Tüm alıştırma soruları

111 soru

Soru 101Soru

The table below presents operating and compliance metrics for seven municipal water treatment plants during 2025:

Plant NameDaily Capacity (MGD)Operating Cost ($/MG)Compliance Rate (%)Energy Consumption (kWh/MG)Staff Count
Clearwater Station4532098.4%1,45028
Blue River Plant8028099.1%1,20042
Oak Valley Works3041097.8%1,68022
Pine Creek Facility11025099.5%1,15050
Cedar Ridge Station6034098.9%1,38035
Ironwood Treatment2545096.5%1,75018
Laurel Heights7031098.7%1,30031

Match each statement on the left with its correct boolean evaluation outcome and quantitative justification on the right.

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Öğeler

Statement I: The plant with the median Staff Count has an Operating Cost greater than the mean Operating Cost of all seven plants.
Statement II: For plants with a Compliance Rate above 98.5%, the ratio of total Daily Capacity to total Staff Count exceeds 2.0 MGD per staff member.
Statement III: When sorted in ascending order by Energy Consumption, the plant with the third-lowest energy usage has a higher Daily Capacity than the plant with the lowest Operating Cost.

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Cevap

Statement I matches with the outcome stating it evaluates to FALSE (310<310 < 337.14); Statement II matches with the outcome stating it evaluates to TRUE (2.03 > 2.0); Statement III matches with the outcome stating it evaluates to FALSE (70 MGD < 110 MGD).
Each statement matches its exact evaluation outcome based on standard tabular data calculations: Statement I evaluates to False because 310perMGisbelowthe310 per MG is below the 337.14 per MG mean; Statement II evaluates to True because 320 divided by 158 equals approximately 2.03, which is greater than 2.0; Statement III evaluates to False because a capacity of 70 MGD is less than 110 MGD.

Adım Adım Çözüm

1
Evaluate Statement I by determining the median staff count and mean operating cost.
Sorted staff counts: 18, 22, 28, 31 (Laurel Heights), 35, 42, 50. Median plant is Laurel Heights with an operating cost of 310.Meanoperatingcost=(320+280+410+250+340+450+310)/7=2360/7310. Mean operating cost = (320 + 280 + 410 + 250 + 340 + 450 + 310) / 7 = 2360 / 7 ≈ 337.14. Since 310islessthan310 is less than 337.14, Statement I is False.
Identify the median position (4th out of 7) and compare its specific column value to the unweighted mean across all rows.
2
Evaluate Statement II by filtering plants with compliance > 98.5% and calculating aggregate capacity to staff ratio.
Qualifying plants are Blue River, Pine Creek, Cedar Ridge, and Laurel Heights. Total capacity = 80 + 110 + 60 + 70 = 320 MGD. Total staff = 42 + 50 + 35 + 31 = 158. Ratio = 320 / 158 ≈ 2.025 MGD/staff. Since 2.025 > 2.0, Statement II is True.
Filter rows according to a conditional threshold, aggregate columns across the selected subset, and perform ratio division.
3
Evaluate Statement III by sorting by Energy Consumption and comparing capacities of the 3rd-lowest energy plant and 1st-lowest cost plant.
Energy ordered plants: Pine Creek (1,150), Blue River (1,200), Laurel Heights (1,300). 3rd-lowest is Laurel Heights (Capacity = 70 MGD). Lowest operating cost plant is Pine Creek ($250, Capacity = 110 MGD). Comparing capacities: 70 MGD < 110 MGD. Statement III is False.
Sort table columns independently to locate specified ordinal positions (3rd-lowest energy, 1st-lowest cost) and compare their respective cross-column properties.

Anahtar Kavram

Multi-Statement Boolean Evaluation using tabular sorting, descriptive statistics, conditional filtering, and cross-column ratios.
Tahmini Süre:2m 0s
Soru 102Soru

An earth-observation satellite payload operator allocates integer numbers of channels to two operational payload modes: High-Resolution Imaging (II) and Atmospheric Sounding (SS). Operational parameters and resource bounds are specified as follows:

- Bandwidth: Each imaging channel requires 15 MHz15\text{ MHz} and each sounding channel requires 10 MHz10\text{ MHz}. Total available payload bandwidth is at most 210 MHz210\text{ MHz}.
- Power: Each imaging channel consumes 40 W40\text{ W} and each sounding channel consumes 15 W15\text{ W}. Total available power budget is at most 440 W440\text{ W}.
- Mission Minimums: The satellite must operate at least 22 imaging channels (I2I \ge 2) and at least 44 sounding channels (S4S \ge 4).
- Buffer Constraint: The number of sounding channels cannot exceed twice the number of imaging channels (S2IS \le 2I).

Match each payload optimization target on the left with its corresponding integer channel count on the right.

