Time and Distance: Notes और MCQ Practice

आसान भाषा में:

Time and Distance का basic फंडा बहुत simple है—अगर आपकी Speed ज्यादा होगी, तो same Distance cover करने में Time कम लगेगा (Inverse relation)। RRB NTPC exam में इस topic से Trains, Relative Speed और Average Speed पर direct formula-based सवाल पूछे जाते हैं। बस Units का ध्यान रखो और full marks पक्के!

Important Formulas & Rules:

• Basic Formula: Distance = Speed × Time
• Time: Time = Distance ÷ Speed
• Speed: Speed = Distance ÷ Time
• Unit Conversion:
• km/h to m/s: Speed × (5 ÷ 18)
• m/s to km/h: Speed × (18 ÷ 5)
• Average Speed (जब Distance equal हो): (2 × S1 × S2) ÷ (S1 + S2)
• Relative Speed:
• Opposite direction (आमने-सामने): S1 + S2
• Same direction (एक ही दिशा): S1 − S2

2 Quick Tricks:

1. Ratio Method: अगर Distance constant है, तो Speed का ratio (a : b) हमेशा Time के ratio (b : a) का उल्टा होगा। इससे equation बनाने की जरूरत नहीं पड़ती।
2. Train Crossing Concept:

• Pole या Man को cross करने में Distance = Train की Length
• Platform या Bridge को cross करने में Distance = Train की Length + Platform की Length

2 Common Mistakes:

1. Unit Mismatch: Speed km/h में और Time seconds में रखकर direct multiply कर देना। Solve करने से पहले units हमेशा match करें (km/h के साथ Hours, m/s के साथ Seconds)।
2. Average Speed का गलत Formula: Average Speed निकालने के लिए कभी भी (S1 + S2) ÷ 2 मत करना! हमेशा Total Distance ÷ Total Time use करें।

Time and Distance

1. Concept

Time and Distance is one of the most fundamental topics in Quantitative Aptitude. It deals with how fast an object moves, how much ground it covers, and the time it takes to do so.

At its core, the relationship between these three elements is direct and intuitive:

• Distance is the total length of the path traveled.
• Time is the duration taken to cover that distance.
• Speed is the rate at which distance is covered per unit of time (how fast an object is moving).

If you move faster (higher speed), you take less time to cover a fixed distance. If you move for a longer time at a steady speed, you cover more distance.

2. Important Formulas, Rules, and Facts

Fundamental Relationships

• Distance = Speed × Time
• Speed = Distance ÷ Time
• Time = Distance ÷ Speed

Unit Conversions

Units must always be consistent before solving a problem.

• km/h to m/s: Multiply by 5/18
• Speed in m/s = Speed in km/h × (5 ÷ 18)
• m/s to km/h: Multiply by 18/5
• Speed in km/h = Speed in m/s × (18 ÷ 5)
• Time conversion:
• 1 hour = 60 minutes = 3600 seconds
• 1 minute = 60 seconds
• Distance conversion:
• 1 km = 1000 meters

Average Speed

Average speed is NOT the simple average of speeds unless time spent at both speeds is exactly equal.

• General Formula:
• Average Speed = Total Distance ÷ Total Time
• Special Case (Equal Distances): If an object travels a certain distance at speed 'x' and returns the same distance at speed 'y':
• Average Speed = (2 × x × y) ÷ (x + y)
• Special Case (Three Equal Distances): If an object covers three equal distances at speeds x, y, and z:
• Average Speed = (3 × x × y × z) ÷ (xy + yz + zx)

Proportionality Rules

• When Distance is constant: Speed is inversely proportional to Time.
• Ratio of Speed = a : b → Ratio of Time = b : a
• When Time is constant: Distance is directly proportional to Speed.
• Ratio of Speed = a : b → Ratio of Distance = a : b
• When Speed is constant: Distance is directly proportional to Time.
• Ratio of Time = a : b → Ratio of Distance = a : b

3. Solved Examples (Exam Style)

Example 1: Basic Unit Conversion

Question: A train covers a distance of 450 km in 5 hours. Find its speed in meters per second (m/s).

