CUET UG Applied Mathematics Booster Test 2 - Scheduling, Races, and Inequalities -

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CUET UG Applied Mathematics Booster Test 2 - Scheduling, Races, and Inequalities

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Scheduling, Races, and Inequalities

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QUESTION 1 OF 20

What is the fundamental objective of applying scheduling principles to activities in a production unit?

Answer Review

1 What is the fundamental objective of applying scheduling principles to activities in a production unit?

A To randomly assign tasks to processors to ensure fairness.

B To maximise the turnaround time of every task.

C To pre-plan and schedule activities to minimise production time and cost.

D To create convoy effects deliberately.

💡 Short Explanation

�� Scheduling organizes tasks efficiently. �� It helps reduce production delays. �� Main aim is minimizing time and cost.

✅ Correct Answer Explanation

Scheduling in production systems refers to: • Planning and organizing activities in advance. • Allocating resources efficiently. • Reducing idle time, delays, and operational cost. Thus option C is correct because the primary goal is optimized productivity with minimum production time and cost. Option A is incorrect because scheduling is systematic, not random. Option B is opposite to the purpose of scheduling. Option D describes a negative scheduling effect.

❓ Why Other Options Are Incorrect

�� Option A → Scheduling follows optimization principles, not random assignment.
�� Option B → Scheduling aims to reduce turnaround time.
�� Option D → Convoy effect is undesirable.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match "scheduling objective" with industrial optimization goals.

Final Logic: Efficient planning minimizes production time and cost.

🧠 Memory Trick

"Schedule smart, save time."

📚 Class XII – Informatics Practices – Chapter: Operating System Concepts – Topic: Scheduling Objectives

2 Match the specific time parameter definitions to their corresponding abbreviations:

List I	List II
1. Completion Time - Arrival Time	a. AWT
2. Turnaround Time - Burst Time	b. TAT
3. Sum of Waiting Times / Number of Processes	c. BT
4. Amount of time required by a process for executing	d. WT

A 1-b, 2-d, 3-a, 4-c

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-c, 3-b, 4-a

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

�� TAT represents Turnaround Time. �� WT represents Waiting Time. �� BT represents Burst Time.

✅ Correct Answer Explanation

Match each term correctly: • 1 → b : Completion Time − Arrival Time = Turnaround Time (TAT) → TAT = CT − AT, so this corresponds to TAT. • 2 → d : Turnaround Time − Burst Time = Waiting Time (WT) → WT = TAT − BT, so this corresponds to WT. • 3 → a : Sum of Waiting Times / Number of Processes = Average Waiting Time (AWT) • 4 → c : Amount of time required by a process for execution = Burst Time (BT) Thus, correct matching is: 1-b, 2-d, 3-a, 4-c Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → WT and TAT mappings are interchanged incorrectly.
�� Option C → All abbreviations are mismatched with definitions.
�� Option D → Incorrect pairing of TAT and WT concepts with definitions.

🎯 Strategy

Used

Option Grouping

Application:

Use standard CPU scheduling formulas:
TAT = CT − AT, WT = TAT − BT, BT = execution time, AWT = average of WT.

Final Logic:

Only Option A correctly matches all scheduling definitions with their standard abbreviations.

🧠 Memory Trick

TAT = Total time, WT = Waiting time, BT = Busy time.

📚 Class XII – Computer Science – Chapter: Operating System – Topic: CPU Scheduling Parameters

3 In analyzing the FCFS scheduling algorithm, which of the following characteristics strictly apply?
1. It is non-pre-emptive.
2. Processes are executed in the order of their burst time.
3. The average waiting time can be high if the first job takes too much CPU time.
4. It can suffer from the Starvation problem.

A 1, 2

B 1, 3

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� FCFS executes processes in arrival order. �� It is non-pre-emptive. �� Long first process increases waiting time.

✅ Correct Answer Explanation

FCFS (First Come First Serve) scheduling works as follows: • Statement 1 → Correct. FCFS is non-pre-emptive; once a process starts, it runs to completion. • Statement 2 → Incorrect. FCFS does NOT use burst time; it uses arrival time. • Statement 3 → Correct. If a long process arrives first, it increases waiting time for others (convoy effect). • Statement 4 → Incorrect. FCFS does not cause starvation because every process eventually gets CPU time. Thus, only statements 1 and 3 are correct. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes statement 2 which is false.
�� Option C → Includes burst-time scheduling and starvation, both incorrect.
�� Option D → Includes starvation, which does not occur in FCFS.

🎯 Strategy

Used

Elimination

Application:

Remove statements that incorrectly assume burst-time ordering or starvation behavior.

