Examkin

Table Analysis

63 soru

Tüm türler63Çoktan seçmeli20Çoklu seçim3Doğru / Yanlış27Sayısal9Eşleştirme4
Soru 21Soru →

The table below displays quarterly logistics metrics for five regional distribution hubs of a global shipping enterprise in 2025:

HubTotal Units ShippedAir Freight Share (%)Q1 Avg Shipping Cost per Unit ($)Q4 Avg Shipping Cost per Unit ($)
North50,00030%120168
South60,00040%200260
East40,00025%150213
West75,00020%160192
Central30,00035%80108

For the hub that experienced the greatest percentage increase in average shipping cost per unit from Q1 to Q4, what is the ratio of its air freight unit volume to the non-air freight unit volume of the hub with the highest total overall expenditure in Q4?

Cevabı ve açıklamayı göster

Cevap: 5:185 : 18

Cevap

The ratio of the air freight unit volume of the hub with the greatest percentage cost increase (East) to the non-air freight unit volume of the hub with the highest Q4 expenditure (South) is 5 : 18.
East has the greatest percentage cost increase (42.0%42.0\%), yielding 10,00010,000 air freight units. South has the highest total Q4 expenditure ($15,600,000\$15,600,000), with a non-air freight unit volume of 36,00036,000. The simplified ratio of these two quantities is 10,000:36,000=5:1810,000 : 36,000 = 5 : 18.

Adım Adım Çözüm

1
Calculate the percentage increase in average shipping cost per unit from Q1 to Q4 for each hub.
North: 168−120120=40.0%\frac{168 - 120}{120} = 40.0\%
South: 260−200200=30.0%\frac{260 - 200}{200} = 30.0\%
East: 213−150150=42.0%\frac{213 - 150}{150} = 42.0\%
West: 192−160160=20.0%\frac{192 - 160}{160} = 20.0\%
Central: 108−8080=35.0%\frac{108 - 80}{80} = 35.0\%
Identify which hub achieved the greatest percentage increase in per-unit cost.
2
Identify the target hub and calculate its air freight unit volume.
The hub with the highest percentage increase is East (42.0%). Its air freight unit volume is 25%25\% of 40,000=10,00040,000 = 10,000 units.
Formulate the numerator for the final ratio.
3
Calculate total Q4 expenditure for each hub to find the hub with the highest overall expenditure.
North: 50,000×$168=$8,400,00050,000 \times \$168 = \$8,400,000
South: 60,000×$260=$15,600,00060,000 \times \$260 = \$15,600,000
East: 40,000×$213=$8,520,00040,000 \times \$213 = \$8,520,000
West: 75,000×$192=$14,400,00075,000 \times \$192 = \$14,400,000
Central: 30,000×$108=$3,240,00030,000 \times \$108 = \$3,240,000
The hub with the highest total Q4 expenditure is South ($15,600,000).
4
Calculate the non-air freight unit volume for South.
South non-air freight share =100%−40%=60%= 100\% - 40\% = 60\%. Non-air freight volume =60%= 60\% of 60,000=36,00060,000 = 36,000 units.
Formulate the denominator for the final ratio.
5
Compute and simplify the final ratio.
10,00036,000=1036=518\frac{10,000}{36,000} = \frac{10}{36} = \frac{5}{18} or 5:185 : 18.
Express the relationship in simplest integer ratio form.

Anahtar Kavram

Multi-Step Tabular Ratio and Percentage Analysis
Soru 22Soru →

The table below details capital allocation, exit valuations, active portfolio companies, and successful exits across five high-technology sectors for a venture capital firm in 2026.

SectorCapital Invested ($M)Realized Exit Valuation ($M)Active Portfolio CompaniesSuccessful Exits
AI Infrastructure4501,350186
Clean Energy320560164
Enterprise SaaS6001,5002510
Advanced Robotics250375103
BioHealth & Genomics400880125

Evaluate the following statement based on the data provided:

The overall return multiple (total realized exit valuation divided by total capital invested across all five sectors combined) is less than the percentage return on investment (net gain divided by capital invested) achieved by the BioHealth & Genomics sector.

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False. The overall return multiple across all five sectors (2.3094×2.3094\times, representing 230.94%230.94\%) is greater than the percentage ROI for BioHealth & Genomics (120%120\%).
The evaluated statement is False. Calculating the total sum of capital invested across all sectors yields $2,020M\$2,020\text{M}, and the total sum of realized exit valuations yields $4,665M\$4,665\text{M}. The overall return ratio is 46652020≈2.3094\frac{4665}{2020} \approx 2.3094 (or 230.94%230.94\%). In contrast, the percentage ROI for BioHealth & Genomics is calculated on its net gain: 880−400400=1.20\frac{880 - 400}{400} = 1.20 (or 120%120\%). Comparing the two values reveals that 2.3094>1.20002.3094 > 1.2000, contradicting the statement's claim that the overall multiple is less.

Adım Adım Çözüm

1
Calculate the total capital invested across all five sectors.
Total Capital Invested = 450+320+600+250+400=$2,020 million450 + 320 + 600 + 250 + 400 = \$2,020\text{ million}.
Aggregating the column provides the denominator for the combined return multiple.
2
Calculate the total realized exit valuation across all five sectors.
Total Exit Valuation = 1,350+560+1,500+375+880=$4,665 million1,350 + 560 + 1,500 + 375 + 880 = \$4,665\text{ million}.
Aggregating the column provides the numerator for the combined return multiple.
3
Compute the overall return multiple across all sectors.
Overall Multiple = $4,665M$2,020M≈2.3094\frac{\$4,665\text{M}}{\$2,020\text{M}} \approx 2.3094 (or 230.94%230.94\% of capital invested).
Determines the benchmark value stated in the prompt.
4
Calculate the percentage ROI for the BioHealth & Genomics sector.
BioHealth Net Gain = $880M−$400M=$480M\$880\text{M} - \$400\text{M} = \$480\text{M}. BioHealth ROI = $480M$400M×100%=120%\frac{\$480\text{M}}{\$400\text{M}} \times 100\% = 120\%.
Percentage ROI measures net profit relative to initial capital invested.
5
Compare the overall return multiple to the BioHealth percentage ROI.
2.3094>1.20002.3094 > 1.2000 (or 230.94%>120%230.94\% > 120\%). The statement claims the overall multiple is less than the BioHealth ROI, which is incorrect.
Direct comparison evaluates the truth value of the target statement.

Anahtar Kavram

Distinguishing Total Return Multiples from Net Percentage ROI and Aggregating Tabular Data
Tahmini Süre:2m 30s
Soru 23Soru →

The table below displays performance metrics for six maritime shipping ports.

PortRegionTonnage (M Tons)On-Time Arrival Rate (%)Operational Cost ($M)
Port AEast458812
Port BWest608815
Port CEast609214
Port DNorth309510
Port EWest458511
Port FEast608516

If the table is sorted first by Tonnage (M Tons) in descending order, and any ties are broken by On-Time Arrival Rate (%) in descending order, Port B will be listed second from the top.

