Two-Part Analysis Practice Questions — GMAT — Page 4

Question 61Question →

An agricultural research station operates two types of automated irrigation pumps: Type Alpha and Type Beta. Operating 4 Type Alpha pumps and 3 Type Beta pumps simultaneously for 5 hours consumes a total of $215\text{ kilowatt-hours (kWh)}$ of electricity. Operating 2 Type Alpha pumps and 5 Type Beta pumps simultaneously for 4 hours consumes a total of $156\text{ kWh}$ of electricity. Based on the information provided, match each given pump metric on the left with its correct hourly electricity consumption value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The hourly consumption rate of one Type Alpha pump is

7\text{ kWh per hour}

, one Type Beta pump is

5\text{ kWh per hour}

, and their combined hourly rate is

12\text{ kWh per hour}

.

Solving the system of simultaneous linear equations

4a + 3b = 43

and

2a + 5b = 39

yields

a = 7

for Type Alpha pumps and

b = 5

for Type Beta pumps. The sum

a + b

equals

12

.

Step-by-Step Solution

1

Set up equations for the total hourly consumption from the given scenario data.

Let

a

be the hourly consumption of Type Alpha in kWh and

b

be the hourly consumption of Type Beta in kWh. From 5 hours of operation:

5(4a + 3b) = 215 \implies 4a + 3b = 43

. From 4 hours of operation:

4(2a + 5b) = 156 \implies 2a + 5b = 39

.

Dividing total energy by total hours converts total consumption into hourly rates for each operating group.

2

Solve the system of simultaneous linear equations for

a

and

b

.

Multiply

2a + 5b = 39

by 2 to get

4a + 10b = 78

. Subtract

4a + 3b = 43

from

4a + 10b = 78

:

(4a + 10b) - (4a + 3b) = 78 - 43 \implies 7b = 35 \implies b = 5

.

Eliminating variable

a

allows direct solution for variable

b

.

3

Substitute

b = 5

back into one of the linear equations to find

a

.

2a + 5(5) = 39 \implies 2a + 25 = 39 \implies 2a = 14 \implies a = 7

.

Determines the single unit rate for Type Alpha pumps.

4

Calculate the combined hourly rate for one Type Alpha and one Type Beta pump.

a + b = 7 + 5 = 12\text{ kWh per hour}

.

Calculates the sum of both individual unit rates.

Key Concept

Formulating and solving a system of two linear equations in two variables from rates and total work/consumption.

Estimated Time:2m 0s

Question 62Question →

A corporate audit department classifies incoming project files using two criteria: Compliance Category and Funding Level.

Files are prioritized according to the following strict hierarchy:
1. First, by Compliance Category: Category 1 (Urgent Audit) > Category 2 (Standard Audit) > Category 3 (Routine Check).
2. Second, within the same Compliance Category, by Funding Level: High Funding > Low Funding.

Based on these rules, what is the correct sequence of the four project files listed below when sorted from highest processing priority (1st) to lowest processing priority (4th)?

Drag items to arrange them in the correct order

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Answer

The correct order from highest to lowest priority is File Beta, File Alpha, File Gamma, and File Delta.

The correct ranking applies the primary criterion (Compliance Category: Category 1 > Category 2 > Category 3) first, which places File Beta and File Alpha in the top two positions, followed by File Gamma in third, and File Delta in fourth. Within Category 1, applying the secondary criterion (Funding Level: High Funding > Low Funding) places File Beta ahead of File Alpha.

Step-by-Step Solution

1

Group the files by the primary precedence criterion (Compliance Category).

Category 1 files (File Beta and File Alpha) have higher priority than Category 2 files (File Gamma), which have higher priority than Category 3 files (File Delta).

Compliance Category is the primary sorting key specified in the rules.

2

Apply the tie-breaker criterion (Funding Level) to files within Category 1.

File Beta (High Funding) ranks ahead of File Alpha (Low Funding).

When Compliance Categories are identical, High Funding takes precedence over Low Funding.

3

Assemble the complete sequence from highest to lowest priority.

1st: File Beta, 2nd: File Alpha, 3rd: File Gamma, 4th: File Delta.

Combining primary category sorting and secondary funding level tie-breaking yields the definitive order.

Key Concept

Multi-attribute categorical classification and hierarchical sorting

Question 63Question →

A smart battery storage unit operates over three consecutive cycles ( $k = 1, 2, 3$ ). At the start of Cycle 1 ( $k = 0$ ), the stored energy ( $E_0$ ) is $50\text{ kWh}$ and the grid demand factor ( $D_0$ ) is $10$ .

For each cycle $k$ :
- Stored energy updates according to the demand factor of the previous cycle: $E_k = E_{k-1} + 20$ if $D_{k-1} \ge 10$ ; otherwise, $E_k = E_{k-1} - 10$ .
- Grid demand factor updates according to: $D_k = D_{k-1} - 2$ .

Which of the following statements regarding the system states must be true? (Select all that apply.)

Select all that apply

At the end of Cycle 1 (

k = 1

), the stored energy is

70\text{ kWh}

.

At the end of Cycle 3 (

k = 3

), the grid demand factor is

4

.

At the end of Cycle 2 (

k = 2

), the stored energy is

90\text{ kWh}

.

At the end of Cycle 3 (

k = 3

), the stored energy is

70\text{ kWh}

.

At the end of Cycle 2 (

k = 2

), the grid demand factor is

8

.

Show answer & explanation

Answer: At the end of Cycle 1 ( $k = 1$ ), the stored energy is $70\text{ kWh}$ .; At the end of Cycle 3 ( $k = 3$ ), the grid demand factor is $4$ .

Answer

The statements confirming that the stored energy at the end of Cycle 1 is 70 kWh and that the grid demand factor at the end of Cycle 3 is 4 are both correct.

Evaluating the transition steps sequentially: At Cycle 1 (

k=1

),

D_0 = 10 \ge 10

, so stored energy becomes

50 + 20 = 70\text{ kWh}

and demand factor becomes

10 - 2 = 8

. This confirms the statement regarding Cycle 1 stored energy. Continuing through Cycle 2 (

D_1 = 8 < 10 \Rightarrow E_2 = 60, D_2 = 6

) to Cycle 3 (

D_2 = 6 < 10 \Rightarrow E_3 = 50, D_3 = 4

), the grid demand factor reaches

4

, confirming the statement regarding Cycle 3 grid demand.