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Öğeler

Maximum possible number of Imaging channels (II)
Maximum possible number of Sounding channels (SS)
Number of Sounding channels (SS) that maximizes daily data return (R=70I+20SR = 70I + 20S Gbit/day)

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Cevap

The maximum possible number of Imaging channels (II) is 9. The maximum possible number of Sounding channels (SS) is 12. The number of Sounding channels (SS) that maximizes daily data return is 5.
Matching the maximum imaging channels (II) to 9 is correct because I=10I = 10 forces S2S \le 2, violating S4S \ge 4, whereas I=9I = 9 is fully feasible with S=5S = 5. Matching the maximum sounding channels (SS) to 12 is correct because at (6,12)(6, 12), bandwidth, power, and buffer constraints are simultaneously satisfied. Matching the sounding channels for maximum data return to 5 is correct because the function R=70I+20SR = 70I + 20S reaches its peak of 730 Gbit/day at (9,5)(9, 5).

Adım Adım Çözüm

1
Formulate the algebraic system of inequality constraints.
Bandwidth: 15I+10S210    3I+2S4215I + 10S \le 210 \implies 3I + 2S \le 42. Power: 40I+15S440    8I+3S8840I + 15S \le 440 \implies 8I + 3S \le 88. Bounds: I2I \ge 2, S4S \ge 4, and S2IS \le 2I.
Converting narrative bounds into simplified algebraic inequalities facilitates evaluating integer extreme points.
2
Find the maximum feasible value for Imaging channels (II).
If I=10I = 10, power requires 8(10)+3S88    3S8    S28(10) + 3S \le 88 \implies 3S \le 8 \implies S \le 2, which violates the minimum requirement S4S \ge 4. If I=9I = 9, power requires 8(9)+3S88    3S16    S58(9) + 3S \le 88 \implies 3S \le 16 \implies S \le 5. With S=4S = 4 or S=5S = 5, bandwidth gives 3(9)+2(5)=37423(9) + 2(5) = 37 \le 42 and S2(9)S \le 2(9) holds. Thus, max I=9I = 9.
Determining the upper bound of II requires testing larger integer values of II while ensuring S4S \ge 4 is not violated.
3
Find the maximum feasible value for Sounding channels (SS).
Since S2IS \le 2I, SS is constrained by II. Testing I=6I = 6: bandwidth gives 3(6)+2S42    2S24    S123(6) + 2S \le 42 \implies 2S \le 24 \implies S \le 12. Power gives 8(6)+3S88    3S40    S13.338(6) + 3S \le 88 \implies 3S \le 40 \implies S \le 13.33. Buffer gives S2(6)=12S \le 2(6) = 12. At (6,12)(6, 12), all constraints hold (18+24=424218+24=42 \le 42, 48+36=848848+36=84 \le 88, 121212 \le 12). Thus, max S=12S = 12.
Maximizing SS requires balancing the ratio constraint S2IS \le 2I against bandwidth and power limits.
4
Maximize the objective function R(I,S)=70I+20SR(I, S) = 70I + 20S over all feasible boundary points.
Evaluating candidate integer vertex points: R(9,5)=70(9)+20(5)=730R(9, 5) = 70(9) + 20(5) = 730; R(9,4)=710R(9, 4) = 710; R(8,8)=720R(8, 8) = 720; R(7,10)=690R(7, 10) = 690; R(6,12)=660R(6, 12) = 660. The maximum data return occurs at (9,5)(9, 5), where S=5S = 5.
Comparing total return at all boundary points identifies the global integer maximum.

Anahtar Kavram

Linear Programming and Integer Bounded Optimization
Soru 103Soru

A municipal water treatment facility operates four filtration basins (Basin A, Basin B, Basin C, and Basin D). Each basin receives raw water at a specified intake rate and operates at a given recovery percentage (the percent of raw water converted to purified water). Match each filtration basin on the left to its corresponding purified water output rate in thousands of gallons per hour (kgal/hr) on the right.

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Öğeler

Basin A: Raw water intake rate of 120 kgal/hr120\text{ kgal/hr} with a 75%75\% recovery rate.
Basin B: Raw water intake rate of 160 kgal/hr160\text{ kgal/hr} with a 65%65\% recovery rate.
Basin C: Raw water intake rate of 200 kgal/hr200\text{ kgal/hr} with a 55%55\% recovery rate.
Basin D: Raw water intake rate of 150 kgal/hr150\text{ kgal/hr} with an 80%80\% recovery rate.