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is True. When sorted primarily by Tonnage (descending) and secondarily by On-Time Arrival Rate (descending), Port B occupies the second row from the top.
Sorting primarily by Tonnage in descending order identifies Ports B, C, and F as the top group with 60 M Tons. Applying the secondary sort by On-Time Arrival Rate in descending order ranks Port C (92%) first and Port B (88%) second.

Adım Adım Çözüm

1
Identify the primary sort column and filter for the highest values.
The primary sort column is Tonnage (M Tons) descending. The maximum tonnage value is 60, shared by Port B, Port C, and Port F.
Primary sort criteria dictate grouping the dataset by the highest primary metric first.
2
Apply the secondary sort criteria (On-Time Arrival Rate descending) to break the three-way tie.
Among Ports B, C, and F, their On-Time Arrival Rates are: Port C (92%), Port B (88%), and Port F (85%). Sorting descending gives the order: Port C, Port B, Port F.
Secondary sort rules determine the internal ordering of rows that share identical primary sort values.
3
Determine the final position of Port B.
Port C is row 1, Port B is row 2, and Port F is row 3.
Comparing the sorted order directly answers the statement's evaluation criteria.

Anahtar Kavram

Multi-Column Table Sorting and Tie-Breaking Logic
Soru 24Soru →

The table below presents operational and maintenance data for seven municipal water pumping stations across four municipal districts.

Station CodeDistrictFacility Age (Years)Energy Consumption (kWh/m³)Maintenance Backlog ($k)Annual Outages
Alpha-1North140.851403
Bravo-4South221.122105
Charlie-2East140.781852
Delta-9North281.25955
Echo-3West220.942101
Foxtrot-7South181.051604
Golf-5West140.851202

Statement: If the table is sorted in ascending order by Facility Age (Years), with any ties broken by sorting in descending order by Maintenance Backlog ($k), then among the stations with a Facility Age of 14 years, the station with the lowest Energy Consumption rating is listed immediately after the station with the highest Annual Outages count.

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is False because Charlie-2 (lowest Energy Consumption among 14-year stations) appears immediately before Alpha-1 (highest Annual Outages among 14-year stations) when applying the secondary descending sort on Maintenance Backlog.
The correct response evaluates the multi-column sort step-by-step: filtering the 14-year-old stations (Alpha-1, Charlie-2, Golf-5), sorting them by descending Maintenance Backlog (yielding Charlie-2 then Alpha-1 then Golf-5), and identifying that Charlie-2 (lowest energy consumption) precedes Alpha-1 (highest annual outages) rather than following it.

Adım Adım Çözüm

1
Filter for the relevant subset of stations
Identify stations with Facility Age = 14: Alpha-1, Charlie-2, Golf-5.
The primary sort isolates all stations with the minimum age (14 years) at the beginning of the table.
2
Apply the tie-breaking multi-column ordering logic
Sort Alpha-1 (140k),Charlie−2(140k), Charlie-2 ( 185k), and Golf-5 (120k)indescendingorderofMaintenanceBacklog:1st=Charlie−2(120k) in descending order of Maintenance Backlog: 1st = Charlie-2 ( 185k), 2nd = Alpha-1 (140k),3rd=Golf−5(140k), 3rd = Golf-5 ( 120k).
Ties in primary sort (Facility Age) must be broken by secondary descending sort on Maintenance Backlog.
3
Evaluate key metrics within the sorted sub-group
Station with lowest Energy Consumption among 14-year stations = Charlie-2 (0.78 kWh/m³). Station with highest Annual Outages count among 14-year stations = Alpha-1 (3 outages).
Identify the target entities to verify their relative vertical ordering.
4
Determine relative position of target stations
Charlie-2 is in 1st position and Alpha-1 is in 2nd position. Therefore, Charlie-2 appears immediately BEFORE Alpha-1, rendering the statement False.
Verify whether the target entity appears immediately after or immediately before.

Anahtar Kavram

Multi-Column Sorting with Primary and Secondary Tie-Breaking Criteria
Soru 25Soru →

The table below presents performance and operational metrics for 8 regional microgrid energy storage facilities.

FacilityTechnologyStorage Capacity (MWh)Round-Trip Efficiency (%)Maintenance Cost ($/MWh)
M-1Solar+Storage4588.514.20
M-2Hybrid6091.016.50
M-3Wind+Storage5088.512.80
M-4Hybrid4091.015.10
M-5Solar+Storage5592.013.80
M-6Hybrid7089.518.00
M-7Wind+Storage6593.511.50
M-8Solar+Storage3588.515.00

Suppose the table is sorted according to the following sequential rules:
1. Primary sort: Technology in ascending alphabetical order.
2. Secondary sort (tie-breaker for identical Technology): Round-Trip Efficiency in descending order (highest to lowest).
3. Tertiary sort (tie-breaker for identical Technology and Round-Trip Efficiency): Maintenance Cost in ascending order (lowest to highest).

After applying this multi-column ordering, what is the Storage Capacity (in MWh) of the facility ranked 5th from the top?

Cevabı ve açıklamayı göster

Cevap: 45

Cevap

45 MWh
To determine the 5th ranked facility, we follow the three-tier sorting hierarchy: first by Technology (alphabetical), then by Round-Trip Efficiency (descending), and finally by Maintenance Cost (ascending). Sorting by Technology creates three groups: Hybrid (ranks 1–3), Solar+Storage (ranks 4–6), and Wind+Storage (ranks 7–8). Within the Solar+Storage group, the highest efficiency is facility M-5 (92.0%), making it rank 4. Facilities M-1 and M-8 both have 88.5% efficiency. Applying the tertiary rule (lower maintenance cost first), facility M-1 (14.20)precedesfacilityM−8(14.20) precedes facility M-8 ( 15.00). Thus, facility M-1 occupies rank 5. Looking across facility M-1's row, its Storage Capacity is 45 MWh.

Adım Adım Çözüm

1
Group and sort facilities by the primary criterion: Technology in alphabetical order.
Group 1: Hybrid (M-2, M-4, M-6); Group 2: Solar+Storage (M-1, M-5, M-8); Group 3: Wind+Storage (M-3, M-7).
Alphabetical order places 'Hybrid' first, followed by 'Solar+Storage', and then 'Wind+Storage'.
2
Sort the 'Hybrid' group (Ranks 1 to 3) using secondary (Efficiency desc) and tertiary (Cost asc) rules.
M-4 and M-2 tie at 91.0% Efficiency; M-4 has lower cost (15.10vs15.10 vs 16.50), placing M-4 1st and M-2 2nd. M-6 (89.5%) is 3rd.
Secondary sort prioritizes higher efficiency, and tertiary sort resolves ties by choosing lower maintenance cost.
3
Sort the 'Solar+Storage' group (Ranks 4 to 6) using secondary and tertiary rules.
M-5 has highest efficiency (92.0%) -> Rank 4. M-1 and M-8 tie at 88.5% Efficiency; M-1 has lower cost (14.20vs14.20 vs 15.00) -> Rank 5: M-1, Rank 6: M-8.
M-1 takes Rank 5 because its 14.20/MWhmaintenancecostbeatsM−8′s14.20/MWh maintenance cost beats M-8's 15.00/MWh cost in the tertiary tie-breaker.
4
Identify the facility in Rank 5 and read its Storage Capacity.
Rank 5 is facility M-1, which has a Storage Capacity of 45 MWh.
Evaluating the 5th entry of the ordered list directly answers the question stem.

