges to

80 + e - x = 500

, which incorrectly subtracts purchases instead of adding them — treating buying as a reduction in supply rather than an addition. B writes

80 + x + e = 500

, which says the ending inventory comes out of available supply without accounting for demand being a separate outflow — it conflates demand with the total rather than treating it as a use of inventory. C writes

x = 80 + 500 + e

, dramatically overstating purchases by adding beginning inventory and demand together as if both were costs to buy against. A reliable study tip: always label your flows explicitly — in (beginning inventory + purchases) and out (demand + ending inventory) — then set them equal. This structure prevents every mistake shown in A, B, and C.

Question 7

A manufacturer can purchase overtime labor in two ordered tiers. The first $100$ overtime hours cost $30$ dollars per hour, and any additional overtime hours cost $45$ dollars per hour. Let $o_1$ and $o_2$ represent hours purchased in the first and second tiers, and let $H$ be the overtime hours required by the production plan. The overall model minimizes cost.

Which linear formulation correctly represents overtime coverage and tiered cost?

$\min 30o_1+45o_2$ subject to $o_1+o_2\ge H$ , $o_1\ge0$ , and $0\le o_2\le100$ .
$\min 30o_1+45o_2$ subject to $o_1+o_2\ge H$ , $0\le o_1\le100$ , and $o_2\ge0$ . (correct answer)
$\min 45o_1+30o_2$ subject to $o_1+o_2\ge H$ , $0\le o_1\le100$ , and $o_2\ge0$ .
$\min 30o_1+30o_2$ subject to $o_1+o_2\ge H$ , $0\le o_1\le100$ , and $o_2\ge0$ .

Explanation: When modeling tiered pricing in linear programming, your goal is to capture two things: (1) the correct cost structure for each tier, and (2) the capacity constraint that forces cheaper hours to be used first before spilling into the more expensive tier. Here, the first

100

hours cost

\$30

each (tier 1, variable

o_1

), and any hours beyond that cost

\$45

each (tier 2, variable

o_2

). Because the model minimizes cost, a cost-minimizing solver will naturally exhaust the cheaper tier before using the expensive one — but only if

o_1

is capped at

100

. That cap is the key structural requirement. Answer B gets this exactly right: the objective

\min 30o_1 + 45o_2

correctly assigns the lower rate to tier 1 and higher rate to tier 2. The coverage constraint

o_1 + o_2 \ge H

ensures total hours meet demand. The bounds

0 \le o_1 \le 100

and

o_2 \ge 0

enforce the tier-1 cap while allowing unlimited tier-2 hours. Answer A swaps the caps — it limits

o_2 \le 100

instead of

o_1

, which incorrectly restricts the expensive tier rather than the cheap one, leaving the cheaper tier unbounded. Answer C reverses the costs in the objective (

45o_1 + 30o_2

), effectively labeling the cheaper tier as expensive and vice versa — a straightforward mix-up of the rate assignments. Answer D assigns

\$30

to both tiers, ignoring the second-tier premium entirely and understating costs. A useful rule of thumb: in any tiered-cost LP, the cheaper tier gets the binding upper bound, while the pricier tier stays uncapped. The minimizing objective does the rest automatically.

Question 8

A subscription company offers standard and premium packages. Let $s$ and $p$ denote the numbers of standard and premium packages sold. To limit support demand, the company may sell at most $3$ premium packages for every $5$ standard packages. It must sell at least $120$ packages in total and at least $20$ premium packages.

Which set of constraints correctly represents these requirements?

$3p\le5s$ , $s+p\ge120$ , $p\ge20$ , and $s,p\ge0$ .
$5p\le3s$ , $s+p\ge120$ , $p\ge20$ , and $s,p\ge0$ . (correct answer)
$5p\ge3s$ , $s+p\ge120$ , $p\ge20$ , and $s,p\ge0$ .
$5p\le3s$ , $s+p\le120$ , $p\ge20$ , and $s,p\ge0$ .

Explanation: When translating word problems into linear constraints, your job is to convert each business rule into a precise mathematical inequality — paying close attention to the direction of the inequality and which variable appears on which side. The trickiest constraint here is the ratio rule: "at most 3 premium packages for every 5 standard packages." This means the ratio

p/s \le 3/5

, or equivalently, multiplying both sides by

5s

5p \le 3s

. Read it plainly — premium is capped relative to standard, so premium gets the larger coefficient on the constrained (left) side. The remaining rules are more direct: total packages must be at least 120, giving

s + p \ge 120

, and premium sales must be at least 20, giving

p \ge 20

. Combined with non-negativity, this is exactly what B states, making it the correct answer. Answer A flips the ratio constraint to

3p \le 5s

, which simplifies to

p \le \frac{5}{3}s

— a much looser restriction that allows far more premium packages than intended. It misrepresents the 3-to-5 ratio entirely. Answer C reverses the inequality to

5p \ge 3s

, which actually requires a high share of premium packages rather than limiting it — the opposite of the business rule. Answer D gets the ratio right (

5p \le 3s

) but changes the total packages constraint to

s + p \le 120

, turning "at least 120" into "at most 120" — a critical sign error. When you see ratio constraints, always set up the fraction first (

p/s \le 3/5

), then cross-multiply carefully. Inequality direction errors are the most common trap on these questions.

