CUET UG Applied Mathematics Booster Test 3 - Scheduling, Races, and Inequalities -

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CUET UG Applied Mathematics Booster Test 3 - 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 3 - Scheduling, Races, and Inequalities

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QUESTION 1 OF 20

Evaluate the difference between Response Time and Waiting Time in a strictly Non-Pre-emptive scheduling system.

Answer Review

1 Evaluate the difference between Response Time and Waiting Time in a strictly Non-Pre-emptive scheduling system.

A Response time is always greater than waiting time.

B Response time is always less than waiting time.

C Response time is exactly the same as waiting time.

D Response time applies only to SJF, while waiting time applies to FCFS.

💡 Short Explanation

�� In non-pre-emptive systems, execution starts only once. �� Response time measures first CPU allocation delay. �� Waiting time equals response time here.

✅ Correct Answer Explanation

In a strictly non-pre-emptive scheduling system: • A process starts execution once and continues until completion. • Response Time = Time from arrival until first CPU allocation. • Waiting Time = Total time spent waiting in ready queue. Since the process is never interrupted, the first response delay equals the total waiting delay. Thus: Hence option C is correct. Option A and B are incorrect because neither quantity exceeds the other in strictly non-pre-emptive scheduling. Option D is wrong because both metrics apply to all scheduling algorithms.

❓ Why Other Options Are Incorrect

�� Option A → Response time does not exceed waiting time in non-pre-emptive systems.
�� Option B → Waiting time is not greater separately here.
�� Option D → Both concepts apply universally in CPU scheduling.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match non-pre-emptive behavior with timing definitions.

Final Logic: No interruption means response delay equals waiting delay.

🧠 Memory Trick

"No pre-emption = Same response & wait."

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

2 Match the specific business partnership data to their resulting profit or investment ratios.

List 1	List 2
1. P invests ₹6000 more than Q; Q invests ₹5000 more than R (Total ₹64000)	a. Equivalent Capital Ratio is 20 : 16 : 15
2. A puts in ₹5000/month (12 months), B puts ₹3000/month (4 months) + ₹4500/month (8 months), C puts ₹4000/month (9 months) + ₹3000/month (3 months)	b. Profit Share Ratio is 6 : 7 : 8
3. Profit ratio is 6 : 7 : 8, Time ratio is 2 : 3 : 4	c. Investment Ratio is 27 : 21 : 16
4. A, B, C share a profit of ₹21000 from capitals ₹36000, ₹42000, ₹48000	d. Investment Ratio is 9 : 7 : 6

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

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

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

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

💡 Short Explanation

�� Partnership profit is proportional to Capital × Time. �� Equivalent capital converts varying investments into a common ratio. �� Profit and investment ratios are obtained using partnership formulas.

✅ Correct Answer Explanation

Partnership is based on the relation: Profit Ratio ∝ Capital × Time 1 → c Let the investment of R be x. Then, Q = x + 5000 P = x + 11000 Given, P + Q + R = 64000 (x + 11000) + (x + 5000) + x = 64000 3x + 16000 = 64000 3x = 48000 x = 16000 Therefore, R = 16000 Q = 21000 P = 27000 Investment Ratio = 27000 : 21000 : 16000 = 27 : 21 : 16 Hence, 1 → c. 2 → a Capital × Time: A = 5000 × 12 = 60000 B = (3000 × 4) + (4500 × 8) = 12000 + 36000 = 48000 C = (4000 × 9) + (3000 × 3) = 36000 + 9000 = 45000 Equivalent Ratio = 60000 : 48000 : 45000 = 20 : 16 : 15 Hence, 2 → a. 3 → d Investment Ratio = Profit Ratio ÷ Time Ratio = 6/2 : 7/3 : 8/4 = 3 : 7/3 : 2 Multiplying by 3, = 9 : 7 : 6 Hence, 3 → d. 4 → b Capital Ratio = 36000 : 42000 : 48000 = 6 : 7 : 8 Since the investment period is the same, the profit-sharing ratio is also 6 : 7 : 8. Hence, 4 → b. Therefore, the correct matching is: 1 → c 2 → a 3 → d 4 → b Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because Question 1 matches Investment Ratio 27 : 21 : 16, not Equivalent Capital Ratio 20 : 16 : 15.
�� Option C → Incorrect because Question 2 gives the Equivalent Capital Ratio 20 : 16 : 15, not Investment Ratio 27 : 21 : 16.
�� Option D → Incorrect because Question 4 corresponds to the Profit Share Ratio 6 : 7 : 8, not Investment Ratio 9 : 7 : 6.