Anahtar Kavram

Multi-column hierarchical table sorting with primary, secondary, and tertiary tie-breaking conditions.
Soru 26Soru →

The table below presents operational metrics and financial allocations for six municipal water treatment facilities in 2025.

FacilityTotal Water Processed (Million Gallons)Recycled Water (Million Gallons)Distribution Loss Rate (%)Annual Operating Budget ($ Millions)
Alton450908%18.0
Belton60015012%21.0
Colton350706%14.0
Dalton70021015%22.4
Easton50013010%17.5
Fulton400805%16.0

For the facility with the lowest operating budget per million gallons of total water processed, what percentage of its total processed water was recycled?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

30%
Evaluating the unit budget per million gallons processed across all facilities reveals that Dalton has the lowest operating rate at 22.4M/700 MG=$0.0320M / MG22.4\text{M} / 700\text{ MG} = \$0.0320\text{M / MG}. Computing the percentage of total processed water that was recycled for Dalton gives (210/700)×100%=30%(210 / 700) \times 100\% = 30\%.

Adım Adım Çözüm

1
Calculate operating budget per million gallons processed for each facility to determine unit cost.
Alton = 0.0400; Belton = 0.0350; Colton = 0.0400; Dalton = 0.0320; Easton = 0.0350; Fulton = 0.0400 ($M per MG).
To identify which facility operates with the lowest financial cost per unit of water processed.
2
Identify the minimum unit cost value among all six facilities.
Dalton has the lowest operating budget per million gallons ($0.0320M per MG).
Dalton satisfies the condition of having the lowest operating budget per million gallons processed.
3
Divide Dalton's recycled water volume by its total processed water volume and express as a percentage.
(210 / 700) * 100% = 30%.
To find the specific proportion of recycled water relative to total processed water at the identified facility.

Anahtar Kavram

Unit Rate Comparison and Percentage Calculation from Tabular Data
Soru 27Soru →

The table below displays operational performance metrics for 10 regional cold-chain distribution centers operated by a pharmaceutical logistics company across two geographic zones.

Distribution CenterZoneCold Storage Volume (103 m310^3\text{ m}^3)Temperature ExcursionsOn-Time Delivery Rate (%)
Center H-1Zone East451294.2
Center H-2Zone West801891.5
Center H-3Zone East60896.8
Center H-4Zone East302289.0
Center H-5Zone West551493.4
Center H-6Zone East75598.1
Center H-7Zone West401692.0
Center H-8Zone East901195.5
Center H-9Zone West65995.0
Center H-10Zone East501592.6

By how many percentage points does the median On-Time Delivery Rate of the distribution centers in Zone East exceed the mean On-Time Delivery Rate of all 10 distribution centers?

Cevabı ve açıklamayı göster

Cevap: 1.04 percentage points

Cevap

The median On-Time Delivery Rate of Zone East distribution centers exceeds the mean On-Time Delivery Rate of all 10 centers by 1.04 percentage points.
The mean of all 10 distribution centers is calculated by summing all on-time delivery rates (938.1%938.1\%) and dividing by 10 to get 93.81%93.81\%. Filtering for Zone East gives 6 centers with delivery rates [89.0%,92.6%,94.2%,95.5%,96.8%,98.1%][89.0\%, 92.6\%, 94.2\%, 95.5\%, 96.8\%, 98.1\%]. Because N=6N = 6 is even, the median is the average of the 3rd (94.2%94.2\%) and 4th (95.5%95.5\%) terms, yielding 94.85%94.85\%. Subtracting 93.81%93.81\% from 94.85%94.85\% gives 1.041.04 percentage points.

Adım Adım Çözüm

1
Calculate the mean On-Time Delivery Rate for all 10 distribution centers.
Sum of all delivery rates = 94.2+91.5+96.8+89.0+93.4+98.1+92.0+95.5+95.0+92.6=938.1%94.2 + 91.5 + 96.8 + 89.0 + 93.4 + 98.1 + 92.0 + 95.5 + 95.0 + 92.6 = 938.1\%. Mean = 938.110=93.81%\frac{938.1}{10} = 93.81\%.
To establish the overall average delivery performance benchmark across all facilities.
2
Filter the dataset to isolate the distribution centers in Zone East.
Zone East consists of 6 centers: H-1 (94.2%94.2\%), H-3 (96.8%96.8\%), H-4 (89.0%89.0\%), H-6 (98.1%98.1\%), H-8 (95.5%95.5\%), and H-10 (92.6%92.6\%).
The question specifically requires evaluating the median of the Zone East subset.
3
Sort the Zone East On-Time Delivery Rates in ascending order.
Sorted list of 6 values: [89.0%,92.6%,94.2%,95.5%,96.8%,98.1%][89.0\%, 92.6\%, 94.2\%, 95.5\%, 96.8\%, 98.1\%].
Sorting is a mandatory prerequisite step before determining the median of any dataset.
4
Calculate the median of the even-count (N = 6) Zone East subset.
The two central values (3rd and 4th) are 94.2%94.2\% and 95.5%95.5\%. Median = 94.2+95.52=94.85%\frac{94.2 + 95.5}{2} = 94.85\%.
For an even number of elements, the median is defined as the exact arithmetic mean of the two middle elements.
5
Compute the difference between the calculated median and mean.
94.85%−93.81%=1.0494.85\% - 93.81\% = 1.04 percentage points.
To answer the specific target question regarding the margin by which the median exceeds the mean.

Anahtar Kavram

Descriptive Statistics Interpretation from Tabular Subsets
Soru 28Soru →

The table below provides operational metrics for 10 international clinical trial sites participating in a multi-center biopharmaceutical study:

Site IDTherapeutic AreaPatients EnrolledDrop-out Rate (%)Average Treatment Duration (Days)
Site AOncology4212.5180
Site BImmunology568.0120
Site COncology3015.0225
Site DNeurology646.2590
Site EOncology7810.0150
Site FImmunology4812.5135
Site GOncology5014.0165
Site HNeurology329.375105
Site IOncology605.0195
Site JImmunology4010.0150

Statement: For the subset of clinical trial sites with a drop-out rate of at least 10.0%, the median Average Treatment Duration exceeds the median Patients Enrolled by more than 110 days.

