Races and Games

1. Core Concepts & Theoretical Blueprint

Races and Games problems apply Time-Speed-Distance logic to a competitive context, where the central objects are a race distance, a head start (either in distance or time), and the gap between competitors at the finish — nearly every problem reduces to comparing how far each competitor travels in the SAME time (since a race runs for a fixed duration or fixed winning distance).

Core Definitions:

• "A gives B a start of x metres" in a race of distance D means: while A runs the full D metres, B only needs to run $(D-x)$ metres to finish at the same time (i.e., they are considered to finish together under this handicap).
• "A beats B by x metres" means: when A finishes the race (distance D), B has covered only $(D-x)$ metres.
• "A beats B by t seconds" means: A finishes t seconds before B does.
• Dead Heat: both competitors finish at EXACTLY the same time (neither "beats" the other).

Core Speed Ratio Formula (from "beats by x metres" in a race of distance D):

(since both run for the same TIME, and A covers D while B covers only $D-x$ in that time)

Core Time Formula (from "beats by t seconds"):

Combined "Start and Beats" Formula:

The Universal Trap: Four persistent traps:

1. Confusing "gives a start of x metres" with "beats by x metres" — a START is a handicap given BEFORE the race begins (B's effective running distance is reduced), while "beats by" describes the GAP AT THE FINISH when A completes the race — these produce structurally different equations and must never be conflated.
2. Sign/direction errors in speed ratio setup — since $A:B$ speed ratio $=D:(D-x)$ for "A beats B by x m," students sometimes invert this, especially when the question asks for B's speed relative to A's instead of A's relative to B's.
3. Forgetting that a "start" in TIME (not distance) requires converting to an effective time advantage, not a distance adjustment, before comparing finishing times.
4. Chain "beats by" problems (A beats B, B beats C) require multiplying RATIOS, not simply adding the individual "beats by" margins — the margins (in metres) do not combine additively across a chain of three or more racers because each margin is measured over the SAME fixed race distance, not cumulatively.

2. Exhaustive Question Typology

                          RACES AND GAMES
                                  |
    -----------------------------------------------------------------------
    |            |              |               |               |          |
Type 1:       Type 2:        Type 3:        Type 4:         Type 5:     Type 6:
Basic Race    A Gives B a    A Beats B by   A Beats B by    Dead Heat   Combined
Finish Time/  Start of x     x Metres       t Seconds       Problems    Start (Both
Speed Given   Metres (Find   (Find Speed                    (Both       Time and
Distance      Winning        Ratio or Time)                 Finish      Distance
              Distance/                                     Together)   Given)
              Ratio of
              Speeds)
    |            |              |
Type 7:       Type 8:        Type 9:
Circular      Game Points    Chain Race
Track Races   Problems       (A Beats B,
(Meeting      ("A Can Give   B Beats C,
Points)       B 20 Points    Find A vs C)
              in a Game of
              100")

Type 1 — Basic race finish time/speed given distance:

• Core Scenario: "A runs a 400 m race in 50 seconds. Find A's speed."
• Governing Equation: $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$ (direct TSD application)

Type 2 — A gives B a start of x metres:

• Core Scenario: "In a race of 800 m, A gives B a start of 100 m and still wins by 50 m. Find the ratio of their speeds."
• Governing Equation: A runs 800 m while B runs $(800-100-50)=650$ m in the same time (B's effective distance covered, given the start AND still losing by the stated margin); speed ratio $A:B=800:650$.

Type 3 — A beats B by x metres:

• Core Scenario: "In a 1000 m race, A beats B by 100 m. Find the ratio of their speeds."
• Governing Equation: $\dfrac{Speed_A}{Speed_B}=\dfrac{D}{D-x}=\dfrac{1000}{900}=\dfrac{10}9$

Type 4 — A beats B by t seconds:

• Core Scenario: "In a race, A finishes in 40 seconds and beats B by 8 seconds. Find B's time."
• Governing Equation: $T_B=T_A+t$

Type 5 — Dead heat problems:

• Core Scenario: "A and B run a race, and it ends in a dead heat. If A gives B a start of 12 m in a 60 m race, find the ratio of their speeds." (Dead heat means both cover their respective distances in the SAME time.)
• Governing Equation: Speed ratio $=$ Distance ratio actually covered by each (since time is equal): $A:B=60:(60-12)=60:48=5:4$

Type 6 — Combined start (both time and distance given):

• Core Scenario: "A gives B a start of 12 seconds in a race of 100 m and still beats B by 40 m. Find A's speed."
• Governing Equation: Combine the time-handicap and distance-margin conditions into a single system relating A's and B's speeds.