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Cevap

Basin A matches with 90 kgal/hr, Basin B matches with 104 kgal/hr, Basin C matches with 110 kgal/hr, and Basin D matches with 120 kgal/hr.
The output rate for each basin is determined by taking the product of the total raw intake rate and the given recovery percentage. Basin A yields 120×0.75=90 kgal/hr120 \times 0.75 = 90\text{ kgal/hr}, Basin B yields 160×0.65=104 kgal/hr160 \times 0.65 = 104\text{ kgal/hr}, Basin C yields 200×0.55=110 kgal/hr200 \times 0.55 = 110\text{ kgal/hr}, and Basin D yields 150×0.80=120 kgal/hr150 \times 0.80 = 120\text{ kgal/hr}.

Adım Adım Çözüm

1
Calculate the purified water output for Basin A
Output = 120×0.75=90 kgal/hr120 \times 0.75 = 90\text{ kgal/hr}
Multiply the raw intake volume by the decimal equivalent of the recovery percentage (75%=0.7575\% = 0.75).
2
Calculate the purified water output for Basin B
Output = 160×0.65=104 kgal/hr160 \times 0.65 = 104\text{ kgal/hr}
Multiply the raw intake volume by the recovery percentage (65%=0.6565\% = 0.65).
3
Calculate the purified water output for Basin C
Output = 200×0.55=110 kgal/hr200 \times 0.55 = 110\text{ kgal/hr}
Multiply the raw intake volume by the recovery percentage (55%=0.5555\% = 0.55).
4
Calculate the purified water output for Basin D
Output = 150×0.80=120 kgal/hr150 \times 0.80 = 120\text{ kgal/hr}
Multiply the raw intake volume by the recovery percentage (80%=0.8080\% = 0.80).

Anahtar Kavram

Calculating net output rate from gross input rate and efficiency percentages
Tahmini Süre:1m 30s
Soru 104Soru

A regional energy commission classifies proposed energy generation projects into four regulatory permitting tiers based on two quantitative metrics: Peak Capacity (in megawatts, MW\text{MW}) and Sanctuary Proximity (distance in kilometers, km\text{km}, from a protected wildlife reserve).

The classification criteria are specified as follows:
- Tier 1 (Comprehensive Review): Peak Capacity >50 MW> 50\text{ MW} AND Sanctuary Proximity <10 km< 10\text{ km}
- Tier 2 (Grid Impact Priority): Peak Capacity >50 MW> 50\text{ MW} AND Sanctuary Proximity 10 km\ge 10\text{ km}
- Tier 3 (Environmental Mitigation Priority): Peak Capacity 50 MW\le 50\text{ MW} AND Sanctuary Proximity <10 km< 10\text{ km}
- Tier 4 (Fast-Track Permitting): Peak Capacity 50 MW\le 50\text{ MW} AND Sanctuary Proximity 10 km\ge 10\text{ km}

Based on the rules provided above, match each project proposal to its designated regulatory permitting tier.

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Öğeler

Project Alpha: Peak Capacity = 75 MW75\text{ MW}, Sanctuary Proximity = 6 km6\text{ km}
Project Beta: Peak Capacity = 60 MW60\text{ MW}, Sanctuary Proximity = 18 km18\text{ km}
Project Gamma: Peak Capacity = 35 MW35\text{ MW}, Sanctuary Proximity = 4 km4\text{ km}
Project Delta: Peak Capacity = 40 MW40\text{ MW}, Sanctuary Proximity = 25 km25\text{ km}

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Cevap

Project Alpha matches Tier 1 (Comprehensive Review), Project Beta matches Tier 2 (Grid Impact Priority), Project Gamma matches Tier 3 (Environmental Mitigation Priority), and Project Delta matches Tier 4 (Fast-Track Permitting).
Each project is mapped to the permitting tier whose dual constraints for Peak Capacity and Sanctuary Proximity are both satisfied.

Adım Adım Çözüm

1
Evaluate Project Alpha metrics against the tier criteria.
Peak Capacity is 75 MW>50 MW75\text{ MW} > 50\text{ MW} and Sanctuary Proximity is 6 km<10 km6\text{ km} < 10\text{ km}. This matches Tier 1.
Both conditions for Tier 1 (High capacity and high environmental sensitivity) are satisfied.
2
Evaluate Project Beta metrics against the tier criteria.
Peak Capacity is 60 MW>50 MW60\text{ MW} > 50\text{ MW} and Sanctuary Proximity is 18 km10 km18\text{ km} \ge 10\text{ km}. This matches Tier 2.
High capacity with lower environmental sensitivity places the project into Tier 2.
3
Evaluate Project Gamma metrics against the tier criteria.
Peak Capacity is 35 MW50 MW35\text{ MW} \le 50\text{ MW} and Sanctuary Proximity is 4 km<10 km4\text{ km} < 10\text{ km}. This matches Tier 3.
Standard capacity combined with high environmental sensitivity requires Tier 3 mitigation.
4
Evaluate Project Delta metrics against the tier criteria.
Peak Capacity is 40 MW50 MW40\text{ MW} \le 50\text{ MW} and Sanctuary Proximity is 25 km10 km25\text{ km} \ge 10\text{ km}. This matches Tier 4.
Standard capacity and low environmental sensitivity qualify for Tier 4 fast-track permitting.