🎯 Strategy

Used: Option Grouping

Application:

Evaluate each partnership case independently using Capital × Time and then compare the complete mapping with the given options.

Final Logic:

Only Option A correctly matches all four business partnership situations.

🧠 Memory Trick

Capital × Time = Profit.

📚 Class XI – Mathematics – Chapter: Ratio and Proportion – Topic: Partnership Applications

3 Consider 4 processes arriving at AT = 0 with Burst Times: P1 = 25, P2 = 4, P3 = 7, P4 = 3. Which of the following evaluations mathematically hold true?
(1) Under FCFS, P1 executes first, and Average Waiting Time = 22.5 units.
(2) Under SJF, P4 executes first (BT = 3), followed by P2, P3, P1.
(3) Under SJF, Average Waiting Time = 6 units.
(4) FCFS produces a smaller Average Turnaround Time than SJF.

A 1, 2

B 1, 2, 3

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� FCFS executes processes in arrival order. �� SJF schedules the process with the smallest burst time first. �� SJF minimizes the average waiting time.

✅ Correct Answer Explanation

Under FCFS: Execution Order: P1 → P2 → P3 → P4 Waiting Times: P1 = 0 P2 = 25 P3 = 29 P4 = 36 Average Waiting Time = (0 + 25 + 29 + 36) / 4 = 90 / 4 = 22.5 units Hence, Statement (1) is correct. Under SJF: Execution Order: P4 → P2 → P3 → P1 Waiting Times: P4 = 0 P2 = 3 P3 = 7 P1 = 14 Average Waiting Time = (0 + 3 + 7 + 14) / 4 = 24 / 4 = 6 units Hence, Statements (2) and (3) are correct. Average Turnaround Time under SJF is also lower than FCFS because shorter jobs finish earlier. Therefore, Statement (4) is false. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because Statement (3) is also correct.
�� Option C → Incorrect because Statement (4) is false.
�� Option D → Incorrect because FCFS does not produce a lower average turnaround time than SJF in this case.

🎯 Strategy

Used: Substitution

Application:

Compute the waiting times directly for FCFS and SJF using the given burst times and compare the results.

Final Logic:

The calculated waiting times confirm that only Statements (1), (2), and (3) are correct.

🧠 Memory Trick

Shortest Job First → Smallest Wait.

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

4 Which of the following statements regarding the non-pre-emptive Shortest Job First (SJF) algorithm is algebraically INCORRECT when assessing arrival times > 0?

A If all processes arrive at time 0, SJF simply sorts them by burst time ascending.

B If a 1-unit burst process arrives at time 2 while a 10-unit process is running (started at 0), the 1-unit process must wait.

C SJF will always preempt the 10-unit process at time 2 to avoid the Convoy Effect.

D A process with identical burst time to another is queued based on FCFS tie-breaking.

💡 Short Explanation

�� Non-pre-emptive SJF does not interrupt running jobs. �� New short process must wait. �� Pre-emption belongs to SRTF.

✅ Correct Answer Explanation

In non-pre-emptive SJF: • Once execution starts, the running process continues until completion. • A later-arriving short process cannot interrupt execution. Therefore, statement C is incorrect because pre-emption is not allowed. Statement A is correct for simultaneous arrivals. Statement B correctly explains waiting behavior. Statement D correctly describes FCFS tie-breaking.

❓ Why Other Options Are Incorrect

�� Option A → Correct ascending burst-time ordering.
�� Option B → Non-pre-emptive scheduling forces waiting.
�� Option D → FCFS resolves equal burst-time conflicts.

🎯 Strategy

Used: Elimination

Application: Identify statement contradicting "non-pre-emptive."

Final Logic: Pre-emption cannot occur in non-pre-emptive SJF.