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True. The median Average Treatment Duration for the filtered subset of 6 sites is 157.5 days, and the median Patients Enrolled is 45 patients. The difference of 112.5 days exceeds 110 days.
The statement is correct (True) because filtering for sites with a drop-out rate of at least 10.0%10.0\% yields exactly 6 sites. The median of Patients Enrolled across these 6 sites is 42+482=45\frac{42+48}{2} = 45, and the median of Average Treatment Duration is 150+1652=157.5\frac{150+165}{2} = 157.5. The difference 157.5−45=112.5157.5 - 45 = 112.5 is strictly greater than 110.

Adım Adım Çözüm

1
Filter the dataset by the given condition
Subset consists of 6 sites with Drop-out Rate ≥10.0%\ge 10.0\%: Site A, Site C, Site E, Site F, Site G, and Site J.
Only sites meeting the threshold of ≥10.0%\ge 10.0\% drop-out rate must be analyzed.
2
Determine the median of Patients Enrolled for the filtered subset
Ordered values: 30,40,42,48,50,7830, 40, 42, 48, 50, 78. Median = 42+482=45\frac{42 + 48}{2} = 45.
Because the subset size N=6N = 6 is even, the median is the arithmetic mean of the two middle elements (3rd and 4th).
3
Determine the median of Average Treatment Duration for the filtered subset
Ordered values: 135,150,150,165,180,225135, 150, 150, 165, 180, 225. Median = 150+1652=157.5\frac{150 + 165}{2} = 157.5.
Because N=6N = 6 is even, the median is the arithmetic mean of the 3rd and 4th ordered values.
4
Calculate the difference between the two medians and evaluate the statement
Difference = 157.5−45=112.5157.5 - 45 = 112.5 days, which is greater than 110 days.
Since 112.5>110112.5 > 110, the statement is True.

Anahtar Kavram

Descriptive Statistics Interpretation on Filtered Even-Count Subsets
Soru 29Soru →

The table below details performance and financial metrics for 10 municipal express bus routes operated by a regional transit authority:

Route IDZone TypeDaily Ridership (Thousands)On-Time Performance (%)Operating Cost per Passenger ($)
Route U1Urban Core14.214.282.5%82.5\%3.403.40
Route U2Urban Core18.518.576.0%76.0\%2.902.90
Route U3Urban Core22.022.088.0%88.0\%4.104.10
Route U4Urban Core11.011.091.5%91.5\%3.703.70
Route U5Urban Core16.416.479.0%79.0\%2.502.50
Route U6Urban Core25.125.184.0%84.0\%4.804.80
Route S1Suburban8.28.294.0%94.0\%5.205.20
Route S2Suburban6.56.592.5%92.5\%5.805.80
Route S3Suburban9.09.089.0%89.0\%4.904.90
Route S4Suburban5.45.495.5%95.5\%6.106.10

Based on the table, what is the median Operating Cost per Passenger ($) for the express bus routes operating in the Urban Core zone?

Cevabı ve açıklamayı göster

Cevap: 3.553.55

Cevap

The median Operating Cost per Passenger for the Urban Core routes is $3.55.
Filtering the table for 'Urban Core' yields 6 routes. Sorting their Operating Cost per Passenger in ascending order gives: 2.50,2.50, 2.90, 3.40,3.40, 3.70, 4.10,and4.10, and 4.80. Because there is an even number of items (6), the median is the arithmetic average of the middle two values (3.40and3.40 and 3.70), which equals $3.55.

Adım Adım Çözüm

1
Filter the table data for routes located in the 'Urban Core' Zone Type.
Identified 6 Urban Core routes: Route U1 (3.40),RouteU2(3.40), Route U2 ( 2.90), Route U3 (4.10),RouteU4(4.10), Route U4 ( 3.70), Route U5 (2.50),andRouteU6(2.50), and Route U6 ( 4.80).
The question specifies finding descriptive statistics exclusively for the Urban Core zone routes.
2
Order the Operating Cost per Passenger values of the filtered Urban Core routes from least to greatest.
Sorted list of costs: 2.50,2.50, 2.90, 3.40,3.40, 3.70, 4.10,4.10, 4.80.
Finding a median requires ascending or descending order of values.
3
Calculate the median for the even dataset size (N=6N = 6).
The two central values are the 3rd (3.40)and4th(3.40) and 4th ( 3.70) elements. Median = $3.40+$3.702=$3.55\frac{\$3.40 + \$3.70}{2} = \$3.55.
When a dataset contains an even number of elements, the median is the arithmetic mean of the two middle elements.

Anahtar Kavram

Descriptive Statistics Interpretation (Median of Filtered Even-Count Datasets)
Soru 30Soru →

The table below presents energy and cost metrics for six commercial data center facilities operated by a cloud computing provider in fiscal year 2025.

FacilityIT Load Capacity (MW)Renewable Energy Used (GWh)Total Energy Consumption (GWh)Operational Cost ($M)
Alpha4018024036.0
Beta6531239054.6
Gamma2510016022.4
Delta8036048072.0
Epsilon5022530042.0
Zeta9048654075.6

Statement: If the operational cost of facility Beta increases by 15% in 2026 while its IT load capacity remains unchanged, its 2026 operational cost per megawatt (MW) of IT load capacity will exceed the 2025 operational cost per megawatt of facility Alpha by more than 8%.

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is False because facility Beta's 2026 operational cost per MW (0.966M/MW)exceedsfacilityAlpha′s2025operationalcostperMW(0.966M/MW) exceeds facility Alpha's 2025 operational cost per MW ( 0.900M/MW) by approximately 7.33%, which is less than 8%.
The correct evaluation is False. Facility Alpha's unit operational cost is 0.900MperMW(0.900M per MW ( 36.0M / 40 MW). Facility Beta's unit cost in 2025 is 0.840MperMW(0.840M per MW ( 54.6M / 65 MW). A 15% increase brings Beta's 2026 unit cost to 0.966MperMW.ComparingBeta′s2026rate(0.966M per MW. Comparing Beta's 2026 rate ( 0.966M) to Alpha's 2025 rate (0.900M)givesarelativeincreaseof(0.900M) gives a relative increase of ( 0.966 - 0.900)/0.900) / 0.900 = 7.33%, which does not exceed 8%.

Adım Adım Çözüm

1
Calculate facility Alpha's 2025 operational cost per MW of IT load capacity.
Facility Alpha cost per MW = 36.0M/40MW=36.0M / 40 MW = 0.900M per MW.
Establishes the base value against which the 2026 rate will be compared.
2
Calculate facility Beta's 2025 operational cost per MW and its updated 2026 rate after a 15% increase.
Facility Beta 2025 rate = 54.6M/65MW=54.6M / 65 MW = 0.840M per MW. Facility Beta 2026 rate = 0.840M×1.15=0.840M × 1.15 = 0.966M per MW.
Applies the 15% growth rate to Beta's cost per unit capacity.
3
Compute the percentage change (excess) of facility Beta's 2026 rate relative to facility Alpha's 2025 rate.
Percentage excess = (0.966M−0.966M - 0.900M) / $0.900M = 0.066 / 0.900 = 7.33%.
Evaluates whether the percentage difference exceeds the 8% threshold requested in the statement.

