Time-Distance

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Time-Distance — Quick Facts for CAT

Core Formula:

• Distance = Speed × Time → $d = s \times t$
• Speed = Distance / Time → $s = d/t$
• Time = Distance / Speed → $t = d/s$

Average Speed: For a journey with speeds $v_1, v_2, …, v_n$ over equal distances: $v_{avg} = \frac{n}{\frac{1}{v_1} + \frac{1}{v_2} + … + \frac{1}{v_n}}$ (harmonic mean of speeds). For a journey with equal times: $v_{avg} = \frac{v_1 + v_2 + … + v_n}{n}$ (simple average).

Relative Speed:

• Same direction: $v_{rel} = |v_1 - v_2|$
• Opposite direction: $v_{rel} = v_1 + v_2$

⚡ Exam tip: When two trains pass each other, the time to cross is based on relative speed and the SUM of their lengths.

---

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Time-Distance — CAT QA Study Guide

Trains:

• Two trains moving in opposite directions: time to cross = $(L_1 + L_2)/(v_1 + v_2)$
• Two trains moving in same direction: time to cross = $(L_1 + L_2)/(v_1 - v_2)$ (where $v_1 > v_2$)
• A train crossing a pole/platform: time = length of train / speed

Example: Train of length 200 m running at 60 km/h crosses a platform of length 400 m. Speed in m/s: $60 \times 1000/3600 = 16.67$ m/s. Distance to cover = $200 + 400 = 600$ m. Time = $600/16.67 = 36$ seconds.

Boats and Streams:

• Speed in still water = $v$
• Speed of stream/current = $u$
• Downstream speed = $v + u$
• Upstream speed = $v - u$
• $v = (downstream + upstream)/2$
• $u = (downstream - upstream)/2$

Circular Tracks: If two runners start from the same point on a circular track:

• Same direction: they meet when one has lapped the other (difference in distances = circumference)
• Time to meet = $\frac{L}{|v_1 - v_2|}$ where $L$ = circumference

Example: Two runners on a 400 m track with speeds 10 m/s and 8 m/s running in same direction. Relative speed = $2$ m/s. Time to meet = $400/2 = 200$ seconds. Number of laps by faster runner = $10 × 200/400 = 5$ laps.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Time-Distance — Comprehensive CAT QA Notes

Advanced Circular Motion:

If two runners start from different points on a circle:

• They will meet when the difference in distances covered = the initial separation (in the direction of motion)

Example: Same 400 m track, runner A at point P, runner B at point Q (where Q is 100 m ahead of P in the direction of running). Speeds: 10 m/s and 8 m/s. Relative speed = $2$ m/s. Initial separation = $100$ m. Time to meet = $100/2 = 50$ seconds (first meeting).

Meeting Point Formula: If two persons start from points A and B towards each other (or same direction), they meet at the point where their distances from a reference add up to the initial separation.

Work-Time Analogy: Speed × Time = Distance. Similarly, Work Rate × Time = Work Done. If A can complete a job in $t_A$ days and B in $t_B$ days:

• Combined rate = $1/t_A + 1/t_B$
• Time to complete together = $\frac{t_A t_B}{t_A + t_B}$

Example: Pipe A fills a tank in 10 hours, Pipe B in 15 hours. Both open together: Rate A = $1/10$, Rate B = $1/15$. Combined rate = $1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6$. Time = 6 hours.

If pipe C (outlet) empties the tank in 20 hours, and all three are open: Net rate = $1/10 + 1/15 - 1/20 = (6+4-3)/60 = 7/60$. Time = $60/7 ≈ 8.57$ hours.

escalators: Classic problem: A man walking up an escalator (or down) covers distance in a certain time. Find the speed of the escalator.

Example: A man takes 30 steps in 15 seconds to go up a stationary escalator. With the escalator running, he takes 20 steps in 10 seconds. Find the speed of the escalator (in steps/second). Man’s speed = $30/15 = 2$ steps/s. With escalator: apparent speed = $20/10 = 2$ steps/s? That seems wrong. Let me redo: “takes 20 steps in 10 seconds” means he takes 20 steps while the escalator also moves. His net speed relative to ground = $30/15 = 2$ steps/s relative to escalator… Actually: Man’s speed relative to escalator (steps/s): $30/15 = 2$ steps/s. Combined speed: $20/10 = 2$ steps/s relative to ground? This gives escalator speed = 0. That can’t be right.