Step-by-Step Solution

1

Calculate state values at the end of Cycle 1 (

k = 1

)

E_1 = 50 + 20 = 70\text{ kWh}

because

D_0 = 10 \ge 10

.

D_1 = 10 - 2 = 8

.

Apply the conditional energy rule and the linear demand decrease rule using initial values.

2

Calculate state values at the end of Cycle 2 (

k = 2

)

E_2 = 70 - 10 = 60\text{ kWh}

because

D_1 = 8 < 10

.

D_2 = 8 - 2 = 6

.

Since

D_1 < 10

, the energy decreases by

10\text{ kWh}

.

3

Calculate state values at the end of Cycle 3 (

k = 3

)

E_3 = 60 - 10 = 50\text{ kWh}

because

D_2 = 6 < 10

.

D_3 = 6 - 2 = 4

.

Since

D_2 < 10

, the energy decreases by

10\text{ kWh}

and demand factor decreases by

2

.

Key Concept

Sequential Processes and State Transitions

Question 64Question →

At the beginning of a fiscal year, an asset management firm allocated funds between two portfolios, Portfolio Alpha and Portfolio Beta, in the ratio of $3:5$ , respectively. Over the course of the year, Portfolio Alpha gained $40\%$ in value while Portfolio Beta lost $10\%$ in value. At the end of the year, a total of $\$150,000$ in profits was transferred from Portfolio Alpha to Portfolio Beta. Following this transfer, the final value of Portfolio Beta was exactly $25\%$ greater than the final value of Portfolio Alpha. Based on the information provided, what were the initial asset values of Portfolio Alpha and Portfolio Beta, respectively, at the start of the fiscal year?

Portfolio Alpha:

\$1,350,000

; Portfolio Beta:

\$2,250,000

Portfolio Alpha:

\$2,250,000

; Portfolio Beta:

\$1,350,000

Portfolio Alpha:

\$1,890,000

; Portfolio Beta:

\$2,025,000

Portfolio Alpha:

\$1,740,000

; Portfolio Beta:

\$2,175,000

Portfolio Alpha:

\$1,200,000

; Portfolio Beta:

\$2,000,000

Show answer & explanation

Answer: Portfolio Alpha: $\$1,350,000$ ; Portfolio Beta: $\$2,250,000$

Answer

Portfolio Alpha had an initial asset value of

\$1,350,000

and Portfolio Beta had an initial asset value of

\$2,250,000

.

The correct pair identifies Portfolio Alpha's initial value as

\$1,350,000

and Portfolio Beta's initial value as

\$2,250,000

. This preserves the initial

3:5

ratio. A

40\%

increase brings Alpha to

\$1,890,000

, and a

10\%

decrease brings Beta to

\$2,025,000

. Transferring

\$150,000

results in final values of

\$1,740,000

for Alpha and

\$2,175,000

for Beta, satisfying the condition that Beta's final value is

25\%

greater than Alpha's final value (

\$1,740,000 \times 1.25 = \$2,175,000

).

Step-by-Step Solution

1

Set up algebraic expressions for initial values based on the given ratio.

Let initial Alpha =

3k

and initial Beta =

5k

for some multiplier

k

.

The initial ratio of Portfolio Alpha to Portfolio Beta is given as

3:5

.

2

Express values after annual percentage changes.

Alpha pre-transfer =

3k \times (1 + 0.40) = 4.2k

. Beta pre-transfer =

5k \times (1 - 0.10) = 4.5k

.

Alpha gained

40\%

and Beta lost

10\%

over the year.

3

Account for the

\$150,000

transfer and construct the final equality equation.

Alpha final =

4.2k - 150,000

and Beta final =

4.5k + 150,000

. Equation:

4.5k + 150,000 = 1.25(4.2k - 150,000)

.

Beta's final value is

25\%

(

1.25

times) greater than Alpha's final value after receiving

\$150,000

.

4

Solve for the multiplier

k

and compute initial values.

4.5k + 150,000 = 5.25k - 187,500 \implies 0.75k = 337,500 \implies k = 450,000

. Thus, Alpha initial =

3(450,000) = \$1,350,000

and Beta initial =

5(450,000) = \$2,250,000

.

Evaluating the linear equation yields the constant multiplier

k

.

Key Concept

Multi-stage ratio and percentage equation systems with transition transfers.

Question 65Question →

An automated data-processing system updates two state metrics, $X$ and $Y$ , through a sequential three-stage pipeline ( $k = 1, 2, 3$ ). Starting from an initial state $(X_0, Y_0)$ , the system updates the state variables at each stage $k$ according to the following transition rules:

1. $X_k = X_{k-1} + 2Y_{k-1}$
2. If $X_{k-1}$ is even, $Y_k = Y_{k-1} + k$ ; if $X_{k-1}$ is odd, $Y_k = Y_{k-1} - k$ .

Match each initial state configuration $(X_0, Y_0)$ on the left to its corresponding final state $(X_3, Y_3)$ after Stage 3 on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Initial State

(2,3)

matches Final State

(28,9)

; Initial State

(3,5)

matches Final State

(25,-1)

; Initial State

(4,1)

matches Final State

(18,7)

; Initial State

(5,2)

matches Final State

(9,-4)

.

Each initial state must be tracked through all three stages (

k=1, 2, 3

) using the conditional rule based on whether the preceding value of

X

is even or odd.

Step-by-Step Solution

1

Evaluate the state transition for Initial State

(2,3)

across stages

k=1, 2, 3

.

Stage 1:

(8,4)

, Stage 2:

(16,6)

, Stage 3:

(28,9)

.

Since

X

remains even at every step (

2 \rightarrow 8 \rightarrow 16

),

Y_k

increases by

k

at each stage.

2

Evaluate the state transition for Initial State

(3,5)

across stages

k=1, 2, 3

.

Stage 1:

(13,4)

, Stage 2:

(21,2)

, Stage 3:

(25,-1)

.

Since

X

remains odd at every step (

3 \rightarrow 13 \rightarrow 21

),

Y_k

decreases by

k

at each stage.