🧠 Memory Trick

"Non-pre-emptive = No interruption."

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

5 In a mixed CPU environment, if process A (Priority 1, high) arrives at 5ms while process B (Priority 5, low) has been running since 0ms and needs 10ms total, what occurs under Non-Pre-emptive Priority Scheduling?

A B is terminated instantly.

B A is rejected.

C B finishes at 10ms, and A executes starting at 10ms.

D B pauses at 5ms, A runs, then B resumes.

💡 Short Explanation

�� Non-pre-emptive systems do not interrupt running tasks. �� Higher priority waits until CPU becomes free. �� Process B completes first.

✅ Correct Answer Explanation

Process B starts at 0 ms and requires 10 ms. At 5 ms, process A arrives with higher priority. In non-pre-emptive priority scheduling: • The currently running process is not interrupted. • B continues until completion at 10 ms. • A starts afterward. Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Running processes are not terminated arbitrarily.
�� Option B → High-priority processes are queued, not rejected.
�� Option D → Pausing is pre-emptive behavior.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Apply non-pre-emptive scheduling rule.

Final Logic: Running process completes before switching.

🧠 Memory Trick

"Finish first, switch later."

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

6 A processor handles constraints. If idle time occurs because P1 completes at 4 units but P2 does not arrive until 5 units, what is the length of the Idle Time constraint region?

A 0 units

B 1 unit

C 4 units

D 5 units

💡 Short Explanation

�� Idle time occurs when CPU waits. �� P1 ends at 4. �� P2 arrives at 5.

✅ Correct Answer Explanation

Idle Time is: Substitute values: Hence idle time = 1 unit.

❓ Why Other Options Are Incorrect

�� Option A → CPU is not continuously busy.
�� Option C → Incorrect subtraction logic.
�� Option D → Total timeline, not idle gap.

🎯 Strategy

Used: Substitution

Application: Compute difference between completion and next arrival.

Final Logic: CPU remains idle for exactly one unit.

🧠 Memory Trick

"Gap between jobs = Idle time."

📚 Class XII – Computer Science – Chapter: Scheduling Metrics – Topic: Idle Time

7 Equating the metrics: Process P3 has AT=1, BT=2, and CT=3. What are its Turnaround Time (TAT) and Waiting Time (WT) respectively?

A TAT = 3, WT = 1

B TAT = 2, WT = 0

C TAT = 1, WT = 2

D TAT = 4, WT = 2

💡 Short Explanation

�� TAT = CT − AT. �� WT = TAT − BT. �� Process begins immediately.

✅ Correct Answer Explanation

Use formulas: Substitute values: TAT = 3 − 1 = 2 WT = 2 − 2 = 0 Thus, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect TAT calculation.
�� Option C → TAT and WT reversed incorrectly.
�� Option D → Values exceed actual execution metrics.

🎯 Strategy

Used: Substitution

Application: Insert values into standard formulas.

Final Logic: TAT = 2 and WT = 0 satisfy scheduling equations.

🧠 Memory Trick

"TAT minus BT gives WT."

📚 Class XII – Computer Science – Chapter: Scheduling Parameters – Topic: TAT and WT

8 Calculate the moving average waiting time for 4 processes scheduled in FCFS: P1(WT=0), P2(WT=2), P3(WT=6), P4(WT=9).

A 4.25 units

B 5.00 units

C 17.00 units

D 3.50 units

💡 Short Explanation

�� Average WT = Sum of WT / Number of processes. �� Add all waiting times. �� Divide by 4.

✅ Correct Answer Explanation

Average Waiting Time: Hence option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect arithmetic averaging.
�� Option C → Represents total waiting time only.
�� Option D → Incorrect division result.

🎯 Strategy

Used: Substitution

Application: Apply average formula directly.

Final Logic: Sum divided by process count gives 4.25.

🧠 Memory Trick

"Average = Total ÷ Count."

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

9 What is the probabilistic formula linking Completion Time (CT), Arrival Time (AT), and Turnaround Time (TAT)?

A AT = CT + TAT

B TAT = CT - AT

C CT = TAT - AT

D TAT = AT / CT

💡 Short Explanation

�� TAT measures total process duration. �� It starts at arrival. �� It ends at completion.

