Time, Work, Speed, Distance and Clocks Previous Year Questions (PYQs) for XAT: 7+ Solved Questions with Step-by-Step Solutions

Solve 7+ Time, Work, Speed, Distance and Clocks previous year questions for XAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Time, Work, Speed, Distance and Clocks

Chapter Roadmap

Time, Work, Speed, Distance and Clocks

Topic 1: Speed, Distance & Travel Problems (You are here)

PYQ Weight: 6 questions | Importance: High

What you will master:

Core speed-distance-time relationships and proportionality
Average speed for equal and unequal distances
Relative speed and meeting point problems
Early/late problems with speed changes
Network and shortest-time path problems

Topic 2: Circular Motion, Work & Clocks

PYQ Weight: 5 questions | Importance: Moderate-High

What you will master:

Circular track meetings (same and opposite directions)
Clock hand angles and time calculations
Work-rate and efficiency problems

By the end of this chapter, you will be able to solve any speed, distance, time, work, or clock problem.

Speed, Distance & Travel Problems: The Big Picture

Speed, distance, and time are three sides of the same coin. They are locked together by one simple relationship:

Distance = Speed \times Time

The exam tests whether you can:

See the relationships when one quantity changes
Set up clean equations from word problems
Choose the right approach for different problem types

Flights & Delays
Speed changes to recover lost time

Buses & Trains
Objects moving toward or away

Paths & Networks
Choosing the fastest route

Average Speed
Speed changes during a journey

Time, Work, Speed, Distance and Clocks: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

The swimming pool in an exclusive club is a circular strip. A water flow is artificially maintained in the strip, along a clockwise direction, at a constant speed of 3km/h(kilometers/hour). One morning, Ayub and Rana start swimming at the same time from two diametrically opposite points on the strip. Ayub swims in a counterclockwise direction, while Rana swims in a clockwise direction. They swim at constant, but different, speeds. When they meet for the first time, Ayub has covered 60m(meters). They meet again after Rana has covered 180m from the first meeting point.

If Rana swims at the speed of 3 km/h in still water, how long does Ayub take to complete one round of the circular strip, if he swims in the clockwise direction?

A.
4 minutes 30 seconds
B.
2 minutes 15 seconds
C.
None of the other options is correct
D.
1 minute 48 seconds
E.
54 seconds

Correct Answer:

Step-by-Step Solution

Key idea: This is a circular motion problem with a current, requiring careful handling of effective speeds and relative distances.

Step 1: Convert all speeds to consistent units. Water flow speed = 3 km/h = 50 m/min.

Step 2: Rana's speed in still water is 3 km/h = 50 m/min. Since Rana swims clockwise (with the flow), Rana's effective speed is $50 + 50 = 100$ m/min.

Step 3: Let Ayub's speed in still water be $V_{A}$ km/h. Ayub swims counterclockwise (against the flow), so Ayub's effective speed is $V_{A} - 3$ km/h.

Step 4: They start from diametrically opposite points, so the initial distance between them is $\frac{C}{2}$ , where $C$ is the circumference.

Step 5: At the first meeting, Ayub covers 60 m. The time taken is $t_{1} = \frac{60}{v _{A}}$ , where $v_{A}$ is Ayub's effective speed in m/min.

Step 6: In this same time, Rana covers $100 \times t_{1} = \frac{6000}{v _{A}}$ meters.

Step 7: The sum of their distances is $\frac{C}{2}$ , so $60 + \frac{6000}{v _{A}} = \frac{C}{2}$ .

Step 8: They meet again after Rana covers 180 m. The time for this second leg is $t_{2} = \frac{180}{100} = 1.8$ minutes.

Step 9: In this time, together they cover the full circumference $C$ . So, $C = (v_{A} + 100) \times 1.8$ .

Step 10: Equate the two expressions for $\frac{C}{2}$ : $60 + \frac{6000}{v _{A}} = (v_{A} + 100) \times 0.9$ .

Step 11: Solving this yields $v_{A} = \frac{50}{3}$ m/min (which corresponds to $V_{A} = 7$ km/h).

Step 12: Substitute $v_{A}$ back to find $C$ : $C = (\frac{50}{3} + 100) \times 1.8 = \frac{350}{3} \times \frac{9}{5} = 210$ m. Wait, $C /2 = 60 + 360 = 420 ⟹ C = 840$ m. Let's re-verify: $C = 1.8 \times (350/3) = 210$ ? No, $1.8 \times 350/3 = 0.6 \times 350 = 210$ . But $C /2 = 420 ⟹ C = 840$ .

Correction: The relative speed is the sum of their still water speeds! $V_{r e l} = V_{A} + V_{R} = V_{A} + 3$ km/h.

