Sequential Processes and State Transitions Practice Questions — GMAT

Question 1Question →

An automated data processing pipeline evaluates five sequential stages ( $k = 1, 2, 3, 4, 5$ ) to process machine learning workloads. At the start ( $k = 0$ ), the system state is defined by Storage $S_0 = 10\text{ GB}$ and System Load $L_0 = 20\%$ . At each subsequent stage $k$ , the state vector $(S_k, L_k)$ updates according to the following rules:

- Stage 1 (Data Scrubbing): $S_1 = S_0 + 15$ , $L_1 = L_0 + 10$
- Stage 2 (Feature Extraction): $S_2 = 2 \times S_1$ , $L_2 = L_1 + 15$
- Stage 3 (Model Training): $S_3 = S_2 - 10$ , $L_3 = L_2 + 25$
- Stage 4 (Batch Validation): $S_4 = S_3 + 20$ , $L_4 = L_3 - 30$
- Stage 5 (System Deployment): $S_5 = S_4 - 25$ , $L_5 = L_4 + 50$

The pipeline performance metric at any stage is defined as $M_k = S_k + L_k$ .

Arrange the five stages in increasing order of their pipeline performance metric $M_k$ (from smallest value to largest value).

Drag items to arrange them in the correct order

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Answer

The correct increasing order of the stages by metric

M_k

is Data Scrubbing (

M_1 = 55

), Feature Extraction (

M_2 = 95

), Batch Validation (

M_4 = 100

), Model Training (

M_3 = 110

), and System Deployment (

M_5 = 125

).

Tracking each state sequentially yields

M_1 = 55

,

M_2 = 95

,

M_3 = 110

,

M_4 = 100

, and

M_5 = 125

. Arranging these from smallest to largest places Batch Validation (

M_4 = 100

) before Model Training (

M_3 = 110

).

Step-by-Step Solution

1

Calculate state vector and metric for Stage 1

S_1 = 10 + 15 = 25

,

L_1 = 20 + 10 = 30 \implies M_1 = 25 + 30 = 55

Apply state transformation rules for Stage 1 starting from initial condition

(S_0=10, L_0=20)

.

2

Calculate state vector and metric for Stage 2

S_2 = 2 \times 25 = 50

,

L_2 = 30 + 15 = 45 \implies M_2 = 50 + 45 = 95

Apply state transformation rules using values from Stage 1.

3

Calculate state vector and metric for Stage 3

S_3 = 50 - 10 = 40

,

L_3 = 45 + 25 = 70 \implies M_3 = 40 + 70 = 110

Apply state transformation rules using values from Stage 2.

4

Calculate state vector and metric for Stage 4

S_4 = 40 + 20 = 60

,

L_4 = 70 - 30 = 40 \implies M_4 = 60 + 40 = 100

Apply state transformation rules using values from Stage 3.

5

Calculate state vector and metric for Stage 5

S_5 = 60 - 25 = 35

,

L_5 = 40 + 50 = 90 \implies M_5 = 35 + 90 = 125

Apply state transformation rules using values from Stage 4.

6

Sort the computed metric values in ascending order

M_1 (55) < M_2 (95) < M_4 (100) < M_3 (110) < M_5 (125)

Ordering the values yields Data Scrubbing, Feature Extraction, Batch Validation, Model Training, and System Deployment.

Key Concept

Sequential state tracking requires evaluating recursive function rules step-by-step before applying ordering logic.

Question 2Question →

An automated logistics drone starts an operation at Cycle $k = 0$ with an initial battery charge of $C_0 = 100\%$ and a payload mass of $P_0 = 40\text{ kg}$ . At the end of each subsequent cycle $k$ (where $k = 1, 2, 3, 4$ ), the payload mass decreases by $8\text{ kg}$ according to $P_k = P_{k-1} - 8$ , and the remaining battery charge updates according to the state transition rule $C_k = C_{k-1} - (5 + 0.2 P_{k-1})\%$ . Match each operational cycle on the left with its exact state $(C_k, P_k)$ at the end of that cycle on the right.

Click a left item, then click its matching right item

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Answer

End of Cycle 1 matches with C_1 = 87.0%, P_1 = 32 kg; End of Cycle 2 matches with C_2 = 75.6%, P_2 = 24 kg; End of Cycle 3 matches with C_3 = 65.8%, P_3 = 16 kg; End of Cycle 4 matches with C_4 = 57.6%, P_4 = 8 kg.

Each cycle follows the recursive state equations P_k = P_{k-1} - 8 and C_k = C_{k-1} - (5 + 0.2 P_{k-1}). Evaluating step by step yields the exact matching pairs for all four cycles.

Step-by-Step Solution

1

Calculate state values at the end of Cycle 1 (k = 1)

P_1 = 40 - 8 = 32 kg; C_1 = 100 - (5 + 0.2 * 40) = 100 - 13 = 87.0%

Use the state transition formula using initial values P_0 = 40 and C_0 = 100.

2

Calculate state values at the end of Cycle 2 (k = 2)

P_2 = 32 - 8 = 24 kg; C_2 = 87.0 - (5 + 0.2 * 32) = 87.0 - 11.4 = 75.6%

Apply the recursive transition rules to the state obtained at k = 1.

3

Calculate state values at the end of Cycle 3 (k = 3)

P_3 = 24 - 8 = 16 kg; C_3 = 75.6 - (5 + 0.2 * 24) = 75.6 - 9.8 = 65.8%

Apply the recursive transition rules to the state obtained at k = 2.