Final Logic:

FCFS is non-pre-emptive and may increase waiting time due to long first process.

🧠 Memory Trick

FCFS = First Come, First Served (arrival order only).

📚 Class XII – Computer Science – Chapter: Operating System – Topic: FCFS Scheduling Algorithm

4 Consider a scenario where processes P1, P2, and P3 arrive at time 0 with burst times 10, 2, and 3 respectively. Identify the INCORRECT statement regarding their scheduling:

A Under FCFS, the average waiting time is high due to the Convoy Effect.

B Under SJF, P2 executes first, followed by P3, then P1.

C SJF provides a higher average waiting time than FCFS in this scenario.

D Under SJF, the waiting time for P2 is 0.

💡 Short Explanation

�� SJF selects shortest burst time first. �� SJF minimizes average waiting time. �� Statement C contradicts SJF property.

✅ Correct Answer Explanation

Burst times: • P1 = 10 • P2 = 2 • P3 = 3 Under SJF: • P2 executes first. • Then P3. • Then P1. SJF is known for minimum average waiting time. Therefore, statement C is incorrect because SJF actually reduces waiting time compared to FCFS.

❓ Why Other Options Are Incorrect

�� Option A → FCFS may create Convoy Effect with long first process.
�� Option B → Correct SJF order based on burst time.
�� Option D → First process in SJF has zero waiting time.

🎯 Strategy

Used: Elimination

Application: Compare known SJF property with given statements.

Final Logic: SJF always reduces average waiting time compared to FCFS here.

🧠 Memory Trick

"Shortest first = shortest wait."

📚 Class XII – Computer Science – Chapter: Operating System – Topic: SJF Scheduling

5 If a system utilizes a mixture of pre-emptive priority protocols, what specific trigger causes the CPU to pre-empt a currently running process?

A The current process completes its burst time.

B A new process arrives at the ready queue with a higher priority than the currently running process.

C A new process arrives with a shorter burst time (ignoring priority tags).

D The processor has been idle for too long.

💡 Short Explanation

�� Pre-emption interrupts execution. �� Higher priority process gets CPU first. �� Priority controls execution order.

✅ Correct Answer Explanation

In pre-emptive priority scheduling: • The CPU always serves the highest priority process. • If a higher priority process arrives, the current process is interrupted immediately. Hence option B is correct. Option A describes normal completion, not pre-emption. Option C relates to SJF. Option D is unrelated.

❓ Why Other Options Are Incorrect

�� Option A → Completion is not pre-emption.
�� Option C → Shorter burst time belongs to SJF logic.
�� Option D → Idle time does not trigger pre-emption.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match "pre-emptive priority" with interruption condition.

Final Logic: Higher priority arrival causes CPU takeover.

🧠 Memory Trick

"Higher priority interrupts."

📚 Class XII – Computer Science – Chapter: CPU Scheduling – Topic: Priority Scheduling

6 For a non-pre-emptive priority scheduling algorithm, what constraint ensures that the currently running low-priority process is NOT disrupted?

A The current process must finish execution before the new high-priority process at the head of the queue can start.

B The high-priority process must be rejected completely.

C The CPU duplicates itself to handle both.

D The low-priority process is terminated entirely without finishing.

💡 Short Explanation

�� Non-pre-emptive means no interruption. �� Running process continues till completion. �� New process waits in queue.

✅ Correct Answer Explanation

In non-pre-emptive scheduling: • A running process cannot be interrupted. • Even if a higher priority process arrives, it must wait. Thus option A correctly describes the scheduling constraint.

❓ Why Other Options Are Incorrect

�� Option B → High-priority processes are not rejected.
�� Option C → CPUs are not duplicated.
�� Option D → Running processes are not terminated arbitrarily.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Understand meaning of "non-pre-emptive."

Final Logic: Running process completes before switching.

🧠 Memory Trick

"Non-pre-emptive = no interruption."

📚 Class XII – Computer Science – Chapter: CPU Scheduling – Topic: Non-Pre-emptive Scheduling

7 Equating the total workflow, if Process A has an Arrival Time of 2 units, a Burst Time of 4 units, and finishes execution at 10 units, what is its Turnaround Time (TAT)?

A 4 units

B 10 units

C 8 units

D 6 units

💡 Short Explanation

�� TAT = Completion Time − Arrival Time. �� Completion Time = 10. �� Arrival Time = 2.

✅ Correct Answer Explanation

Turnaround Time formula: Substitute values: Therefore TAT = 8 units. Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equals burst time, not TAT.
�� Option B → Uses completion time only.
�� Option D → Incorrect subtraction.

