Races and Games of Skill — Formulas, Shortcuts, and Worked Examples | TNPSC, SSC, Bank

Quick Concept Summary

Races and Games of Skill questions test your ability to compare speeds or skill levels between competitors using ratios and relative performance data.

Key terms you must know:

• Race: A competition to cover a fixed distance. The winner is who completes it first.
• Start / Head start: A distance advantage given to a weaker competitor. If A gives B a start of 10 m in a 100 m race, B begins 10 m ahead.
• Beat by distance: A wins and B is still d metres behind the finish line when A finishes.
• Beat by time: A finishes t seconds before B crosses the finish line.
• Dead heat: Both finish at the same time (a tie).
• Game of n: Scored game (like billiards) where the winner reaches n points. If A gives B m points in a game of n, A scores n when B scores only (n − m).

The core insight: Speed ∝ Distance when time is constant. So comparing distances covered by two runners in the same time gives their speed ratio directly.

Key Formulas

Worked Examples

Example 1 — Beat by Distance

Problem: In a 100 m race, A beats B by 10 m. Find the ratio of speeds of A and B.

Example 2 — Head Start Problem

Problem: In a 200 m race, A gives B a start of 20 m and still beats B by 10 m. Find their speed ratio.

Example 3 — Beat by Time

Problem: A runs 100 m in 10 seconds. A beats B by 2 seconds. How far behind was B when A finished?

Example 4 — Three-Competitor Race

Problem: In a 500 m race, A beats B by 50 m and A beats C by 100 m. By how much does B beat C in the same 500 m race?

Example 5 — Game of Points

Problem: In a game of 100, A can give B 20 points and A can give C 32 points. How many points can B give C in a game of 80?

Shortcuts and Tricks

Trick 1 — Direct Speed Ratio from Race Data

Whenever two runners complete a race where one person finishes and the other hasn't, their speed ratio = distances covered by each in the same duration.

A beats B by 15 m in 100 m → Speed ratio = 100 : 85 = 20 : 17

Trick 2 — Combining "Start + Beat"

If A gives B a start of x metres AND beats B by y metres in a race of L metres:

• Actual distance B runs = L − x − y
• Speed ratio A : B = L : (L − x − y)

Trick 3 — Three-Person Race Shortcut

A beats B by x, A beats C by y (race of L):

• B beats C by = L − (L−x)(L−y)/L

Example: L=100, x=10, y=20: B beats C = 100 − (90×80/100) = 100 − 72 = 28 m — but only in a 100 m race. Scale if needed.

Trick 4 — Game of Points Conversion

To go from "A gives B m points in game of n" to "B gives C p points in game of q": 1. Find common base (when A scores n, B scores n−m, C scores n−p) 2. Use cross multiplication to find B vs C ratio 3. Scale to required game size

Trick 5 — Beat by Time Conversion

If you know A's speed or time, convert "beat by time" to "beat by distance" using: Beat distance = Speed of B × time by which B is beaten

TNPSC Exam Pattern

Frequency in TNPSC exams: Races and Games appears occasionally in Group II and Group IV exams — typically 1 question per paper. It is more commonly tested in SSC CGL, SSC CHSL, and Bank PO exams where 2–3 questions appear.

What TNPSC tests:

• Simple "A beats B by x metres" → find speed ratio
• Basic head start problems
• Very rarely: three-person races or games of points

What SSC/Bank tests additionally:

• Combined start + beat by time problems
• Three-competitor race linking
• Games of points with multiple players

Expected marks in TNPSC Group II/IV: 1 question (2 marks in Group II, 1.5 marks in Group IV).

Strategy for TNPSC: Learn the core speed ratio = distance ratio concept and the two basic scenarios (beat by distance, give a start). That covers ~90% of TNPSC race questions.

Practice Questions

Q1. In a 100 m race, A beats B by 10 m. What is the ratio of speeds of A and B?

Q2. In a 200 m race, A gives B a start of 40 m and wins by 20 m. Find the ratio of their speeds.

Q3. A runs 100 m in 20 sec. B runs 100 m in 25 sec. By how many metres does A beat B?

Q4. In a race of 600 m, A beats B by 60 m and B beats C by 30 m. By how many metres does A beat C?

• When A runs 600 m, B runs 540 m.
• When B runs 600 m, C runs 570 m → when B runs 540 m, C runs = (540 × 570)/600 = 513 m.
• A beats C by 600 − 513 = 87 m.

Q5. A and B run a race of 1 km. A gives B a start of 100 m. If A's speed is 10 m/s and B's speed is 8 m/s, who wins?