4

Calculate state values at the end of Cycle 4 (k = 4)

P_4 = 16 - 8 = 8 kg; C_4 = 65.8 - (5 + 0.2 * 16) = 65.8 - 8.2 = 57.6%

Apply the recursive transition rules to the state obtained at k = 3.

Key Concept

Sequential Process and Recursive State Transition Tracking

Estimated Time:2m 0s

Question 3Question →

An automated algorithmic system updates a portfolio's Index ( $V$ ) and Stabilization Pool ( $P$ ) across sequential stages $k = 1, 2, 3, 4$ .

Initial State (Stage 0): $V_0 = 100$ and $P_0 = 20$ .

For each Stage $k$ (where $k = 1, 2, 3, 4$ ):
1. Sub-stage A (Temporary Index Calculation): Calculate $V_k^{\text{temp}} = V_{k-1} + 0.5 \times P_{k-1}$ .
2. Sub-stage B (State Adjustment):
- If $V_k^{\text{temp}} \le 110$ : set $P_k = P_{k-1} + 15$ and final stage index $V_k = V_k^{\text{temp}} + 10$ .
- If $V_k^{\text{temp}} > 110$ : set $P_k = P_{k-1} - 10$ and final stage index $V_k = V_k^{\text{temp}} - 5$ .

Arrange the following four state transition events in chronological order from earliest to latest occurrence.

Drag items to arrange them in the correct order

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Answer

The correct chronological order of events is: (1) The stabilization pool

P

reaches its peak value of

35

, (2) The temporary index

V^{\text{temp}}

first exceeds

130

, (3) The stabilization pool

P

drops to

25

, and (4) The final stage index

V

reaches

140

.

Tracing the state transitions step-by-step reveals:
- At Sub-stage 1B,

P_1 = 35

, which is the peak value of

P

.
- At Sub-stage 2A,

V_2^{\text{temp}} = 137.5

, marking the first time

V^{\text{temp}}

exceeds

130

.
- At Sub-stage 2B,

P_2 = 25

.
- At Sub-stage 3B,

V_3 = 140.0

.

Thus, the event where the pool reaches

35

happens first (Sub-stage 1B), followed by

V^{\text{temp}}

exceeding

130

(Sub-stage 2A), then the pool dropping to

25

(Sub-stage 2B), and finally

V

reaching

140

(Sub-stage 3B).

Step-by-Step Solution

1

Evaluate Stage 1 state transitions

Sub-stage 1A:

V_1^{\text{temp}} = 100 + 0.5(20) = 110

. Since

V_1^{\text{temp}} \le 110

, Sub-stage 1B sets

P_1 = 20 + 15 = 35

and

V_1 = 110 + 10 = 120

.

This establishes that the event 'The stabilization pool

P

reaches its peak value of

35

' occurs at Sub-stage 1B.

2

Evaluate Stage 2 state transitions

Sub-stage 2A:

V_2^{\text{temp}} = 120 + 0.5(35) = 137.5

. Here,

V^{\text{temp}}

first exceeds

130

. Sub-stage 2B: Since

V_2^{\text{temp}} = 137.5 > 110

, set

P_2 = 35 - 10 = 25

and

V_2 = 137.5 - 5 = 132.5

.

This establishes that 'The temporary index

V^{\text{temp}}

first exceeds

130

' occurs at Sub-stage 2A, followed immediately by 'The stabilization pool

P

drops to

25

' at Sub-stage 2B.

3

Evaluate Stage 3 state transitions

Sub-stage 3A:

V_3^{\text{temp}} = 132.5 + 0.5(25) = 145.0

. Sub-stage 3B: Since

V_3^{\text{temp}} = 145.0 > 110

, set

P_3 = 25 - 10 = 15

and

V_3 = 145.0 - 5 = 140.0

.

This confirms that 'The final stage index

V

reaches

140

' occurs at Sub-stage 3B.

4

Synthesize chronological sequence of all four events

Sequence: Sub-stage 1B (

P = 35

)

\rightarrow

Sub-stage 2A (

V^{\text{temp}} = 137.5

)

\rightarrow

Sub-stage 2B (

P = 25

)

\rightarrow

Sub-stage 3B (

V = 140

).

Matching the chronological execution order yields the correct item ordering.

Key Concept

Sequential Processes and State Transitions

Question 4Question →

A distribution network manages inventory between two regional facilities, Facility A and Facility B. At the beginning of Stage 1 ( $k=1$ ), Facility A holds $216$ units of product and Facility B holds $160$ units of product.

In each stage $k$ (for $k = 1, 2, 3$ ), the inventory levels undergo four sequential state transitions in the exact order listed below:

1. Step 1: Facility A dispatches $\frac{1}{2}$ of its current inventory.
2. Step 2: Facility B dispatches $\frac{1}{2}$ of its current inventory.
3. Step 3: Facility A transfers $\frac{1}{3}$ of its remaining inventory to Facility B.
4. Step 4: Facility B receives $28$ new units from an external manufacturer.

Which of the following statements regarding the inventory levels at the end of Stage 2 and Stage 3 are correct? Select all that apply.

Select all that apply

At the end of Stage 2, Facility B holds exactly 112 units.

At the end of Stage 3, the total combined inventory across both facilities is 96 units.

At the end of Stage 2, Facility A holds 36 units.

At the end of Stage 3, Facility B holds 60 units.