✅ Correct Answer Explanation

Turnaround Time is defined as total elapsed time from arrival to completion. Formula: Thus option B is correct. Other options incorrectly rearrange or misuse variables.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect addition relation.
�� Option C → Wrong rearrangement.
�� Option D → Division has no scheduling meaning here.

🎯 Strategy

Used: Elimination

Application: Identify standard scheduling equation.

Final Logic: TAT equals completion minus arrival.

🧠 Memory Trick

"Completion minus Arrival = Turnaround."

📚 Class XII – Computer Science – Chapter: Scheduling Metrics – Topic: Turnaround Time

10 If the vector of execution is purely FCFS with processes P1(AT=0, BT=2), P2(AT=1, BT=2), P3(AT=5, BT=3). What is the CT of P3?

A 5

B 7

C 8

D 10

💡 Short Explanation

�� FCFS executes by arrival order. �� P1 finishes at 2. �� P3 arrives later at 5.

✅ Correct Answer Explanation

Execution sequence: • P1: 0 → 2 • P2: 2 → 4 • CPU idle: 4 → 5 • P3: 5 → 8 Therefore, Completion Time of P3 = 8. Hence option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Equals arrival time only.
�� Option B → Ignores full burst duration.
�� Option D → Adds unnecessary delay.

🎯 Strategy

Used: Substitution

Application: Construct FCFS execution timeline stepwise.

Final Logic: P3 starts at 5 and runs 3 units, finishing at 8.

🧠 Memory Trick

"FCFS follows arrival queue."

📚 Class XII – Computer Science – Chapter: FCFS Scheduling – Topic: Completion Time Calculation

11 If A runs 3 times as fast as B, and A gives B a head start of 40 metres, what must be the total linear area (length) of the racecourse so that A and B reach the goal at the exact same time?

A 120 metres

B 60 metres

C 80 metres

D 100 metres

💡 Short Explanation

�� Speed ratio A:B = 3:1 �� B gets 40 m head start �� Equal finishing time forms proportional distance relation

✅ Correct Answer Explanation

Let total race length be x metres. → A runs x metres while B runs only (x − 40) metres. → Since A is 3 times as fast as B: → Therefore, the total racecourse length is 60 metres. → Option B is correct. → Option A, C, and D do not satisfy the proportional speed condition.

❓ Why Other Options Are Incorrect

�� Option A → 120 m gives distance ratio 120:80 = 3:2, not 3:1.
�� Option C → 80 m gives ratio 80:40 = 2:1, incorrect.
�� Option D → 100 m gives ratio 100:60 = 5:3, incorrect.

🎯 Strategy

Used: Substitution

Application: Convert the race condition into a speed-distance proportional equation and solve algebraically.

Final Logic: Equal finishing time implies distance ratio equals speed ratio.

🧠 Memory Trick

"Same Time ⇒ Speed Ratio = Distance Ratio"

📚 Class XI – Mathematics – Chapter: Linear Inequalities – Topic: Ratio-Based Applications

12 Integrating the ratios in a 1000m race: A defeats B by 100m, and B defeats C by 100m. What is the integrated ratio of the distances covered by A, B, and C in the same amount of time?

A 100 : 90 : 80

B 10 : 9 : 8

C 100 : 90 : 81

D 1000 : 900 : 800

💡 Short Explanation

�� A:B = 1000:900 �� B:C = 1000:900 �� Multiply proportional relations carefully

✅ Correct Answer Explanation

If A defeats B by 100 m in a 1000 m race: → If B defeats C by 100 m: → Combine both: → Hence, Option C is correct. → Option A ignores compounded proportional effect. → Option B is incomplete scaling. → Option D incorrectly assumes linear subtraction.

❓ Why Other Options Are Incorrect

�� Option A → Uses simple subtraction instead of chained ratios.
�� Option B → Does not preserve the second proportional relation properly.
�� Option D → Treats differences directly rather than using ratios.

🎯 Strategy

Used: Substitution

Application: Convert race margins into speed ratios and integrate sequentially.

Final Logic: Combined race ratios multiply proportionally.