Anahtar Kavram

Unit Rate Calculations and Percentage Change Base Selection
Soru 31Soru →

The table below details 2025 performance and operational metrics for seven metropolitan transit corridors:

Corridor CodeVehicle Fleet Size (Buses)Zero-Emission Vehicles (%)On-Time Departure Rate (%)Annual Passenger Trips (Millions)Operating Cost per Passenger Trip ($)
C-North12045%92%18.0$2.40
C-South20060%85%32.0$1.95
C-East8025%94%10.0$3.10
C-West15040%88%22.5$2.20
C-Central25070%81%45.0$1.80
C-Harbor9055%90%12.0$2.85
C-Airport11030%96%15.0$2.60

Match each of the three statements on the left with its correct evaluation description on the right based on the data provided.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Statement 1: The median operating cost per passenger trip across all seven corridors is greater than the weighted average operating cost per passenger trip for the entire transit system.
Statement 2: For corridors where the Zero-Emission Vehicles percentage is at least 45%45\%, the mean vehicle fleet size is strictly less than the mean vehicle fleet size of corridors with a Zero-Emission Vehicles percentage below 45%45\%.
Statement 3: Among corridors with an On-Time Departure Rate of at least 88%88\%, the corridor with the highest ratio of annual passenger trips per vehicle is C-South.

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Statement 1 matches the evaluation confirming it is True (2.40>2.20262.40 > 2.2026); Statement 2 matches the evaluation confirming it is False because 165>113.33165 > 113.33; Statement 3 matches the evaluation confirming it is False due to C-South failing the conditional threshold.
Each statement matches its precise evaluation. Statement 1 correctly identifies that the median cost per trip (2.402.40) exceeds the weighted average system cost (2.20262.2026). Statement 2 identifies that the mean fleet size for high zero-emission corridors (165165) is higher, not lower, than for low zero-emission corridors (113.33113.33). Statement 3 correctly identifies that C-South is excluded altogether because its on-time departure rate is 85%85\%, below the required 88%88\% threshold.

Adım Adım Çözüm

1
Calculate the median operating cost and system-wide weighted average operating cost for Statement 1.
Corridor operating costs in ascending order: $1.80,$1.95,$2.20,$2.40,$2.60,$2.85,$3.10\$1.80, \$1.95, \$2.20, \$2.40, \$2.60, \$2.85, \$3.10. Median = $2.40\$2.40. Total system operating cost = $340.3 million\$340.3\text{ million}; total trips = 154.5 million154.5\text{ million}. Weighted average = $340.3/154.5≈$2.2026\$340.3 / 154.5 \approx \$2.2026. Median ($2.40\$2.40) > Weighted Average ($2.2026\$2.2026). Statement 1 is True.
Evaluates median vs. weighted average accurately across unequal trip volumes.
2
Filter and calculate subset fleet means for Statement 2.
Corridors with Zero-Emission ≥45%\ge 45\%: C-North (120120), C-South (200200), C-Central (250250), C-Harbor (9090). Mean = 660/4=165660 / 4 = 165. Corridors with Zero-Emission <45%< 45\%: C-East (8080), C-West (150150), C-Airport (110110). Mean = 340/3≈113.33340 / 3 \approx 113.33. 165>113.33165 > 113.33, so the statement claiming 165<113.33165 < 113.33 is False.
Compares conditional subgroup averages correctly.
3
Apply conditional filtering for Statement 3 based on On-Time Departure Rate.
Corridors with On-Time Departure Rate ≥88%\ge 88\% include C-North, C-East, C-West, C-Harbor, and C-Airport. C-South has an On-Time Departure Rate of 85%<88%85\% < 88\%, so it is excluded from the comparison set. Statement 3 is False.
Prevents evaluating metrics on elements that do not satisfy pre-requisite conditional criteria.

Anahtar Kavram

Multi-Statement Boolean Evaluation on Tabular Data
Soru 32Soru →

The table below details semiconductor fabrication performance metrics across eight production lines during Q2 2026:

Fab LineProcess Node (nm)Monthly Wafers (thousands)Defect Density (defects/cm²)Yield Rate (%)R&D Intensity ($/wafer)
Line A3450.2878.5%420
Line B5800.2284.0%310
Line C71200.1591.2%180
Line D141500.0895.6%95
Line E3350.3472.0%460
Line F5900.1886.5%290
Line G71100.1293.0%210
Line H282000.0498.2%50

Evaluate the truth value of the following claim regarding three boolean statements about the dataset:

Statement I: Among the Fab Lines with a Process Node of 7 nm or smaller, the median Defect Density is greater than the mean Defect Density of lines producing more than 100 thousand Monthly Wafers.
Statement II: The weighted average Yield Rate across all 3 nm and 5 nm Fab Lines combined (weighted by Monthly Wafers) is less than 82.0%.
Statement III: When the table is sorted in descending order of R&D Intensity ($/wafer), the Fab Line in the third row has a higher Yield Rate than the Fab Line with the lowest overall Defect Density.

Claim: Evaluating statements I, II, and III yields True, False, and False, respectively.

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True
The claim stating that evaluating statements I, II, and III yields True, False, and False, respectively, is correct because mathematical evaluation confirms Statement I is True (0.20>0.09750.20 > 0.0975), Statement II is False (82.23%>82.0%82.23\% > 82.0\%), and Statement III is False (84.0%<98.2%84.0\% < 98.2\%).

Adım Adım Çözüm

1
Filter table for Process Node ≤ 7 nm and calculate median Defect Density, then compare with mean Defect Density of lines with Monthly Wafers > 100k.
Process Node ≤ 7 nm values: {0.12, 0.15, 0.18, 0.22, 0.28, 0.34} → Median = 0.20 defects/cm². Monthly Wafers > 100k values: {0.15, 0.08, 0.12, 0.04} → Mean = 0.0975 defects/cm². 0.20 > 0.0975 → Statement I is True.
Evaluates conditional filtering and statistical medians versus means across filtered subsets.
2
Compute the weighted average Yield Rate for 3 nm and 5 nm lines combined.
Total Wafers = 250k; Total Conforming Wafers = 205.575k → Weighted Yield = 82.23%. Since 82.23% ≥ 82.0%, Statement II is False.
Evaluates multi-row weighted average calculation versus simple unweighted averaging.
3
Sort table by R&D Intensity descending, identify 3rd row line, and compare its Yield Rate with the line having lowest overall Defect Density.
3rd row line by R&D Intensity is Line B (Yield Rate = 84.0%). Lowest Defect Density line is Line H (Yield Rate = 98.2%). 84.0% < 98.2% → Statement III is False.
Tests column sorting, positional indexing, and cross-column correlation comparison.
4
Combine the evaluations of Statements I, II, and III.
Evaluation sequence is (True, False, False), matching the claim exactly.
Confirms the ultimate boolean evaluation of the composite claim.

