← All topics🚀 Head Start & Dead HeatMaths
Linear & Circular Races · Topic 2 of 5

🚀 Head Start & Dead Heat

1 exam question type, fully solved

A head start handicaps the faster runner so the race is fair. The start can be given two ways: a few metres ahead (the slower runner has less ground to cover) or a few seconds earlier (the slower runner runs for more time). Reading which kind you have is half the battle.

Turn the words into one equation

• Start of metres. B begins m ahead, so B only runs while A runs the full .

• “Finish together / dead heat” ⇒ equal times: .

• “A beats B by seconds” ⇒ the loser takes more time: .

• Ratio + start. When speeds are a ratio (say A is times as fast), let the finish/meeting distance be ; A covers , B covers , and equal time gives . The faster runner's distance is — and “distance from the start” means exactly this , measured from the common line.

🏃Linear Race SimulatorAnimate the race
AB
A: 0.0 mB: 0.0 mResult: A wins by 20.0 m = 2.5 s

B runs a 500 m race at 5 m/s. If A finishes 10 s sooner, A's time is s.

📝Practice Questions

Q1X gives Y a 20 s head start over 600 m and they finish together. Y runs at 5 m/s. X's time is:

Q2A is twice as fast as B and gives B a 50 m start. Both finish together. The race length is:

Q3In an 800 m race A runs at 8 m/s, gives B a 40 m start and beats B by 5 s. B's speed is about:

📚

Real exam questions — Head Start & Dead Heat

1 question types · 5 solved examples from real SSC papers

A start can be in metres (loser runs less) or in seconds (loser takes more time). Translate the words into "equal time" or "loser's time is bigger" and the rest is one equation.

How to solve this type
A "head start" or "start of ss metres" means the slower runner B begins ss metres ahead, so B only has to cover (d−s)(d-s) metres while A covers the full dd.

The key sentence tells you how their TIMES compare:
• "Both finish together / dead heat" ⇒ A's time = B's time. Set dvA=d−svB\dfrac{d}{v_A}=\dfrac{d-s}{v_B}.
• "A beats B by tt seconds" ⇒ the loser B takes tt seconds MORE, so tB=tA+tt_B=t_A+t.

When speeds are given as a ratio like "A is 54\dfrac{5}{4} times as fast", let the meeting/finish distance be DD: A covers DD, B covers D−sD-s, and equal time gives DvA=D−svB\dfrac{D}{v_A}=\dfrac{D-s}{v_B}. The "distance from the start" they ask for is A's distance DD — measured from the common starting line, not from B's advanced position.

A gives B a 10 s head start in a 1500 m race and they finish together. B's speed is 6 m/s. How long (in minutes) does A take to finish?

A8 minB3 minC5 minD4 min

In a 600 m race, X runs at 12 m/s, gives Y a 30 m start and beats Y by 5 s. Find Y's speed.

A12 m/sB9 m/sC11 m/sD10 m/s

In a 400 m race, A gives B a 15 m start and beats B by 10 s. A's speed is 16 m/s. Find B's speed.

A11 m/sB9 m/sC10 m/sD13 m/s

Ram runs 3/2 times as fast as Mohan. Ram gives Mohan a 200 m head start, and they reach the winning post together. How far is the winning post?

A575 mB550 mC625 mD600 m

Reema runs 5/4 times as fast as Rekha. Reema gives Rekha a 50 m head start. At what distance from the starting point do they meet?

A225 mB240 mC200 mD250 m