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Question 381Question →

A cloud computing infrastructure employs two types of processing nodes: Server-Type X and Server-Type Y. A configuration consisting of 3 Server-Type X nodes and 5 Server-Type Y nodes processes 4,100 data transactions per second. A second configuration consisting of 5 Server-Type X nodes and 2 Server-Type Y nodes processes 4,300 data transactions per second. Assuming each node of a given type operates at a constant rate, what is the processing capacity, in data transactions per second, of a single Server-Type X node?

Show answer & explanation

Answer: 700

Answer

700 data transactions per second

Let

x

be the capacity of Server-Type X and

y

be the capacity of Server-Type Y. The system

3x + 5y = 4100

and

5x + 2y = 4300

simplifies to

19x = 13300

after eliminating

y

, yielding

x = 700

.

Step-by-Step Solution

1

Set up the linear equations representing the two node configurations.

3x + 5y = 4100

and

5x + 2y = 4300

, where

x

and

y

represent the processing rates of Server-Type X and Server-Type Y, respectively.

Translating word problems into algebraic equations allows systematic solving for unknown rates.

2

Eliminate the variable

y

by multiplying the equations by appropriate constants.

6x + 10y = 8200

and

25x + 10y = 21500

.

Aligning the coefficients of

y

enables elimination through subtraction.

3

Subtract the first modified equation from the second and solve for

x

.

19x = 13300 \implies x = 700

.

Solving the single-variable equation yields the processing capacity for Server-Type X.

Key Concept

Solving simultaneous linear equations in two variables using elimination

Question 382Question →

The table below details physical and operational performance metrics for 10 commercial office properties managed by a real estate investment trust:

Property ID	Building Grade	Leasable Area (sq ft)	Occupancy Rate (%)	Energy Intensity (kWh/sq ft)
P-101	Class A	120,000	92%	18.4
P-102	Class B	85,000	88%	22.1
P-103	Class A	210,000	95%	14.8
P-104	Class A	165,000	84%	19.2
P-105	Class B	95,000	91%	24.5
P-106	Class A	140,000	96%	16.0
P-107	Class B	110,000	79%	21.0
P-108	Class A	180,000	90%	15.6
P-109	Class A	250,000	89%	17.2
P-110	Class B	75,000	94%	23.8

What is the median annual energy intensity, in kWh/sq ft, of the properties that have an occupancy rate of at least 90%?

Show answer & explanation

Answer: 17.2

Answer

17.2

Filtering the table for properties with an Occupancy Rate of at least 90% selects 6 properties: P-101 (18.4), P-103 (14.8), P-105 (24.5), P-106 (16.0), P-108 (15.6), and P-110 (23.8). Sorting these 6 values gives [14.8, 15.6, 16.0, 18.4, 23.8, 24.5]. Since the count is even (

N=6

), the median is the average of the two middle values, 16.0 and 18.4, which equals 17.2.

Step-by-Step Solution

1

Filter the dataset based on the occupancy rate criteria

Properties meeting the condition (Occupancy Rate

\ge 90\%

) are P-101 (92%), P-103 (95%), P-105 (91%), P-106 (96%), P-108 (90%), and P-110 (94%). This yields a subset of 6 properties.

Only properties meeting the threshold of at least 90% occupancy should be included in the statistical calculation.

2

Extract the corresponding Energy Intensity values for the filtered subset

The corresponding Energy Intensity values (in kWh/sq ft) are: 18.4, 14.8, 24.5, 16.0, 15.6, and 23.8.

Target descriptive metric calculation requires the specific Energy Intensity values for the 6 selected properties.

3

Order the extracted values from smallest to largest

Ordered values: 14.8, 15.6, 16.0, 18.4, 23.8, 24.5.

Finding the median requires arranging numerical data in sequential order.

4

Compute the median for the even-count dataset (

N=6

)

The 3rd value is 16.0 and the 4th value is 18.4. The median is

\frac{16.0 + 18.4}{2} = \frac{34.4}{2} = 17.2

.

When a dataset contains an even number of elements, the median is the arithmetic mean of the two middle elements.

Key Concept

Descriptive Statistics Interpretation

Question 383Question →

A commercial roastery blends two batches of coffee beans, Batch A and Batch R. Batch A contains whole beans and cracked beans in a mass ratio of $7:3$ . Batch R contains whole beans and cracked beans in a mass ratio of $3:2$ . A roaster combines $x$ kilograms of Batch A with $y$ kilograms of Batch R to create an unroasted blend in which the overall mass ratio of whole beans to cracked beans is $13:7$ .