Question 9

A company is selecting the footprint of a new distribution center. Because of zoning categories, the facility must be either no larger than $40$ thousand square feet or at least $70$ thousand square feet. Let $x$ denote the footprint in thousands of square feet.

Which statement best describes the condition $x\le40$ or $x\ge70$ in a linear optimization model?

It is equivalent to imposing both $x\le40$ and $x\ge70$ in one continuous linear program.
It is equivalent to the pair of constraints $x\ge40$ and $x\le70$ in one continuous linear program.
It is a disjunctive condition that generally requires a binary variable or two separately solved linear programs. (correct answer)
It is an ordinary continuous linear constraint because each side of the word "or" is individually linear.

Explanation: When you encounter an "or" condition in optimization — especially one where a variable must satisfy one range or another — you're entering the territory of disjunctive constraints. The key question to ask is: can a single, continuous linear program capture this "either/or" logic? The answer is almost never yes. Here,

x \le 40

x \ge 70

means the feasible region is split into two disconnected intervals:

[0, 40]

and

[70, \infty)

. A standard linear program requires a convex feasible region — one continuous, unbroken set. Two disconnected intervals violate this requirement. To handle the condition properly, you typically introduce a binary variable

y \in \{0, 1\}

and reformulate: if

y = 0

, enforce

x \le 40

; if

y = 1

, enforce

x \ge 70

. Alternatively, you solve two separate LPs — one for each case — and compare. Either way, choice C correctly identifies this as a disjunctive condition requiring special treatment. Choice A is wrong because combining

x \le 40

and

x \ge 70

simultaneously in one LP would require

x

to be both at most 40 and at least 70 at the same time — an impossible, infeasible contradiction. Choice B flips the inequalities entirely, describing

40 \le x \le 70

, which is actually the excluded middle range — the opposite of the intended condition. Choice D is a tempting trap. Yes, each individual constraint is linear, but "or" between two linear constraints creates a non-convex feasible region, which breaks standard LP assumptions. Study tip: Anytime you see "or" connecting two inequality constraints in optimization, flag it immediately as disjunctive — it almost always signals the need for binary (integer) variables or scenario-based LP comparisons.

Question 10

A beverage company must purchase exactly $1{,}000$ liters of concentrate from two suppliers. Supplier A's concentrate has a quality score of $92\%$ , and supplier B's concentrate has a quality score of $78\%$ . The combined purchase must have an average quality score of at least $86\%$ . Let $a$ and $b$ be liters purchased from suppliers A and B.

Which constraints correctly impose the required quantity and average quality?

$a+b\le1000$ and $0.92a+0.78b\ge0.86(1000)$ , with $a,b\ge0$ .
$a+b=1000$ and $0.92a+0.78b\ge0.86(a-b)$ , with $a,b\ge0$ .
$a+b=1000$ and $0.92a+0.78b\le0.86(1000)$ , with $a,b\ge0$ .
$a+b=1000$ and $0.92a+0.78b\ge0.86(1000)$ , with $a,b\ge0$ . (correct answer)

Explanation: When translating a word problem into linear constraints, your job is to convert every "must" or "at least" statement into precise mathematical language — one constraint per condition. Here, two conditions exist: a fixed total quantity and a minimum quality threshold. The total purchase must be exactly 1,000 liters, which means

a + b = 1000

(an equality, not an inequality). For the quality constraint, think of it as a weighted average: each liter from Supplier A contributes 0.92 quality units, and each liter from Supplier B contributes 0.78 quality units. The total quality points collected must meet or exceed what 1,000 liters at 86% would produce, giving

0.92a + 0.78b \ge 0.86(1000)

. Combined with non-negativity, answer D captures both conditions correctly. Answer A fails on the quantity constraint — using

a + b \le 1000

allows purchasing fewer than 1,000 liters, which violates the "exactly 1,000" requirement. Answer B uses the correct equality for quantity, but the right-hand side of the quality constraint is

0.86(a - b)

, which is meaningless here — the target quality should be applied to the total volume, not the difference between suppliers. Answer C gets the quantity right but flips the inequality on quality:

\le 0.86(1000)

would enforce a maximum quality score, the opposite of what the problem requires. Study tip: Always map each word carefully — "exactly" means

=

, "at least" means

\ge

, and "at most" means

\le

. When you see weighted average constraints, ask yourself: quality points earned ≥ quality points required at the threshold.