🧠 Memory Trick

"Race Defeat → Convert to Speed Ratio First"

📚 Class XI – Mathematics – Chapter: Ratio and Proportion Applications – Topic: Relative Speed Ratios

13 Based on the integrated ratio 100:90:81, by how many metres does A defeat C in a 1000m race?

A 200 metres

B 190 metres

C 181 metres

D 100 metres

💡 Short Explanation

�� Ratio A:C = 100:81 �� A finishes 1000 m first �� C covers proportional reduced distance

✅ Correct Answer Explanation

From the ratio: → When A runs 1000 m, C runs: → Therefore, A defeats C by: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Overestimates the margin.
�� Option C → Numerical confusion from ratio term 81.
�� Option D → Ignores compounded race ratio.

🎯 Strategy

Used: Substitution

Application: Use integrated ratio directly to compute remaining distance.

Final Logic: Distance lag = Total race distance − proportional covered distance.

🧠 Memory Trick

"100:81 ⇒ 19% behind"

📚 Class XI – Mathematics – Chapter: Ratio and Proportion – Topic: Competitive Race Ratios

14 In a 500m race, A defeats B by 60 metres OR 12 seconds. What is the time taken by A to complete the race?

A 88 seconds

B 100 seconds

C 112 seconds

D 76 seconds

💡 Short Explanation

�� B covers 60 m in 12 sec �� B's speed = 5 m/s �� Use race ratio for A's speed

✅ Correct Answer Explanation

B covers remaining 60 m in 12 sec: → When A finishes 500 m, B covers 440 m. → Therefore: → A's speed: → Time taken by A: → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Assumes incorrect equal speed relation.
�� Option C → Overestimates completion time.
�� Option D → Gives faster speed than ratio permits.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Convert metres and seconds into speed, then compute race completion time.

Final Logic: Use "remaining distance ÷ extra time" to derive speed.

🧠 Memory Trick

"Beat by metres + seconds ⇒ Find speed first"

📚 Class XI – Mathematics – Chapter: Motion and Relative Speed – Topic: Race Problems

15 A, B, and C enter a partnership. B contributes 1/3 of the capital. A contributes as much as B and C together. What is the ratio of their capitals (A:B:C)?

A 1:2:3

B 3:2:1

C 2:1:1

D 4:2:1

💡 Short Explanation

�� Let total capital = 1 �� B contributes 1/3 �� A = B + C

✅ Correct Answer Explanation

Let total capital be 1. → B contributes: → Remaining for A and C: → Given: → Since: then: Also: Substituting: Then: So: Multiply by 6: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Does not satisfy A = B + C.
�� Option C → Makes A equal to B + C false.
�� Option D → Produces incorrect total proportion.

🎯 Strategy

Used: Substitution

Application: Form equations from partnership conditions and simplify ratios.

Final Logic: Translate verbal relations into algebraic equations.

🧠 Memory Trick

"A equals B+C → Largest share"

📚 Class XI – Mathematics – Chapter: Ratio and Proportion – Topic: Partnership Applications

16 Priya invests Rs 40,000. After 4 months, Rekha joins with Rs 50,000. They earn Rs 220,000 profit at the end of the year. What is the exact profit share of Rekha?

A Rs 1,00,000

B Rs 1,20,000

C Rs 1,10,000

D Rs 80,000

💡 Short Explanation

�� Profit ∝ Capital × Time �� Priya invests for 12 months �� Rekha invests for 8 months

✅ Correct Answer Explanation

Priya's investment-time: → Rekha's investment-time: → Ratio: → Total parts: → Rekha's share: → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Assumes larger contribution than actual ratio.
�� Option C → Incorrect proportional division.
�� Option D → Underestimates Rekha's investment-time effect.

🎯 Strategy

Used: Substitution

Application: Use Capital × Time proportionality for partnership profit sharing.

Final Logic: Profit share follows effective investment ratio.

🧠 Memory Trick

"Profit ∝ Money × Months"

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

17 Evaluate the inequality: (y - 1) / 3 + (y - 5) / 5 > (y + 3) / 6 + (y + 2) / 2. Which numerical inequality operation must be carefully monitored if multiplying by the LCM across the inequality?

A Changing the inequality sign if the LCM is positive.