Type 7 — Circular track races (meeting points):

• Core Scenario: "A and B run around a circular track of length 400 m at speeds of 10 m/s and 8 m/s respectively, starting together in the same direction. Find when they will next meet at the starting point," or "find how many times they meet on the track (not just at start) in a given duration."
• Governing Equation: Same direction: relative speed $=S_A-S_B$; time for first meeting anywhere on track $=\dfrac{\text{track length}}{\text{relative speed}}$; meeting at START specifically uses LCM logic as in the Time and Distance chapter.

Type 8 — Game points problems ("A can give B 20 points in a game of 100"):

• Core Scenario: "In a game of 100 points, A can give B 20 points and still win. Find how many points B can give C, if B can give C 10 points in a game of 100, and determine A vs C combined."
• Governing Equation: "A gives B x points in a game of P" means: when A scores P, B scores only $(P-x)$ — structurally identical to the "start" concept in races, just measured in points instead of metres.

Type 9 — Chain race (A beats B, B beats C, find A vs C):

• Core Scenario: "In a race of 1000 m, A beats B by 100 m, and B beats C by 100 m (in a race of the same original distance run by each pair). Find by how much A beats C."
• Governing Equation: Multiply the SPEED RATIOS (not distances) along the chain: $\dfrac{Speed_A}{Speed_C}=\dfrac{Speed_A}{Speed_B}\times\dfrac{Speed_B}{Speed_C}$, then reapply the race-distance formula using this combined ratio.

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Basic Race Finish Time/Speed Given Distance

MCQ 1. A runs a 400 m race in 50 seconds. Find A's speed in m/s. (A) 8 m/s (B) 7 m/s (C) 9 m/s (D) 6 m/s

Correct Answer: (A) Solution: Speed$=400/50=8$ m/s.

MCQ 2. B covers a 900 m race in 1 minute 30 seconds. Find B's speed in m/s. (A) 10 m/s (B) 9 m/s (C) 12 m/s (D) 8 m/s

Correct Answer: (A) Solution: Time$=90$ s. Speed$=900/90=10$ m/s.

MCQ 3. A runner completes a 1500 m race at a speed of 6 m/s. Find the time taken. (A) 250 seconds (B) 240 seconds (C) 260 seconds (D) 230 seconds

Correct Answer: (A) Solution: Time$=1500/6=250$ seconds.

Type 2 — A Gives B a Start of x Metres

MCQ 1. In a race of 800 m, A gives B a start of 100 m and still wins by 50 m. Find the ratio of their speeds. (A) 16:13 (B) 8:7 (C) 13:16 (D) 7:8

Correct Answer: (A) Solution: A runs full 800 m; B (with 100 m start) needs to run only 700 m to finish, but A still wins by 50 m, meaning B has covered only $700-50=650$ m when A finishes. Speed ratio $A:B=800:650=16:13$.

MCQ 2. In a race of 500 m, A gives B a start of 40 m and beats B by 10 m. Find the ratio of their speeds. (A) 10:9 (B) 9:10 (C) 5:4 (D) 4:5

Correct Answer: (A) Solution: B's effective required distance$=500-40=460$ m; B actually covers $460-10=450$ m when A finishes 500 m. Ratio$=500:450=10:9$.

MCQ 3. A can run 200 m in 25 seconds. In a race of 200 m, A gives B a start of 20 m. If both finish together (dead heat with the start), find B's speed. (A) 7.2 m/s (B) 8 m/s (C) 7.5 m/s (D) 6.8 m/s

Correct Answer: (A) Solution: A's speed$=200/25=8$ m/s, so A takes 25 seconds for 200 m. With the start, B only needs to cover $200-20=180$ m in the same 25 seconds (since they finish together). B's speed$=180/25=7.2$ m/s.

Type 3 — A Beats B by x Metres

MCQ 1. In a 1000 m race, A beats B by 100 m. Find the ratio of their speeds. (A) 10:9 (B) 9:10 (C) 11:10 (D) 10:11

Correct Answer: (A) Solution: Ratio$=1000:(1000-100)=1000:900=10:9$.

MCQ 2. In a 500 m race, A beats B by 50 m. If A's speed is 10 m/s, find B's speed. (A) 9 m/s (B) 8.5 m/s (C) 9.5 m/s (D) 8 m/s

Correct Answer: (A) Solution: Ratio$=500:450=10:9$. Since A's speed$=10$, and ratio$A:B=10:9$: B's speed$=9$ m/s.

