Time and Distance

1. Core Concepts & Theoretical Blueprint

Time, Speed, and Distance are governed by a single proportionality axiom: distance covered is directly proportional to speed when time is constant, and to time when speed is constant.

Absolute Core Formula:

Unit Conversion:

Inverse Proportionality Between Speed and Time:

Average Speed:

For two equal distances at speeds $S_1, S_2$:

The Universal Trap: The most heavily tested trap: averaging speeds directly when the distances — not the times — are equal. Second trap: sign errors in "reaches T minutes late/early" problems.

2. Exhaustive Question Typology

                          TIME AND DISTANCE
                                 |
    -------------------------------------------------------------
    |            |              |               |               |
Type 1:       Type 2:        Type 3:        Type 4:         Type 5:
Basic D=S×T   Speed-Time     Average        Late/Early      Relative
computation   Inverse        Speed          Arrival         Speed —
              Proportion     (unequal       (change in       Two Bodies
              (Same D,       speed          speed affects    (opposite/
              find new T)    scenarios)     arrival time)     same dir)
    |            |              |
Type 6:       Type 7:        Type 8:
Circular      Escalator/     Races &
track meeting Moving         proportional
problems      walkway        speed
(same/opp.    problems       comparisons
direction)

Type 1 — Basic distance/speed/time computation:

• Core Scenario: Direct application, find any one variable given the other two.
• Governing Equation: $D = S \times T$

Type 2 — Speed changes, same distance, find new time:

• Core Scenario: "If speed is increased by x%, find the % change in time taken."
• Governing Equation: $S_1T_1 = S_2T_2 \Rightarrow \dfrac{T_2}{T_1} = \dfrac{S_1}{S_2}$

Type 3 — Average speed over a journey with unequal speeds:

• Core Scenario: Person travels part of the journey at $S_1$, another part at $S_2$.
• Governing Equation: Equal distances → $\dfrac{2S_1S_2}{S_1+S_2}$; Equal times → $\dfrac{S_1+S_2}{2}$

Type 4 — Late/early arrival due to speed change:

• Core Scenario: "A person walking at $S_1$ reaches t minutes late; at $S_2$ reaches t' minutes early."
• Governing Equation:
  D = \frac{S_1 S_2 (t+t')}{S_1 - S_2}

Type 5 — Two bodies moving toward/away from each other:

• Core Scenario: Two people start from points A and B, moving toward or in the same direction.
• Governing Equation: Opposite: $t = \dfrac{D}{S_1+S_2}$; Same direction: $t = \dfrac{D}{S_1-S_2}$

Type 6 — Circular track meeting problems:

• Core Scenario: Two runners start together on a circular track of circumference C.
• Governing Equation: Opposite: $t = \dfrac{C}{S_1+S_2}$; Same direction: $t = \dfrac{C}{S_1-S_2}$

Type 7 — Escalator/moving walkway problems:

• Core Scenario: A person walks up/down a moving escalator.
• Governing Equation: With escalator: $S_{eff} = S_{person} + S_{escalator}$; Against: $S_{eff} = S_{person} - S_{escalator}$

Type 8 — Races and proportional speed comparison:

• Core Scenario: "A can run x meters while B runs y meters in the same time."
• Governing Equation:
  $$\frac{S_A}{S_B} = \frac{x}{y}$$

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Basic D=S×T Computation

MCQ 1. A car travels at 60 km/hr for 3.5 hours. Find the distance covered. (A) 210 km (B) 200 km (C) 220 km (D) 205 km

Correct Answer: (A) Solution: $D=60\times3.5=210$ km.

MCQ 2. A cyclist covers 90 km in 4 hours. Find his speed. (A) 22.5 km/hr (B) 20 km/hr (C) 25 km/hr (D) 21 km/hr

Correct Answer: (A) Solution: $S=90/4=22.5$ km/hr.

MCQ 3. How long will it take to cover 250 km at a speed of 62.5 km/hr? (A) 4 hours (B) 3.5 hours (C) 4.5 hours (D) 5 hours

Correct Answer: (A) Solution: $T=250/62.5=4$ hours.

Type 2 — Speed-Time Inverse Proportion

MCQ 1. If a car's speed is increased by 25%, find the % decrease in time taken to cover the same distance. (A) 20% (B) 25% (C) 18% (D) 22%

Correct Answer: (A) Solution: Speed ratio 5:4 (new:old). Time ratio (inverse) 4:5. Decrease$=\dfrac{1}{5}\times100=20\%$.

MCQ 2. A train's speed is reduced to 3/4 of its usual speed. Find the % increase in time taken. (A) 33.33% (B) 25% (C) 30% (D) 35%

Correct Answer: (A) Solution: Speed ratio 3:4 (new:old). Time ratio (inverse) 4:3. Increase$=\dfrac13\times100=33.33\%$.

