Time, Work, Speed, Distance and Clocks Practice Questions for XAT: 183+ Solved Questions with Step-by-Step Solutions

Solve 183+ Time, Work, Speed, Distance and Clocks practice 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

A person travels from city A to city B. If he travels the entire distance at 40 km/h, he reaches 1 hour late. If he travels the entire distance at 60 km/h, he reaches 1 hour early. He starts at his usual scheduled time. He travels at 40 km/h for the first 1.5 hours. Then, he increases his speed by 50% for the next 1 hour. After that, he takes a 30-minute stoppage. To reach city B exactly at his scheduled time, what must be his speed for the remaining distance?

A.
50 km/h
B.
70 km/h
C.
80 km/h
D.
60 km/h

Correct Answer:

Step-by-Step Solution

Key idea: This is an early/late framework problem combined with piecewise motion and a stoppage. You must first find the total distance and scheduled time, then calculate the remaining distance and time to find the required speed.

Step 1: Find total distance (D) and scheduled time (T).

Let D be the distance.

At 40 km/h, time = D/40 = T + 1.

At 60 km/h, time = D/60 = T - 1.

Subtract the two equations: D/40 - D/60 = 2.

(3D - 2D) / 120 = 2 => D / 120 = 2 => D = 240 km.

Substitute D back: 240 / 40 = T + 1 => 6 = T + 1 => T = 5 hours.

Step 2: Calculate distance and time for the first two segments.

Segment 1: 1.5 hours at 40 km/h. Distance = 1.5 * 40 = 60 km.

Segment 2: Speed increases by 50% -> 40 * 1.5 = 60 km/h.

Time = 1 hour. Distance = 1 * 60 = 60 km.

Total distance covered so far = 60 + 60 = 120 km.

Total time used so far = 1.5 + 1 = 2.5 hours.

Step 3: Account for the stoppage and find remaining requirements.

Stoppage = 30 minutes = 0.5 hours.

Total time used including stoppage = 2.5 + 0.5 = 3 hours.

Time remaining to reach on time = T - 3 = 5 - 3 = 2 hours.

Distance remaining = D - 120 = 240 - 120 = 120 km.

Step 4: Calculate required speed for the remaining distance.

Required speed = Distance remaining / Time remaining = 120 / 2 = 60 km/h.

Answer: 60 km/h

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

Two runners, A and B, start from the same point on a circular track of circumference 1200 meters and run in opposite directions. A's speed is 5 m/s, and B's speed is 7 m/s. Every time they meet, they must stop and rest. B always rests for exactly 18 seconds. A rests for 12 seconds during their first meeting, and his rest time increases by 2 seconds for each subsequent meeting. They resume running only when both have finished resting. How many times do they meet within the first 3600 seconds?

A.
24
B.
28
C.
26
D.
30

Correct Answer:

Step-by-Step Solution

Key idea: This is a circular motion problem with an arithmetic progression in rest times. The key is to calculate the total time for each cycle (run + rest) and sum them up until the total exceeds 3600 seconds.

Step 1: Calculate the time to run between meetings. Relative speed = 5 + 7 = 12 m/s. Distance = 1200 m. Time to run = 1200 / 12 = 100 seconds.

Step 2: Determine the rest time for each cycle. They resume when both are ready, so the rest time for the cycle is the maximum of A's and B's rest times. B always rests 18s. A rests 12, 14, 16, 18, 20...

Step 3: Calculate cycle times.

Cycles 1 to 4: A's rest is 12, 14, 16, 18. Max rest is always 18s (since B rests 18s). Total time per cycle = 100 + 18 = 118s.

Total time for 4 meetings = 4 * 118 = 472 seconds.

Step 4: For cycle $k$ (where $k \geq 5$ ), A's rest is $12 + 2 (k - 1) = 10 + 2 k$ . Since $k \geq 5$ , A's rest is $\geq 20$ , which is greater than B's 18s. So the max rest is $10 + 2 k$ .

Total time for cycle $k$ = 100 + 10 + 2k = 110 + 2k.

Step 5: Sum the times for $n$ meetings ( $n > 4$ ).

Total time = 472 + $\sum_{k = 5}^{n} (110 + 2 k)$

= 472 + 110(n - 4) + 2 * $\frac{( n + 5 ) ( n - 4 )}{2}$

= 472 + 110n - 440 + n^2 + n - 20

= n^2 + 111n + 12.

Step 6: Find the maximum $n$ such that total time $\leq 3600$ .

n^2 + 111n + 12 \le 3600

n^2 + 111n - 3588 \le 0

For n = 26: 26^2 + 111(26) + 12 = 676 + 2886 + 12 = 3574 \le 3600.

For n = 27: 27^2 + 111(27) + 12 = 729 + 2997 + 12 = 3738 > 3600.

Answer: 26

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

Which of the following is the correct rearrangement of the fundamental formula $Distance = Speed \times Time$ to solve for Time?