3

Evaluate the state transition for Initial State

(4,1)

across stages

k=1, 2, 3

.

Stage 1:

(6,2)

, Stage 2:

(10,4)

, Stage 3:

(18,7)

.

Since

X

remains even at every step (

4 \rightarrow 6 \rightarrow 10

),

Y_k

increases by

k

at each stage.

4

Evaluate the state transition for Initial State

(5,2)

across stages

k=1, 2, 3

.

Stage 1:

(9,1)

, Stage 2:

(11,-1)

, Stage 3:

(9,-4)

.

Since

X

remains odd at every step (

5 \rightarrow 9 \rightarrow 11

),

Y_k

decreases by

k

at each stage.

Key Concept

Multi-stage recursive state update logic and conditional branch evaluation.

Question 66Question →

A manufacturing facility uses two processing lines, Line A and Line B, to produce liquid chemical compounds. Line A operates at a constant rate of $r_A$ liters per hour and yields a mixture containing $15\%$ active solute by volume. Line B operates at a constant rate of $r_B$ liters per hour and yields a mixture containing $40\%$ active solute by volume.

When Line A operates for 4 hours and Line B operates for 6 hours, the combined output is $1,600$ liters of a solution containing $30\%$ active solute by volume.

If Line A and Line B are run simultaneously for 5 hours, how many liters of active solute will be produced in total?

Show answer & explanation

Answer: 440

Answer

440 liters of active solute will be produced in total.

By setting up a system of two linear equations relating the individual line flow rates (

r_A

and

r_B

) to total volume and total solute, we find

r_A = 160

L/hr and

r_B = 160

L/hr. Running Line A and Line B together yields

0.15(160) + 0.40(160) = 88

liters of active solute per hour. Over 5 hours, this produces

88 \times 5 = 440

liters of active solute.

Step-by-Step Solution

1

Formulate total volume equation

4r_A + 6r_B = 1600 \implies 2r_A + 3r_B = 800

Line A runs for 4 hours and Line B runs for 6 hours to produce a total volume of 1,600 liters.

2

Formulate active solute equation

0.15(4r_A) + 0.40(6r_B) = 0.30(1600) \implies 0.6r_A + 2.4r_B = 480 \implies r_A + 4r_B = 800

The total solute produced is

30\%

of the 1,600-liter mixture, which equals 480 liters.

3

Solve the system of linear equations

r_A = 160

liters/hour and

r_B = 160

liters/hour

Substituting

r_A = 800 - 4r_B

into

2r_A + 3r_B = 800

yields

5r_B = 800 \implies r_B = 160

and

r_A = 160

.

4

Calculate solute rate per hour when both run simultaneously

0.15(160) + 0.40(160) = 24 + 64 = 88

liters of solute per hour

Each line produces solute at its respective concentration percentage of its hourly fluid rate.

5

Compute total solute produced in 5 hours

88 \times 5 = 440

liters

Multiplying the combined solute production rate of 88 liters per hour by 5 hours gives the final quantity.

Key Concept

Simultaneous equations combining volumetric rates and mixture percentage concentrations.

Estimated Time:2m 30s

Question 67Question →

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 68Question →

A electronics retail warehouse categorizes customer returns into three disposition tiers based on two attributes: Package Seal Condition (Intact vs. Broken) and Resale Value ( $100 or higher vs. Under$ 100).

- Class 1 (Immediate Direct Restock): Requires an Intact seal AND a Resale Value of $100 or higher. - **Class 2 (Inspect and Repackage)**: Assigned to any item with a Broken seal, regardless of Resale Value. - **Class 3 (Discount Outlet Transfer)**: Assigned to any item with an Intact seal BUT a Resale Value under$ 100.

Match each returned product item description to its correct disposition tier.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The returned wireless speaker (

140 value, Intact seal) corresponds to Class 1; the returned tablet (

250 value, Broken seal) corresponds to Class 2; and the returned gaming mouse ($45 value, Intact seal) corresponds to Class 3.

Each returned product is classified by applying the explicit conditional rules governing seal integrity and monetary threshold. The wireless speaker meets both Class 1 requirements. The tablet meets the single condition (Broken seal) for Class 2. The gaming mouse meets the criteria for Class 3 (Intact seal with value under $100).

Step-by-Step Solution

1

Evaluate the returned wireless speaker ($140 value, Intact seal).

Matches Class 1 (Immediate Direct Restock).

It satisfies both requirements: Intact seal and Resale Value

\ge \$100

.

2

Evaluate the returned tablet ($250 value, Broken seal).

Matches Class 2 (Inspect and Repackage).

Any item with a Broken seal is automatically classified as Class 2.

3

Evaluate the returned gaming mouse ($45 value, Intact seal).

Matches Class 3 (Discount Outlet Transfer).

It has an Intact seal but its value is strictly less than $100.

Key Concept

Multi-Attribute Categorical Sorting and Classification

Question 69Question →

A chemical synthesis apparatus monitors two reactant volumes, $A_k$ and $B_k$ (in liters), across three sequential stages ( $k = 1, 2, 3$ ). At the initial state ( $k = 0$ ), the apparatus contains $A_0 = 100$ liters and $B_0 = 40$ liters.

For each subsequent stage

k

, the volumes are updated sequentially according to the following state transition equations:

A_k = A_{k-1} - B_{k-1} + 40

B_k = B_{k-1} + \frac{A_{k-1}}{2} - 20

Based on the process above, which of the following statements regarding the reactant volumes are correct? Select all that apply.

Select all that apply

At the end of Stage 2 (

k = 2

), the volume of reactant A (

A_2

) is 70 liters.

At the end of Stage 3 (

k = 3

), the volume of reactant B (

B_3

) is 115 liters.

At the end of Stage 2 (

k = 2

), the volume of reactant B (

B_2

) is 70 liters.

At the end of Stage 3 (

k = 3

), the volume of reactant A (

A_3

) is 70 liters.

At the end of Stage 1 (

k = 1

), the volume of reactant B (

B_1

) is 100 liters.

Show answer & explanation

Answer: At the end of Stage 2 ( $k = 2$ ), the volume of reactant A ( $A_2$ ) is 70 liters.; At the end of Stage 3 ( $k = 3$ ), the volume of reactant B ( $B_3$ ) is 115 liters.

