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Races and Relative Speeds

Introduction

Races and Relative Speeds का कॉन्सेप्ट time-speed-distance की बेसिक्स पर बना है, जहाँ दो या अधिक प्रतिभागी एक साथ चल रहे होते हैं। यह pattern इसलिए ज़रूरी है क्योंकि यह आपको उनकी speeds compare करने, head starts (leads) निकालने, और यह तय करने में मदद करता है कि कौन जीतेगा या कब मिलेंगे।

ऐसे सवाल aptitude tests में बहुत आते हैं और इनमें relative speed की साफ समझ चाहिए - खासकर जब प्रतियोगी same या opposite directions में चलते हैं।

Pattern: Races and Relative Speeds

Pattern

Key concept: Distance या time को compare करें formula Speed = Distance ÷ Time और relative speed की idea से।

  • जब same direction में चलते हैं: Relative speed = (Speed₁ - Speed₂)
  • जब opposite directions में चलते हैं: Relative speed = (Speed₁ + Speed₂)
  • अगर दोनों same distance पूरी करते हैं: Time speed के उल्टा होता है → T₁ : T₂ = S₂ : S₁
  • अगर किसी को head start (lead) मिला है: Winner या margin निकालने के लिए times या distances का difference यूज़ करें।

Step-by-Step Example

Question

500 m की race में A की speed 5 m/s और B की speed 4 m/s है। A जब finish करता है, तब उसके पास कितना lead होता है?

Solution

  1. Step 1: A का finish करने का time निकालें

    Time = Distance ÷ Speed = 500 ÷ 5 = 100 seconds.

  2. Step 2: उसी time में B कितनी दूरी तय करेगा

    Distance = Speed × Time = 4 × 100 = 400 m.

  3. Step 3: Lead निकालें

    Lead = 500 - 400 = 100 m.

  4. Final Answer:

    A, B को 100 m से हरा देता है।

  5. Quick Check:

    Speed ratio = 5 : 4 → difference = 1/5 of 500 = 100 ✅

Quick Variations

1. Lead निकालना (distance difference)।

2. Tie के लिए required head start।

3. Distance की जगह time-based races।

4. Ratio-based सवाल (Speed ratio ↔ Distance ratio ↔ Time ratio)।

5. A, B, C जैसे multiple participants की comparison।

Trick to Always Use

  • Step 1: Speed × Time या Distance ÷ Speed से पहले तय करें कि कौन पहले finish करेगा।
  • Step 2: Lead के लिए, same time में तय की गई distances का फर्क निकालें।
  • Step 3: Head start वाले cases में distances बराबर करके required lead निकालें।
  • Step 4: Units हमेशा consistent रखें (m/s या km/h).

Summary

Summary

  • जब speeds अलग हों, तो faster व्यक्ति का lead = (Speed diff) × Time.
  • Equal distances पर Time ∝ 1/Speed.
  • Head start = times equal करने के लिए जरूरी दूरी।
  • पूरे calculation में units एक जैसे रखें।

Practice

(1/5)
1. In a 400 m race, A runs at 8 m/s and B at 6 m/s. How much lead does A have when A finishes the race?
easy
A. 50 m
B. 150 m
C. 200 m
D. 100 m

Solution

  1. Step 1: Time for A to finish

    Time = Distance ÷ Speed = 400 ÷ 8 = 50 s.

  2. Step 2: Distance B covers in same time

    Distance = Speed × Time = 6 × 50 = 300 m.

  3. Step 3: Lead = Remaining distance

    Lead = 400 - 300 = 100 m.

  4. Final Answer:

    A beats B by 100 m → Option D.

  5. Quick Check:

    Speed ratio 8:6 = 4:3 → 1/4 of 400 = 100 m ✅
Hint: Lead = Distance × (1 - slower/faster).
Common Mistakes: Mixing up time and speed ratios.
2. A runs at 10 m/s and B at 8 m/s in a 500 m race. If B is given a 50 m head start, who wins and by how much?
easy
A. A wins by 50 m
B. A wins by 25 m
C. B wins by 25 m
D. B wins by 50 m

Solution

  1. Step 1: Time for A to finish

    Time = 500 ÷ 10 = 50 s.

  2. Step 2: Distance B covers in same time

    Distance = 8 × 50 = 400 m; plus head start = 400 + 50 = 450 m.

  3. Step 3: Difference

    A = 500 m, B = 450 m → A wins by 50 m.

  4. Final Answer:

    A wins by 50 m → Option A.

  5. Quick Check:

    Without head start A would be 100 m ahead (speed ratio), head start reduces it to 50 m ✅
Hint: Compute both distances in the same time, include head start before comparing.
Common Mistakes: Forgetting to add the head start for B.
3. A and B start together in a 600 m race. A’s speed is 9 m/s and B’s is 8 m/s. When A finishes, how far is B from the finish line?
medium
A. 50.67 m
B. 66.67 m
C. 75.67 m
D. 80.67 m

Solution

  1. Step 1: Time for A to finish

    Time = 600 ÷ 9 = 66.666... s.

  2. Step 2: Distance B covers in same time

    Distance = 8 × 66.666... = 533.333... m.

  3. Step 3: Remaining distance to finish

    600 - 533.333... = 66.666... m (≈ 66.67 m).

  4. Final Answer:

    B is about 66.67 m from the finish → Option B.

  5. Quick Check:

    Speed ratio 9:8 → (1/9) of 600 = 66.666... m ✅
Hint: Remaining gap = Distance × (1 - slower/faster).
Common Mistakes: Rounding too early; keep fractions until final step.
4. In a 1 km race, A can run 1 km in 180 seconds and B in 200 seconds. How much start can A give B to finish together?
medium
A. 100 m
B. 80 m
C. 90 m
D. 120 m

Solution

  1. Step 1: Speeds

    A's speed = 1000 ÷ 180 ≈ 5.555... m/s; B's speed = 1000 ÷ 200 = 5 m/s.

  2. Step 2: Let head start = x (m) for B

    Then B runs (1000 - x) m in same time A runs 1000 m:
    (1000 - x) ÷ 5 = 1000 ÷ 5.555... → 1000 - x = 900 → x = 100 m.

  3. Final Answer:

    A can give B a 100 m head start → Option A.

  4. Quick Check:

    If B starts 100 m ahead, B runs 900 m at 5 m/s = 180 s same as A ✅
Hint: Set times equal and solve for head start: (D - x)/v_slow = D/v_fast.
Common Mistakes: Comparing times directly without forming equation for head start.
5. In an 800 m race, A beats B by 80 m and B beats C by 40 m. By how much does A beat C?
medium
A. 100 m
B. 110 m
C. 116 m
D. 120 m

Solution

  1. Step 1: Derive speed ratios

    When A runs 800 m, B runs 720 m → A:B = 800:720 = 10:9.
    When B runs 800 m, C runs 760 m → B:C = 800:760 = 20:19.

  2. Step 2: Combine ratios to get A:C

    A:C = (10/9) × (20/19) = 200:171.

  3. Step 3: When A runs 800 m, C runs (171/200)×800 = 684 m → Lead = 800 - 684 = 116 m.

  4. Final Answer:

    A beats C by 116 m → Option C.

  5. Quick Check:

    Chaining ratios is equivalent to composing relative distances → 116 m is consistent with given pairwise leads ✅
Hint: Multiply A:B and B:C ratios to get A:C; then compute remaining distance.
Common Mistakes: Adding pairwise leads instead of using speed/distance ratios.

Mock Test

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