Linear & Circular Races · Topic 5 of 5

🧩 Last Meeting & Dead-Heat Systems

2 exam question types, fully solved

The toughest race questions stack two conditions together — a handicap plus a beat, or a “last meeting before the winner finishes”. None of them need new theory: set up the speed ratio, then track distances at the exact moment the question asks about.

Pin the moment, then compare distances

• Two conditions on one race. “A beats B by 10 m and by 2 s” means B covers that last 10 m in 2 s, so m/s. One sentence usually hands you a speed directly.

• Dead-heat with a handicap. Convert “can give a start of m” into the speed ratio , then apply it to the new race length.

• Last meeting before the finish. Find the winner's finishing time, then see where the slower runner is at that instant — the final gap, not a fresh lap.

Always work in one consistent unit (m and s), and keep speeds as a clean ratio so the arithmetic cancels.

🏃Linear Race SimulatorAnimate the race

Speed A (m/s): 10Speed B (m/s): 8Distance (m): 100B's head start (m): 0

A: 0.0 mB: 0.0 mResult: A wins by 20.0 m = 2.5 s

A beats B by 10 m and by 2 s in a race. B's speed is m/s.

Q1On a 1000 m loop A laps in 100 s, B in 200 s. In a 4 km race, their last meeting is at:

Q2In a 400 m race a 30 m start gives a 6 s win, a 50 m start a dead heat. B's speed is:

Q3In a 300 m race a 20 m start gives a 4 s win, a 40 m start a dead heat. A's time for 300 m is:

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Real exam questions — Last Meeting & Dead-Heat Systems

2 question types · 6 solved examples from real SSC papers

Two finishers for the chapter: long circular races where you cap meetings at the winner's finish time, and two-condition dead-heat puzzles you crack by subtracting the two scenarios.

How to solve this type

A race of total distance

R

is run on a loop of length

L

. They meet (same direction) every

\dfrac{L}{|v_1-v_2|}

seconds: at

1,2,3,\dots

times that interval.

But meetings only count while BOTH are still on the track. The faster runner finishes at

\dfrac{R}{v<sub>\text{fast</sub>}}

— read the question carefully, it usually asks "before the FASTER runner finishes". The last meeting is the largest multiple of the interval that is

\le

that finish time.

So: get both speeds from

\dfrac{L}{\text{lap time}}

, find the meeting interval, find the winner's finish time, and take the biggest interval-multiple under it.

A and B run a 12 km race on a 1200 m circular track. A laps in 300 s, B in 400 s. When is their last meeting before the faster runner finishes?

A2400 sB9600 sC6000 sD10800 s

A laps a 1000 m track in 200 s, B in 250 s. They run a 10 km race. When is their last meeting?

A3000 sB4000 sC2000 sD2500 s

A and B run an 18 km race on an 800 m circular track. A laps in 200 s, B in 250 s. When is their last meeting before the faster finishes?

A1800 sB2700 sC2250 sD4000 s