Answer

The correct statements are that at the end of Stage 2, the volume of reactant A is 70 liters, and at the end of Stage 3, the volume of reactant B is 115 liters.

Evaluating the process iteratively stage by stage shows that at

k=2

,

A_2 = 70

liters, and at

k=3

,

B_3 = 115

liters. Both of these statements match the calculated values.

Step-by-Step Solution

1

Calculate state values at the end of Stage 1 (

k=1

)

A_1 = 100 - 40 + 40 = 100

liters;

B_1 = 40 + \frac{100}{2} - 20 = 70

liters.

Apply state equations using initial inputs

A_0 = 100

and

B_0 = 40

.

2

Calculate state values at the end of Stage 2 (

k=2

)

A_2 = 100 - 70 + 40 = 70

liters;

B_2 = 70 + \frac{100}{2} - 20 = 100

liters.

Apply state equations using inputs

A_1 = 100

and

B_1 = 70

.

3

Calculate state values at the end of Stage 3 (

k=3

)

A_3 = 70 - 100 + 40 = 10

liters;

B_3 = 100 + \frac{70}{2} - 20 = 115

liters.

Apply state equations using inputs

A_2 = 70

and

B_2 = 100

.

4

Evaluate the proposed options against calculated values

The statement specifying

A_2 = 70

liters is correct. The statement specifying

B_3 = 115

liters is correct.

Match calculated state outputs with the offered statements.

Key Concept

Sequential Process Execution and Coupled State Transitions

Question 70Question →

An automated municipal water treatment facility utilizes two primary filtration systems, System Alpha and System Beta, to treat urban wastewater. Let $x$ represent the daily processing rate of System Alpha in megaliters (ML) per day, and let $y$ represent the daily processing rate of System Beta in megaliters (ML) per day, where both $x$ and $y$ are positive values.

Operating System Alpha for 2 days and System Beta for 3 days yields a combined total throughput of 85 megaliters. Additionally, the difference between the square of System Alpha's daily processing rate and the square of System Beta's daily processing rate is equal to 175.

In the table below, select the value for the daily processing rate of System Alpha and the value for the daily processing rate of System Beta that are consistent with the information provided.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The daily processing rate for System Alpha is 20 ML and the daily processing rate for System Beta is 15 ML.

Solving the system of simultaneous equations

2x + 3y = 85

and

x^2 - y^2 = 175

yields

x = 20

ML for System Alpha and

y = 15

ML for System Beta. This satisfies both throughput and squared difference conditions while maintaining positive rates.

Step-by-Step Solution

1

Formulate the system of simultaneous equations from the problem statement.

Equation 1:

2x + 3y = 85

; Equation 2:

x^2 - y^2 = 175

, with constraints

x > 0

and

y > 0

.

Operating Alpha for 2 days and Beta for 3 days totals 85 ML (

2x + 3y = 85

), and the difference of their squared rates is 175 (

x^2 - y^2 = 175

).

2

Express

x

in terms of

y

using Equation 1.

x = \frac{85 - 3y}{2}

Isolating

x

allows substitution into the non-linear equation.

3

Substitute

x = \frac{85 - 3y}{2}

into Equation 2 and simplify the quadratic equation.

\left(\frac{85 - 3y}{2}\right)^2 - y^2 = 175 \implies \frac{7225 - 510y + 9y^2}{4} - y^2 = 175 \implies 5y^2 - 510y + 6525 = 0 \implies y^2 - 102y + 1305 = 0

.

Expanding and clearing denominators yields a standard single-variable quadratic equation in terms of

y

.

4

Solve the quadratic equation

y^2 - 102y + 1305 = 0

for

y

.

(y - 15)(y - 87) = 0 \implies y = 15

or

y = 87

.

Factoring the quadratic gives the two mathematical roots for

y

.

5

Evaluate the corresponding values of

x

for each root of

y

to verify positivity constraints.

If

y = 87

,

x = \frac{85 - 3(87)}{2} = -88

(rejected since

x > 0

). If

y = 15

,

x = \frac{85 - 3(15)}{2} = 20

(valid since

x > 0

).

Physical processing rates must be positive, making

x = 20

and

y = 15

the unique valid solution.

Key Concept

Solving non-linear systems of simultaneous equations using algebraic substitution and quadratic factoring under real-world domain constraints.

Estimated Time:3m 0s

Question 71Question →

A commercial real estate development firm categorizes building proposals using two regulatory compliance rules:

1. Zoning Board Review: A proposal is classified as *Required* if the building height exceeds $50\text{ m}$ OR if the building footprint exceeds $2{}000\text{ m}^2$ . Otherwise, it is classified as *Exempt*.
2. Traffic Impact Assessment: A proposal is classified as *Required* if the building height exceeds $40\text{ m}$ AND the building footprint exceeds $1{}500\text{ m}^2$ . Otherwise, it is classified as *Exempt*.

Building Proposal X specifies a building height of $45\text{ m}$ and a building footprint of $1{}800\text{ m}^2$ .

Which of the following correctly pairs the classification for Zoning Board Review and Traffic Impact Assessment for Building Proposal X?

Zoning Board Review: Exempt; Traffic Impact Assessment: Required

Zoning Board Review: Required; Traffic Impact Assessment: Exempt

Zoning Board Review: Required; Traffic Impact Assessment: Required

Zoning Board Review: Exempt; Traffic Impact Assessment: Exempt

Zoning Board Review: Required; Traffic Impact Assessment: Conditional

Show answer & explanation

Answer: Zoning Board Review: Exempt; Traffic Impact Assessment: Required

Answer

Zoning Board Review: Exempt; Traffic Impact Assessment: Required

Evaluating Proposal X (

45\text{ m}

height,

1{}800\text{ m}^2

footprint) under Rule 1 shows that neither

45 > 50

nor

1{}800 > 2{}000

holds true, making Zoning Board Review Exempt. Under Rule 2, both

45 > 40

and

1{}800 > 1{}500

hold true, making Traffic Impact Assessment Required. The option pairing 'Exempt' for Zoning Board Review and 'Required' for Traffic Impact Assessment is correct.