Solution:

• Step 1: Calculate speed in km/h.
• Speed = Distance ÷ Time
• Speed = 450 ÷ 5 = 90 km/h
• Step 2: Convert km/h into m/s by multiplying by 5/18.
• Speed in m/s = 90 × (5 ÷ 18)
• Speed in m/s = 5 × 5 = 25 m/s

Answer: 25 m/s

Example 2: Average Speed for Equal Distance

Question: A motorist travels from Town A to Town B at a speed of 60 km/h and returns from Town B to Town A at a speed of 40 km/h. Find the average speed for the entire journey.

Solution:

• Step 1: Identify that the distance going and returning is equal.
• Step 2: Apply the equal-distance average speed formula:
• Average Speed = (2 × x × y) ÷ (x + y)
• Here, x = 60 km/h, y = 40 km/h
• Step 3: Substitute the values:
• Average Speed = (2 × 60 × 40) ÷ (60 + 40)
• Average Speed = 4800 ÷ 100 = 48 km/h

Answer: 48 km/h

Example 3: Speed Ratio and Time Difference (Late/Early Type)

Question: Walking at 3/4 of his usual speed, a student reaches his school 15 minutes late. What is his usual time to reach the school?

Solution:

• Step 1: Understand the inverse relationship.
• Distance is constant (home to school).
• Ratio of speeds (Usual : New) = 1 : (3/4) = 4 : 3
• Step 2: Determine ratio of time.
• Ratio of time taken (Usual : New) = 3 : 4
• Step 3: Compare unit difference.
• Difference in time units = 4 - 3 = 1 unit
• Given that 1 unit difference = 15 minutes
• Step 4: Calculate usual time.
• Usual time = 3 units = 3 × 15 = 45 minutes

Answer: 45 minutes

Example 4: Late and Early Conditions

Question: If a person travels at 4 km/h, he reaches his office 10 minutes late. If he travels at 5 km/h, he reaches 5 minutes early. Find the distance to his office.

Solution:

• Step 1: Find the total difference in time.
• Late by 10 min = +10
• Early by 5 min = -5
• Total difference = 10 - (-5) = 15 minutes
• Convert minutes into hours: 15 ÷ 60 = 1/4 hour
• Step 2: Use the distance formula for different speeds:
• Distance = (Product of Speeds ÷ Difference of Speeds) × Difference in Time
• Distance = [(4 × 5) ÷ (5 - 4)] × (1 ÷ 4)
• Distance = (20 ÷ 1) × (1 ÷ 4) = 5 km

Answer: 5 km

4. Shortcuts and Tricks

1. Multiples of 18 and 5 Rule

Instead of calculating 5/18 every time, memorize common benchmarks:

• 18 km/h = 5 m/s
• 36 km/h = 10 m/s
• 54 km/h = 15 m/s
• 72 km/h = 20 m/s
• 90 km/h = 25 m/s
• 108 km/h = 30 m/s

Trick: If speed is a multiple of 18 in km/h, divide by 18 and multiply by 5 to get m/s instantly.

2. Direct Formula for Late/Early Problems

When speed s1 causes late/early time t1 and speed s2 causes late/early time t2:

• Distance = [ (s1 × s2) ÷ |s1 - s2| ] × (Change in Time in hours)
• If one case is 'late' and the other is 'early', Add the times: Change in Time = t1 + t2
• If both cases are 'late' or both are 'early', Subtract the times: Change in Time = |t1 - t2|

3. Change in Speed as a Fraction

If speed becomes (a ÷ b) of original speed, then the new time becomes (b ÷ a) of original time.

• Extra time taken = [(b - a) ÷ a] × Original Time (when b > a)

5. Common Mistakes to Avoid

1. Arithmetic Mean for Average Speed: Never calculate average speed as (s1 + s2) ÷ 2 when distance is constant. Always use total distance ÷ total time or 2xy ÷ (x + y).
2. Ignoring Units: Adding hours and minutes directly without converting (e.g., adding 2 hours + 20 minutes as 2.2 hours instead of 2 + 20/60 = 2.33 hours).
3. Mixing Metric Systems: Multiplying speed in km/h directly with time in seconds to find distance in meters without converting km/h to m/s.
4. Confusion with Time Differences:

• Late by 10 min and Late by 5 min → Difference is 5 min (10 - 5).
• Late by 10 min and Early by 5 min → Difference is 15 min (10 + 5).