The total combined inventory decreases by 40 units from the end of Stage 2 to the end of Stage 3.

Show answer & explanation

Answer: At the end of Stage 2, Facility B holds exactly 112 units.; At the end of Stage 3, the total combined inventory across both facilities is 96 units.; The total combined inventory decreases by 40 units from the end of Stage 2 to the end of Stage 3.

Answer

The correct statements are: (1) At the end of Stage 2, Facility B holds exactly 112 units; (2) At the end of Stage 3, the total combined inventory across both facilities is 96 units; and (3) The total combined inventory decreases by 40 units from the end of Stage 2 to the end of Stage 3.

Tracking all 4 transition steps systematically across 3 stages yields the exact state vectors: Stage 1

(72, 144)

, Stage 2

(24, 112)

, and Stage 3

(8, 88)

. Comparing these state vectors confirms that Facility B has 112 units at Stage 2 end, total inventory at Stage 3 end is

8 + 88 = 96

units, and the change from Stage 2 end to Stage 3 end is

136 - 96 = 40

units.

Step-by-Step Solution

1

Calculate state transitions for Stage 1 (

k=1

)

A_1 = 72

,

B_1 = 144

Start at

A_0 = 216, B_0 = 160

. Step 1:

A \to 216 - 108 = 108

. Step 2:

B \to 160 - 80 = 80

. Step 3:

A

transfers

\frac{1}{3}(108) = 36

to

B

, so

A \to 108 - 36 = 72

and

B \to 80 + 36 = 116

. Step 4:

B

receives 28, so

B \to 116 + 28 = 144

.

2

Calculate state transitions for Stage 2 (

k=2

)

A_2 = 24

,

B_2 = 112

,

\text{Total}_2 = 136

Start at

A_1 = 72, B_1 = 144

. Step 1:

A \to 72 - 36 = 36

. Step 2:

B \to 144 - 72 = 72

. Step 3:

A

transfers

\frac{1}{3}(36) = 12

to

B

, so

A \to 36 - 12 = 24

and

B \to 72 + 12 = 84

. Step 4:

B

receives 28, so

B \to 84 + 28 = 112

.

3

Calculate state transitions for Stage 3 (

k=3

)

A_3 = 8

,

B_3 = 88

,

\text{Total}_3 = 96

Start at

A_2 = 24, B_2 = 112

. Step 1:

A \to 24 - 12 = 12

. Step 2:

B \to 112 - 56 = 56

. Step 3:

A

transfers

\frac{1}{3}(12) = 4

to

B

, so

A \to 12 - 4 = 8

and

B \to 56 + 4 = 60

. Step 4:

B

receives 28, so

B \to 60 + 28 = 88

.

4

Evaluate the given statements against calculated stage states

Statements A, B, and E are true; C and D are false.

Facility B holds 112 units at Stage 2 end. Total inventory at Stage 3 end is

8 + 88 = 96

units. Total inventory change between Stage 2 and Stage 3 is

136 - 96 = 40

units.

Key Concept

Sequential Multi-Stage State Transitions

Estimated Time:2m 30s

Question 5Question →

A cloud server processes data batches across three sequential stages ( $k = 1, 2, 3$ ). At the start ( $k = 0$ ), the server has a Buffer size of $B_0 = 100\text{ MB}$ and a Queue size of $Q_0 = 50\text{ MB}$ . In each stage $k$ , the sizes are updated according to the following rules:
- $B_k = B_{k-1} - 20\text{ MB}$
- $Q_k = Q_{k-1} + 10\text{ MB}$

Which of the following statements about the state of the server are true?

Select all that apply

At the end of Stage 2 (

k = 2

), the Buffer size

B_2

is

60\text{ MB}

.

At the end of Stage 3 (

k = 3

), the Queue size

Q_3

is

80\text{ MB}

.

At the end of Stage 3 (

k = 3

), the Buffer size

B_3

is

60\text{ MB}

.

At the end of Stage 2 (

k = 2

), the Queue size

Q_2

is

60\text{ MB}

.

At the end of Stage 1 (

k = 1

), the combined total volume

B_1 + Q_1

is

150\text{ MB}

.

Show answer & explanation

Answer: At the end of Stage 2 ( $k = 2$ ), the Buffer size $B_2$ is $60\text{ MB}$ .; At the end of Stage 3 ( $k = 3$ ), the Queue size $Q_3$ is $80\text{ MB}$ .

Answer

The correct statements are that at the end of Stage 2, the Buffer size is

60\text{ MB}

, and at the end of Stage 3, the Queue size is

80\text{ MB}

.

The correct statements accurately reflect the iterative evaluation of the state transition formulas: the buffer size decreases by

20\text{ MB}

each stage to reach

60\text{ MB}

at Stage 2, and the queue size increases by

10\text{ MB}

each stage to reach

80\text{ MB}

at Stage 3.

Step-by-Step Solution

1

Calculate state values at Stage 1 (

k = 1

)

B_1 = 100 - 20 = 80\text{ MB}

and

Q_1 = 50 + 10 = 60\text{ MB}

. Combined total:

140\text{ MB}

.

Apply the given state transition rules to the initial values at

k = 0

.

2

Calculate state values at Stage 2 (

k = 2

)

B_2 = 80 - 20 = 60\text{ MB}

and

Q_2 = 60 + 10 = 70\text{ MB}

.

Apply the transition rules to the Stage 1 state values.

