Races and Circular Tracks
Races and Circular Tracks
A race question compares two runners over the same distance. "A beats B by x metres" means when A finishes, B is x metres behind; "by t seconds" means B finishes t seconds after A. On a circular track, runners meet when the faster gains (or together they cover) one full lap.
1. Rules
| Statement | Meaning |
|---|---|
| A beats B by x m in a d m race | When A runs d, B runs d − x; speed ratio A : B = d : (d − x) |
| A beats B by t s | B takes t seconds more than A for d metres |
| A beats B by x m or t s | B covers x metres in t seconds: B's speed = x/t |
| A gives B a start of x m | B starts x metres ahead; a dead heat means both finish together |
| Chain: A beats B, B beats C | Multiply ratios: C's distance = (B's distance when A finishes) × (C's per B) |
Circular track of length L, speeds u > v:
Opposite directions: first meeting after L/(u + v) Same direction: first meeting after L/(u − v)
First meeting at the starting point: LCM of the lap times L/u and L/v
Distinct meeting points (speeds in ratio a : b in lowest terms): opposite a + b, same direction a − b
✗ A beats B by 20 m and B beats C by 25 m in 500 m, so A beats C by 45 m | ✓ C runs 475/500 of B's distance: when B is at 480, C is at 456, so A beats C by 44 m
हिंदी नोट: "A ने B को x मीटर से हराया" का अर्थ है जब A दौड़ पूरी करता है तब B, x मीटर पीछे होता है। वृत्ताकार पथ पर विपरीत दिशा में चलने पर चालें जुड़ती हैं, एक दिशा में घटती हैं।
Exam Pointer: Verified NTPC patterns: A beats B by metres and seconds (Graduate CBT-1, March 2026) and runners on a circular track, first meeting and number of meeting points (Graduate CBT-1, June 2025 and March 2026). Head-start questions had no verified NTPC shift and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: A beats B by metres or seconds
[PYQ: NTPC Graduate CBT-1 22-Mar-2026 Shift-2]
EXAM LEVEL
Q. In a 1 km race A finishes in 190 seconds and beats B by 50 m. By how many seconds does A beat B?
In 190 seconds B covers 950 m, so B's speed = 5 m/s and B needs 200 seconds for 1,000 m. A wins by 10 seconds.
Answer: 10 seconds
EXAMATLAS LEVEL
Q. In a 1,000 m race A beats B by 40 m or 8 seconds. Find A's time.
B covers the 40 m gap in 8 seconds, so B's speed = 5 m/s and B's time = 200 seconds. A finishes 8 seconds earlier: 192 seconds.
Answer: 192 seconds
Pattern 2: Head starts and chains of races
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. A runs 1.25 times as fast as B. In a 500 m race, what start should A give B so that the race ends in a dead heat?
While A runs 500 m, B runs 500/1.25 = 400 m. So B needs a start of 100 m.
Answer: 100 m
EXAMATLAS LEVEL
Q. In a 500 m race A beats B by 20 m and B beats C by 25 m. By how many metres does A beat C?
When A finishes, B is at 480 m. C runs 475 m while B runs 500 m, so when B is at 480, C is at 480 × 475/500 = 456 m. A beats C by 44 m.
Answer: 44 m
Pattern 3: Runners on a circular track
[PYQ: NTPC Graduate CBT-1 6-Jun-2025 Shift-3 | NTPC Graduate CBT-1 27-Mar-2026 Shift-3]
EXAM LEVEL
Q. Two runners start together from the same point of a 600 m circular track at 5 m/s and 7 m/s. When do they first meet if they run in opposite directions, and if they run in the same direction?
Opposite: 600/(5 + 7) = 50 seconds. Same direction: the faster must gain a lap, 600/(7 − 5) = 300 seconds.
Answer: 50 seconds; 300 seconds
EXAMATLAS LEVEL
Q. For the same two runners (5 m/s and 7 m/s on a 600 m track), at how many distinct points do they meet when running in opposite directions, and when do they first meet at the starting point?
Speed ratio 5 : 7 in lowest terms, so in opposite directions they meet at 5 + 7 = 12 distinct points. Lap times are 120 s and 600/7 s; their LCM = 600/HCF(5, 7) = 600 seconds, the first time both are back at the start together.
Answer: 12 points; after 600 seconds
Pattern 4: Dead heats and starts from speeds or times
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. A runs 400 m in 50 seconds and B in 60 seconds. How many seconds' start can A give B so that they finish together?
B needs 10 seconds more for the same distance, so A can give B a 10-second start. In metres: in 50 seconds B covers 400 × 50/60 = 333 1/3 m, so a start of 66 2/3 m.
Answer: 10 seconds (66 2/3 m)
EXAMATLAS LEVEL
Q. In a 1 km race A can give B a 100 m start and C a 150 m start. How much start can B give C in a 1 km race?
When A runs 1,000 m, B runs 900 m and C runs 850 m, so C runs 850/900 of B's distance. When B runs 1,000 m, C runs 1,000 × 850/900 = 944 4/9 m. B can give C a start of 55 5/9 m, not 50 m.
Answer: 55 5/9 m
60-Second Revision
- Beats by x m in d m: speed ratio d : (d − x); by x m or t s: loser's speed = x/t.
- Head start for a dead heat: start = d − (slower's distance while faster runs d).
- Chains multiply ratios; never add the margins.
- Circular track: opposite L/(u + v), same L/(u − v); points a + b or a − b.
Next Step: Time, Speed and Distance done. Practise trains, boats and races in the ExamAtlas RRB NTPC 2026 mock tests; convert every speed to m/s before you start.