🎯 Strategy

Used: Substitution

Application: Apply TAT formula directly.

Final Logic: TAT equals completion minus arrival time.

🧠 Memory Trick

"TAT = Total Active Time."

📚 Class XII – Computer Science – Chapter: Operating System – Topic: Turnaround Time

8 A process has a Turnaround Time of 12 units and a Burst Time of 5 units. Moving through its lifecycle, what is the calculated Waiting Time (WT) for this process?

A 17 units

B 7 units

C 60 units

D 2.4 units

💡 Short Explanation

�� WT = TAT − BT. �� TAT = 12. �� BT = 5.

✅ Correct Answer Explanation

Waiting Time formula: Substitute values: Thus Waiting Time = 7 units. Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Adds instead of subtracting.
�� Option C → Incorrect multiplication result.
�� Option D → Incorrect division operation.

🎯 Strategy

Used: Substitution

Application: Insert values into standard WT formula.

Final Logic: Waiting time equals turnaround minus burst time.

🧠 Memory Trick

"Wait = Total − Execution."

📚 Class XII – Computer Science – Chapter: CPU Scheduling – Topic: Waiting Time

9 What sequence of arrivals maximizes the probability of the Convoy Effect occurring in a non-pre-emptive FCFS schedule?

A Several short processes followed by a long process.

B A massive, CPU-intensive process followed by multiple very short processes.

C All processes having the exact same burst time.

D Processes arriving at intervals perfectly matching their burst times.

💡 Short Explanation

�� FCFS follows arrival order. �� Long first process delays all others. �� This creates Convoy Effect.

✅ Correct Answer Explanation

Convoy Effect occurs when: • A long CPU-bound process arrives first. • Many short jobs wait behind it. • System efficiency decreases because short jobs remain blocked. Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Long process at end does not block earlier jobs.
�� Option C → Equal burst times do not create convoy effect strongly.
�� Option D → Matching intervals reduce waiting problems.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify the condition that maximizes waiting behind one large task.

Final Logic: Large first process traps many short processes.

🧠 Memory Trick

"Big job blocks convoy."

📚 Class XII – Computer Science – Chapter: Scheduling Problems – Topic: Convoy Effect

10 Starvation directs smaller jobs to the front in SJF. If a long job (P1) is waiting, what specific continuous vector of events will cause P1 to starve infinitely?

A No new processes arrive.

B Processes with increasingly longer burst times arrive.

C An infinite stream of new processes with shorter burst times than P1 keeps arriving.

D P1 is moved to a pre-emptive queue.

💡 Short Explanation

�� SJF prefers shortest burst time. �� Long jobs remain postponed. �� Continuous short arrivals cause starvation.

✅ Correct Answer Explanation

In SJF scheduling: • The shortest available process always executes first. • If shorter processes continuously arrive, a long process may never get CPU time. This indefinite postponement is called starvation. Thus option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Without new arrivals, P1 eventually executes.
�� Option B → Longer jobs do not block P1 in SJF.
�� Option D → Queue type alone does not cause starvation.

🎯 Strategy

Used: Elimination

Application: Remove scenarios where P1 would eventually run.

Final Logic: Endless shorter jobs prevent P1 execution forever.

🧠 Memory Trick

"Short jobs starve long jobs."

📚 Class XII – Computer Science – Chapter: CPU Scheduling – Topic: Starvation in SJF

11 An athlete is running on a circular 400m racecourse. If "A beats B by 40 metres", what distance has B covered in the area of the track by the time A crosses the finishing point?

A 400m

B 440m

C 360m

D 40m

💡 Short Explanation

�� A finishes the full race distance. �� B is behind by 40 m. �� So B covers 400 − 40 = 360 m.

✅ Correct Answer Explanation

If A beats B by 40 metres in a 400 m race: • A completes the entire 400 m race. • At that exact moment, B is still 40 m behind. Therefore: Hence B has covered 360 m. Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → 400 m means B also finished, which is incorrect.
�� Option B → 440 m exceeds the total race distance.
�� Option D → 40 m is only the remaining distance.

🎯 Strategy

Used: Substitution

Application: Use the race definition directly with total distance minus lead distance.

Final Logic: Distance covered by loser = Total distance − Beating margin.

🧠 Memory Trick

"Beaten by x → subtract x."

📚 Class XI – Mathematics – Chapter: Time, Speed and Distance – Topic: Races and Games

12 Integrating the race conditions, what does the phrase "A beats B by x metres and t minutes" mathematically establish about B's speed?

A B's speed is strictly (x * t).

B B covers the remaining x metres in an extra t minutes, making B's speed = x / t.