Anahtar Kavram

Multi-attribute categorical classification using strict logical boundary constraints.
Soru 105Soru

A commercial bakery operates three distinct production units—Unit X, Unit Y, and Unit Z—to fulfill daily shipments of bread loaves:

- Unit X produces 150150 loaves per hour, with 4%4\% discarded due to minor defects.
- Unit Y produces 200200 loaves per hour, with 5%5\% discarded due to minor defects.
- Unit Z produces 250250 loaves per hour, with 2%2\% discarded due to minor defects.

Match each production unit on the left with its corresponding net yield of non-defective loaves produced during an 88-hour shift on the right.

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Öğeler

Unit X
Unit Y
Unit Z

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Cevap

Unit X matches with 1,152 non-defective loaves; Unit Y matches with 1,520 non-defective loaves; Unit Z matches with 1,960 non-defective loaves.
Each unit's net non-defective yield is derived by calculating gross output over 8 hours (Hourly Rate × 8) and multiplying by (1Defect Rate)(1 - \text{Defect Rate}). Unit X yields 1,200×0.96=1,1521,200 \times 0.96 = 1,152, Unit Y yields 1,600×0.95=1,5201,600 \times 0.95 = 1,520, and Unit Z yields 2,000×0.98=1,9602,000 \times 0.98 = 1,960.

Adım Adım Çözüm

1
Calculate total gross production for each unit over an 8-hour shift.
Unit X produces 150×8=1,200150 \times 8 = 1,200 loaves; Unit Y produces 200×8=1,600200 \times 8 = 1,600 loaves; Unit Z produces 250×8=2,000250 \times 8 = 2,000 loaves.
Gross production equals hourly rate multiplied by duration.
2
Calculate the non-defective percentage for each unit.
Unit X non-defective rate is 100%4%=96%100\% - 4\% = 96\%; Unit Y non-defective rate is 100%5%=95%100\% - 5\% = 95\%; Unit Z non-defective rate is 100%2%=98%100\% - 2\% = 98\%.
The non-defective rate is the complement of the discard percentage.
3
Multiply gross output by the non-defective rate to determine net yield.
Unit X yield is 1,200×0.96=1,1521,200 \times 0.96 = 1,152; Unit Y yield is 1,600×0.95=1,5201,600 \times 0.95 = 1,520; Unit Z yield is 2,000×0.98=1,9602,000 \times 0.98 = 1,960.
Applying the non-defective fraction gives the final usable loaf count for each unit.

Anahtar Kavram

Work rate and net yield calculation using percentages
Soru 106Soru

A museum conservation board classifies historical artifacts into three management streams based on two parameters: Material Sensitivity (High or Moderate) and Structural Fragility Index (a numerical score from 11 to 100100). The classification rules are defined as follows:

- Track Y (Long-Term Preservation): Assigned if an artifact has High Material Sensitivity AND a Structural Fragility Index score of at least 7070.
- Track X (Immediate Stabilization): Assigned if an artifact meets EXACTLY ONE of the two conditions (either High Material Sensitivity OR a Structural Fragility Index score of at least 7070, but not both).
- Routine Storage: Assigned if an artifact meets NEITHER condition (Moderate Material Sensitivity AND a Structural Fragility Index score below 7070).

Match each of the following three artifacts to its correct management stream.

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Öğeler

Artifact Alpha: High Material Sensitivity; Structural Fragility Index of 75
Artifact Beta: Moderate Material Sensitivity; Structural Fragility Index of 82
Artifact Gamma: Moderate Material Sensitivity; Structural Fragility Index of 50

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Cevap

Artifact Alpha matches with Track Y (Long-Term Preservation); Artifact Beta matches with Track X (Immediate Stabilization); Artifact Gamma matches with Routine Storage.
Artifact Alpha satisfies both criteria (High Sensitivity and Fragility Index 70\ge 70), placing it in Track Y. Artifact Beta satisfies only the Fragility Index criterion (70\ge 70), placing it in Track X. Artifact Gamma satisfies neither criterion, placing it in Routine Storage.

Adım Adım Çözüm

1
Evaluate Artifact Alpha against the classification rules.
Material Sensitivity is High (True) and Fragility Index is 757075 \ge 70 (True).
Both conditions are satisfied simultaneously, which matches the criteria for Track Y (Long-Term Preservation).
2
Evaluate Artifact Beta against the classification rules.
Material Sensitivity is Moderate (False) and Fragility Index is 827082 \ge 70 (True).
Exactly one condition is satisfied, which matches the criteria for Track X (Immediate Stabilization).
3
Evaluate Artifact Gamma against the classification rules.
Material Sensitivity is Moderate (False) and Fragility Index is 50<7050 < 70 (False).
Neither condition is satisfied, which matches the criteria for Routine Storage.