Step-by-Step Solution

1

Evaluate the condition for Zoning Board Review

Building height is

45\text{ m}

(not

> 50\text{ m}

) and footprint is

1{}800\text{ m}^2

(not

> 2{}000\text{ m}^2

). Since neither condition of the OR clause is satisfied, Zoning Board Review is Exempt.

An OR logical rule requires at least one condition to be true for the designation to apply.

2

Evaluate the condition for Traffic Impact Assessment

Building height is

45\text{ m}

(

> 40\text{ m}

) and footprint is

1{}800\text{ m}^2

(

> 1{}500\text{ m}^2

). Since both conditions of the AND clause are satisfied, Traffic Impact Assessment is Required.

An AND logical rule requires all specified conditions to be met simultaneously.

3

Combine the evaluations into the paired result

Zoning Board Review: Exempt; Traffic Impact Assessment: Required

Matching both calculated categorical outcomes with the choices provided.

Key Concept

Categorical classification using logical conjunctions (AND) and disjunctions (OR)

Question 72Question →

An investment fund allocates capital between two asset classes: Asset Class $P$ and Asset Class $Q$ . The annual percentage yields earned by these asset classes remained constant over a two-year period.

- In Year 1, an investment of $\$3,000,000$ in Asset Class $P$ and $\$2,000,000$ in Asset Class $Q$ produced a total return of $\$410,000$ .
- In Year 2, an investment of $\$2,000,000$ in Asset Class $P$ and $\$5,000,000$ in Asset Class $Q$ produced a total return of $\$640,000$ .

Based on the information provided, select the annual percentage yield earned by Asset Class $P$ and the annual percentage yield earned by Asset Class $Q$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

The annual percentage yield for Asset Class P is 7%, and the annual percentage yield for Asset Class Q is 10%.

Solving the system of simultaneous linear equations

3r_P + 2r_Q = 41

and

2r_P + 5r_Q = 64

yields

r_P = 7\%

for Asset Class P and

r_Q = 10\%

for Asset Class Q.

Step-by-Step Solution

1

Define variables for the unknown annual return rates.

Let

r_P

be the annual percentage yield of Asset Class

P

and

r_Q

be the annual percentage yield of Asset Class

Q

.

Establishing explicit variables allows translation of the financial statements into a system of linear equations.

2

Set up the simultaneous equations for Year 1 and Year 2.

Year 1 equation:

3,000,000 \cdot \frac{r_P}{100} + 2,000,000 \cdot \frac{r_Q}{100} = 410,000 \implies 30,000 r_P + 20,000 r_Q = 410,000 \implies 3r_P + 2r_Q = 41

.
Year 2 equation:

2,000,000 \cdot \frac{r_P}{100} + 5,000,000 \cdot \frac{r_Q}{100} = 640,000 \implies 20,000 r_P + 50,000 r_Q = 640,000 \implies 2r_P + 5r_Q = 64

.

Simplifying by dividing both sides by 10,000 reduces the coefficients to manageable integers.

3

Solve the system of linear equations using the elimination method.

Multiply the first equation by 5:

15r_P + 10r_Q = 205

.
Multiply the second equation by 2:

4r_P + 10r_Q = 128

.
Subtract the second modified equation from the first:

(15r_P - 4r_P) + (10r_Q - 10r_Q) = 205 - 128 \implies 11r_P = 77 \implies r_P = 7

.

Eliminating

r_Q

allows direct solution for the value of

r_P

.

4

Substitute

r_P = 7

back into the first simplified equation to solve for

r_Q

.

3(7) + 2r_Q = 41 \implies 21 + 2r_Q = 41 \implies 2r_Q = 20 \implies r_Q = 10

.

Substituting the known variable yields the remaining unknown.

Key Concept

Simultaneous Linear Equations in Two Variables

Estimated Time:2m 0s

Question 73Question →

A solar micro-grid facility utilizes two distinct types of energy storage battery modules: Type-X and Type-Y. On Monday, a configuration of 4 Type-X modules and 3 Type-Y modules stored a total of $180\text{ kWh}$ of energy. On Tuesday, a configuration of 2 Type-X modules and 5 Type-Y modules stored a total of $160\text{ kWh}$ of energy. Assuming each battery module of a given type stores a constant amount of energy, what is the individual storage capacity of one Type-X module and one Type-Y module, respectively?

Type-X =

30\text{ kWh}

, Type-Y =

20\text{ kWh}

Type-X =

20\text{ kWh}

, Type-Y =

30\text{ kWh}

Type-X =

27\text{ kWh}

, Type-Y =

24\text{ kWh}

Type-X =

25\text{ kWh}

, Type-Y =

25\text{ kWh}

Type-X =

35\text{ kWh}

, Type-Y =

15\text{ kWh}

Show answer & explanation

Answer: Type-X = $30\text{ kWh}$ , Type-Y = $20\text{ kWh}$

Answer

Type-X =

30\text{ kWh}

and Type-Y =

20\text{ kWh}

The correct option correctly states that Type-X has a capacity of

30\text{ kWh}

and Type-Y has a capacity of

20\text{ kWh}

. Substituting these values into both equations yields

4(30) + 3(20) = 180

and

2(30) + 5(20) = 160

, confirming both daily constraints are satisfied.

Step-by-Step Solution

1

Formulate a system of two linear equations in two variables.

Let

x

represent the capacity of one Type-X module and

y

represent the capacity of one Type-Y module. The given conditions translate to:
Equation (1):

4x + 3y = 180

Equation (2):

2x + 5y = 160

Simultaneous word problems require establishing algebraic equations representing total quantities for each scenario.

2

Eliminate variable

x

by multiplying Equation (2) by 2.

Equation (2) multiplied by 2 becomes:
Equation (3):

4x + 10y = 320

Aligning coefficients of

x

enables linear elimination.

3

Subtract Equation (1) from Equation (3) to solve for

y

.

(4x + 10y) - (4x + 3y) = 320 - 180 \implies 7y = 140 \implies y = 20

Subtracting the equations eliminates

x

, producing a single variable equation in

y

.

4

Substitute

y = 20

back into Equation (1) to find

x

.

4x + 3(20) = 180 \implies 4x + 60 = 180 \implies 4x = 120 \implies x = 30

Determining the remaining unknown variable.

