Time, Work, Speed, Distance and Clocks Short Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Time, Work, Speed, Distance and Clocks short notes for XAT: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Summary: The Speed-Distance-Time Toolkit

Fundamental:

D = S \times T

| Proportionality:

S \propto \frac{1}{T}

(constant D)

Average Speed: Equal distances

\to \frac{2 ab}{a + b}

. General

\to \frac{Total D}{Total T}

Relative Speed: Opposite

\to S_{1} + S_{2}

. Same

\to ∣ S_{1} - S_{2} ∣

Meeting: Time

= \frac{Gap}{Rel. Speed}

. Early/Late: Two equations, solve simultaneously.

Memory Hook

Distance equals speed times time. When one is constant, the other two are locked together. Master the relationships, and every problem becomes straightforward.

Summary: The Circular, Work & Clock Toolkit

Circular Meeting Anywhere: $T = \frac{C}{S _{relative}}$
Circular Meeting at Start: $T = LCM (T_{1}, T_{2})$
Distinct Meeting Points: Opposite $= a + b$ , Same $= ∣ a - b ∣$
Work LCM Method: Total Work = LCM of times. Rate = Total Work / Time.
Efficiency: Inverse of time ratio.
Universal Work Formula: $\frac{M _{1} D _{1} H _{1}}{W _{1}} = \frac{M _{2} D _{2} H _{2}}{W _{2}}$
Pipes: Inlet = Positive, Outlet = Negative.
Clock Speeds: Minute = $6^{\circ}$ /min, Hour = $0. 5^{\circ}$ /min, Relative = $5. 5^{\circ}$ /min.
Clock Angle: $∥30 H - 5.5 M ∥$ (Take acute angle).

Memory Hook: Circular motion is just straight lines wrapping around. Work is just rates and LCMs. Clocks are just circular tracks with fixed speeds. Map the cycles, and the math will follow.

Time, Work, Speed, Distance and Clocks: Solved Questions with Step-by-Step Explanations (2 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

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Time, Work, Speed, Distance and Clocks Short Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Time, Work, Speed, Distance and Clocks short notes for XAT: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice question

A question from this chapter

Question 1

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?

Question 2

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?

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