Let's use km/h. $t_{1} = \frac{0.06}{V _{A} - 3}$ . Rana covers $6 \times t_{1} = \frac{0.36}{V _{A} - 3}$ .

$\frac{C}{2} = 0.06 + \frac{0.36}{V _{A} - 3}$ .

$t_{2} = \frac{0.18}{6} = 0.03$ hours. $C = (V_{A} + 3) \times 0.03$ .

$0.06 + \frac{0.36}{V _{A} - 3} = (V_{A} + 3) \times 0.015 ⟹ 20 + \frac{120}{V _{A} - 3} = 5 (V_{A} + 3)$ .

Let $x = V_{A} - 3$ . $20 x + 120 = 5 x (x + 6) ⟹ 5 x^{2} + 10 x - 120 = 0 ⟹ x^{2} + 2 x - 24 = 0 ⟹ x = 4$ .

So $V_{A} = 7$ km/h.

Step 13: If Ayub swims clockwise, his effective speed is $V_{A} + 3 = 10$ km/h = $\frac{500}{3}$ m/min.

Step 14: $C = 0.03 \times 10 = 0.3$ km = 300 m.

Step 15: The time to complete one round is $\frac{300}{500/3} = 1.8$ minutes, which is 1 minute 48 seconds.

Answer: 1 minute 48 seconds

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Some members of a social service organization in Kolkata decide to prepare 2400 laddoos to gift to children in various orphanages and slums in the city, during Durga puja. The plan is that each of them makes the same number of laddoos. However, on the laddoo-making day, ten members are absent, thus each remaining member makes 12 laddoos more than earlier decided.

How many members actually make the laddoos?

A.
100
B.
50
C.
90
D.
24
E.
40

Correct Answer:

Step-by-Step Solution

Key idea: This is a work-rate problem involving inverse proportionality between the number of workers and the work done per worker.

Step 1: Let the initial number of members be $N$ . The total work is 2400 laddoos.

Step 2: Initially, each member makes $\frac{2400}{N}$ laddoos.

Step 3: On the day, 10 members are absent, so the number of working members is $N - 10$ .

Step 4: Each remaining member makes $\frac{2400}{N - 10}$ laddoos.

Step 5: The problem states that each remaining member makes 12 laddoos more than planned. So, $\frac{2400}{N - 10} - \frac{2400}{N} = 12$ .

Step 6: Simplify the equation: $2400 (\frac{1}{N - 10} - \frac{1}{N}) = 12 ⟹ 200 (\frac{N - ( N - 10 )}{N ( N - 10 )}) = 1$ .

Step 7: $200 \times \frac{10}{N ^{2} - 10 N} = 1 ⟹ 2000 = N^{2} - 10 N ⟹ N^{2} - 10 N - 2000 = 0$ .

Step 8: Factor the quadratic: $(N - 50) (N + 40) = 0$ . Since $N$ must be positive, $N = 50$ .

Step 9: The question asks for the number of members who actually make the laddoos, which is $N - 10 = 50 - 10 = 40$ .

Answer: 40

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

A ight, traveling to a destination 11,200 kms away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the ight got delayed by three hours. The pilot increased the average speed of the airplane by 100 km/hr from the initially planned average speed, to reduce the overall delay to one hour.
Had the pilot increased the average speed by 350 km/hr from the initially planned average speed, when would have the ight reached its destination?

A.
11:30 PM
B.
7:50 PM
C.
5:10 PM
D.
8:10 PM
E.
10:36 PM

Correct Answer:

Step-by-Step Solution

Key idea: This is a time-speed-distance problem involving a change in speed to compensate for a departure delay.

Step 1: Let the initially planned average speed be $S$ km/hr. The planned travel time is $\frac{11200}{S}$ hours.

Step 2: The flight departs 3 hours late but arrives with an overall delay of only 1 hour. This means the actual travel time is $3 - 1 = 2$ hours less than the planned time.

Step 3: The new speed is $S + 100$ km/hr. So, the equation is $\frac{11200}{S + 100} = \frac{11200}{S} - 2$ .

Step 4: Simplify the equation: $\frac{5600}{S} - \frac{5600}{S + 100} = 1 ⟹ 5600 (100) = S (S + 100) ⟹ S^{2} + 100 S - 560000 = 0$ .

Step 5: Factoring gives $(S - 700) (S + 800) = 0$ . Since speed must be positive, $S = 700$ km/hr.

Step 6: If the speed is increased by 350 km/hr, the new speed is $700 + 350 = 1050$ km/hr.

Step 7: The travel time at this new speed is $\frac{11200}{1050} = \frac{32}{3}$ hours = 10 hours 40 minutes.

Step 8: The planned departure was 6:30 AM. The actual departure is 3 hours later, at 9:30 AM.

Step 9: Adding 10 hours 40 minutes to 9:30 AM gives an arrival time of 8:10 PM.