Key Concept

Solving Systems of Linear Equations in Two Variables

Question 74Question →

A municipal utility operates three water desalination plants—Plant Alpha, Plant Beta, and Plant Gamma—to process seawater into purified drinking water. Each plant operates at a constant total intake rate of seawater and yields a specific percentage of brine byproduct, with the remainder converted to purified drinking water:

- Plant Alpha: Total intake rate of $40,000$ gallons per hour; $15\%$ of intake becomes brine byproduct.
- Plant Beta: Total intake rate of $60,000$ gallons per hour; $10\%$ of intake becomes brine byproduct.
- Plant Gamma: Total intake rate of $50,000$ gallons per hour; $20\%$ of intake becomes brine byproduct.

Match each operational metric on the left with its corresponding calculated value on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The operational metrics match as follows: the combined output rate of Plant Alpha and Plant Beta is

88

thousand gallons per hour; the ratio of brine byproduct volume produced by Plant Alpha (in 5 hours) to Plant Gamma (in 4 hours) is

3 : 4

; and the time required for Plant Beta to equal Plant Gamma's 9-hour purified output is

\frac{20}{3}

hours.

Each calculation requires decomposing total plant intake into its constituent components (purified water and brine byproduct) using given percentages, followed by executing standard work-rate and ratio operations (

R \times T = W

). The resulting values match

88

thousand gallons per hour,

3 : 4

, and

\frac{20}{3}

hours respectively.

Step-by-Step Solution

1

Calculate individual rates for purified drinking water and brine byproduct for each plant.

Plant Alpha: Purified =

0.85 \times 40 = 34

k gal/hr, Brine =

0.15 \times 40 = 6

k gal/hr. Plant Beta: Purified =

0.90 \times 60 = 54

k gal/hr, Brine =

0.10 \times 60 = 6

k gal/hr. Plant Gamma: Purified =

0.80 \times 50 = 40

k gal/hr, Brine =

0.20 \times 50 = 10

k gal/hr.

Deconstructing the total intake rates into purified water rates and byproduct rates allows direct substitution into each multi-part question.

2

Sum the purified drinking water rates of Plant Alpha and Plant Beta.

Combined purified rate =

34 + 54 = 88

thousand gallons per hour.

When two plants operate simultaneously, their effective production rates add together.

3

Compute total brine output for Plant Alpha in 5 hours and Plant Gamma in 4 hours, then simplify the resulting ratio.

Alpha brine volume =

6 \times 5 = 30

thousand gallons. Gamma brine volume =

10 \times 4 = 40

thousand gallons. Ratio =

30 : 40 = 3 : 4

.

Total volume equals hourly rate multiplied by operational duration.

4

Find total purified output of Plant Gamma in 9 hours, and divide by Plant Beta's hourly purified rate.

Gamma purified total =

40 \times 9 = 360

thousand gallons. Required time for Beta =

\frac{360}{54} = \frac{20}{3}

hours.

Time equals total required volume divided by the rate of production.

Key Concept

Work rates and percent composition decomposition in multi-unit processing systems.

Estimated Time:2m 30s

Question 75Question →

A freight logistics firm installed automated route-optimization software across its fleet in Region X while simultaneously introducing a new financial bonus for drivers who voluntarily reduce engine idling time. Over the following six months, fuel consumption per delivery dropped by 18%, and fleet idle time decreased significantly. Executive management concluded that the installation of the route-optimization software was the primary cause of the reduction in fuel consumption. Which of the following statements, if true, provides the strongest counterargument to management's conclusion by identifying an alternative causal explanation?

The financial bonus incentivized drivers to turn off engines during waiting periods, accounting for the reduced idle time and lower fuel consumption independently of the software.

The reduction in fuel consumption led management to install the route-optimization software across all remaining fleet vehicles.

Drivers who received the route-optimization software reported higher overall job satisfaction after the fuel consumption dropped.

The route-optimization software was programmed specifically to minimize total driving mileage rather than monitoring engine idling hours.

Fuel prices in Region X fluctuated throughout the six-month period, altering the total monetary expenditure on fuel for the logistics firm.

Show answer & explanation

Answer: The financial bonus incentivized drivers to turn off engines during waiting periods, accounting for the reduced idle time and lower fuel consumption independently of the software.

Answer

The statement that driver financial bonuses incentivized engine shutdown provides the strongest counterargument by introducing an alternative cause for the reduced fuel consumption.

The argument concludes that the software caused the drop in fuel consumption. However, the stem notes that financial bonuses for reducing idling time were introduced simultaneously. Pointing out that these bonuses directly incentivized lower idling and lower fuel use presents a compelling alternative explanation, showing that the outcome could have occurred regardless of the software.

Step-by-Step Solution

1

Identify the causal claim in the argument stem.

Management claims that introducing route-optimization software (Cause) directly led to an 18% reduction in fuel consumption (Effect).

Understanding the precise cause-and-effect structure is essential before evaluating potential flaws or weak points in the reasoning.

2

Analyze concurrent events and potential confounding variables.

The company introduced two simultaneous changes: route-optimization software AND financial bonuses for reducing engine idling time.

When two potential causes are introduced at the same time, attributing the outcome solely to one cause without isolating variables commits a causal flaw.

3

Select the option that offers a valid alternative explanation for the observed outcome.

Attributing the reduced fuel consumption to the idling bonuses presents a direct, plausible alternative cause, neutralizing the conclusion that the software was the primary driver.

A causal argument is weakened when the observed outcome can be fully explained by an alternate independent factor.

Key Concept

Evaluating Alternative Causal Explanations and Confounding Variables

Question 76Question →

A semiconductor manufacturing facility operates two models of photolithography equipment: Model D and Model E. Over a 10-hour operational cycle, each Model D scanner consumes $400\text{ kWh}$ of energy and processes $600$ silicon wafers, while each Model E scanner consumes $1,200\text{ kWh}$ of energy and processes $1,000$ silicon wafers. If the combined fleet consumed a total of $9,600\text{ kWh}$ of energy and processed $10,400$ wafers over the cycle, which of the following correctly identifies the number of Model D scanners and Model E scanners operated by the facility?