MCQ 3. If speed increases in the ratio 5:7, find the ratio of time taken. (A) 7:5 (B) 5:7 (C) 5:5 (D) 7:7

Correct Answer: (A) Solution: Time ratio is inverse of speed ratio: $7:5$.

Type 3 — Average Speed

MCQ 1. A car travels the first half of a journey at 40 km/hr and the second half at 60 km/hr. Find the average speed. (A) 48 km/hr (B) 50 km/hr (C) 45 km/hr (D) 52 km/hr

Correct Answer: (A) Solution: Equal distances: $\dfrac{2\times40\times60}{100}=48$ km/hr.

MCQ 2. A man travels for equal TIMES at 20 km/hr and 30 km/hr. Find his average speed. (A) 25 km/hr (B) 24 km/hr (C) 26 km/hr (D) 24.5 km/hr

Correct Answer: (A) Solution: Equal times: simple average $=\dfrac{20+30}{2}=25$ km/hr.

MCQ 3. A cyclist covers a distance at 15 km/hr and returns at 10 km/hr. Find the average speed for the whole journey. (A) 12 km/hr (B) 12.5 km/hr (C) 11.5 km/hr (D) 13 km/hr

Correct Answer: (A) Solution: $\dfrac{2\times15\times10}{25}=12$ km/hr.

Type 4 — Late/Early Arrival

MCQ 1. Walking at 3/4 of his usual speed, a man reaches his office 20 minutes late. Find his usual time. (A) 60 minutes (B) 55 minutes (C) 65 minutes (D) 50 minutes

Correct Answer: (A) Solution: Speed ratio 3:4 → time ratio 4:3. Difference (1 unit)=20 min. Usual time (3 units)$=60$ min.

MCQ 2. A student walking at 5 km/hr reaches school 6 minutes late; walking at 6 km/hr, he reaches 2 minutes early. Find the distance to school. (A) 4 km (B) 3.5 km (C) 4.5 km (D) 3.8 km

Correct Answer: (A) Solution: D=\dfrac{S_1S_2(t+t')}{S_1-S_2}=\dfrac{5\times6\times(8/60)}{1}=\dfrac{30\times8/60}{1}=4 km.

MCQ 3. Walking at 4 km/hr, a man is late by 10 minutes; walking at 5 km/hr he is early by 5 minutes. Find the distance. (A) 5 km (B) 4.5 km (C) 5.5 km (D) 4.8 km

Correct Answer: (A) Solution: $D=\dfrac{4\times5\times(15/60)}{1}=\dfrac{20\times0.25}{1}=5$ km.

Type 5 — Two Bodies Moving Toward/Away

MCQ 1. Two persons start from points 120 km apart, moving toward each other at 25 km/hr and 35 km/hr. Find the time to meet. (A) 2 hours (B) 2.5 hours (C) 1.8 hours (D) 2.2 hours

Correct Answer: (A) Solution: $t=120/(25+35)=120/60=2$ hours.

MCQ 2. Two cars start from the same point, moving in the same direction at 50 km/hr and 65 km/hr. Find how long it takes for them to be 60 km apart. (A) 4 hours (B) 3.5 hours (C) 4.5 hours (D) 3.8 hours

Correct Answer: (A) Solution: $t=60/(65-50)=60/15=4$ hours.

MCQ 3. Two trains 300 km apart move toward each other at 40 km/hr and 60 km/hr. Find the time to meet. (A) 3 hours (B) 2.5 hours (C) 3.5 hours (D) 2.8 hours

Correct Answer: (A) Solution: $t=300/(40+60)=3$ hours.

Type 6 — Circular Track Meeting Problems

MCQ 1. Two runners on a 400 m circular track run at 8 m/s and 6 m/s in the same direction, starting together. Find the time for the first meeting. (A) 200 seconds (B) 180 seconds (C) 220 seconds (D) 190 seconds

Correct Answer: (A) Solution: $t=400/(8-6)=200$ seconds.

MCQ 2. Two runners on a 300 m track run in opposite directions at 5 m/s and 7 m/s, starting together. Find the time for the first meeting. (A) 25 seconds (B) 30 seconds (C) 20 seconds (D) 28 seconds

Correct Answer: (A) Solution: $t=300/(5+7)=25$ seconds.

MCQ 3. Two athletes run around a 500 m circular track at speeds of 10 m/s and 15 m/s in opposite directions. Find how many times they meet in 5 minutes (300 seconds). (A) 15 times (B) 12 times (C) 18 times (D) 10 times

Correct Answer: (A) Solution: Time for one meeting$=500/(10+15)=20$ s. In 300 s: $300/20=15$ meetings.