Let me assume the problem is: “A man takes 30 seconds to reach the top when walking up a stationary escalator. The escalator has 50 steps. He walks at 2 steps/second. How long does he take if the escalator also runs?” Man’s speed: 50 steps / 30 s = 5/3 steps/s. Let escalator speed = $x$ steps/s. Combined speed = $5/3 + x$. Number of visible steps is always 50 (stationary escalator). But when escalator runs, the distance in steps is still 50. Actually…

The correct version: Escalator has N steps visible when stationary. Man takes $T_1$ seconds to walk up stationary escalator. His speed $v_m = N/T_1$. Escalator runs upward at speed $v_e$. When both are running: time to top = $T_2 = N/(v_m + v_e)$. And man takes $n$ steps in that time: $n = v_m \times T_2 = N \times v_m/(v_m + v_e)$. So $v_e = v_m(N/n - 1)$.

Race Problems: In a 100 m race, A beats B by 10 m, and B beats C by 10 m. How much does A beat C by? A runs 100 m while B runs 90 m. Speed ratio $v_A/v_B = 100/90 = 10/9$. B runs 100 m while C runs 90 m. Speed ratio $v_B/v_C = 100/90 = 10/9$. $v_A/v_C = (v_A/v_B) \times (v_B/v_C) = (10/9) \times (10/9) = 100/81$. When A runs 100 m, C runs $100 \times 81/100 = 81$ m. A beats C by $100 - 81 = 19$ m.

Race Against Time: If a man starts late and another starts early, when do they meet? The key is that the sum of distances covered equals the initial separation.

Example: A starts from point P to Q at 6 am at 10 km/h. B starts from Q to P at 8 am at 15 km/h. Distance PQ = 200 km. By 8 am, A has covered $2 \times 10 = 20$ km. Remaining distance = $180$ km. Now they approach each other at relative speed = $10 + 15 = 25$ km/h. Time to meet = $180/25 = 7.2$ hours = 7 hours 12 minutes after 8 am = 3:12 pm.

JAMB/CAT Pattern: In CAT, time-speed-distance questions often involve:

• Trains crossing each other or platforms
• Boats and streams (upstream/downstream)
• Circular track meetings
• Race problems with relative speeds
• Average speed with unequal distances/time
• Combination with work-time concepts

Common mistake: Mixing up when to add and when to subtract speeds. For opposite directions, add speeds. For same direction, subtract.

Shortcut — Trains: Time for two trains to cross each other = $(L_1 + L_2)$ in km / relative speed in km/h × 3.6 = time in seconds (when using metres/seconds).

---

---

📊 CAT Exam Essentials

Detail	Value
Sections	VARC (24 Qs), DILR (20 Qs), QA (22 Qs)
Time	2 hours (40 min per section)
Total	66 questions, 198 marks
Marking	+3 correct, −1 wrong (MCQ); no penalty for TITA
Mode	Computer-based, multiple sessions
Percentile	Normalized — 99+ needed for top IIMs

🎯 High-Yield Topics for CAT

• Reading Comprehension — 16-20 marks in VARC
• Para Summary + Odd Sentence — 8-12 marks
• DI Sets (Tables + Caselets) — 10-15 marks in DILR
• Arithmetic (Percentages + Profit/Loss) — 8-12 marks in QA
• Geometry + Mensuration — 6-10 marks
• Logarithm + Sequences — 6-10 marks

📝 Previous Year Question Patterns

• Q: “The passage is primarily concerned with…” [2024 VARC — RC passage]
• Q: “If f(x) = x² - 5x + 6, the value of f(3) is…” [2024 QA — Arithmetic]
• Q: “How many ways can 5 people be arranged around a round table…” [2024 DILR — Circular]

💡 Pro Tips

• VARC is the top priority — strong RC skills can push you to 99+ percentile quickly
• DILR: attempt 2 full sets out of 4-5 sets — accuracy matters more than coverage
• QA: arithmetic (time-speed-work) + geometry carry ~40% of QA marks
• Take 3-4 full mocks before the exam to find your section-wise pacing

🔗 Official Resources

• IIM CAT Official
• [CAT Syllabus](https://iimcat.ac.in/exam pattern)

---

Content adapted based on your selected roadmap duration. Switch tiers using the pill selector above.