C B's speed is y / (x + t).

D A and B have the same speed, but different starting times.

💡 Short Explanation

�� B is behind by x metres. �� B takes t more time to finish. �� Speed = distance ÷ time.

✅ Correct Answer Explanation

If A beats B by x metres and t minutes: • When A finishes, B still has x metres remaining. • B requires an additional t minutes to complete that distance. Thus B's speed becomes: Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Speed is distance divided by time, not multiplied.
�� Option C → No such standard race relation exists.
�� Option D → Statement does not imply equal speeds.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Speed must always have units of distance/time.

Final Logic: x metres in t minutes gives speed x/t.

🧠 Memory Trick

"Speed = remaining distance/time."

📚 Class XI – Mathematics – Chapter: Time, Speed and Distance – Topic: Race Relations

13 In a 500 metres race, A gives B a head start of 60 metres. How much distance does B actually have to run to complete the race?

A 560 m

B 500 m

C 440 m

D 60 m

💡 Short Explanation

�� Head start reduces race distance. �� B starts 60 m ahead. �� So B runs 500 − 60 = 440 m.

✅ Correct Answer Explanation

Total race distance = 500 m. B gets a head start of 60 m, meaning B begins 60 m ahead of the starting line. Thus B actually runs: Therefore option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Distance is not increased.
�� Option B → B does not run full distance due to head start.
�� Option D → 60 m is only the advantage distance.

🎯 Strategy

Used: Substitution

Application: Directly subtract head start from total race length.

Final Logic: Actual distance = Total distance − Head start.

🧠 Memory Trick

"Head start means less running."

📚 Class XI – Mathematics – Chapter: Races and Games – Topic: Head Start Problems

14 A and B complete a 100m race in 36 seconds and 48 seconds respectively. A defeats B by 12 seconds. By how many metres does A defeat B?

A 12 metres

B 25 metres

C 36 metres

D 20 metres

💡 Short Explanation

�� A finishes in 36 s. �� In 36 s, B covers only part of race. �� Remaining distance is the defeat margin.

✅ Correct Answer Explanation

A completes 100 m in 36 s. B takes 48 s for 100 m. So in 36 s, B covers: Thus B is behind by: Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect proportional calculation.
�� Option C → This equals A's time, not distance margin.
�� Option D → Does not satisfy speed ratio.

🎯 Strategy

Used: Substitution

Application: Use proportional speed-distance relation.

Final Logic: Distance covered by B in A's finishing time determines defeat margin.

🧠 Memory Trick

"Same time → proportional distance."

📚 Class XI – Mathematics – Chapter: Time, Speed and Distance – Topic: Races and Relative Speed

15 A, B, and C invested capital for 2, 3, and 4 months respectively. Their profit ratio is 6:7:8. What is the fundamental formula linking their investments, time, and profit?

A Ratio of Investment + Time = Ratio of Profit

B Ratio of Investment × Time = Ratio of Profit

C Ratio of Profit / Investment = Time Squared

D Investment = Profit + Time

💡 Short Explanation

�� Partnership profit depends on money and time. �� Higher capital or time gives larger profit. �� Profit ∝ Capital × Time.

✅ Correct Answer Explanation

In partnership: Thus: • Profit share depends on both investment and duration. • Partners investing more money for more time earn more profit. Hence option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Profit is not based on simple addition.
�� Option C → No squared relation exists.
�� Option D → Investment does not equal profit plus time.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match standard partnership formula with profit-sharing rule.

Final Logic: Profit is proportional to capital multiplied by time.

🧠 Memory Trick

"Partnership = Money × Time."

📚 Class XI – Mathematics – Chapter: Commercial Mathematics – Topic: Partnership Formula

16 P, Q, and R jointly invested Rs 64,000 in a compound partnership. P invested Rs 6000 more than Q, and Q invested Rs 5000 more than R. If R's investment is x, what algebraic expression represents P's investment?

A x + 5000

B x + 6000

C x + 11000

D 3x + 11000

💡 Short Explanation

�� R invests x. �� Q invests 5000 more than R. �� P invests 6000 more than Q.

✅ Correct Answer Explanation

Given: • R = x • Q = x + 5000 • P = (x + 5000) + 6000 Therefore: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents Q's relation with R only.
�� Option B → Ignores Q's additional investment.
�� Option D → Incorrect algebraic expansion.

🎯 Strategy

Used: Substitution

Application: Sequentially translate verbal relations into algebra.

Final Logic: Add both increases step-by-step.

🧠 Memory Trick

"5000 + 6000 = 11000."