Anahtar Kavram

Multi-Statement Boolean Evaluation using tabular data filtering, weighted averages, and multi-column sorting.
Tahmini Süre:3m 0s
Soru 33Soru →

The table below presents quarterly operational metrics for 10 semiconductor fabrication plants:

Plant IDRegionWafer SizeDefect Density (defects/cm²)Operational Yield (%)
Fab 101Asia-Pacific300 mm1.894.2
Fab 102Asia-Pacific300 mm2.492.6
Fab 103Asia-Pacific300 mm3.189.5
Fab 104Asia-Pacific300 mm3.788.0
Fab 105Asia-Pacific300 mm4.386.4
Fab 106Asia-Pacific300 mm5.183.1
Fab 107North America300 mm2.293.0
Fab 108North America200 mm4.885.0
Fab 109Europe300 mm3.587.5
Fab 110Europe200 mm5.581.0

Statement: For semiconductor fabrication plants with a 300 mm300\text{ mm} wafer size located in the Asia-Pacific region, the median defect density is greater than 3.3 defects/cm23.3\text{ defects/cm}^2.

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True
The correct response is True because the filtered subset of 6 plants in the Asia-Pacific region with 300 mm wafer sizes has defect densities of 1.8, 2.4, 3.1, 3.7, 4.3, and 5.1 defects/cm². Averaging the two middle values (3.1 and 3.7) yields a median of 3.4 defects/cm², which exceeds 3.3 defects/cm².

Adım Adım Çözüm

1
Filter the dataset according to specified criteria
Identified 6 relevant plants: Fab 101, Fab 102, Fab 103, Fab 104, Fab 105, and Fab 106 (Region = Asia-Pacific AND Wafer Size = 300 mm).
The question requires analyzing descriptive statistics specifically for 300 mm wafer plants in the Asia-Pacific region.
2
Extract and sort the defect density data for the filtered subset
Sorted list of defect densities (N=6N = 6): 1.8,2.4,3.1,3.7,4.3,5.11.8, 2.4, 3.1, 3.7, 4.3, 5.1.
Calculating a median requires arranging the data points in ascending order.
3
Calculate the median of the even-count dataset
The 3rd element is 3.13.1 and the 4th element is 3.73.7. The median is 3.1+3.72=3.4 defects/cm2\frac{3.1 + 3.7}{2} = 3.4\text{ defects/cm}^2.
When NN is even, the median is defined as the mean of the two middle numbers at positions N2\frac{N}{2} and N2+1\frac{N}{2} + 1.
4
Evaluate the statement
The calculated median of 3.4 defects/cm23.4\text{ defects/cm}^2 is strictly greater than 3.3 defects/cm23.3\text{ defects/cm}^2. Therefore, the statement is True.
Comparing the calculated value (3.43.4) directly against the threshold in the stem (3.33.3) confirms the statement.

Anahtar Kavram

Descriptive Statistics Interpretation (Median of Even-Count Subsets)
Soru 34Soru →

The table below presents operational patient data across five regional health centers for fiscal year 2025.

Regional Health CenterTotal Inpatient AdmissionsEmergency Department VisitsOutpatient Telehealth Consultations30-Day Inpatient Readmissions
Central Metro12,00045,00018,0001,440
North Ridge8,00025,00015,000880
Eastside15,00060,00020,0001,500
Valley Pines6,00018,00012,000540
West Coast10,00032,00014,0001,100

Across all five regional health centers combined, what percentage of total inpatient admissions resulted in a 30-day inpatient readmission?

Cevabı ve açıklamayı göster

Cevap: 10.7%

Cevap

10.7%
To calculate the aggregate percentage across all five regional health centers, sum the total number of readmissions (1,440+880+1,500+540+1,100=5,4601,440 + 880 + 1,500 + 540 + 1,100 = 5,460) and divide by the sum of total inpatient admissions (12,000+8,000+15,000+6,000+10,000=51,00012,000 + 8,000 + 15,000 + 6,000 + 10,000 = 51,000). Dividing 5,4605,460 by 51,00051,000 yields approximately 0.107060.10706, or 10.7%10.7\%.

Adım Adım Çözüm

1
Calculate the total number of 30-day inpatient readmissions across all five regional health centers.
1,440+880+1,500+540+1,100=5,4601,440 + 880 + 1,500 + 540 + 1,100 = 5,460
To find aggregate percentage, the combined numerator must be determined.
2
Calculate the total number of inpatient admissions across all five regional health centers.
12,000+8,000+15,000+6,000+10,000=51,00012,000 + 8,000 + 15,000 + 6,000 + 10,000 = 51,000
The question specifically asks for readmissions as a percentage of total inpatient admissions, establishing the base denominator.
3
Divide total readmissions by total inpatient admissions and convert to a percentage.
5,46051,000×100%≈10.70588%≈10.7%\frac{5,460}{51,000} \times 100\% \approx 10.70588\% \approx 10.7\%
Determines the combined weighted percentage of readmissions.

Anahtar Kavram

Weighted aggregate percentage calculation from tabular data
Tahmini Süre:1m 30s
Soru 35Soru →

The table lists performance and cost metrics for 10 municipal water treatment facilities during the 2025 fiscal year:

Facility IDRegionDaily Capacity (MGD)Average Turbidity (NTU)Operating Cost ($/thousand gallons)
W-01North24.00.121.42
W-02South18.00.251.85
W-03North32.00.081.28
W-04Central12.00.191.65
W-05South28.00.151.56
W-06Central40.00.101.18
W-07North15.00.221.70
W-08South22.00.141.50
W-09Central35.00.111.34
W-10North10.00.281.92

Based on the data provided, what is the median operating cost, in dollars per thousand gallons, for the facilities that have a daily capacity greater than 20.0 MGD?

Cevabı ve açıklamayı göster

Cevap: 1.38

Cevap

The median operating cost for facilities with a daily capacity greater than 20.0 MGD is $1.38 per thousand gallons.
Filtering the table for facilities with Daily Capacity > 20.0 MGD yields 6 facilities (W-01, W-03, W-05, W-06, W-08, W-09). Sorting their operating costs in ascending order produces: 1.18,1.18, 1.28, 1.34,1.34, 1.42, 1.50,1.50, 1.56. Because the count of elements is even (6), the median is the average of the 3rd element (1.34)andthe4thelement(1.34) and the 4th element ( 1.42), which equals $1.38.

Adım Adım Çözüm

1
Filter the dataset by the given capacity constraint.
Identified 6 facilities with Daily Capacity > 20.0 MGD: W-01 (24.0 MGD), W-03 (32.0 MGD), W-05 (28.0 MGD), W-06 (40.0 MGD), W-08 (22.0 MGD), and W-09 (35.0 MGD).
Only facilities meeting the criterion 'Daily Capacity > 20.0 MGD' must be evaluated.
2
Extract and order the operating costs for the filtered subset.
Operating costs in ascending order: 1.18,1.18, 1.28, 1.34,1.34, 1.42, 1.50,1.50, 1.56.
Determining the median of a dataset requires arranging values sequentially.
3
Calculate the median for an even-numbered dataset.
Average of the 3rd and 4th values: \(\frac{1.34 + 1.42}{2} = 1.38\).
When a dataset contains an even number of elements \(N = 6\), the median is the arithmetic mean of the two central numbers at positions \(N/2 = 3\) and \(N/2 + 1 = 4\).