Anahtar Kavram

Categorical classification using multi-attribute boolean logical operators (AND, XOR, NOR).
Soru 107Soru

A regional logistics operator manages cargo transit on a river corridor using two types of vessels: Express Barges (EE) and Heavy-Haul Barges (HH). Weekly operational constraints are defined as follows:

- Fuel Allowance: Each Express Barge consumes 88 metric tons of fuel per trip, and each Heavy-Haul Barge consumes 1212 metric tons. Total weekly fuel consumption cannot exceed 144144 metric tons.
- Crew Availability: Each Express Barge requires 33 crew shifts, and each Heavy-Haul Barge requires 22 crew shifts. Total available crew shifts per week cannot exceed 3030.
- Service Minimum: The operator must deploy at least 22 Heavy-Haul Barges (H2H \ge 2) per week to maintain baseline commercial obligations.
- Profit Structure: Each Express Barge generates a net profit of $5,000\$5,000, and each Heavy-Haul Barge generates a net profit of $6,000\$6,000.

Match each operational metric on the left with its correct optimal value under the profit-maximizing schedule of integer barge deployments.

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Öğeler

Number of Express Barges (EE) in the profit-maximizing schedule
Number of Heavy-Haul Barges (HH) in the profit-maximizing schedule
Maximum total weekly profit achieved (in thousands of dollars)

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Cevap

The profit-maximizing deployment requires 33 Express Barges and 1010 Heavy-Haul Barges, yielding a total maximum profit of 7575 thousand dollars ($75,000\$75,000).
The correct matches establish that deploying 33 Express Barges and 1010 Heavy-Haul Barges yields the maximum possible weekly profit of $75,000\$75,000. This combination uses 2(3)+3(10)=362(3) + 3(10) = 36 tons of fuel (using 100%100\% of fuel capacity) and 3(3)+2(10)=293(3) + 2(10) = 29 crew shifts (out of 3030 available), satisfying all joint constraints.

Adım Adım Çözüm

1
Formulate the linear programming inequalities from the narrative constraints.
Fuel constraint: 8E+12H144    2E+3H368E + 12H \le 144 \implies 2E + 3H \le 36.
Crew constraint: 3E+2H303E + 2H \le 30.
Bounds: H2H \ge 2, E0E \ge 0, with E,HZE, H \in \mathbb{Z}.
Profit function: P(E,H)=5E+6HP(E, H) = 5E + 6H (in thousands of dollars).
Converting real-world bounds into simplified algebraic inequalities is required to determine the feasible integer region.
2
Identify boundary integer candidate points near the intersection of the constraints.
Solving 2E+3H=362E + 3H = 36 and 3E+2H=303E + 2H = 30 gives 5H=48    H=9.6,E=3.65H = 48 \implies H = 9.6, E = 3.6.
Testing adjacent integer points (E,H)(E, H) within the feasible region:
- Case 1: H=10    2E+3036    E3H = 10 \implies 2E + 30 \le 36 \implies E \le 3. Check crew for (3,10)(3, 10): 3(3)+2(10)=29303(3) + 2(10) = 29 \le 30 (Feasible). Profit P=5(3)+6(10)=75P = 5(3) + 6(10) = 75.
- Case 2: H=9    2E+2736    E4H = 9 \implies 2E + 27 \le 36 \implies E \le 4. Check crew for (4,9)(4, 9): 3(4)+2(9)=30303(4) + 2(9) = 30 \le 30 (Feasible). Profit P=5(4)+6(9)=74P = 5(4) + 6(9) = 74.
- Case 3: H=12    2E+3636    E=0H = 12 \implies 2E + 36 \le 36 \implies E = 0. Check crew for (0,12)(0, 12): 3(0)+2(12)=24303(0) + 2(12) = 24 \le 30 (Feasible). Profit P=6(12)=72P = 6(12) = 72.
- Case 4: H=11    2E+3336    E1H = 11 \implies 2E + 33 \le 36 \implies E \le 1. Check crew for (1,11)(1, 11): 3(1)+2(11)=25303(1) + 2(11) = 25 \le 30 (Feasible). Profit P=5(1)+6(11)=71P = 5(1) + 6(11) = 71.
Since integer constraints apply, evaluating feasible grid points along the constraint boundary identifies the global maximum.
3
Compare profit outcomes and confirm optimal values.
The maximum profit occurs at (E,H)=(3,10)(E, H) = (3, 10), giving a profit of $75,000\$75,000.
Comparing all candidate integer points verifies that E=3E=3, H=10H=10, and P=75P=75 is the unique optimal solution.

