Time, Speed & Distance

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high importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Races & Handicaps

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⏱ 5 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

In a race of L metres, 'A beats B by x metres' means when A finishes L, B has covered only $L - x$ .

Same running time, so $\dfrac{S_A}{S_B} = \dfrac{L}{L - x}$ .

'A beats B by t seconds': B needs t more seconds after A finishes. A dead heat means they finish together. 'A gives B a start of 20 m' shortens B's race to $L - 20$ .

The language of races

Decode the phrase first; the arithmetic is then plain.

"A beats B by x metres" — when A covers the full distance D, B has covered $D - x$ .
"A beats B by t seconds" — B needs t more seconds after A finishes.
"A gives B a start of x metres" — B begins x m ahead, running only $D - x$ .
"A gives B a start of t seconds" — B starts running t seconds early.

A dead heat is a tie: both finish together.

Rule: Whoever "gives" the start is the faster runner and still runs the FULL distance.

Beats by metres gives the speed ratio

When A finishes, both have run for the same time, so distances carry the ratio:

\dfrac{S_A}{S_B} = \frac{D}{D - x}

In a 100 m race, beating B by 10 m makes the speeds $100 : 90 = 10 : 9$ . That single ratio answers most follow-ups: longer races, new margins, the start needed for a tie.

Tip: Compute the ratio once, reuse it everywhere in the question.

Beats by seconds, or by both

Beaten by t seconds: $t_B = t_A + t$ .
"A beats B by 20 m or 5 s" links both views: B runs the final 20 m in 5 s → $S_B = \frac{20}{5} = 4$ m/s. B's full time is $\dfrac{200}{4} = 50$ s, so A's is 45 s.
Winner's times given: A runs 100 m in 12 s, B in 15 s. When A finishes, B has covered $\dfrac{100}{15} \times 12 = 80$ m → beaten by 20 m.

Starts and handicaps

A start of x m means B runs $D - x$ . A dead heat then gives $\dfrac{S_A}{S_B} = \frac{D}{D-x}$ — the same equation as "beats by x", read the other way.

If A gives the start AND still wins by y m, B effectively ran $D - x - y$ . A gives B 30 m in a 300 m race and wins by 45 m: B ran $300 - 30 - 45 = 225$ → speeds $300 : 225 = 4 : 3$ .

Watch: Apply the start to the RECEIVER, and subtract the winning margin too when there is one.

Games of one hundred

"A can give B 20 points in a game of 100" means $S_A : S_B = 100 : 80$ . Chain two statements by multiplying ratios.

A gives B 20, B gives C 10: when A scores 100, C scores $80 \times \frac{90}{100} = 72$ → A can give C 28 points.

Chained races

"A beats B by 5 m in 100 m; B beats C by 8 m in 100 m" does NOT give "A beats C by 13 m". Convert to ratios and multiply: when A runs 100, C runs $95 \times \frac{92}{100} = 87.4$ → A beats C by 12.6 m.

Careful: Margins multiply through ratios; they never add.

Circular tracks

Two runners on a circular track of length L meet again when the FASTER gains exactly one full lap:

\text{same way: } t = \frac{L}{S_1 - S_2}, \qquad \text{opposite ways: } t = \frac{L}{S_1 + S_2}

On a 400 m track at 6 m/s and 4 m/s: same way they meet every $\dfrac{400}{2} = 200$ s; opposite ways every $\dfrac{400}{10} = 40$ s.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Race won by x metres

How to spot it:

'A beats B by x metres in a D-metre race' — the speed ratio, a margin in a longer race, or a speed is asked.

\frac{S_A}{S_B} = \frac{D}{D - x}

Method

At the moment A finishes, B has covered exactly D − x in the same time.
Equal times cancel: speeds are in the ratio D : (D − x).
Reuse the ratio for every follow-up.

Why it works:

'Beats by x' is two runners observed over one common time interval.

Try this

In a 1 km race A beats B by 100 m. If A runs at 11 m/s, find B's speed.

Show solution

$\dfrac{S_B}{S_A} = \frac{900}{1000}$ .
$S_B = 11 \times \frac{9}{10} = 9.9$ m/s.

Answer

9.9 m/s

Type 2very common3 practice Q

Race won by t seconds or by x m or t s

How to spot it:

'A beats B by t seconds', or the combined 'by x metres or t seconds' — a finishing time or a speed is asked.

t_B = t_A + t; \qquad S_B = \frac{x}{t}

Method

By t s: the loser needs t more seconds after the winner finishes.
By x m or t s: the loser covers the last x m in t s → $S_B = \frac{x}{t}$ .
Convert to full times: $t_B = \frac{D}{S_B}$ , then $t_A = t_B - t$ .

Why it works:

x metres and t seconds are two views of the loser's final stretch.

Try this

In a 200 m race A beats B by 20 metres or 5 seconds. Find A's time for the race.

Show solution

$S_B = \frac{20}{5} = 4$ m/s.
B's full time = $\dfrac{200}{4} = 50$ s.
A's time = $50 - 5 = 45$ s.