B Ensuring the LCM does not equal 1.

C The inequality sign remains unchanged because the LCM of strictly positive denominators is positive.

D Subtracting fractions flips the sign.

💡 Short Explanation

�� Multiplying by positive numbers preserves inequality �� Denominators are all positive �� No sign reversal occurs

✅ Correct Answer Explanation

In inequalities, multiplying or dividing by a positive number keeps the inequality direction unchanged. → Here denominators are: All are positive. Their LCM is also positive. → Therefore, multiplying throughout by the LCM does NOT reverse the sign. → Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Sign changes only for negative multiplication/division.
�� Option B → LCM being 1 is irrelevant.
�� Option D → Subtraction does not automatically flip inequality signs.

🎯 Strategy

Used: Elimination

Application: Remove options violating standard inequality rules.

Final Logic: Positive multiplication preserves inequality direction.

🧠 Memory Trick

"Positive keeps direction; negative flips."

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

18 If A = 2, B = 4, evaluate the quantities AB and A+B. Why can we NOT definitively establish a universal static inequality like AB > A+B for all positive integers?

A Because if A=1, B=4, AB (4) is less than A+B (5), but if A=2, B=4, AB (8) is greater than A+B (6).

B Because AB is always equal to A+B.

C Because inequalities only apply to fractions.

D Because B must be negative.

💡 Short Explanation

�� Product and sum vary with values �� No universal inequality exists �� Counterexamples disprove universal claims

✅ Correct Answer Explanation

For A=2 and B=4: So: → But for A=1 and B=4: So: → Since both outcomes occur, no universal inequality is always true. → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Product and sum are not always equal.
�� Option C → Inequalities apply broadly, not only to fractions.
�� Option D → B need not be negative.

🎯 Strategy

Used: Substitution

Application: Use numerical examples to test universal claims.

Final Logic: One counterexample disproves a universal inequality.

🧠 Memory Trick

"Counterexample breaks universality."

📚 Class XI – Mathematics – Chapter: Inequalities – Topic: Verification Using Examples

📄 Passage

In a race, competitors cover a specific distance on a racecourse. Certain terms define their relative performance. If 'A gives B a start of x metres', A runs the full distance y, while B runs (y - x). If 'A beats B by x metres and t minutes', it means while A reaches the finish, B is x metres behind and takes an additional t minutes to finish. Hence, B covers the remaining x metres in t minutes, making B's speed x/t. A 'Dead Heat' occurs when all participants finish exactly at the same time.

Question:
According to the passage, how is B's speed strictly derived if A beats B by 60 metres and 12 seconds?

A 60 * 12 m/s

B 60 / 12 m/s

C (60 + 12) / 2 m/s

D 12 / 60 m/s

💡 Short Explanation

�� Remaining distance = 60 m �� Extra time = 12 sec �� Speed = Distance ÷ Time

✅ Correct Answer Explanation

When A finishes, B is 60 m behind. → B covers this remaining distance in 12 seconds. Therefore: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Multiplies instead of dividing.
�� Option C → No averaging principle exists here.
�� Option D → Reverses speed formula.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Apply the standard speed formula carefully with correct units.

Final Logic: Speed always equals distance divided by time.

🧠 Memory Trick

"Speed = D/T"

📚 Class XI – Physics – Chapter: Motion in a Straight Line – Topic: Speed Formula

📄 Passage

Question:
Using the derived speed (5 m/s) from the passage scenario, how much time does B take to complete a full 500m racecourse?

A 88 seconds

B 100 seconds

C 120 seconds

D 90 seconds

💡 Short Explanation

�� Speed of B = 5 m/s �� Distance = 500 m �� Time = Distance ÷ Speed

✅ Correct Answer Explanation

Given: → Therefore: → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Corresponds to another race scenario.
�� Option C → Assumes slower speed.
�� Option D → Gives incorrect division result.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Use standard motion formula directly.

Final Logic: Time equals distance divided by speed.

🧠 Memory Trick

"T = D/S"

📚 Class XI – Physics – Chapter: Motion in a Straight Line – Topic: Speed, Distance and Time

CUET UG Applied Mathematics Booster Test 3 - Scheduling, Races, and Inequalities

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