A.
$e x t T im e = e x t D i s t an ce im ese x t S p ee d$
B.
$ext{Time} = rac{ ext{Speed}}{ ext{Distance}}$
C.
$e x t T im e = e x t D i s t an ce + e x t S p ee d$
D.
$ext{Time} = rac{ ext{Distance}}{ ext{Speed}}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct recall question testing algebraic manipulation of the core equation.

Step 1: Start with the fundamental formula: $Distance = Speed \times Time$ ( $D = S \times T$ ).

Step 2: To isolate Time ( $T$ ), divide both sides of the equation by Speed ( $S$ ).

Step 3: This yields: $\frac{D}{S} = T$ , or $Time = \frac{Distance}{Speed}$ .

Answer: $Time = \frac{Distance}{Speed}$

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

A and B can complete a work in 20 hours and 30 hours respectively. They start working together at 12:00 PM on a faulty clock that gains 12 minutes every hour. After 4 true hours, A leaves. B continues alone. However, after 8 true hours of total work (from the start), B's efficiency drops by 20%. They work until the task is finished. What time does the faulty clock show when the work is completed?

A.
9:24 PM
B.
9:48 PM
C.
9:36 PM
D.
10:00 PM

Correct Answer:

Step-by-Step Solution

Key idea: This is a synthesis of work rates, staggered efficiency, and faulty clock time mapping. You must calculate true time elapsed, then map it to the faulty clock's time.

Step 1: Calculate work done in the first phase (0 to 4 true hours).

A's rate = 1/20, B's rate = 1/30. Combined rate = 1/20 + 1/30 = 5/60 = 1/12 per hour.

Work done in 4 hours = 4 * (1/12) = 1/3.

Remaining work = 1 - 1/3 = 2/3.

Step 2: Calculate work done in the second phase (4 to 8 true hours).

A has left. B works alone at 1/30 per hour.

Duration = 8 - 4 = 4 hours.

Work done = 4 * (1/30) = 4/30 = 2/15.

Remaining work = 2/3 - 2/15 = 10/15 - 2/15 = 8/15.

Step 3: Calculate work done in the third phase (after 8 true hours).

B's efficiency drops by 20%. New rate = (1/30) * 0.8 = 1/37.5 = 2/75 per hour.

Time to finish remaining 8/15 work = (8/15) / (2/75) = (8/15) * (75/2) = 20 hours.

Step 4: Calculate total true time elapsed.

Total true time = 4 + 4 + 20 = 28 hours.

Step 5: Map true time to faulty clock time.

The clock gains 12 minutes every hour, meaning it covers 72 minutes for every 60 true minutes.

Ratio = 72 / 60 = 1.2.

Time on faulty clock = 28 * 1.2 = 33.6 hours = 33 hours and 36 minutes.

Step 6: Determine the final clock time.

Start time = 12:00 PM.

Add 24 hours -> 12:00 PM (next day).

Add remaining 9 hours 36 minutes -> 9:36 PM.

Answer: 9:36 PM

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

An inter-city express train is scheduled to cover a distance of 480 km in exactly 8 hours. Due to a signaling malfunction, the train is forced to reduce its speed to 75% of its scheduled speed for the first 120 km. After clearing the affected zone, the train accelerates to a new constant speed for the remaining journey so as to arrive exactly on schedule.

What is the percentage increase in speed required for the remaining journey compared to the original scheduled speed?

Correct Answer:

12.5

Step-by-Step Solution

Key idea: This is a Flight Delay / Partial Journey Recovery problem. The core task is balancing lost time in one segment with gained speed in another to maintain total time.

Step 1: Determine Scheduled Parameters.

Total Distance $D = 480$ km.

Scheduled Time $T = 8$ hrs.

Scheduled Speed $S = 480/8 = 60$ km/h.

Step 2: Analyze the Affected Segment.

Distance $d_{1} = 120$ km.

Reduced Speed $S_{1} = 0.75 \times 60 = 45$ km/h.

Time Taken $t_{1} = 120/45 = 8/3$ hours ( $2$ hrs $40$ mins).

Step 3: Calculate Requirements for Remaining Segment.

Remaining Distance $d_{2} = 480 - 120 = 360$ km.

Scheduled Time for $d_{2}$ would be $360/60 = 6$ hrs.

Available Time $t_{2} = T o t a l T im e - t_{1} = 8 - 8/3 = 16/3$ hours.

Step 4: Find New Speed and Percentage Increase.

Required Speed $S_{2} = d_{2} / t_{2} = 360/ (16/3) = (360 \times 3) /16 = 1080/16 = 67.5$ km/h.

Percentage Increase:

$\frac{S _{2} - S}{S} \times 100 = \frac{67.5 - 60}{60} \times 100$

$= \frac{7.5}{60} \times 100 = \frac{1}{8} \times 100 = 12.5%$

Answer: 12.5

Time, Work, Speed, Distance and Clocks Practice Questions for XAT: 183+ 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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