9 Model D scanners and 5 Model E scanners

5 Model D scanners and 9 Model E scanners

12 Model D scanners and 4 Model E scanners

6 Model D scanners and 6 Model E scanners

15 Model D scanners and 3 Model E scanners

Show answer & explanation

Answer: 9 Model D scanners and 5 Model E scanners

Answer

9 Model D scanners and 5 Model E scanners

The correct answer states 9 Model D scanners and 5 Model E scanners. Setting up the system

d + 3e = 24

(energy) and

3d + 5e = 52

(production) and solving by substitution gives

e = 5

and

d = 9

, which uniquely satisfies both constraints.

Step-by-Step Solution

1

Define variables and establish equations based on given rates and totals.

Let

d

be the number of Model D scanners and

e

be the number of Model E scanners.
Energy equation:

400d + 1200e = 9600

Wafer equation:

600d + 1000e = 10400

Translate the physical constraints into a system of two linear equations.

2

Simplify both linear equations by dividing by their greatest common divisors.

Dividing the energy equation by

400

gives:

d + 3e = 24 \quad \text{--- (Equation 1)}

Dividing the wafer equation by

200

gives:

3d + 5e = 52 \quad \text{--- (Equation 2)}

Simplifying constants reduces calculation complexity and prevents arithmetic errors.

3

Solve the system using substitution or elimination.

From Equation 1, express

d

in terms of

e

:

d = 24 - 3e

Substitute this expression into Equation 2:

3(24 - 3e) + 5e = 52

72 - 9e + 5e = 52

72 - 4e = 52

4e = 20 \implies e = 5

Now calculate

d

:

d = 24 - 3(5) = 24 - 15 = 9

Isolating one variable yields exact values for both equipment models.

4

Verify values against both constraints.

Energy:

400(9) + 1200(5) = 3600 + 6000 = 9600\text{ kWh}

Wafers:

600(9) + 1000(5) = 5400 + 5000 = 10400\text{ wafers}

Ensures the solution satisfies both independent constraints simultaneously.

Key Concept

Simultaneous Linear Equations in Two Variables

Question 77Question →

A regional logistics node routes cargo containers through a three-stage automated sorting process ( $k = 1, 2, 3$ ). At the start of Stage 1 (state $k = 0$ ), the node holds $V_0 = 120$ heavy units and $W_0 = 80$ light units. For each stage $k \in \{1, 2, 3\}$ , the counts of heavy units ( $V_k$ ) and light units ( $W_k$ ) transition according to the following rules:

- $V_k = \frac{1}{2} V_{k-1} + 20$
- $W_k = W_{k-1} + \frac{1}{4} V_{k-1} - 10$

Which of the following statements regarding the state of the cargo units across these three stages are correct?

Select all that apply

At the end of Stage 2 (

k = 2

), the number of light units is

110

.

The total number of units (heavy plus light) at the end of Stage 3 (

k = 3

) is

165

.

At the end of Stage 1 (

k = 1

), the number of heavy units exceeds the number of light units by

20

.

The number of heavy units at the end of Stage 3 (

k = 3

) is

60

.

At the end of Stage 2 (

k = 2

), the total number of units is

180

.

Show answer & explanation

Answer: At the end of Stage 2 ( $k = 2$ ), the number of light units is $110$ .; The total number of units (heavy plus light) at the end of Stage 3 ( $k = 3$ ) is $165$ .

Answer

The correct statements are that at the end of Stage 2 (

k = 2

), the number of light units is

110

, and the total number of units (heavy plus light) at the end of Stage 3 (

k = 3

) is

165

.

Evaluating the multi-stage updates step by step shows that at Stage 2 (

k=2

),

W_2 = 110

, making the statement about 110 light units at Stage 2 correct. At Stage 3 (

k=3

),

V_3 = 50

and

W_3 = 115

, giving a total of

50 + 115 = 165

units, making the total unit statement at Stage 3 correct.

Step-by-Step Solution

1

Calculate state values at the end of Stage 1 (

k = 1

)

V_1 = \frac{1}{2}(120) + 20 = 80

;

W_1 = 80 + \frac{1}{4}(120) - 10 = 100

. Total units

T_1 = 80 + 100 = 180

.

Apply the given transition rules to initial values

V_0 = 120

and

W_0 = 80

.

2

Calculate state values at the end of Stage 2 (

k = 2

)

V_2 = \frac{1}{2}(80) + 20 = 60

;

W_2 = 100 + \frac{1}{4}(80) - 10 = 110

. Total units

T_2 = 60 + 110 = 170

.

Apply transition rules to

V_1 = 80

and

W_1 = 100

.

3

Calculate state values at the end of Stage 3 (

k = 3

)

V_3 = \frac{1}{2}(60) + 20 = 50

;

W_3 = 110 + \frac{1}{4}(60) - 10 = 115

. Total units

T_3 = 50 + 115 = 165

.

Apply transition rules to

V_2 = 60

and

W_2 = 110

.

4

Evaluate each candidate statement against the computed values

Statement regarding light units being 110 at Stage 2 is TRUE (

W_2 = 110

). Statement regarding total units being 165 at Stage 3 is TRUE (

T_3 = 165

). All other statements are false.

Compare calculated stage states directly with the statement claims.

Key Concept

Sequential state transitions involve iteratively updating state variables through defined algebraic relationships across multiple stages.

Question 78Question →

A large public university introduced interactive virtual laboratory simulations across its online introductory physics courses to address high course drop rates. In the semester following the introduction of these simulations, overall course completion rates increased by $22\%$ , and average student ratings for course engagement doubled. The university administration concluded that the interactive virtual laboratory simulations directly caused the increase in student retention by helping students visualize complex physical concepts.

Select the statement that most strongly supports the university administration's causal conclusion, and select the statement that most strongly undermines it. Make exactly one selection in each column.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The statement that most strongly supports the conclusion is the concurrent trial showing higher retention in sections with virtual simulations compared to identical sections without them taught by the same instructors. The statement that most strongly undermines the conclusion is that the university lowered the minimum passing score from

70\%

to

55\%

at the start of the semester.

To support the causal claim, we need evidence isolating the virtual labs as the factor driving retention. The controlled comparison between sections with and without simulations taught by the same instructors directly reinforces the causal connection. To undermine the causal claim, we need an alternative factor that explains the higher completion rates; lowering the passing grade threshold from

70\%

to

55\%

provides a direct alternative explanation for why more students completed the course.