Answer: 8:10 PM

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

There are five dustbins along a circular path at different places. Ramesh takes multiple rounds of the path every morning, always at the same speed. He noticed that it took him a different number of steps to walk between any two consecutive dustbins. Ramesh also noticed that starting from any of the dustbins, it took a minimum 800 steps to reach every second dustbin. On the other hand, starting from any of the dustbins, it took a maximum 1260 steps to reach every third dustbin.

If Ramesh’s one step is 0.77 metre, and the width of the path is negligible, which the following can be the radius of the circular path?

A.
230 metres
B.
240 metres
C.
260 metres
D.
250 metres
E.
220 metres

Correct Answer:

Step-by-Step Solution

Key idea: This is a circular motion problem involving the complementary sums of segments on a circular track.

Step 1: Understand the segments. There are 5 dustbins, creating 5 segments between consecutive dustbins. Let their lengths in steps be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ .

Step 2: The total number of steps in one full round is $S = x_{1} + x_{2} + x_{3} + x_{4} + x_{5}$ .

Step 3: The distances between "every second dustbin" are the sums of 2 consecutive segments. There are 5 such sums.

Step 4: The distances between "every third dustbin" are the sums of 3 consecutive segments. There are 5 such sums.

Step 5: Notice that the sum of 2 consecutive segments and the sum of the remaining 3 consecutive segments always add up to $S$ . Thus, the set of 3-segment sums is exactly the set of complements of the 2-segment sums.

Step 6: Therefore, the maximum of the 3-segment sums is equal to $S$ minus the minimum of the 2-segment sums.

Step 7: We are given the minimum of the 2-segment sums is 800, and the maximum of the 3-segment sums is 1260.

Step 8: So, $1260 = S - 800 ⟹ S = 2060$ steps.

Step 9: The total distance of the circular path is $2060 \times 0.77 = 1586.2$ metres.

Step 10: The circumference is $2 π R = 1586.2$ . Using $π \approx 22/7$ , $R = \frac{1586.2 \times 7}{44} \approx 252.4$ metres.

Step 11: The closest option to 252.4 metres is 250 metres.

Answer: 250 metres

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Two buses travel between Jamshedpur and Kolkata in the opposite directions, on the same road. On that road, the maximum allowed speeds are different(but constant) for the opposite directions. Usually, both buses travel at the respective maximum allowed speeds to their respective destinations: the bus from Jamshedpur to Kolkata takes 4 hours, while the bus from Kolkata to Jamshedpur takes 3 hours.

One day, the two buses start at the same time. However, one hour after starting, the bus from Jamshedpur to Kolkata reduces its speed to half of its maximum allowed speed due to congestion on the road.

If both buses do not stop anywhere in between, how many hours after starting do they meet?

A.
2 hours
B.
23/12 hours
C.
12/5 hours
D.
21/11 hours
E.
10/11 hours

Correct Answer:

Step-by-Step Solution

Key idea: This is a relative speed and meeting point problem with a mid-journey speed change.

Step 1: Let the distance between Jamshedpur and Kolkata be $D$ .

Step 2: The speed of the bus from Jamshedpur to Kolkata (J to K) is $\frac{D}{4}$ . The speed of the bus from Kolkata to Jamshedpur (K to J) is $\frac{D}{3}$ .

Step 3: After 1 hour, the J to K bus has covered $\frac{D}{4}$ , and the K to J bus has covered $\frac{D}{3}$ .

Step 4: The remaining distance between them is $D - \frac{D}{4} - \frac{D}{3} = D - \frac{7 D}{12} = \frac{5 D}{12}$ .

Step 5: After 1 hour, the J to K bus reduces its speed to half, so its new speed is $\frac{1}{2} \times \frac{D}{4} = \frac{D}{8}$ .

Step 6: The K to J bus maintains its speed of $\frac{D}{3}$ . Their relative speed is now $\frac{D}{8} + \frac{D}{3} = \frac{11 D}{24}$ .

Step 7: The time taken to cover the remaining distance is $\frac{Remaining Distance}{Relative Speed} = \frac{5 D /12}{11 D /24} = \frac{5}{12} \times \frac{24}{11} = \frac{10}{11}$ hours.

Step 8: The total time after starting is the initial 1 hour plus the additional $\frac{10}{11}$ hours, which equals $1 + \frac{10}{11} = \frac{21}{11}$ hours.

Answer: 21/11 hours

Time, Work, Speed, Distance and Clocks Previous Year Questions (PYQs) for XAT: 7+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Time, Work, Speed, Distance and Clocks

Time, Work, Speed, Distance and Clocks

Speed, Distance & Travel Problems: The Big Picture

Speed, Distance & Travel Problems: The Big Picture

Time, Work, Speed, Distance and Clocks: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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