MCQ 3. In a 200 m race, A beats B by 20 m or 4 seconds. Find A's time to complete the race. (A) 36 seconds (B) 40 seconds (C) 32 seconds (D) 44 seconds

Correct Answer: (A) Solution: B covers the last 20 m (the margin) in 4 seconds (since this is the distance B still needs when A finishes, covered at B's own speed). B's speed$=20/4=5$ m/s. B's time for full 200 m$=200/5=40$ s. Since A beats B by 4 seconds, A's time$=40-4=36$ seconds.

Type 4 — A Beats B by t Seconds

MCQ 1. In a race, A finishes in 40 seconds and beats B by 8 seconds. Find B's time. (A) 48 seconds (B) 44 seconds (C) 50 seconds (D) 46 seconds

Correct Answer: (A) Solution: $T_B=T_A+t=40+8=48$ seconds.

MCQ 2. A and B run a 100 m race. A's speed is 5 m/s, and A beats B by 5 seconds. Find B's speed. (A) 4 m/s (approximately, exact: 4 m/s) (B) 4.5 m/s (C) 3.5 m/s (D) 4.2 m/s

Correct Answer: (A) Solution: A's time$=100/5=20$ s. B's time$=20+5=25$ s. B's speed$=100/25=4$ m/s.

MCQ 3. In a 400 m race, A's time is 50 seconds, and B's time is 55 seconds. By how many seconds does A beat B, and by how many metres (at B's own pace)? (A) 5 seconds, and B covers $400\times\frac{50}{55}\approx36.36$ m in that time gap — i.e., A beats B by approximately 36.36 m (B) 5 seconds, 40 m (C) 5 seconds, 30 m (D) 5 seconds, 45 m

Correct Answer: (A) Solution: Time difference$=55-50=5$ seconds (direct). For the DISTANCE margin: B's speed$=400/55=7.27$ m/s; in the 5-second gap, B covers $7.27\times5\approx36.36$ m — this is the distance margin by which A beats B.

Type 5 — Dead Heat Problems

MCQ 1. A and B run a race, and it ends in a dead heat. If A gives B a start of 12 m in a 60 m race, find the ratio of their speeds. (A) 5:4 (B) 4:5 (C) 6:5 (D) 5:6

Correct Answer: (A) Solution: Since it's a dead heat (equal time), speed ratio$=$ distance ratio actually run: $A:B=60:(60-12)=60:48=5:4$.

MCQ 2. In a race of 90 m, A gives B a start of 15 m, and the race ends in a dead heat. Find the ratio of A's speed to B's speed. (A) 6:5 (B) 5:6 (C) 3:2 (D) 2:3

Correct Answer: (A) Solution: $A:B=90:(90-15)=90:75=6:5$.

MCQ 3. A and B's speeds are in the ratio 7:6. In a race, if it ends in a dead heat, find the start (in metres) A must give B, in a race of 210 m. (A) 30 m (B) 25 m (C) 35 m (D) 20 m

Correct Answer: (A) Solution: Since it's a dead heat, $\dfrac{A}{B}=\dfrac{210}{210-x}=\dfrac76\Rightarrow7(210-x)=6\times210\Rightarrow1470-7x=1260\Rightarrow7x=210\Rightarrow x=30$ m.

Type 6 — Combined Start (Both Time and Distance Given)

MCQ 1. A gives B a start of 12 seconds in a race of 100 m and still beats B by 40 m. If A's speed is 5 m/s, find B's speed. (A) 2.5 m/s (B) 3 m/s (C) 2 m/s (D) 3.5 m/s

Correct Answer: (A) Solution: A's time for 100 m$=100/5=20$ s. B starts 12 seconds early, so B has been running for $20+12=32$ seconds when A finishes. In that time, B covers $100-40=60$ m (since A beats B by 40 m). B's speed$=60/32=1.875$ m/s. (Recheck: gives 1.875, not matching option A cleanly; correcting the option set.)

MCQ 1 (verified). Correct Answer: (E)/restated as 1.875 m/s Solution: As derived: B's speed = 60/32 = 1.875 m/s.

MCQ 2. In a race of 300 m, A gives B a start of 10 seconds and beats B by 15 m. If B's speed is 6 m/s, find A's speed. (A) 8 m/s (B) 7.5 m/s (C) 8.5 m/s (D) 9 m/s

Correct Answer: (A) Solution: B covers $300-15=285$ m when A finishes. B's time for this$=285/6=47.5$ s. Since B started 10 s early, A's time$=47.5-10=37.5$ s. A's speed$=300/37.5=8$ m/s.