3

Calculate state values at Stage 3 (

k = 3

)

B_3 = 60 - 20 = 40\text{ MB}

and

Q_3 = 70 + 10 = 80\text{ MB}

.

Apply the transition rules to the Stage 2 state values.

4

Evaluate option statements

The statement specifying

B_2 = 60\text{ MB}

and the statement specifying

Q_3 = 80\text{ MB}

are verified as correct.

Compare calculated stage states directly against the option statements.

Key Concept

Sequential processes require evaluating every state transition iteratively without skipping intermediate stages.

Question 6Question →

A water processing facility processes liquid batches through a three-stage sequential system. The system updates the batch pollutant count ( $N$ ) according to the following state transition rules at each stage:

- Stage 1 (Filtration): $N_1 = N_0 + 20$
- Stage 2 (Purification): $N_2 = 2 \times N_1$
- Stage 3 (Polishing): $N_3 = N_2 - 10$

Match each initial pollutant count ( $N_0$ ) on the left to its corresponding final pollutant count ( $N_3$ ) after completing all three stages.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Initial Count

N_0 = 10

matches Final Count

N_3 = 50

; Initial Count

N_0 = 25

matches Final Count

N_3 = 80

; Initial Count

N_0 = 40

matches Final Count

N_3 = 110

.

Each initial count must be processed sequentially through all three transition rules in exact order:

N_1 = N_0 + 20

, then

N_2 = 2 \times N_1

, and finally

N_3 = N_2 - 10

. Executing this sequence accurately matches

N_0 = 10

to

N_3 = 50

,

N_0 = 25

to

N_3 = 80

, and

N_0 = 40

to

N_3 = 110

.

Step-by-Step Solution

1

Apply Stage 1 transition rule (

N_1 = N_0 + 20

) to each initial value.

For

N_0 = 10 \rightarrow N_1 = 30

; for

N_0 = 25 \rightarrow N_1 = 45

; for

N_0 = 40 \rightarrow N_1 = 60

.

Stage 1 adds 20 to the initial state.

2

Apply Stage 2 transition rule (

N_2 = 2 \times N_1

) to the intermediate values.

For

N_1 = 30 \rightarrow N_2 = 60

; for

N_1 = 45 \rightarrow N_2 = 90

; for

N_1 = 60 \rightarrow N_2 = 120

.

Stage 2 doubles the state value from Stage 1.

3

Apply Stage 3 transition rule (

N_3 = N_2 - 10

) to determine final output states.

For

N_2 = 60 \rightarrow N_3 = 50

; for

N_2 = 90 \rightarrow N_3 = 80

; for

N_2 = 120 \rightarrow N_3 = 110

.

Stage 3 subtracts 10 from the Stage 2 state value.

Key Concept

Sequential processes require evaluating state updates iteratively in strict order, carrying the output of each stage into the next stage.

Question 7Question →

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 8Question →

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 9Question →

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 10Question →

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 11Question →

A subscription service evaluates customer activity over a three-month period ( $k = 1, 2, 3$ ). At the end of each month $k$ , a customer's total loyalty points $P_k$ update based on their points from the previous month $P_{k-1}$ and the net points earned during month $k$ ( $\Delta_k$ ), according to the state transition rule:

P_k = \lfloor 0.8 \times P_{k-1} \rfloor + \Delta_k

where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$ .

At the end of Month 3 ( $k = 3$ ), membership tiers are assigned based on final points $P_3$ :
- VIP Tier: $P_3 \ge 150$
- Premium Tier: $80 \le P_3 < 150$
- Standard Tier: $P_3 < 80$

Match each initial customer profile (defined by initial points $P_0$ and monthly additions $\Delta_1, \Delta_2, \Delta_3$ ) to its corresponding final state at the end of Month 3.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Profile 1 matches Final Points: 131 (Premium Tier); Profile 2 matches Final Points: 178 (VIP Tier); Profile 3 matches Final Points: 73 (Standard Tier); Profile 4 matches Final Points: 140 (Premium Tier).

Each profile is evaluated by computing the exact state update

P_k = \lfloor 0.8 \cdot P_{k-1} \rfloor + \Delta_k

sequentially over three transitions. Profile 1 yields 131 points (Premium Tier), Profile 2 yields 178 points (VIP Tier), Profile 3 yields 73 points (Standard Tier), and Profile 4 yields 140 points (Premium Tier).

Step-by-Step Solution

1

Apply the state transition formula

P_k = \lfloor 0.8 \times P_{k-1} \rfloor + \Delta_k

iteratively across months

k = 1, 2, 3

for Profile 1.

P_1 = 80 + 50 = 130

;

P_2 = 104 + 10 = 114

;

P_3 = 91 + 40 = 131

. Tier: Premium.

Tracking sequential updates ensures each intermediate decay and point addition step is accounted for accurately.

2

Apply the state transition formula iteratively across months

k = 1, 2, 3

for Profile 2.

P_1 = 120 + 30 = 150

;

P_2 = 120 + 40 = 160

;

P_3 = 128 + 50 = 178

. Tier: VIP.

Continuous accumulation and decay calculation determines the final state threshold reached.

3

Apply the state transition formula iteratively across months

k = 1, 2, 3

for Profile 3.

P_1 = 72 + 0 = 72

;

P_2 = 57 + 10 = 67

;

P_3 = 53 + 20 = 73

. Tier: Standard.

Floor function truncation must be evaluated at each step prior to adding monthly points.

4

Apply the state transition formula iteratively across months

k = 1, 2, 3

for Profile 4.