Answer

45 seconds

Type 3common3 practice Q

Starts and handicaps

How to spot it:

A runner gets a head start of x metres or t seconds; the dead heat, the win margin or the needed start is asked.

B \text{ runs } D - x; \quad \text{giver still wins by } y \Rightarrow S_A:S_B = D:(D - x - y)

Method

Write how far each runner actually runs — the receiver runs D − x.
Dead heat → equal times → ratio D : (D − x).
Giver still wins by y → the receiver ran D − x − y.

Why it works:

A start shortens one course; a winning margin shortens it again at the finish.

Try this

In a 300 m race A gives B a start of 30 m and still beats him by 45 m. Find the ratio of their speeds.

Show solution

B runs $300 - 30 - 45 = 225$ m.
$S_A : S_B = 300 : 225 = 4 : 3$ .

Answer

4 : 3

Type 4common2 practice Q

Games of points and chained races

How to spot it:

'A can give B p points in a game of N' chains, or two race margins (A over B, B over C) combine into A over C.

\frac{S_A}{S_B} = \frac{N}{N - p}; \qquad \frac{S_A}{S_C} = \frac{S_A}{S_B} \times \frac{S_B}{S_C}

Method

Turn every statement into a ratio: giving p in a game of N means N : (N − p).
Multiply the ratios along the chain.
Read the combined margin from the final ratio.

Why it works:

Margins compose multiplicatively because each link compares distances over a common time.

Try this

In a game of 100, A can give B 20 points and B can give C 10 points. How many points can A give C in a game of 100?

Show solution

$\dfrac{A}{B} = \frac{100}{80}$ , $\dfrac{B}{C} = \frac{100}{90}$ .
When A has 100, C has $80 \times \frac{90}{100} = 72$ .
A gives C $100 - 72 = 28$ points.

Answer

28 points

Type 5occasional

Meeting on a circular track

How to spot it:

Two runners start together on a circular track; when they next meet (at the start point or elsewhere) is asked.

t = \frac{L}{S_1 - S_2} \ (\text{same way}), \qquad t = \frac{L}{S_1 + S_2} \ (\text{opposite})

Method

Same direction: the faster gains one full lap L at the difference of speeds.
Opposite directions: together they cover one lap L at the sum of speeds.
Multiply the gap time by either speed for where they meet.

Why it works:

Meeting again needs exactly one lap of separation, gained or covered together.

Try this

Two runners start together on a 400 m circular track at 6 m/s and 4 m/s, in the same direction. After how long do they meet for the first time? And if they run in opposite directions?

Show solution

Same way: $\dfrac{400}{6 - 4} = 200$ s.
Opposite: $\dfrac{400}{6 + 4} = 40$ s.

Answer

200 s same way; 40 s opposite ways

Formula sheet

Beating margin in metres

\frac{S_A}{S_B} = \frac{D}{D - x}

Beating margin in time

t_B = t_A + t; \quad S_B = \frac{x}{t}

Start handicap

B \text{ runs } D - \text{start} \ ( - \text{win margin})

Circular track meeting

t = \frac{L}{S_1 \mp S_2}

Shortcuts that save time

⚡ Translate beats by x m into a speed ratio

Same finishing time — distances are in the speed ratio.

Example

In a 200 m race A beats B by 20 metres. Find the ratio of their speeds.

Show solution

$S_A : S_B = 200 : 180$ .
= $10 : 9$ .

Answer

10 : 9

⚡ By x metres or t seconds reveals both speeds

B's last x metres took t seconds — that is B's speed for free.

Example

In a 200 m race A beats B by 20 m or 5 seconds. Find B's speed.

Show solution

$S_B = \frac{20}{5} = 4$ m/s.
(B's full time = $\dfrac{200}{4} = 50$ s; A's = 45 s.)

Answer

4 m/s

⚡ Handle the handicap

A start shortens one runner's distance — recompute the ratio.

Example

A gives B a start of 20 m in a 200 m race and they finish together. Find the ratio of their speeds.

Show solution

B runs $200 - 20 = 180$ m while A runs 200 m.
$S_A : S_B = 200 : 180 = 10 : 9$ .

Answer

10 : 9

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Reading beats by 20 m as B running 20 m.

B runs D − 20 when A runs D; B is 20 m SHORT at the finish.

Mistake 02

Confusing a time margin with a distance margin.

By t seconds is about finishing times; by x metres is about positions.

Mistake 03

Applying the start to the giver.

The RECEIVER of the start runs the shorter distance D − x.

Mistake 04

Adding margins in a chained race.

Multiply ratios: A over C comes from (A/B) × (B/C).

Mistake 05

Using the winner's time for the loser's distance.

Match each distance with its own runner's time.

Quick revision

Read this the night before the exam.

Beats by x m → S_A : S_B = D : (D − x).
Beats by t s → t_B = t_A + t; by x m or t s → S_B = x/t.
Start of x m → B runs D − x; if the giver still wins by y, B ran D − x − y.
Games of 100: giving p points → N : (N − p); chain by multiplying.
Circular track: same way L/(difference), opposite ways L/(sum).

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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