Anahtar Kavram

Calculating the median of a filtered even-count subset from tabular data.

Alternatif Yöntem

As a verification step, sum all operating costs in the filtered set: 1.18+1.28+1.34+1.42+1.50+1.56=8.281.18 + 1.28 + 1.34 + 1.42 + 1.50 + 1.56 = 8.28. The mean is 8.28/6=1.388.28 / 6 = 1.38. In this symmetric subset, the mean and median coincide at 1.381.38.
Tahmini Süre:2m 0s
Soru 36Soru →

The table below presents subscriber numbers and annual operating revenue data for five regional divisions of a telecommunications provider in fiscal year 2025.

RegionTotal Broadband Subscribers (thousands)Fiber-Optic Subscribers (thousands)Total Operating Revenue ($ millions)
North25015045.0
South40020064.0
East1809027.0
West32022457.6
Central50035080.0

Across the two regional divisions with the highest total operating revenue combined, fiber-optic subscribers represent what percentage of total broadband subscribers? Express your answer to the nearest tenth of a percent.

Cevabı ve açıklamayı göster

Cevap: 61.1

Cevap

61.1%
To find the proportion of fiber-optic subscribers across the top two revenue-generating divisions, first rank the divisions by Total Operating Revenue: Central (80.0M)andSouth(80.0M) and South ( 64.0M) are the highest. Combine their subscriber counts to find the appropriate base: Central has 500 thousand total broadband subscribers and South has 400 thousand, giving a total base of 500+400=900500 + 400 = 900 thousand. Next, combine their fiber-optic subscribers: Central has 350 thousand and South has 200 thousand, giving 350+200=550350 + 200 = 550 thousand fiber-optic subscribers. The combined percentage is 550900×100%≈61.111%\frac{550}{900} \times 100\% \approx 61.111\%, which rounds to 61.1%61.1\%.

Adım Adım Çözüm

1
Identify the top two divisions by operating revenue
Central (80.0M)andSouth(80.0M) and South ( 64.0M)
The question specifically restricts the calculation to the two divisions with the highest operating revenue.
2
Sum total broadband subscribers for Central and South
500 + 400 = 900 thousand subscribers
This establishes the correct aggregate base value for the percentage calculation.
3
Sum fiber-optic subscribers for Central and South
350 + 200 = 550 thousand subscribers
This gives the total combined numerator value for the fiber-optic subgroup.
4
Calculate percentage and round to the nearest tenth
(550 / 900) * 100% = 61.111...% -> 61.1%
Dividing the target subgroup total by the base total yields the requested proportion.

Anahtar Kavram

Ratio, Proportion, and Percent Calculations across Filtered Aggregate Groups
Tahmini Süre:1m 30s
Soru 37Soru →

The table below displays operational metrics for eight semiconductor fabrication facilities (fabs).

Fab IDRegionWafer Yield Rate (%)Defect Density (defects/cm²)Monthly Capacity (k Wafers)Operating Cost per Wafer ($)
Fab AlphaAsia-Pacific94.20.08120450
Fab BetaNorth America91.50.1285520
Fab GammaEurope95.80.0560490
Fab DeltaAsia-Pacific91.50.10140410
Fab EpsilonEurope93.00.0995475
Fab ZetaNorth America94.20.07110510
Fab EtaAsia-Pacific89.80.15150390
Fab ThetaEurope95.80.0670465

Evaluate the following statement as True or False:

If the table is sorted primarily by Wafer Yield Rate (%) in descending order, and secondarily by Defect Density (defects/cm²) in ascending order to break ties, the facility ranked 4th has a lower Monthly Capacity than the facility ranked 3rd.

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is False.
Sorting primarily by Wafer Yield Rate (%) descending places facilities with 95.8%95.8\% yield in positions 1–2 and facilities with 94.2%94.2\% yield in positions 3–4. To break the tie for positions 3 and 4, we sort Defect Density in ascending order. Fab Zeta (0.070.07 defects/cm²) precedes Fab Alpha (0.080.08 defects/cm²). Thus, Fab Zeta is ranked 3rd with a capacity of 110110 k wafers, and Fab Alpha is ranked 4th with a capacity of 120120 k wafers. Since 120120 is greater than 110110, the statement claiming the 4th-ranked facility has a lower capacity is False.

Adım Adım Çözüm

1
Identify the primary sorting criterion and group facilities by Wafer Yield Rate (%) in descending order.
Group 1 (95.8%95.8\%): Fab Gamma, Fab Theta
Group 2 (94.2%94.2\%): Fab Alpha, Fab Zeta
Group 3 (93.0%93.0\%): Fab Epsilon
Group 4 (91.5%91.5\%): Fab Beta, Fab Delta
Group 5 (89.8%89.8\%): Fab Eta
Primary sort requires ordering by highest Wafer Yield Rate first.
2
Apply the secondary sorting criterion (Defect Density in ascending order) to break ties in the top groups to determine ranks 1 through 4.
For Group 1 (95.8%95.8\%):
- Fab Gamma (0.050.05) ranks 1st
- Fab Theta (0.060.06) ranks 2nd

For Group 2 (94.2%94.2\%):
- Fab Zeta (0.070.07) ranks 3rd
- Fab Alpha (0.080.08) ranks 4th
Ascending order means lower numerical defect density comes first.
3
Compare the Monthly Capacity values of the 3rd and 4th ranked facilities.
- 3rd Ranked (Fab Zeta): 110110 k Wafers
- 4th Ranked (Fab Alpha): 120120 k Wafers
Since 120>110120 > 110, the 4th-ranked facility has a higher capacity than the 3rd-ranked facility.
Verifies whether the condition '4th-ranked facility has a lower Monthly Capacity than 3rd-ranked facility' holds true.

Anahtar Kavram

Multi-Column Sorting and Secondary Tie-Breaking Logic
Soru 38Soru →

The table below presents operational and usage metrics for eight branch locations within a regional public library network during 2025:

Branch CodeTotal Volumes (thousands)Active Cardholders (thousands)Annual Circulations (thousands)Operating Expenses ($ thousands)Digital Resource Accesses (thousands)
CEN450609003,600300
NOR120253001,000150
EAS801616064080
WES15020240960120
SOU200405001,500250
RIV901822572090
HAR11022220880110
UPL160324801,440160

Consider the following three statements regarding the data:

I. The median number of Annual Circulations among the branches with more than 20,000 Active Cardholders is greater than 350,000.
II. For Eastgate (EAS), the ratio of Annual Circulations per Total Volume is higher than that for West End (WES).
III. Exactly three branches satisfy both of the following conditions: Operating Expenses per Circulation are strictly less than $4.00, AND Digital Resource Accesses exceed 140,000.