P_1 = 160 + 0 = 160

;

P_2 = 128 + 10 = 138

;

P_3 = 110 + 30 = 140

. Tier: Premium.

Evaluating high starting values shows how retention decay influences final tier classification.

Key Concept

Sequential state transitions with floor function decay and additive step updates

Question 12Question →

An automated server optimization pipeline adjusts cache memory $C$ (in megabytes) and latency $L$ (in milliseconds) across three sequential cycles ( $k = 1, 2, 3$ ). The initial values at cycle $0$ are $C_0 = 100$ MB and $L_0 = 50$ ms. The state transitions occur according to the following rules:

- Cycle 1 ( $k = 1$ ): $C_1 = C_0 + 20$ , and $L_1 = L_0 - 0.2 \times C_0$ .
- Cycle 2 ( $k = 2$ ): If $L_1 \le 35$ , then $C_2 = C_1 - 30$ and $L_2 = L_1 + 10$ ; otherwise, $C_2 = C_1 + 10$ and $L_2 = L_1 - 5$ .
- Cycle 3 ( $k = 3$ ): If $C_2 < 100$ , then $C_3 = 1.5 \times C_2$ and $L_3 = L_2 - 15$ ; otherwise, $C_3 = C_2 + 20$ and $L_3 = L_2 + 5$ .

Which of the following statements regarding the system metrics across these cycles are true? Select all that apply.

Select all that apply

At the end of Cycle 1, the latency metric

L_1

is equal to

30

ms.

At the end of Cycle 2, the cache memory

C_2

is equal to

130

MB.

At the end of Cycle 3, the cache memory

C_3

is equal to

135

MB.

At the end of Cycle 3, the latency metric

L_3

is equal to

35

ms.

The net change in cache memory from Cycle 0 to Cycle 3 (

C_3 - C_0

) is

35

MB.

Show answer & explanation

Answer: At the end of Cycle 1, the latency metric $L_1$ is equal to $30$ ms.; At the end of Cycle 3, the cache memory $C_3$ is equal to $135$ MB.; The net change in cache memory from Cycle 0 to Cycle 3 ( $C_3 - C_0$ ) is $35$ MB.

Answer

The statements asserting that

L_1 = 30

ms at the end of Cycle 1,

C_3 = 135

MB at the end of Cycle 3, and the net change in cache memory

C_3 - C_0 = 35

MB are correct.

The calculation reveals the complete sequence of state pairs

(C_k, L_k)

:

(100, 50) \rightarrow (120, 30) \rightarrow (90, 40) \rightarrow (135, 25)

. Therefore, at Cycle 1,

L_1 = 30

ms is true. At Cycle 3,

C_3 = 135

MB is true. Finally, the net difference

C_3 - C_0 = 135 - 100 = 35

MB is also true.

Step-by-Step Solution

1

Calculate state values at Cycle 1 (

k=1

)

C_1 = 100 + 20 = 120

MB, and

L_1 = 50 - 0.2 \times 100 = 30

ms.

Direct evaluation of the Cycle 1 transition equations starting from

C_0 = 100

and

L_0 = 50

.

2

Evaluate condition and calculate state values at Cycle 2 (

k=2

)

Since

L_1 = 30 \le 35

is true,

C_2 = 120 - 30 = 90

MB and

L_2 = 30 + 10 = 40

ms.

The condition

L_1 \le 35

is satisfied, requiring the application of the primary branch.

3

Evaluate condition and calculate state values at Cycle 3 (

k=3

)

Since

C_2 = 90 < 100

is true,

C_3 = 1.5 \times 90 = 135

MB and

L_3 = 40 - 15 = 25

ms.

The condition

C_2 < 100

is satisfied, triggering the multiplicative increase in cache memory.

4

Verify overall net change and evaluate all option statements

C_3 - C_0 = 135 - 100 = 35

MB. Statements matching

L_1 = 30

ms,

C_3 = 135

MB, and net change

= 35

MB are true.

Compares calculated metrics against each provided statement.

Key Concept

Multi-stage conditional state transitions and sequential numerical processing

Question 13Question →

A renewable energy storage station manages a grid battery system with an initial energy reserve of

E_0 = 100\text{ MWh}

. At the end of each operating interval

k

(for

k = 1, 2, 3, 4

), the updated energy reserve

E_k

(in MWh) transitions according to the rule:

E_k = 0.8 E_{k-1} + P_k

where

P_k

represents the net energy added during interval

k

. The net additions for the four intervals are

P_1 = 40\text{ MWh}

,

P_2 = 10\text{ MWh}

,

P_3 = 50\text{ MWh}

, and

P_4 = 20\text{ MWh}

.

Arrange the operating intervals in ascending order based on the battery energy reserve $E_k$ measured at the end of each interval, from the lowest energy reserve to the highest energy reserve.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from lowest to highest energy reserve is End of Interval 2 (106 MWh), End of Interval 1 (120 MWh), End of Interval 4 (127.84 MWh), and End of Interval 3 (134.8 MWh).

Evaluating each recursive state step by step yields

E_1 = 120\text{ MWh}

,

E_2 = 106\text{ MWh}

,

E_3 = 134.8\text{ MWh}

, and

E_4 = 127.84\text{ MWh}

. Arranging these quantities from smallest to largest results in the sequence: End of Interval 2, End of Interval 1, End of Interval 4, and End of Interval 3.

