Races and Games

This topic deals with competition scenarios where participants race (typically in running, cycling, swimming, etc.) or play games (like tournaments), with a focus on time, distance, speed, head starts, leads, and eliminations.

It is essentially an application of Speed, Distance, and Time, along with ratios and relative speeds.

2. Key Formulas & Shortcuts

Let:

  • dd = total distance of race
  • SA,SBS_A, S_B = speeds of A and B
  • tA,tBt_A, t_B = time taken by A and B respectively
  • LL = Lead (in time or distance)
  • R=SASBR = \frac{S_A}{S_B} = speed ratio

A. If A gives B a head start of ‘x’ meters in a race of ‘d’ meters:

It means A covers d while B covers only d – x

SASB=dd−x⇒x=d(1−SBSA)\frac{S_A}{S_B} = \frac{d}{d - x} \quad\Rightarrow\quad x = d \left(1 - \frac{S_B}{S_A} \right)

B. If A beats B by ‘x’ meters in a race of ‘d’ meters:

It means when A finishes d, B covers d – x

SASB=dd−x⇒x=d(1−SBSA)\frac{S_A}{S_B} = \frac{d}{d - x} \quad\Rightarrow\quad x = d \left(1 - \frac{S_B}{S_A} \right)

C. If A beats B by ‘t’ seconds:

Then use speeds and time:

x=SB×tx = S_B \times t

D. If two persons run at speeds in the ratio m:nm:n:

Then their time taken will be in the inverse ratio:

Time ratio=n:m\text{Time ratio} = n : m

E. Start of the race: Head Start / Start Advantage

If A gives B a start of xx meters, it means B starts at xx meters ahead, but the race length is same for both.

F. Games / Tournaments (Round Robins)

  • Single Round Robin (each plays all once):
Total matches=n(n−1)2\text{Total matches} = \frac{n(n - 1)}{2}
  • Double Round Robin (each plays all twice):
Total matches=n(n−1)\text{Total matches} = n(n - 1)
  • Knockout format:
Total matches=n−1\text{Total matches} = n - 1

3. Conceptual Tips & Common Mistakes

  • Always compare speeds using ratios if exact values aren’t known.

  • Never confuse “beats by x meters” with “gives a start of x meters.”

    • Beat = both run the same race, but one finishes earlier.
    • Start = unequal starting point.

  • If time lead is given, use the loser’s speed to calculate the distance gap.

  • In knockout formats, number of matches = number of eliminations = n – 1

  • In round-robin formats, don’t forget to divide by 2 (for single round robin).

4. Visual Explanation

Race Track Visualization

|<----------- d = 100 m ----------->|
A starts at 0      →→→→→→→→→→→ Finish
B starts at 0      →→→→→→→→     (loses by x m)

If A beats B by 20 meters, then:

  • A runs 100 m in t seconds
  • B runs 80 m in t seconds ⇒ Relative speed concept

Tournament Match Chart (Round Robin, 4 players)

Match #Player 1Player 2
1AB
2AC
3AD
4BC
5BD
6CD

→ Total matches = 4×32=6\frac{4 \times 3}{2} = 6

5. Solved Examples

Example 1: Distance-based Lead

Q: A beats B by 20 m in a 100 m race. Find the ratio of their speeds.

A:
When A runs 100 m, B runs 80 m

SASB=10080=54\frac{S_A}{S_B} = \frac{100}{80} = \frac{5}{4}

Example 2: Time Lead

Q: A beats B by 10 seconds in a race. If B’s speed is 6 m/s, how many meters did A beat B by?

A:
Lead in distance = 6×10=606 \times 10 = 60 meters

Example 3: Head Start

Q: A gives B a start of 20 m in a 200 m race. A's speed is 10 m/s. B’s speed?

A:

SASB=200180=109⇒SB=910×10=9 m/s\frac{S_A}{S_B} = \frac{200}{180} = \frac{10}{9} \quad\Rightarrow\quad S_B = \frac{9}{10} \times 10 = 9 \text{ m/s}

Example 4: Tournament Matches

Q: In a single round-robin tournament with 8 teams, how many matches are played?

A:

Total matches=8×72=28\text{Total matches} = \frac{8 \times 7}{2} = 28

Example 5: Knockout Matches

Q: In a knockout chess tournament with 256 players, how many matches will be played?

A:

Matches=256−1=255\text{Matches} = 256 - 1 = 255
PRACTICE QUESTIONS
PREVIOUSNEXT