Type 7 — Escalator/Moving Walkway Problems

MCQ 1. A man walks up a moving escalator at 3 steps/sec, and it takes him 20 seconds. If the escalator alone (without walking) takes 50 seconds to carry a stationary person, find the total number of steps. (A) approximately 71.4 steps (B) 70 steps (C) 75 steps (D) 68 steps

Correct Answer: (A) Solution: Escalator's own rate$=N/50$ steps/sec, where N=total steps. Combined rate (walking + escalator)$=N/20$. Man's own walking rate$=3$ steps/sec (given, in escalator-relative terms this represents his contribution). Set up: $N/20=3+N/50$. Multiply by 100: $5N=300+2N\Rightarrow3N=300\Rightarrow N=100$. (Recheck: this gives 100, not matching option A; correcting.)

MCQ 1 (verified). Correct Answer: (E)/restated as 100 steps Solution: As derived: total visible steps N=100.

MCQ 2. A boy walking at 3 km/hr crosses a moving walkway in 2 minutes. If the walkway alone moves at 2 km/hr, find the length of the walkway. (A) 166.67 m (B) 150 m (C) 170 m (D) 160 m

Correct Answer: (A) Solution: Effective speed$=3+2=5$ km/hr$=5000/60$ m/min$=83.33$ m/min. Length$=83.33\times2=166.67$ m.

MCQ 3. A man walks against a moving walkway (speed 1.5 km/hr) at his own walking speed of 4 km/hr, covering 50 m. Find the time taken. (A) 72 seconds (B) 75 seconds (C) 70 seconds (D) 68 seconds

Correct Answer: (A) Solution: Effective speed$=4-1.5=2.5$ km/hr$=2500/3600$ m/s$=0.694$ m/s. Time$=50/0.694=72$ s.

Type 8 — Races and Proportional Speed Comparison

MCQ 1. A can run 100 m while B runs 90 m in the same time. Find the ratio of their speeds. (A) 10:9 (B) 9:10 (C) 5:4 (D) 4:5

Correct Answer: (A) Solution: $S_A:S_B=100:90=10:9$.

MCQ 2. In the time A covers 400 m, B covers 350 m. If A's speed is 8 m/s, find B's speed. (A) 7 m/s (B) 7.5 m/s (C) 6.5 m/s (D) 6 m/s

Correct Answer: (A) Solution: Ratio$=400:350=8:7$. Since A's speed=8: B's speed$=7$ m/s.

MCQ 3. A runs 5/4 times as fast as B. In a race, if A gives B a start of 60 m, find the length of the race so that both reach the finish together. (A) 300 m (B) 280 m (C) 320 m (D) 260 m

Correct Answer: (A) Solution: Speed ratio $A:B=5:4$. Let race length=D. A covers D, B covers D-60 in the same time. $\dfrac{D}{D-60}=\dfrac54\Rightarrow4D=5D-300\Rightarrow D=300$ m.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — Direct Ratio Method

• Application: Whenever a percentage/ratio change in speed is given and the question asks for % change in time, with distance constant.
• Mental Model: Since $S \propto \frac{1}{T}$ for constant D, directly invert the ratio.

Shortcut 2 — The Late/Early Arrival Direct Formula

• Application: Classic "reaches late at one speed, early at another" problems.
• Mental Model: Plug directly into D = \frac{S_1S_2(t+t')}{S_1-S_2}.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): Walking at 3/4 of his usual speed, a man reaches his office 20 minutes late. Find his usual time to cover the distance.

Traditional Method (Slow): Let usual speed = S, usual time = T. New speed=(3/4)S, new time=(4/3)T. $(4/3)T-T=20\Rightarrow(1/3)T=20\Rightarrow T=60$ minutes. (~40 seconds.)

Exam Shortcut (Fast): Speed ratio 3:4 → Time ratio 4:3. Difference in ratio units=1 unit=20 minutes. Usual time (3 units)=$3\times20=$ 60 minutes. (Under 15 seconds.)

Problem 2 (UPSC/Banking Advanced): Two trains start simultaneously from stations A and B towards each other. After crossing, they take 4 hours and 9 hours respectively to reach B and A. If the speed of the first train is 45 km/hr, find the speed of the second train.

Step-by-Step Breakdown:

1. $\dfrac{S_1}{S_2} = \sqrt{\dfrac{t_2}{t_1}} = \sqrt{\dfrac94} = \dfrac32$
2. $\dfrac{45}{S_2} = \dfrac32 \Rightarrow S_2 = 30$
3. Answer: 30 km/hr.

6. Chapter Checklist for Students

• I never average two speeds directly unless the TIME (not distance) for each leg is explicitly equal.
• I apply the inverse ratio method instantly for "speed changes, distance constant" problems.
• I have memorized the late/early arrival formula and can apply it directly.
• I correctly assign "+" for opposite-direction meeting and "−" for same-direction catch-up problems.
• I can apply the $\sqrt{t_2/t_1}$ speed-ratio shortcut for post-crossing time problems.