MCQ 3. A gives B a start of 5 seconds in a 100 m race and beats him by 20 m. A's time to run 100 m is 10 seconds. Find B's speed. (A) 5.33 m/s (approx) (B) 5 m/s (C) 6 m/s (D) 5.5 m/s

Correct Answer: (A) Solution: B has been running for $10+5=15$ seconds when A finishes, covering $100-20=80$ m. B's speed$=80/15=5.33$ m/s.

Type 7 — Circular Track Races (Meeting Points)

MCQ 1. A and B run around a circular track of length 400 m at speeds of 10 m/s and 8 m/s respectively, starting together in the same direction. Find the time for them to meet again for the first time (anywhere on the track). (A) 200 seconds (B) 180 seconds (C) 220 seconds (D) 190 seconds

Correct Answer: (A) Solution: Same direction: relative speed$=10-8=2$ m/s. Time to meet (covering one full lap gap)$=400/2=200$ seconds.

MCQ 2. Two runners run around a circular track of 600 m at speeds of 15 m/s and 9 m/s in OPPOSITE directions, starting together. Find the time for their first meeting. (A) 25 seconds (B) 30 seconds (C) 20 seconds (D) 35 seconds

Correct Answer: (A) Solution: Opposite directions: relative speed$=15+9=24$ m/s. Time$=600/24=25$ seconds.

MCQ 3. A and B run around a circular track of 800 m, A at 20 m/s and B at 12 m/s, in the same direction, starting together. Find how many times they meet in 10 minutes (600 seconds). (A) 15 times (B) 12 times (C) 18 times (D) 10 times

Correct Answer: (A) Solution: Relative speed$=20-12=8$ m/s. Time for one meeting$=800/8=100$ seconds. In 600 seconds: $600/100=6$ meetings after the start... (Recheck: this gives 6 meetings, not matching option A; correcting.)

MCQ 3 (verified). Correct Answer: (E)/restated as 6 times Solution: As derived: meetings = 600/100 = 6 times (in addition to the initial start, which is not counted as a "meeting" for this type of question).

Type 8 — Game Points Problems

MCQ 1. In a game of 100 points, A can give B 20 points and still win. Find the ratio of A's score to B's score when A finishes. (A) 100:80 = 5:4 (B) 80:100 (C) 20:100 (D) 100:20

Correct Answer: (A) Solution: When A scores 100 (full game), B has only scored $100-20=80$. Ratio$=100:80=5:4$.

MCQ 2. A can give B 20 points in a game of 100, and B can give C 10 points in a game of 100. Find how many points A can give C in a game of 100. (A) 28 points (B) 30 points (C) 25 points (D) 32 points

Correct Answer: (A) Solution: When A scores 100, B scores 80. When B scores 100, C scores 90 — so when B scores 80, C scores $\dfrac{90}{100}\times80=72$. So when A scores 100, C scores 72, meaning A can give C $100-72=28$ points.

MCQ 3. In a game of 150 points, A can give B a start of 30 points and still win. Find the ratio of their scores when A finishes. (A) 150:120=5:4 (B) 120:150 (C) 30:150 (D) 5:6

Correct Answer: (A) Solution: When A scores 150, B scores $150-30=120$. Ratio$=150:120=5:4$.

Type 9 — Chain Race (A Beats B, B Beats C)

MCQ 1. In a race of 1000 m, A beats B by 100 m, and B beats C by 100 m (each margin measured in a full 1000 m race for that pair). Find by how many metres A beats C. (A) 190 m (B) 200 m (C) 180 m (D) 210 m

Correct Answer: (A) Solution: Speed ratio $A:B=1000:900=10:9$. Speed ratio $B:C=1000:900=10:9$. Combined $A:C=(10\times10):(9\times9)=100:81$. In a 1000 m race, when A covers 1000 m, C covers $\dfrac{81}{100}\times1000=810$ m. A beats C by $1000-810=190$ m.

MCQ 2. In a race, A beats B by 20 m, and B beats C by 10 m, in a race of 200 m each. Find by how many metres A beats C. (A) 29 m (B) 30 m (C) 28 m (D) 31 m

Correct Answer: (A) Solution: $A:B=200:180=10:9$. $B:C=200:190=20:19$. Combined $A:C=(10\times20):(9\times19)=200:171$. When A covers 200 m, C covers $\dfrac{171}{200}\times200=171$ m. A beats C by $200-171=29$ m.

MCQ 3. A beats B by 10 seconds, and B beats C by 15 seconds, in a race where each runs the same fixed distance. If A's time is 50 seconds, find C's time. (A) 75 seconds (B) 70 seconds (C) 65 seconds (D) 80 seconds

Correct Answer: (A) Solution: B's time=A's time+10=50+10=60 s. C's time=B's time+15=60+15=75 s (time-based "beats by" margins DO add directly along a chain, unlike distance margins, since time is a direct additive quantity along the same timeline).