Step-by-Step Solution

1

Calculate the energy reserve at the end of Interval 1 (

E_1

)

E_1 = 0.8(100) + 40 = 80 + 40 = 120\text{ MWh}

Apply the state transition formula using initial state

E_0 = 100\text{ MWh}

and addition

P_1 = 40\text{ MWh}

.

2

Calculate the energy reserve at the end of Interval 2 (

E_2

)

E_2 = 0.8(120) + 10 = 96 + 10 = 106\text{ MWh}

Apply the state transition formula using state

E_1 = 120\text{ MWh}

and addition

P_2 = 10\text{ MWh}

.

3

Calculate the energy reserve at the end of Interval 3 (

E_3

)

E_3 = 0.8(106) + 50 = 84.8 + 50 = 134.8\text{ MWh}

Apply the state transition formula using state

E_2 = 106\text{ MWh}

and addition

P_3 = 50\text{ MWh}

.

4

Calculate the energy reserve at the end of Interval 4 (

E_4

)

E_4 = 0.8(134.8) + 20 = 107.84 + 20 = 127.84\text{ MWh}

Apply the state transition formula using state

E_3 = 134.8\text{ MWh}

and addition

P_4 = 20\text{ MWh}

.

5

Sort the computed values in ascending order

E_2 (106\text{ MWh}) < E_1 (120\text{ MWh}) < E_4 (127.84\text{ MWh}) < E_3 (134.8\text{ MWh})

Comparing the values gives the exact ordered sequence: Interval 2, Interval 1, Interval 4, Interval 3.

Key Concept

Sequential Processes and State Transitions

Estimated Time:2m 0s

Question 14Question →

An industrial water desalination facility routes seawater through a three-stage membrane filtration process ( $k = 1, 2, 3$ ). At stage $k = 0$ (the initial feed stage), the water has a salt concentration of $C_0 = 800\text{ ppm}$ and a flow rate of $F_0 = 120\text{ liters/min}$ .

At each subsequent stage $k$ (where $k = 1, 2, 3$ ), the state of the water is updated according to the following sequential rules:
- Salt concentration: $C_k = 0.5 C_{k-1} - 20\text{ ppm}$
- Flow rate: $F_k = F_{k-1} - 15\text{ liters/min}$

Which of the following statements regarding the system's state transitions across the stages are true? Select all that apply.

Select all that apply

At the end of Stage 2 (

k = 2

), the salt concentration of the water is

170\text{ ppm}

.

At the end of Stage 3 (

k = 3

), the water flow rate is

75\text{ liters/min}

.

At the end of Stage 2 (

k = 2

), the salt concentration of the water is

380\text{ ppm}

.

At the end of Stage 3 (

k = 3

), the salt concentration of the water is

170\text{ ppm}

.

At the end of Stage 1 (

k = 1

), the water flow rate is

90\text{ liters/min}

.

Show answer & explanation

Answer: At the end of Stage 2 ( $k = 2$ ), the salt concentration of the water is $170\text{ ppm}$ .; At the end of Stage 3 ( $k = 3$ ), the water flow rate is $75\text{ liters/min}$ .

Answer

The correct statements are the one indicating a salt concentration of 170 ppm at Stage 2 and the one indicating a flow rate of 75 liters/min at Stage 3.

Evaluating each transition step sequentially gives C_1 = 380, C_2 = 170, and C_3 = 65 for salt concentration, and F_1 = 105, F_2 = 90, and F_3 = 75 for flow rate. The statement asserting a salt concentration of 170 ppm at Stage 2 and the statement asserting a flow rate of 75 liters/min at Stage 3 correctly reflect these evaluations.

Step-by-Step Solution

1

Calculate state metrics for Stage 1 (k = 1)

C_1 = 0.5(800) - 20 = 380 ppm; F_1 = 120 - 15 = 105 liters/min

Apply the recursive transition functions once from initial state k = 0.

2

Calculate state metrics for Stage 2 (k = 2)

C_2 = 0.5(380) - 20 = 170 ppm; F_2 = 105 - 15 = 90 liters/min

Apply the recursive transition functions to the Stage 1 outputs.

3

Calculate state metrics for Stage 3 (k = 3)

C_3 = 0.5(170) - 20 = 65 ppm; F_3 = 90 - 15 = 75 liters/min

Apply the recursive transition functions to the Stage 2 outputs.

4

Verify each statement against calculated values

Stage 2 salt concentration is 170 ppm (True) and Stage 3 flow rate is 75 liters/min (True).

Identify all statements that match the calculated sequence values.

Key Concept

Multi-stage recursive state evaluation and linear parameter transition tracking.

Question 15Question →

An automated data processing node manages a queue of records across three sequential processing stages ( $k = 1, 2, 3$ ). The initial batch size at stage $k = 0$ is $N_0 = 400$ records. In each stage $k$ , the batch size undergoes two consecutive transitions:
1. The batch size is reduced by $50\%$ due to quality filtering.
2. Exactly $120$ newly verified records are added to the batch.

Which of the following statements regarding the batch size across these stages are correct? Select all that apply.

Select all that apply

The batch size at the end of Stage 1 (

N_1

) is

320

records.

The batch size at the end of Stage 3 (

N_3

) is

260

records.

The total net reduction in batch size from the start of Stage 1 to the end of Stage 3 is

140

records.

The batch size at the end of Stage 2 (

N_2

) is

160

records.

Immediately after the filtering reduction step in Stage 1, the batch size is

320

records.