Anahtar Kavram

Integer Optimization under Simultaneous Bounded Constraints
Soru 108Soru

A venture capital firm classifies its portfolio companies into four strategic management categories based on two quarterly performance metrics: Burn Multiple (BB) and Customer Retention Rate (RR).

Classification Criteria:
- Category 1 (Prime Growth): B1.5B \le 1.5 and R85%R \ge 85\%
- Category 2 (Capital Efficient): B1.5B \le 1.5 and R<85%R < 85\%
- Category 3 (Expansion Focused): B>1.5B > 1.5 and R85%R \ge 85\%
- Category 4 (High Risk): B>1.5B > 1.5 and R<85%R < 85\%

Match each portfolio company to its corresponding strategic management category.

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Öğeler

Company P: B=1.2B = 1.2, R=90%R = 90\%
Company Q: B=1.4B = 1.4, R=80%R = 80\%
Company R: B=1.8B = 1.8, R=88%R = 88\%
Company S: B=2.1B = 2.1, R=75%R = 75\%

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Cevap

Company P matches Category 1 (Prime Growth); Company Q matches Category 2 (Capital Efficient); Company R matches Category 3 (Expansion Focused); Company S matches Category 4 (High Risk).
Each company is mapped to its category by evaluating its two metrics against the threshold boundary conditions (B=1.5B = 1.5 and R=85%R = 85\%). Company P satisfies low burn (B1.5B \le 1.5) and high retention (R85%R \ge 85\%), placing it in Category 1. Company Q meets low burn but has lower retention (R<85%R < 85\%), placing it in Category 2. Company R exceeds the burn threshold (B>1.5B > 1.5) but maintains high retention (R85%R \ge 85\%), placing it in Category 3. Company S exceeds the burn threshold and has lower retention, placing it in Category 4.

Adım Adım Çözüm

1
Evaluate Company P (B=1.2,R=90%B = 1.2, R = 90\%) against threshold rules.
B=1.21.5B = 1.2 \le 1.5 is true, and R=90%85%R = 90\% \ge 85\% is true. This satisfies Category 1 (Prime Growth).
Both metric conditions match the exact criteria for Category 1.
2
Evaluate Company Q (B=1.4,R=80%B = 1.4, R = 80\%) against threshold rules.
B=1.41.5B = 1.4 \le 1.5 is true, but R=80%<85%R = 80\% < 85\% is true. This satisfies Category 2 (Capital Efficient).
Low burn multiple combined with retention rate below 85%85\% places the entity in Category 2.
3
Evaluate Company R (B=1.8,R=88%B = 1.8, R = 88\%) against threshold rules.
B=1.8>1.5B = 1.8 > 1.5 is true, and R=88%85%R = 88\% \ge 85\% is true. This satisfies Category 3 (Expansion Focused).
Burn multiple exceeding 1.51.5 combined with high retention places the entity in Category 3.
4
Evaluate Company S (B=2.1,R=75%B = 2.1, R = 75\%) against threshold rules.
B=2.1>1.5B = 2.1 > 1.5 is true, and R=75%<85%R = 75\% < 85\% is true. This satisfies Category 4 (High Risk).
Both metrics fail the favorable thresholds, placing the entity in Category 4.

Anahtar Kavram

Categorical Classification and Sorting
Soru 109Soru

A municipal transit agency is installing two types of electric vehicle (EV) charging stations at a new central station: Level 2 Chargers (LL) and Level 3 Fast Chargers (DD). Each Level 2 charger costs $4,000\$4,000 to install and draws 10 kW10\text{ kW} of grid power. Each Level 3 charger costs $15,000\$15,000 to install and draws 50 kW50\text{ kW} of grid power. The agency has a total installation budget of $100,000\$100,000 and a maximum available grid power allocation of 320 kW320\text{ kW}. To meet minimum service grant requirements, the hub must install at least 5 Level 2 chargers (L5L \ge 5) and at least 3 Level 3 chargers (D3D \ge 3). Each Level 2 charger can serve up to 8 sessions per day, and each Level 3 charger can serve up to 30 sessions per day. Match each operational optimization metric on the left with its corresponding correct value on the right that maximizes the total daily charging sessions served by the hub.

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Öğeler

Optimal number of Level 2 chargers (LL)
Optimal number of Level 3 chargers (DD)
Maximum total daily charging sessions

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Cevap

The optimal operational configuration consists of 10 Level 2 chargers and 4 Level 3 chargers, achieving a maximum total of 200 daily charging sessions.
To maximize daily sessions S=8L+30DS = 8L + 30D under budget 4L+15D1004L + 15D \le 100, power L+5D32L + 5D \le 32, and lower bounds L5,D3L \ge 5, D \ge 3, testing valid integer points shows that setting Level 3 chargers to 4 and Level 2 chargers to 10 satisfies all joint constraints while achieving the peak capacity of 200 total daily sessions.