📚 Class XI – Mathematics – Chapter: Algebraic Expressions – Topic: Linear Expressions

17 Evaluate the inequality 4x - 2 < 8. What is the boundary condition for the value of x?

A x < 10

B x > 2.5

C x < 2.5

D x = 2.5

💡 Short Explanation

�� Solve step-by-step. �� Isolate x. �� Final inequality gives x boundary.

✅ Correct Answer Explanation

Given: Add 2 to both sides: Divide by 4: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → 10 is intermediate value before division.
�� Option B → Inequality direction is incorrect.
�� Option D → Equality is not included.

🎯 Strategy

Used: Substitution

Application: Solve algebraically using inequality rules.

Final Logic: Isolate x carefully without changing inequality sign.

🧠 Memory Trick

"Move constants, divide coefficient."

📚 Class XI – Mathematics – Chapter: Linear Inequalities – Topic: Solving Inequalities

18 If a > b, and c < 0, which of the following operations correctly preserves the mathematical relationship?

A a * c > b * c

B a / c > b / c

C a * c < b * c

D a + c < b + c

💡 Short Explanation

�� Negative multiplication flips inequality. �� c is negative here. �� So direction reverses.

✅ Correct Answer Explanation

Given: Multiplying by a negative number reverses inequality: Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Sign must reverse for negative multiplication.
�� Option B → Division by negative also reverses sign.
�� Option D → Addition preserves same direction, not reverse.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Apply standard inequality operation rule.

Final Logic: Negative operations reverse inequality direction.

🧠 Memory Trick

"Negative flips sign."

📚 Class XI – Mathematics – Chapter: Linear Inequalities – Topic: Operations on Inequalities

📄 Passage

Scheduling algorithms manage the execution of processes to optimize performance. A GANTT chart is a type of bar chart that illustrates a project schedule, helping to map out Turnaround Time (TAT) and Waiting Time (WT). First Come First Serve (FCFS) executes processes strictly by Arrival Time (AT). Shortest Job First (SJF) selects the process with the minimum Burst Time (BT) next. While SJF provides an optimal Average Waiting Time, it risks starving longer processes. Conversely, FCFS guarantees no starvation but can suffer from the Convoy Effect if a massive process arrives early. Both non-pre-emptive FCFS and SJF do not interrupt a process once it has started executing.

Question:
According to the passage, what is the primary consequence of using the SJF scheduling algorithm?

A It provides the maximum possible turnaround time.

B It suffers from the Convoy Effect.

C It provides an optimal Average Waiting Time but risks starving longer processes.

D It pre-empts currently running tasks immediately.

💡 Short Explanation

�� SJF selects shortest jobs first. �� It minimizes average waiting time. �� Long jobs may starve indefinitely.

✅ Correct Answer Explanation

Shortest Job First (SJF): • Executes the process with minimum burst time first. • Produces minimum average waiting time. • Can cause starvation for longer processes if short jobs keep arriving. Thus, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → SJF minimizes waiting time, not maximizes it.
�� Option B → Convoy Effect is mainly linked with FCFS.
�� Option D → Standard SJF here is non-pre-emptive.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match passage statement directly with SJF property.

Final Logic: SJF optimizes waiting but risks starvation.

🧠 Memory Trick

"Short first, long starve."

📚 Class XII – Computer Science – Chapter: Operating System Concepts – Topic: SJF Scheduling

📄 Passage

Question:
In analyzing scheduling effects, why does FCFS guarantee no starvation but risk the Convoy Effect?

A Because it strictly executes jobs by Arrival Time, meaning short jobs might get trapped behind a massive early job.

B Because it executes the shortest job first, leaving long jobs behind.

C Because it constantly interrupts running jobs.

D Because the GANTT chart limits the number of processes to four.

💡 Short Explanation

�� FCFS follows arrival order. �� No process is skipped permanently. �� Long jobs can delay many short jobs.

✅ Correct Answer Explanation

FCFS scheduling: • Executes processes strictly according to arrival time. • Ensures every process eventually executes, so starvation does not occur. • However, a very long process arriving first can delay all smaller jobs behind it, creating the Convoy Effect. Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → This describes SJF, not FCFS.
�� Option C → FCFS is non-pre-emptive.
�� Option D → GANTT charts do not limit process count.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Link FCFS properties with arrival-order execution.

Final Logic: Arrival-order execution prevents starvation but creates convoy delays.

🧠 Memory Trick

"First come, everyone waits."

📚 Class XII – Computer Science – Chapter: Operating System Concepts – Topic: FCFS Scheduling and Convoy Effect

CUET UG Applied Mathematics Booster Test 2 - Scheduling, Races, and Inequalities

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