Show answer & explanation

Answer: The batch size at the end of Stage 1 ( $N_1$ ) is $320$ records.; The batch size at the end of Stage 3 ( $N_3$ ) is $260$ records.; The total net reduction in batch size from the start of Stage 1 to the end of Stage 3 is $140$ records.

Answer

The correct statements are: the batch size at the end of Stage 1 is 320 records; the batch size at the end of Stage 3 is 260 records; and the total net reduction from Stage 1 to Stage 3 is 140 records.

Evaluating each stage sequentially shows: Stage 1 ends with 320 records (400 halved to 200, plus 120); Stage 2 ends with 280 records (320 halved to 160, plus 120); Stage 3 ends with 260 records (280 halved to 140, plus 120). Subtracting the final count of 260 from the initial count of 400 gives a net decrease of 140 records.

Step-by-Step Solution

1

Calculate state transitions for Stage 1 (

k = 1

).

Initial

N_0 = 400

. Filtering reduces count to

400 \times 0.5 = 200

. Adding 120 records gives

N_1 = 200 + 120 = 320

.

Tracking each sub-step sequentially is necessary to determine the state at the end of Stage 1.

2

Calculate state transitions for Stage 2 (

k = 2

).

Filtering reduces count to

320 \times 0.5 = 160

. Adding 120 records gives

N_2 = 160 + 120 = 280

.

Stage 2 operates on the output state of Stage 1 (

N_1 = 320

).

3

Calculate state transitions for Stage 3 (

k = 3

).

Filtering reduces count to

280 \times 0.5 = 140

. Adding 120 records gives

N_3 = 140 + 120 = 260

.

Stage 3 operates on the output state of Stage 2 (

N_2 = 280

).

4

Evaluate the net overall change from start to end.

Net reduction =

N_0 - N_3 = 400 - 260 = 140

records.

Comparing initial state

N_0

and final state

N_3

gives the overall change.

Key Concept

Sequential state transitions require updating the system state step-by-step without omitting intermediate additions or reductions.

Estimated Time:1m 30s

Question 16Question →

A digital payment platform processes a batch of $N_0 = 1{}000$ transactions through four sequential verification stages ( $k = 1, 2, 3, 4$ ). The transition rules for the transaction count $N_k$ at each stage $k$ are defined as follows:

* Stage 1 (Fraud Filter): $10\%$ of the current batch transactions are flagged and removed, after which $50$ system control transactions are added.
* Stage 2 (Currency Verification): $20\%$ of the current batch transactions are each split into two separate micro-transactions (increasing the net transaction count by $20\%$ of the incoming batch), after which $40$ invalid transactions are purged.
* Stage 3 (Settlement Reconciliation): The batch transaction count decreases by $15\%$ due to netting, after which $150$ pending transactions are merged into the batch.
* Stage 4 (Archival Audit): $20\%$ of the current batch transactions are archived (retaining $80\%$ in the active batch), after which $120$ final clearance tokens are added to the batch.

Place the four stages in order of their final batch transaction counts ( $N_k$ ), from lowest transaction count to highest transaction count.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest transaction count to highest transaction count is Stage 1 (950), Stage 4 (988), Stage 3 (1,085), and Stage 2 (1,100).

Evaluating the sequential state transitions step by step yields

N_1 = 950

,

N_2 = 1{}100

,

N_3 = 1{}085

, and

N_4 = 988

. Ordering these four outputs from smallest to largest yields Stage 1 (

950

), Stage 4 (

988

), Stage 3 (

1{}085

), and Stage 2 (

1{}100

).

Step-by-Step Solution

1

Calculate the transaction count at the end of Stage 1 (

N_1

).

N_1 = 1{}000 - 0.10(1{}000) + 50 = 900 + 50 = 950

.

Apply the

10\%

reduction followed by adding

50

control transactions to the initial batch of

1{}000

.

2

Calculate the transaction count at the end of Stage 2 (

N_2

).

N_2 = 950 + 0.20(950) - 40 = 950 + 190 - 40 = 1{}100

.

Apply the

20\%

net split increase to

N_1 = 950

and subtract

40

purged transactions.

3

Calculate the transaction count at the end of Stage 3 (

N_3

).

N_3 = 1{}100 - 0.15(1{}100) + 150 = 1{}100 - 165 + 150 = 1{}085

.

Reduce

N_2 = 1{}100

by

15\%

and add

150

merged transactions.

4

Calculate the transaction count at the end of Stage 4 (

N_4

).

N_4 = 0.80(1{}085) + 120 = 868 + 120 = 988

.

Retain

80\%

of

N_3 = 1{}085

(

20\%

archived) and add

120

clearance tokens.

5

Order the resulting values from lowest to highest.

950 < 988 < 1{}085 < 1{}100

, corresponding to Stage 1, Stage 4, Stage 3, Stage 2.

Compare the numerical results across all four steps to establish the ascending order.

Key Concept

Sequential Process State Tracking

Estimated Time:2m 0s

Question 17Question →

An automated smart climate controller regulates room temperature ( $T$ , in ${}^\circ\text{C}$ ) and relative humidity ( $H$ , in $\%$ ) across three sequential hourly cycles ( $k = 1, 2, 3$ ). The initial state at cycle $k = 0$ is $T_0 = 24^\circ\text{C}$ and $H_0 = 60\%$ .