Adım Adım Çözüm

1
Formulate the linear system of constraints and objective function.
Budget constraint: 4,000L+15,000D100,000    4L+15D1004,000L + 15,000D \le 100,000 \implies 4L + 15D \le 100.
Power constraint: 10L+50D320    L+5D3210L + 50D \le 320 \implies L + 5D \le 32.
Minimum bounds: L5L \ge 5 and D3D \ge 3.
Objective function to maximize: S=8L+30DS = 8L + 30D.
Define the mathematical boundaries and optimization goal based on problem parameters.
2
Evaluate feasible integer values for Level 3 chargers (DD).
Case 1: If D=3D = 3:
- Budget bound: 4L+15(3)100    4L55    L134L + 15(3) \le 100 \implies 4L \le 55 \implies L \le 13.
- Power bound: L+5(3)32    L17L + 5(3) \le 32 \implies L \le 17.
- Max feasible L=13L = 13.
- Total sessions S=8(13)+30(3)=104+90=194S = 8(13) + 30(3) = 104 + 90 = 194.

Case 2: If D=4D = 4:
- Budget bound: 4L+15(4)100    4L40    L104L + 15(4) \le 100 \implies 4L \le 40 \implies L \le 10.
- Power bound: L+5(4)32    L12L + 5(4) \le 32 \implies L \le 12.
- Max feasible L=10L = 10.
- Total sessions S=8(10)+30(4)=80+120=200S = 8(10) + 30(4) = 80 + 120 = 200.

Case 3: If D=5D = 5:
- Budget bound: 4L+15(5)100    4L25    L64L + 15(5) \le 100 \implies 4L \le 25 \implies L \le 6.
- Power bound: L+5(5)32    L7L + 5(5) \le 32 \implies L \le 7.
- Max feasible L=6L = 6.
- Total sessions S=8(6)+30(5)=48+150=198S = 8(6) + 30(5) = 48 + 150 = 198.

Case 4: If D=6D = 6:
- Budget bound: 4L+15(6)100    4L10    L24L + 15(6) \le 100 \implies 4L \le 10 \implies L \le 2.
- Violates minimum constraint L5L \ge 5.
Systematically check integer combinations of variables within the feasible region to identify maximum daily sessions.
3
Compare total sessions across valid integer pairs.
The maximum session count occurs when D=4D = 4 and L=10L = 10, producing S=200S = 200 daily sessions.
Determine the unique optimal combination satisfying all joint constraints.

Anahtar Kavram

Linear Optimization under Bounded Integer Constraints
Tahmini Süre:2m 0s
Soru 110Soru

A printing company produces two types of customized marketing materials: Standard brochures (SS) and Premium brochures (PP). Each Standard brochure requires 22 minutes of printing time and 11 minute of binding time. Each Premium brochure requires 33 minutes of printing time and 33 minutes of binding time. The production facility has a maximum daily capacity of 240240 minutes for printing and 180180 minutes for binding. Due to a recurring client agreement, the facility must produce at least 2020 Standard brochures per day. The profit is $4\$4 per Standard brochure and $9\$9 per Premium brochure. Match each optimization variable or outcome on the left with its correct value on the right that maximizes daily total profit while satisfying all operational constraints.

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Öğeler

Optimal number of Standard brochures (SS)
Optimal number of Premium brochures (PP)
Maximum total daily profit (ZZ)

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Cevap

The optimal production values are 6060 Standard brochures and 4040 Premium brochures, resulting in a maximum daily profit of $600\$600.
Solving the system of simultaneous constraints yields the optimal vertex at S=60S = 60 Standard brochures and P=40P = 40 Premium brochures. At this point, both printing (2(60)+3(40)=2402(60) + 3(40) = 240) and binding (60+3(40)=18060 + 3(40) = 180) capacities are fully utilized, achieving the maximum daily profit of 4(60)+9(40)=$6004(60) + 9(40) = \$600.