Step-by-Step Solution

1

Identify the author's core causal claim

Claim: Interactive virtual laboratory simulations directly caused the

22\%

increase in student course completion rates.

Evaluating support or weakness requires isolating the explicit cause (virtual labs) and effect (completion rate increase).

2

Evaluate candidate statements to find the strong supporter

The statement comparing identical sections with and without simulations taught by the same instructors isolates the virtual lab variable, eliminating instructor bias and external temporal factors.

Controlled comparison directly strengthens a causal link by demonstrating that the effect occurs when the cause is present and does not occur to the same degree when the cause is absent.

3

Evaluate candidate statements to find the strong underminer

The statement regarding lowering the passing grade from

70\%

to

55\%

introduces an obvious alternative explanation for why completion rates rose.

Introducing a viable alternative cause for the observed outcome weakens the argument that the virtual labs were responsible for the increase.

Key Concept

Causal Arguments and Alternative Explanations in Two-Part Analysis

Question 79Question →

A institutional review board classifies biomedical research protocols into three oversight streams based on two parameters: Participant Risk Level (High, Medium, or Low) and Data Integrity Score ( $S$ , where $0 \le S \le 100$ ). The classification is governed by the following strict hierarchy of rules:

1. Full Board Audit: Assigned if the protocol has a High Participant Risk Level OR a Data Integrity Score of $S < 60$ .
2. Expedited Committee Review: Assigned if the protocol does not require a Full Board Audit AND meets at least one of the following conditions: Medium Participant Risk Level OR a Data Integrity Score in the range $60 \le S \le 80$ .
3. Administrative Exemption: Assigned to any protocol that does not meet the criteria for either Full Board Audit or Expedited Committee Review (i.e., Low Participant Risk Level AND $S > 80$ ).

Four trial protocols are currently under review:
- Protocol 101: Medium Participant Risk; $S = 55$
- Protocol 102: High Participant Risk; $S = 85$
- Protocol 103: Low Participant Risk; $S = 75$
- Protocol 104: Low Participant Risk; $S = 90$

Match each research protocol to its specific oversight stream classification.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Protocol 101 matches Full Board Audit triggered by

S < 60

; Protocol 102 matches Full Board Audit triggered by High Risk; Protocol 103 matches Expedited Committee Review; Protocol 104 matches Administrative Exemption.

Each protocol must be evaluated strictly sequentially according to the rule hierarchy: Full Board Audit criteria first, then Expedited Committee Review criteria, and finally Administrative Exemption criteria.

Step-by-Step Solution

1

Evaluate Rule 1 (Full Board Audit criteria) for all protocols

Protocol 101 meets

S < 60

(

S = 55

). Protocol 102 meets High Participant Risk. Both are assigned to Full Board Audit.

Rule 1 is evaluated first in the hierarchy. Any protocol meeting either High Risk or

S < 60

is classified under Full Board Audit.

2

Evaluate Rule 2 (Expedited Committee Review criteria) for remaining protocols

Protocol 103 has Low Risk and

S = 75

, which satisfies

60 \le S \le 80

. It is assigned to Expedited Committee Review.

Protocol 103 did not meet Rule 1 criteria, so Rule 2 is evaluated next.

3

Evaluate Rule 3 (Administrative Exemption criteria) for remaining protocols

Protocol 104 has Low Risk and

S = 90 > 80

, satisfying Administrative Exemption.

Protocol 104 does not meet Rule 1 or Rule 2 criteria, defaulting to Rule 3.

Key Concept

Multi-attribute categorical classification governed by strict hierarchical decision rules.

Question 80Question →

A hybrid microgrid power station generates electricity using Solar Array S and Wind Turbine W. In Month 1, the ratio of the energy output produced by Solar Array S to that produced by Wind Turbine W was $4 : 3$ .

In Month 2, the energy output of Solar Array S increased by $30\%$ relative to Month 1, while the energy output of Wind Turbine W increased by $10\%$ relative to Month 1. The total energy output produced by both sources combined in Month 2 was $1,700\text{ MWh}$ .

Based on the information provided, which of the following options correctly pairs the energy output of Solar Array S in Month 1 (Column 1) with the energy output of Wind Turbine W in Month 2 (Column 2)?

Column 1: 800 MWh; Column 2: 660 MWh

Column 1: 660 MWh; Column 2: 800 MWh

Column 1: 800 MWh; Column 2: 600 MWh

Column 1: 1,040 MWh; Column 2: 660 MWh

Column 1: 600 MWh; Column 2: 1,040 MWh

Show answer & explanation

Answer: Column 1: 800 MWh; Column 2: 660 MWh

Answer

Column 1: 800 MWh; Column 2: 660 MWh

The option specifying Column 1 as 800 MWh and Column 2 as 660 MWh correctly identifies the Month 1 Solar Array S output (

4 \times 200 = 800\text{ MWh}

) and the Month 2 Wind Turbine W output (

3.3 \times 200 = 660\text{ MWh}

).

Step-by-Step Solution

1

Define variables based on Month 1 ratio

Let Month 1 output for Solar Array S be

S_1 = 4x

and for Wind Turbine W be

W_1 = 3x

.

The given ratio of Solar output to Wind output in Month 1 is

4 : 3

.

2

Express Month 2 outputs in terms of

x

S_2 = 4x \times (1 + 0.30) = 5.2x

and

W_2 = 3x \times (1 + 0.10) = 3.3x

.

Solar output increased by

30\%

and Wind output increased by

10\%

in Month 2.

3

Set up and solve the total output equation for Month 2

5.2x + 3.3x = 8.5x = 1,700 \implies x = 200\text{ MWh}

.

The combined Month 2 output is given as

1,700\text{ MWh}

.

4

Calculate the target quantities for Column 1 and Column 2

Column 1 (

S_1

) =

4 \times 200 = 800\text{ MWh}

; Column 2 (

W_2

) =

3.3 \times 200 = 660\text{ MWh}

.

Column 1 asks for Solar output in Month 1, and Column 2 asks for Wind output in Month 2.

Key Concept

Two-Part Rate and Ratio Analysis with Multi-Stage Percentage Changes

Estimated Time:2m 30s

Two-Part Analysis

Answer

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