During the roasting process, moisture evaporation causes whole beans to lose $10\%$ of their mass and cracked beans to lose $20\%$ of their mass. If the total mass of the roasted blend is $346$ kilograms, what was the initial mass $x$ of Batch A, in kilograms?

Show answer & explanation

Answer: 200

Answer

The initial mass of Batch A (

x

) is 200 kg.

The correct calculation shows that equal masses of Batch A and Batch R (

x = y

) satisfy the combined unroasted ratio of

13:7

. Applying the respective

10\%

and

20\%

mass losses for whole and cracked beans results in a total roasted mass factor of

1.73x

. Equating

1.73x = 346

gives

x = 200

kg.

Step-by-Step Solution

1

Set up expressions for initial whole and cracked bean masses

Batch A has

0.7x

kg whole and

0.3x

kg cracked beans; Batch R has

0.6y

kg whole and

0.4y

kg cracked beans.

Ratios

7:3

and

3:2

convert directly to component fractions

7/10 = 0.7

,

3/10 = 0.3

for A, and

3/5 = 0.6

,

2/5 = 0.4

for R.

2

Equate the total component ratio to

13:7

to find the relationship between

x

and

y

x = y

Setting

\frac{0.7x + 0.6y}{0.3x + 0.4y} = \frac{13}{7}

and cross-multiplying yields

4.9x + 4.2y = 3.9x + 5.2y

, simplifying to

x = y

.

3

Apply roasting mass loss percentages to compute total roasted mass in terms of

x

Total roasted mass =

1.73x

Whole beans retain

90\%

mass (

0.90 \times 1.3x = 1.17x

) and cracked beans retain

80\%

mass (

0.80 \times 0.7x = 0.56x

). Summing yields

1.17x + 0.56x = 1.73x

.

4

Solve for

x

given total roasted mass of

346

kg

x = 200

Dividing

346

by

1.73

yields

200

kg.

Key Concept

Weighted average ratios combined with proportional percentage change

Question 384Question →

A logistics company operates two types of delivery drones: Model X and Model Y. Working continuously at their respective constant hourly rates, a fleet of 3 Model X drones and 5 Model Y drones delivers a total of 190 packages per hour. Alternatively, a fleet of 4 Model X drones and 2 Model Y drones delivers a total of 160 packages per hour. What is the hourly package delivery capacity of a single Model X drone?

Show answer & explanation

Answer: 30

Answer

The hourly package delivery capacity of a single Model X drone is 30 packages per hour.

By translating the problem into simultaneous linear equations,

3x + 5y = 190

and

4x + 2y = 160

, simplifying the second equation gives

2x + y = 80

, so

y = 80 - 2x

. Substituting this into the first equation yields

3x + 5(80 - 2x) = 190

, which reduces to

-7x = -210

, giving

x = 30

packages per hour.

Step-by-Step Solution

1

Set up a system of two linear equations based on the rates of Model X (

x

) and Model Y (

y

).

System of equations:
(1)

3x + 5y = 190

(2)

4x + 2y = 160

Each statement provides a linear combination of the hourly output of the two drone types.

2

Express

y

in terms of

x

using the second equation.

2x + y = 80 \implies y = 80 - 2x

Simplifying the second equation isolates

y

easily for substitution.

3

Substitute the expression for

y

into the first equation and solve for

x

.

3x + 5(80 - 2x) = 190 \implies 3x + 400 - 10x = 190 \implies -7x = -210 \implies x = 30

Solving the single-variable linear equation yields the precise capacity of Model X.

Key Concept

Solving Systems of Two Linear Equations with Two Unknowns

Estimated Time:1m 30s

Question 385Question →

The table below presents annual production and resource consumption metrics for four semiconductor fabrication plants in 2025.

Fab Facility	Silicon Wafers Processed (in thousands)	Solvent Used per Wafer (in mL)	Total Electricity Consumed (in MWh)
Fab Alpha	450	800	3,600
Fab Beta	600	500	4,200
Fab Gamma	250	1,200	2,750
Fab Delta	800	450	5,200

Note: $1\text{ kiloliter (kL)} = 1,000\text{ liters (L)} = 1,000,000\text{ milliliters (mL)}$ .

Based on the table, what was the total volume of solvent used by Fab Alpha in 2025, expressed in kiloliters (kL)?

Show answer & explanation

Answer: 360

Answer

The total volume of solvent used by Fab Alpha in 2025 was 360 kL.