Adım Adım Çözüm

1
Formulate the linear programming model with constraints and objective function.
Constraints: 2S+3P2402S + 3P \le 240 (printing time), S+3P180S + 3P \le 180 (binding time), S20S \ge 20 (contractual minimum), and S,P0S, P \ge 0. Objective: Maximize Z=4S+9PZ = 4S + 9P.
Clear mathematical formulation is required to define the feasible region bounded by resource limits.
2
Calculate the intersection points (vertices) of the boundary lines of the feasible region.
Subtracting S+3P=180S + 3P = 180 from 2S+3P=2402S + 3P = 240 yields S=60S = 60. Substituting S=60S = 60 into S+3P=180S + 3P = 180 gives 3P=120    P=403P = 120 \implies P = 40. Corner point 1: (60,40)(60, 40).
The fundamental theorem of linear programming states that maximum profit occurs at a vertex of the feasible region.
3
Find remaining boundary vertices satisfying all constraints.
Corner point 2: (120,0)(120, 0) from printing limit when P=0P=0 (satisfies binding limit 120180120 \le 180). Corner point 3: (20,53)(20, 53) when S=20S=20 (integer constraint on PP where 20+3P180    P53.3320 + 3P \le 180 \implies P \le 53.33). Corner point 4: (20,0)(20, 0).
All feasible vertices must be identified to compare their objective function values.
4
Evaluate the profit function Z=4S+9PZ = 4S + 9P at each vertex point.
At (60,40)(60, 40): Z=4(60)+9(40)=240+360=$600Z = 4(60) + 9(40) = 240 + 360 = \$600. At (120,0)(120, 0): Z=4(120)=$480Z = 4(120) = \$480. At (20,53)(20, 53): Z=4(20)+9(53)=80+477=$557Z = 4(20) + 9(53) = 80 + 477 = \$557. At (20,0)(20, 0): Z=4(20)=$80Z = 4(20) = \$80.
Direct substitution reveals which corner point yields the global maximum value inside the bounded constraints.

Anahtar Kavram

Linear programming optimization under joint resource and integer bound constraints
Soru 111Soru

A regional emergency health network deploys two types of mobile medical units for event coverage: Rapid Response Units (RR) and Heavy Support Units (HH).

The deployment is governed by the following staffing and operational constraints:
- Each Rapid Response Unit (RR) requires 11 paramedic and 22 EMTs.
- Each Heavy Support Unit (HH) requires 33 paramedics and 11 EMT.
- On any given shift, a maximum of 2121 paramedics and 2222 EMTs are available.
- Operational policy mandates deploying at least 33 Rapid Response Units (R3R \ge 3) and at least 22 Heavy Support Units (H2H \ge 2).

Each Rapid Response Unit can treat 1010 patients per hour, and each Heavy Support Unit can treat 1515 patients per hour.

Match each operational metric on the left to its corresponding optimal value on the right that maximizes total patient treatment capacity per hour.

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Öğeler

Optimal number of Rapid Response Units (RR)
Optimal number of Heavy Support Units (HH)
Maximum total patient treatment capacity per hour

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Cevap

The optimal operational configuration pairs 9 Rapid Response Units with 4 Heavy Support Units, achieving a maximum treatment capacity of 150 patients per hour.
The optimal pairing matches 9 Rapid Response Units, 4 Heavy Support Units, and a total capacity of 150 patients per hour. The optimal point (9,4)(9, 4) fully utilizes both available resource pools (1(9)+3(4)=211(9) + 3(4) = 21 paramedics and 2(9)+1(4)=222(9) + 1(4) = 22 EMTs), maximizing total throughput 10R+15H10R + 15H.

Adım Adım Çözüm

1
Set up the linear inequality constraints and objective function.
Paramedic resource constraint: R+3H21R + 3H \le 21; EMT resource constraint: 2R+H222R + H \le 22; Operational bounds: R3,H2R \ge 3, H \ge 2; Objective function to maximize: C=10R+15HC = 10R + 15H.
Translating the narrative conditions into algebraic forms defines the feasible region of operation.
2
Calculate the corner vertices of the bounded feasible polygon.
Solving R+3H=21R + 3H = 21 and 2R+H=222R + H = 22 simultaneously yields R=9R = 9 and H=4H = 4. Other extreme vertices defined by lower bounds are (3,6)(3, 6), (3,2)(3, 2), and (10,2)(10, 2).
Linear programming optimization principles guarantee that the maximum objective value occurs at one of the corner vertices of the feasible region.
3
Evaluate the capacity objective function C(R,H)=10R+15HC(R, H) = 10R + 15H at each vertex.
At (3,2):10(3)+15(2)=60(3, 2): 10(3) + 15(2) = 60; At (3,6):10(3)+15(6)=120(3, 6): 10(3) + 15(6) = 120; At (10,2):10(10)+15(2)=130(10, 2): 10(10) + 15(2) = 130; At (9,4):10(9)+15(4)=150(9, 4): 10(9) + 15(4) = 150.
Comparing capacities across all vertices establishes (9,4)(9, 4) as the global maximum.

Anahtar Kavram

Linear Programming Optimization under Bounded System Constraints
ÖncekiSayfa 6 / 6
Tüm alıştırma soruları — GMAT | Examkin