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The "Same Time, Different Distance" Ratio Reflex

• Application: Every "beats by x metres" and "gives a start of x metres" problem (Types 2, 3, 5) — arguably the single most-used insight in this chapter.
• Mental Model: Whenever a race scenario doesn't explicitly involve TIME, default to the reflex that both racers ran for the EXACT SAME DURATION — so their DISTANCES covered in that shared time directly give the SPEED RATIO. This eliminates the need to introduce a time variable at all for the majority of "beats by" and "start" problems.

Shortcut 2 — Multiply Speed Ratios (Never Add Distance Margins) for Chain Races

• Application: Every Type 9 chain-race problem.
• Mental Model: When given "A beats B by x m" and "B beats C by y m" (each in their own separate race of the SAME fixed distance D), never simply add x+y to guess "A beats C by x+y m" — this is almost always wrong. Instead, always convert each margin to a SPEED RATIO first ($A:B$ and $B:C$), MULTIPLY these ratios to get $A:C$, and only then reapply the race-distance formula to find the actual metre margin between A and C.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): In a 500 m race, A gives B a start of 50 m and beats B by 5 seconds. If A's speed is 10 m/s, find B's speed.

Traditional Method (Slow): A's time for 500 m$=500/10=50$ s. B's effective running distance (with the start)$=500-50=450$ m. Since A beats B by 5 seconds, B's time$=50+5=55$ s. B's speed$=450/55=8.18$ m/s. (Requires computing A's time, adjusting for the start distance, then adjusting for the time margin, then dividing — ~35-40 seconds.)

Exam Shortcut (Fast): Set up the SAME reasoning but streamline the arithmetic: A's time$=50$s (instant division). B's required distance$=450$m (instant subtraction, "start" reduces the distance). B's time$=50+5=55$s (instant addition, "beats by seconds" adds directly to A's time). B's speed$=450/55$, simplified by dividing both by 5 first: $=90/11\approx8.18$ m/s. Answer: ≈8.18 m/s (exactly 90/11 m/s), with the KEY speed insight being to perform each of the three quick sub-steps (A's time, B's distance, B's time) as instant one-line operations in sequence, rather than pausing to re-derive the logic at each stage — under 20 seconds once the sequence is internalized as a fixed template.

Problem 2 (UPSC/Banking Advanced): A, B, and C participate in a 1200 m race. A finishes the race in 100 seconds. A beats B by 200 m, and A beats C by 375 m. Find the distance by which B beats C, and B's time to complete the race.

Step-by-Step Breakdown:

1. When A covers 1200 m (finishing in 100 s), B has covered $1200-200=1000$ m, and C has covered $1200-375=825$ m.
2. Find B's speed: since B covers 1000 m in the SAME 100 seconds that A takes to finish, B's speed$=1000/100=10$ m/s.
3. Find C's speed: similarly, C's speed$=825/100=8.25$ m/s.
4. Find B's time to complete the FULL 1200 m race: $T_B=1200/10=120$ seconds.
5. Now find how far C has traveled in B's total time (120 seconds): distance$=8.25\times120=990$ m.
6. Distance by which B beats C (at the point B finishes the race) $=1200-990=210$ m.
7. Answer: B's time to complete the race is 120 seconds, and B beats C by 210 m. This demonstrates the standard advanced technique for THREE-WAY race comparison problems: first extract each runner's actual SPEED using the "same time, different distance" reflex relative to the fastest runner's fixed finishing time, then use those speeds to independently answer any pairwise comparison question (B vs C here) by computing each runner's distance covered over whatever NEW time interval is relevant to that specific sub-question (B's own finishing time, in this case) — treating speed as the stable intermediate quantity that unlocks every subsequent comparison.

6. Chapter Checklist for Students

• I clearly distinguish "gives a start of x metres" (reduces the opponent's effective required distance BEFORE the race) from "beats by x metres" (the gap AT THE FINISH when the winner completes the full distance).
• I default to the "same time, different distance = speed ratio" reflex for any race problem that doesn't explicitly require introducing a time variable.
• I add time-based "beats by" margins directly along a chain (A beats B by t1, B beats C by t2 → C's time = A's time + t1 + t2), since time is directly additive.
• I multiply SPEED RATIOS (never simply add distance margins) when solving chain races involving distance-based "beats by" conditions.
• I extract each runner's actual SPEED as the first step in any multi-runner comparison problem, then use that speed to answer whatever specific pairwise question is asked, rather than trying to compare distances/times directly across mismatched time intervals.