To find the total solvent used by Fab Alpha in kiloliters, first account for the scale factor in the wafer column header: Fab Alpha processed 450 thousand wafers, which is

450 \times 1,000 = 450,000

wafers. Next, calculate the volume in milliliters:

450,000 \text{ wafers} \times 800 \text{ mL/wafer} = 360,000,000 \text{ mL}

. Finally, convert to kiloliters by dividing by

1,000,000

(

1 \text{ kL} = 1,000,000 \text{ mL}

), which gives

360 \text{ kL}

.

Step-by-Step Solution

1

Extract wafer volume with scale multiplier

Fab Alpha processed 450,000 wafers.

The table column header specifies 'in thousands', requiring a multiplier of 1,000.

2

Calculate total solvent volume in milliliters

Total solvent volume = 360,000,000 mL.

Multiply 450,000 wafers by 800 mL per wafer.

3

Convert volume to target unit (kiloliters)

360 kL.

Divide 360,000,000 mL by 1,000,000 mL per kL.

Key Concept

Unit Conversion and Scale Interpretation

Question 386Question →

An e-commerce fulfillment center uses two automated sorting machines, Machine X and Machine Y, to sort packages. Machine X processes packages at a constant rate of $150$ packages per hour, with an error rate of $4\%$ . Machine Y processes packages at a constant rate of $200$ packages per hour, with an error rate of $1.5\%$ . If both machines operate simultaneously and continuously to process a total batch of $2,100$ packages, what is the total number of incorrectly processed packages in this batch?

Show answer & explanation

Answer: 54

Answer

54 packages

Working together at a combined rate of

350

packages per hour, the machines take

6

hours to complete

2,100

packages. Machine X processes

900

packages with

36

errors (

4\%

), while Machine Y processes

1,200

packages with

18

errors (

1.5\%

). Adding these yields

54

total errors.

Step-by-Step Solution

1

Determine the combined processing rate of the two machines

350

packages per hour

Machine X and Machine Y work concurrently, so their rates add up (

150 + 200 = 350

).

2

Determine total operating time needed for the batch

6

hours

Dividing total packages (

2,100

) by the combined rate (

350

) gives the total operating duration.

3

Calculate individual volume handled by each machine

Machine X:

900

packages; Machine Y:

1,200

packages

Multiplying the duration (

6

hours) by each machine's hourly rate yields their individual output.

4

Calculate the error quantity for each machine and total errors

36 + 18 = 54

packages

4\%

of

900 = 36

, and

1.5\%

of

1,200 = 18

. The sum of errors is

54

.

Key Concept

Combined Rates and Weighted Percentage Errors

Question 387Question →

A pharmaceutical laboratory uses two automated centrifuge units, Unit Alpha and Unit Beta, to process liquid compound solutions into a central storage tank.

- Unit Alpha processes a solution containing $15\%$ active compound at a constant rate of $40$ liters per hour.
- Unit Beta processes a solution containing $35\%$ active compound at a constant rate of $60$ liters per hour.

If both units operate simultaneously for $5$ hours, what is the percentage of active compound in the total combined solution collected in the tank?

Show answer & explanation

Answer: 27

Answer

The active compound concentration in the combined solution is

27\%

.

The correct answer of

27\%

is obtained by determining the overall mass balance: Unit Alpha contributes

30

liters of active compound from

200

liters of total liquid, while Unit Beta contributes

105

liters of active compound from

300

liters of total liquid. Dividing the total active compound (

135

liters) by the total liquid (

500

liters) gives

0.27

, or

27\%

.

Step-by-Step Solution

1

Determine the total volume and active compound volume produced by Unit Alpha.

Unit Alpha produces

200

liters of total liquid (

40 \times 5

) containing

30

liters of active compound (

200 \times 0.15

).

Volume equals rate multiplied by time, and active compound volume equals total volume multiplied by its concentration percentage.

2

Determine the total volume and active compound volume produced by Unit Beta.

Unit Beta produces

300

liters of total liquid (

60 \times 5

) containing

105

liters of active compound (

300 \times 0.35

).

Applied the same volume and percentage relationship to Unit Beta.

3

Compute the combined total volume and total active compound volume.

Total combined liquid volume =

200 + 300 = 500

liters. Total active compound =

30 + 105 = 135

liters.

Combining the output of both units sums their respective liquid and active compound volumes.

4

Calculate the combined percentage concentration.

\frac{135}{500} \times 100\% = 27\%.

The final concentration is the ratio of total active compound to total liquid volume expressed as a percentage.

Key Concept

Weighted Average Percentage Concentration

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Step-by-Step Solution

Key Concept

Common Mistakes