Which of the following correctly identifies the truth values (True or False) of Statements I, II, and III, respectively?

Cevabı ve açıklamayı göster

Cevap: Statement I: True; Statement II: True; Statement III: True

Cevap

Statement I: True; Statement II: True; Statement III: True
The correct option is the one identifying all three statements as True. Statement I is True because the median circulation of the 5 filtered branches is 480,000, which exceeds 350,000. Statement II is True because EAS has a circulation-to-volume ratio of 2.0 compared to 1.6 for WES. Statement III is True because NOR, SOU, and UPL are the exactly three branches that satisfy both expenses per circulation < $4.00 and digital accesses > 140,000.

Adım Adım Çözüm

1
Evaluate Statement I by filtering and sorting.
Branches with Active Cardholders > 20 (in thousands) are CEN (60), NOR (25), SOU (40), HAR (22), and UPL (32), total 5 branches. Their Annual Circulations (in thousands) are 900, 300, 500, 220, and 480. Sorting these values yields 220, 300, 480, 500, 900. The median (3rd value) is 480 thousand (480,000). Since 480,000 > 350,000, Statement I is True.
Filtering required strict application of the cardholder threshold before ordering the circulation data.
2
Evaluate Statement II by calculating row ratios.
For EAS, Circulations / Volumes = 160 / 80 = 2.0. For WES, Circulations / Volumes = 240 / 150 = 1.6. Since 2.0 > 1.6, Statement II is True.
Ratio comparison requires placing Circulations in the numerator and Volumes in the denominator.
3
Evaluate Statement III using dual-condition logical filtering.
Calculate Operating Expenses (thousands)/AnnualCirculations(thousands)toget thousands) / Annual Circulations (thousands) to get per circulation: CEN (3,600 / 900 = 4.00),NOR(1,000/300=4.00), NOR (1,000 / 300 = 3.33), EAS (640 / 160 = 4.00),WES(960/240=4.00), WES (960 / 240 = 4.00), SOU (1,500 / 500 = 3.00),RIV(720/225=3.00), RIV (720 / 225 = 3.20), HAR (880 / 220 = 4.00),UPL(1,440/480=4.00), UPL (1,440 / 480 = 3.00). Branches with expense per circulation strictly less than $4.00 are NOR, SOU, RIV, and UPL. Among these, Digital Resource Accesses (> 140 thousand) are NOR (150), SOU (250), and UPL (160). RIV has only 90. Exactly 3 branches meet both conditions. Statement III is True.
Strict inequality rules out branches with expenses per circulation equal to $4.00.

Anahtar Kavram

Multi-Statement Boolean Evaluation on Tabular Data
Soru 39Soru →

The table below details operational metrics for 10 wind turbines managed by a renewable energy utility during the previous calendar year:

Turbine IDRegionCapacity (MW)Availability (%)Unplanned Maintenance (hours)
T-01Offshore6.094.248
T-02Onshore4.097.518
T-03Offshore8.091.084
T-04Onshore3.598.112
T-05Offshore6.095.836
T-06Offshore8.093.562
T-07Onshore4.096.024
T-08Offshore6.092.470
T-09Onshore3.598.88
T-10Offshore8.096.528

Based on the table, what is the median number of unplanned maintenance hours for turbines located in the Offshore region?

Cevabı ve açıklamayı göster

Cevap: 55

Cevap

The median number of unplanned maintenance hours for turbines in the Offshore region is 55 hours.
Filtering the table for Offshore turbines yields 6 entries with unplanned maintenance hours of 48, 84, 36, 62, 70, and 28. Arranging these values in ascending order gives {28, 36, 48, 62, 70, 84}. Because the count is even (N = 6), the median is the average of the 3rd and 4th elements: (48 + 62) / 2 = 55 hours.

Adım Adım Çözüm

1
Filter the dataset by Region
Identified 6 turbines in the Offshore region: T-01 (48 hrs), T-03 (84 hrs), T-05 (36 hrs), T-06 (62 hrs), T-08 (70 hrs), and T-10 (28 hrs).
The question specifically restricts the calculation to turbines operating in the Offshore region.
2
Sort the filtered unplanned maintenance hours in ascending order
Ordered set: 28, 36, 48, 62, 70, 84.
Finding the median requires data elements to be arranged sequentially.
3
Calculate the median of the even-count dataset (N = 6)
The two middle values are the 3rd element (48) and the 4th element (62). Mean = (48 + 62) / 2 = 55.
When a dataset contains an even number of observations, the median is the arithmetic average of the two central terms.

Anahtar Kavram

Descriptive Statistics Interpretation (Median of Even-Count Filtered Subsets)
Tahmini Süre:1m 30s
Soru 40Soru →

The table below provides operational performance metrics for 10 electric vehicle (EV) charging hubs operated by a regional clean energy authority during the second quarter of 2026:

Hub IDRegionFast ChargersDaily Utilization Rate (%)Average Session Duration (min)
H-101North1272%28.5
H-102Metro1681%42.0
H-103South858%35.0
H-104Metro1068%31.0
H-105Metro1488%46.5
H-106West662%22.0
H-107Metro2075%38.0
H-108North1070%44.0
H-109Metro874%34.0
H-110Metro1276%52.0

What is the median Average Session Duration, in minutes, for the subset of charging hubs with a Daily Utilization Rate exceeding 71%?

Cevabı ve açıklamayı göster

Cevap: 40.0

Cevap

40.0 minutes
Filtering the table for hubs with a Daily Utilization Rate strictly greater than 71% yields exactly 6 hubs: H-101 (28.5 min), H-102 (42.0 min), H-105 (46.5 min), H-107 (38.0 min), H-109 (34.0 min), and H-110 (52.0 min). Arranging these 6 duration values in ascending order gives: 28.5, 34.0, 38.0, 42.0, 46.5, 52.0. Because the dataset contains an even number of values (6), the median is calculated by averaging the middle two values: (38.0 + 42.0) / 2 = 40.0 minutes.

Adım Adım Çözüm

1
Filter the table rows by the condition: Daily Utilization Rate > 71%.
Six hubs meet the threshold: H-101 (72%), H-102 (81%), H-105 (88%), H-107 (75%), H-109 (74%), and H-110 (76%).
Filtering isolates the relevant subset of hubs specified in the question stem.
2
Extract and order the Average Session Duration values for these 6 hubs in ascending numerical order.
Ordered values: 28.5, 34.0, 38.0, 42.0, 46.5, 52.0.
To determine a median, the data points must be arranged sequentially.
3
Calculate the median for an even dataset size (N=6N = 6).
Median = 38.0+42.02=40.0\frac{38.0 + 42.0}{2} = 40.0 minutes.
When NN is even, the median is the arithmetic mean of the N2\frac{N}{2}-th element (3rd element = 38.0) and the (N2+1)(\frac{N}{2} + 1)-th element (4th element = 42.0).

Anahtar Kavram

Descriptive Statistics Interpretation
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Table Analysis Alıştırma Soruları — GMAT — Sayfa 2 | Examkin