07_time_speed_distance_work.md

Time, Speed & Distance / Time & Work

Section A · Futures First International — Market Trading Intern OA

📌 What Has Been Asked (OA & Rounds)

FormatTypical Question Style
OA (MCQ)"A train 300m long crosses a platform 200m long in 25 seconds. Find speed."
OA"Two trains running in opposite directions at 50 & 70 km/h cross each other in 12 sec. Sum of lengths?"
OA"A can do a job in 12 days, B in 18 days. Together, how many days?"
OA"Pipe A fills tank in 6 hrs, Pipe B empties in 8 hrs. Both open — how long to fill?"
OA"A man walks 4 km/h, reaches 5 min late. At 5 km/h, reaches 5 min early. Find distance."
Buzzer"60 km/h = ? m/s" → 16.67 → instant conversion

PART 1: TIME, SPEED & DISTANCE

🧠 Core Concepts

Speed = Distance / Time Distance = Speed × Time Time = Distance / Speed

Unit Conversions (Memorize Cold)

km/h → m/s: × 5/18 m/s → km/h: × 18/5
km/hm/s
3610
5415
7220
9025
10830

Key Formulas

1. Relative Speed:

Same direction: |S₁ − S₂| Opposite direction: S₁ + S₂

2. Trains Crossing:

Train crossing a pole: Time = Length of train / Speed Train crossing platform: Time = (Length of train + Length of platform) / Speed Two trains crossing: Time = (L₁ + L₂) / Relative speed

3. Average Speed (two trips):

If same distance traveled at speeds S₁ and S₂: Average speed = 2 × S₁ × S₂ / (S₁ + S₂) [Harmonic mean]

⚠️ NOT (S₁ + S₂)/2 — this is a classic trap!

4. Late/Early problems:

"At S₁, arrives t₁ late. At S₂, arrives t₂ early." Distance = S₁ × S₂ × (t₁ + t₂) / (S₂ − S₁)

Example: At 4 km/h, 5 min late. At 5 km/h, 5 min early. D = 4 × 5 × (10/60) / (5−4) = 20 × 1/6 = 10/3 km ≈ 3.33 km

Wait — let me redo: t₁ + t₂ = 10 min = 1/6 hr. D = (4×5×1/6)/(5−4) = 20/6 = 10/3 km

5. Boats & Streams:

Downstream speed = Boat + Stream Upstream speed = Boat − Stream Boat speed = (Downstream + Upstream) / 2 Stream speed = (Downstream − Upstream) / 2

PART 2: TIME & WORK

🧠 Core Concepts

If A does a job in 'a' days → A's 1-day work = 1/a If A and B work together → combined 1-day work = 1/a + 1/b Time together = ab / (a + b)

LCM Method (Fastest)

Instead of fractions, take LCM of days as "total work units":

A finishes in 12 days, B in 18 days. LCM(12,18) = 36 units of work. A's rate = 36/12 = 3 units/day B's rate = 36/18 = 2 units/day Together = 5 units/day → Time = 36/5 = 7.2 days

Pipes & Cisterns

Same as Time & Work, but:

  • Filling pipe = positive work
  • Emptying pipe = negative work
Pipe A fills in 6 hrs (rate = +1/6) Pipe B empties in 8 hrs (rate = −1/8) Together: 1/6 − 1/8 = (4−3)/24 = 1/24 Time to fill = 24 hrs

⚡ Speed Tricks

Trick 1: "Two trains" → just add lengths and use relative speed

Two trains of 200m and 300m, speeds 50 & 70 km/h, opposite direction:

Relative speed = 120 km/h = 120 × 5/18 = 100/3 m/s Time = 500 / (100/3) = 15 seconds

Trick 2: Average speed for equal distances

Going at 40, returning at 60:

Avg = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h

Trick 3: LCM method for Work problems

Always faster than fraction method. Just:

  1. LCM of days → total work
  2. Divide by each person's days → their rate
  3. Add rates → combined rate
  4. Total work / combined rate → answer

Trick 4: "A is twice as efficient as B"

If B takes 20 days → A takes 10 days → Together = 20×10/30 = 6.67 days

Trick 5: "Work-wage" proportionality

Wages divided in ratio of work done (= ratio of efficiencies, NOT time taken)

🎯 What Can Be Asked (Variations)

  1. "A car covers first half at 40 km/h and second half at 60 km/h. Average speed?" → 2×40×60/100 = 48 km/h

  2. "A and B together can do in 12 days. A alone in 20. B alone?" → 1/B = 1/12 − 1/20 = (5−3)/60 = 2/60 = 1/30 → B alone = 30 days

  3. "Pipe A fills in 4 hrs, B fills in 6 hrs, C empties in 12 hrs. All open, time to fill?" → 1/4 + 1/6 − 1/12 = (3+2−1)/12 = 4/12 = 1/3 → 3 hours

  4. "Two men and three boys can do in 10 days. Three men and two boys in 8 days. Five men and five boys?"

📝 Practice Questions (with Answers)

Q1. A train 250m long crosses a bridge 150m long in 20 seconds. Find speed in km/h.

A: Speed = (250+150)/20 = 400/20 = 20 m/s = 20 × 18/5 = 72 km/h

Q2. A man rows upstream at 6 km/h and downstream at 10 km/h. Find his speed in still water.

A: (10+6)/2 = 8 km/h

Q3. A can finish a job in 15 days, B in 10 days. They work together for 4 days. How much work is left?

A: LCM(15,10) = 30. A = 2/day, B = 3/day. Together = 5/day × 4 = 20 done. Left = 10/30 = 1/3

Q4. Two cars start from same point in opposite directions at 50 km/h and 60 km/h. After how many hours are they 330 km apart?

A: Relative speed = 110 km/h. Time = 330/110 = 3 hours

Q5. Pipe A fills in 10 min, B fills in 15 min, C empties in 30 min. All open, time to fill?

A: LCM = 30. Rates: A=3, B=2, C=−1. Net = 4/min. Time = 30/4 = 7.5 min

Q6. A goes from X to Y at 40 km/h and returns at 60 km/h. Total distance is 240 km. Total time?

A: 120/40 + 120/60 = 3 + 2 = 5 hours

🔑 Exam Strategy

  • Convert km/h ↔ m/s instinctively — memorize the table above
  • For work problems, always use LCM method — no fractions, no errors
  • Average speed for equal distances = harmonic mean (NOT arithmetic mean) — common trap
  • Train problems: just sum lengths + use relative speed — done in 15 seconds
  • Pipes & Cisterns = Time & Work with negatives — same logic