• A's time to run 1000 m = 1000/10 = 100 sec.
• B runs 1000 − 100 = 900 m; B's time = 900/8 = 112.5 sec.
• A finishes in 100 sec; B takes 112.5 sec. A wins by 12.5 sec.

Q6. In a game of 90, A can give B 15 points. In a game of 100, A can give C 25 points. How many points can B give C in a game of 80?

• When A=90, B=75 → A:B = 90:75 = 6:5
• When A=100, C=75 → A:C = 100:75 = 4:3
• B:C = (B/A) × (A/C) = (5/6) × (4/3) = 20/18 = 10:9
• In a game of 80 (B scores 80), C scores = 80 × 9/10 = 72.
• B gives C = 8 points.

Q7. A beats B by 30 m in a 120 m race. In a race of 200 m on the same track, how many metres start should A give B for a dead heat?

• Speed ratio A:B = 120:90 = 4:3
• In 200 m, when A runs 200 m, B runs 200 × (3/4) = 150 m.
• B needs to run only 200 − start distance = 150 m → start = 200 − 150 = 50 m.

Q8. In a 400 m race, A beats B by 5 seconds. B's speed is 8 m/s. What is A's speed?

• B's time = 400/8 = 50 sec.
• A's time = 50 − 5 = 45 sec.
• A's speed = 400/45 = 80/9 m/s ≈ 8.89 m/s.

Q9. Three runners A, B, C run a 500 m race. A beats B by 25 m and beats C by 50 m. In a 250 m race, by how many metres does B beat C?

• In 500 m: A=500, B=475, C=450.
• B:C speed ratio = 475:450 = 19:18.
• When B runs 250 m, C runs = 250 × 18/19 = 4500/19 ≈ 236.84 m.
• B beats C by ≈ 13.16 m (or 250/19 m).

Q10. In a race, A can give B a start of 20 m or 5 seconds. Find the length of the race.

• B covers 20 m in 5 seconds → B's speed = 20/5 = 4 m/s.
• A runs the full race while B runs (race − 20) metres, and the time difference is 5 sec.
• Let race = L. Time for A = (L−20)/4 (since A covers L at speed L/(L-20) × 4... )
• Simpler: B runs at 4 m/s. In 5 sec, B covers 20 m. So 20 m = 5 sec for B → B's speed = 4 m/s.
• The race length = ? A gives B a 20 m head start in a race, and A finishes 5 sec before B.
• Let A's speed = v. Then L/v = A's time. B runs (L−20) m at 4 m/s → time = (L−20)/4.
• A beats B by 5 sec: (L−20)/4 − L/v = 5.
• Using head start: speed ratio A:B = L:(L−20). So v/4 = L/(L−20) → v = 4L/(L−20).
• Substitute: (L−20)/4 − (L−20)/4 = 5 → This approach needs one more condition.
• Direct approach: 20 m start = 5 sec for B. B's speed = 4 m/s. A's time = B's time − 5.
• In the race, A runs L in time t; B runs (L−20) in time (t+5).
• (L−20)/4 = t + 5 → t = (L−20)/4 − 5.
• A and B have the same finish time when start is given: A runs L in time t, B runs L−20 in same time t.
• So (L−20)/(L) = B's speed / A's speed. But separately, B covers 20 m in 5 sec.
• A's speed = L/t; B's speed = (L−20)/t = 4 m/s → t = (L−20)/4.
• A's speed = L / [(L−20)/4] = 4L/(L−20).
• Time for B to run full race L = L/4. A beats B by = L/4 − t = L/4 − (L−20)/4 = 20/4 = 5 sec. ✓
• This is consistent for any L unless we use the "5 seconds" condition differently.
• Standard interpretation: The 5-second start means B gets a 5-sec head start OR A beats B by 5 sec.
• If "A gives B a start of 5 seconds" = B starts running 5 sec before A:
• In those 5 sec, B covers 5 × 4 = 20 m. This matches the 20 m start.
• The race length cannot be determined from this alone — any length is consistent.
• Re-read: "A gives B a start of 20 m OR 5 seconds" — these are equivalent starts.
• B's speed = 20 m / 5 sec = 4 m/s. Length determination requires one more piece of data.
• Standard answer for this classic problem type: Race length = 100 m (if additionally A's speed is given or a third condition exists). Check: if L=100, A:B = 100:80 = 5:4. B speed=4 → A speed=5. In 100m race A takes 20 sec, B takes (80/4)=20 sec — they tie, which contradicts. The problem as stated only gives B's speed = 4 m/s. The race distance is indeterminate without more data.