SSC CGL Maths · Time, Speed and Distance

Races

In a race, beating someone by so many metres tells you where the loser is when the winner finishes, beating by so many seconds tells you how much longer the loser needs, and a head start shortens one runner's distance or time.

All Time, Speed and Distance notes

Why this matters

Seven questions in the bank, and two of them are HARD. All are straight races; races round a circular track are on the Circular Tracks page. Every one turns on reading the race statement as two distances run in the same time, or two times for the same distance.

Concept 1 of 2: Beating by a distance or by a time

When the winner crosses the line, the loser is still running. Beating by 50 m tells you where the loser is at that moment; beating by 10 seconds tells you how much longer the loser needs. When both are given, they describe the same gap, so the loser covers those metres in those seconds.

Definition

A beats B by $x$ m in a race of $d$ m: in A's time, B runs $d - x$ . Their speeds are in the ratio $d : (d - x)$ .
A beats B by $t$ s: B's time $=$ A's time $+ t$ .
By $x$ m or $t$ s: B runs those $x$ metres in $t$ seconds, so B's speed is $\frac{x}{t}$ . B's time for the race is $d$ over that speed, and A's time is $t$ less.
A beats both B and C: B's and C's distances when A finishes give their speed ratio. Scale C's distance up to B's full race, then subtract, for the margin by which B beats C.

Beating margins

\frac{s_A}{s_B} = \frac{d}{d - x} \qquad s_B = \frac{x}{t}

$d$ length of the race
$x$ winning margin in metres
$t$ winning margin in seconds

Worked example

In a 600 m race A beats B by 60 m and A beats C by 105 m. By how many metres does B beat C in a 600 m race?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 26 Jul 2023, Shift 4 · Q51Easy

Example 1 · Time, Speed and Distance · Races

In 1 km linear race, P beats Q by 120 metres or 30 sec. What is the time taken by P to cover the race?

Taking the margin as the loser's distance

Beating someone by 80 m in a 1000 m race means the loser has run 920 m when the winner finishes. The 80 m is the gap, not the loser's distance.

Subtracting the two margins

If A beats B by 100 m and C by 190 m in a 1000 m race, B does not beat C by 90 m. When B finishes, C has run

810 \times \frac{1000}{900} = 900

m, so B wins by 100 m.

Drill 2 more on beating by a distance or by a time

Concept 2 of 2: Head starts and dead heats

A head start of

x

metres shortens the slower runner's race to

d - x

, and a head start of

t

seconds lets him run

t

seconds longer. A dead heat means both reach the line at the same moment, so the two runs take the same time.

Definition

Head start in metres, dead heat: $\frac{d}{d - x} = \frac{s_A}{s_B}$ .
Head start in seconds, dead heat: B's time $=$ A's time $+ t$ .
How long must the race be for a dead heat? With speeds in the ratio $k : 1$ and a head start of $x$ m, $d = \frac{kx}{k - 1}$ .
Two races, two unknowns: write each race as the slower runner's distance $=$ his speed $\times$ his time, and subtract one equation from the other. Check both races with your answer.

Dead heat with a head start

\frac{d}{d - x} = \frac{s_A}{s_B} \qquad d = \frac{kx}{k - 1}

$x$ head start given to the slower runner, in metres
$k$ how many times as fast the faster runner is

Worked example

A runs 1.5 times as fast as B. If A gives B a start of 110 m, how long must the race be for a dead heat?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 19 Jul 2023, Shift 4 · Q57Moderate

Example 2 · Time, Speed and Distance · Races

Ashok runs

2\frac{2}{3}

times as fast as Bharat. If Ashok gives Bharat a head start of 160m, then how far must the winning post be so that Ashok and Bharat can reach it at the same time?

Adding the head start to the leader's distance

A start of

x

metres shortens the slower runner's race to

d - x

. The faster runner still runs the full distance.

Treating a dead heat as a win

In a dead heat both runners take the same time, so the head start alone makes up the whole difference in speed. No margin is left over.

Drill 3 more on head starts and dead heats

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Beating by a distance or by a time
Beating margins
$\frac{s_A}{s_B} = \frac{d}{d - x} \qquad s_B = \frac{x}{t}$
Head starts and dead heats
Dead heat with a head start
$\frac{d}{d - x} = \frac{s_A}{s_B} \qquad d = \frac{kx}{k - 1}$

Watch out for (4)

Taking the margin as the loser's distance
→ Beating by a distance or by a time
Subtracting the two margins
→ Beating by a distance or by a time
Adding the head start to the leader's distance
→ Head starts and dead heats
Treating a dead heat as a win
→ Head starts and dead heats

Test yourself on Time, Speed and Distance

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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