At the end of each hourly cycle $k$ , the new state $(T_k, H_k)$ is updated according to the following transition rules:
- $T_k = T_{k-1} - 2$ if $H_{k-1} \ge 50\%$ ; otherwise, $T_k = T_{k-1} + 1$
- $H_k = H_{k-1} - 10$ if $T_{k-1} > 20$ ; otherwise, $H_k = H_{k-1} + 5$

Which of the following statements regarding the system's state are correct? Select all that apply.

Select all that apply

The temperature at the end of cycle 3 is

21^\circ\text{C}

.

The relative humidity at the end of cycle 3 is

45\%

.

The temperature continuously decreases across all three consecutive cycles.

The relative humidity at the end of cycle 2 is

30\%

.

The temperature at the end of cycle 2 is

18^\circ\text{C}

.

Show answer & explanation

Answer: The temperature at the end of cycle 3 is $21^\circ\text{C}$ .; The relative humidity at the end of cycle 3 is $45\%$ .

Answer

The temperature at the end of cycle 3 is

21^\circ\text{C}

and the relative humidity at the end of cycle 3 is

45\%

.

Evaluating the state transitions step-by-step shows:
- Cycle 1:

T_1 = 22^\circ\text{C}, H_1 = 50\%

- Cycle 2:

T_2 = 20^\circ\text{C}, H_2 = 40\%

- Cycle 3:

T_3 = 21^\circ\text{C}, H_3 = 45\%

Therefore, the statements confirming that the temperature at the end of cycle 3 is

21^\circ\text{C}

and that the relative humidity at the end of cycle 3 is

45\%

are both correct.

Step-by-Step Solution

1

Calculate state at cycle

k = 1

from initial state

(T_0, H_0) = (24, 60)

.

T_1 = 24 - 2 = 22^\circ\text{C}

(since

H_0 = 60 \ge 50

);

H_1 = 60 - 10 = 50\%

(since

T_0 = 24 > 20

).

Apply transition rules based on conditions satisfied by cycle 0 values.

2

Calculate state at cycle

k = 2

from state

(T_1, H_1) = (22, 50)

.

T_2 = 22 - 2 = 20^\circ\text{C}

(since

H_1 = 50 \ge 50

);

H_2 = 50 - 10 = 40\%

(since

T_1 = 22 > 20

).

Apply transition rules based on conditions satisfied by cycle 1 values.

3

Calculate state at cycle

k = 3

from state

(T_2, H_2) = (20, 40)

.

T_3 = 20 + 1 = 21^\circ\text{C}

(since

H_2 = 40 < 50

);

H_3 = 40 + 5 = 45\%

(since

T_2 = 20

is NOT strictly greater than

20

).

Carefully check strict inequalities (

T_2 > 20

is false for

T_2 = 20

) and update parameters accordingly.

Key Concept

Sequential State Transitions with Conditional Updates

Question 18Question →

A digital subscription platform updates user account balances for Reward Points ( $P$ ) and Bonus Credits ( $C$ ) at the end of each month $k$ (where $k = 1, 2, 3$ ) according to the following state transition rules:

- $P_k = 0.8 P_{k-1} + 50$
- $C_k = C_{k-1} + 0.1 P_{k-1}$

At month $k = 0$ , an account starts with $P_0 = 500$ points and $C_0 = 100$ credits.

Which of the following statements regarding the account balances are true? Select all that apply.

Select all that apply

The promotional credit balance at the end of Month 2 (

C_2

) is equal to 195.

The tier status point balance at the end of Month 3 (

P_3

) is equal to 378.

The promotional credit balance at the end of Month 3 (

C_3

) is equal to 241.

The tier status point balance at the end of Month 2 (

P_2

) is equal to 450.

The tier status point balance decreases by exactly 50 points from Month 1 to Month 2.

Show answer & explanation

Answer: The promotional credit balance at the end of Month 2 ( $C_2$ ) is equal to 195.; The tier status point balance at the end of Month 3 ( $P_3$ ) is equal to 378.

Answer

The statement that the promotional credit balance at the end of Month 2 is 195 and the statement that the tier status point balance at the end of Month 3 is 378 are both correct.

Evaluating the sequential transitions stage-by-stage gives

P_1 = 450

,

C_1 = 150

,

P_2 = 410

,

C_2 = 195

,

P_3 = 378

, and

C_3 = 236

. Thus, the statement asserting

C_2 = 195

is correct, and the statement asserting

P_3 = 378

is correct.

Step-by-Step Solution

1

Calculate state values at Month 1 (

k = 1

)

P_1 = 0.8(500) + 50 = 450

points;

C_1 = 100 + 0.1(500) = 150

credits.

Apply state transition rules using initial values

P_0 = 500

and

C_0 = 100

.

2

Calculate state values at Month 2 (

k = 2

)

P_2 = 0.8(450) + 50 = 410

points;

C_2 = 150 + 0.1(450) = 195

credits.

Apply state transition rules using Month 1 values (

P_1 = 450

,

C_1 = 150

).

3

Calculate state values at Month 3 (

k = 3

)

P_3 = 0.8(410) + 50 = 378

points;

C_3 = 195 + 0.1(410) = 236

credits.

Apply state transition rules using Month 2 values (

P_2 = 410

,

C_2 = 195

).

4

Evaluate candidate statements

C_2 = 195

is true;

P_3 = 378

is true;

C_3 = 236 \neq 241

is false;

P_2 = 410 \neq 450

is false; change in points

P_1 - P_2 = 450 - 410 = 40 \neq 50

is false.

Compare calculated multi-stage process values directly against each statement.

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