Time, Speed and Distance — SSC CGL Formulas and Traps

Speed, Time and Distance Basics

Distance equals speed times time

Speed, distance and time

d = s \times t \qquad s = \frac{d}{t} \qquad t = \frac{d}{s}

$d$ distance covered
$s$ speed, kept constant
$t$ time taken

Changing km/h to m/s and back

Changing units of speed

1 \text{ km/h} = \frac{5}{18} \text{ m/s} \qquad 1 \text{ m/s} = \frac{18}{5} \text{ km/h}

Speed ratios when the time or the distance is fixed

Ratio of two speeds

\frac{s_1}{s_2} = \frac{d_1}{d_2} \times \frac{t_2}{t_1}

$d_1, t_1$ distance and time of the first mover
$d_2, t_2$ distance and time of the second mover

Common traps

Reading 2 hours 30 minutes as 2.3 hours

30 minutes is half an hour, so 2 hours 30 minutes is 2.5 hours. Change minutes to hours by dividing by 60, never by writing them after the decimal point.

Counting clock time across noon

From 9:30 a.m. to 1:00 p.m. is 3 hours 30 minutes. Count up to 12 noon first, then add the afternoon part, instead of subtracting 9 from 1.

Turning the fraction upside down

If changing a speed from km/h to m/s gives a bigger number, the fraction is upside down. 72 km/h is 20 m/s, a smaller number, because a metre is much shorter than a kilometre.

Mixing seconds and minutes with m/s

A speed in m/s needs the time in seconds. 3 minutes at 4 m/s is 180 seconds, so 720 m, not 12 m.

Using the speed ratio as the time ratio

For the same distance the times are in the inverse ratio. Speeds 2 : 3 give times 3 : 2, so the faster car takes the smaller share of time.

Multiplying the two ratios instead of dividing

Covering a quarter of the distance in three times the time gives a speed ratio of

\frac{1}{4} \div 3 = \frac{1}{12}

, not

\frac{3}{4}

. Divide the distance ratio by the time ratio.

Average Speed

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Average speed of a journey in parts

Average speed

\bar{s} = \frac{d_1 + d_2 + d_3}{t_1 + t_2 + t_3} \qquad \bar{s} = \frac{s_1t_1 + s_2t_2}{t_1 + t_2}

$\bar{s}$ average speed for the whole journey
$d_1, t_1$ distance and time of the first part
$s_1$ speed during the first part

Average speed over equal distances

\bar{s} = \frac{2ab}{a + b} \qquad \bar{s} = \frac{3abc}{ab + bc + ca}

$a, b, c$ the speeds, each kept for the same distance

Common traps

Averaging the speeds

The mean of 40 and 80 is 60, but the average speed is 60 km/h only if equal TIMES were spent at each speed. For equal distances it is lower, because more time goes on the slower part.

Leaving a halt out of the time

A stop adds time and no distance, so it lowers the average speed. Include it in the total time unless the question asks for the speed while moving.

Using 2ab/(a + b) when the times are equal

The formula is for equal distances. When equal times are spent at each speed, the average speed is the plain mean (a + b) ÷ 2.

Taking the average speed as the one-way distance

With a total time given, 2ab/(a + b) gives a speed. The whole round trip is that speed times the total time, and the one-way distance is half of it.

Speed Changes and Equations

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Speed and time on the same route

Fixed distance

s_1 t_1 = s_2 t_2 \qquad \frac{t_2}{t_1} = \frac{s_1}{s_2}

$s_1, t_1$ the usual speed and time
$s_2, t_2$ the new speed and time

Two speeds over the same distance and the time between them

Distance from a time gap

d = \frac{s_1 s_2 \, \Delta t}{s_2 - s_1}

$s_1$ the slower speed
$s_2$ the faster speed
$\Delta t$ how much longer the slower trip takes

Halts, rests and the speed needed to arrive on time

Rests after every k km

\text{number of rests} = \frac{d}{k} - 1 \quad (k \text{ divides } d)

Common traps

Taking a 25% rise in speed as a 25% cut in time

On the same route the time falls in the inverse ratio: speed

\times \frac{5}{4}

makes time

\times \frac{4}{5}

, so the cut is 20%, not 25%.

Giving the new speed when the increase is asked

When a question asks by how much the speed must rise, find the new speed and then subtract the old one. The new speed itself is usually a printed wrong option.

Leaving the time gap in minutes

With speeds in km/h the time gap must be in hours. 20 minutes is

\frac{1}{3}

hour; putting 20 into the formula makes the distance 60 times too big.

Subtracting a late time from an early time

Late by 15 minutes at one speed and early by 5 minutes at the other is a gap of 20 minutes, not 10. Add the two when one trip is late and the other is early.

Counting a rest at the finish

A rest after every 8 km on a 40 km trip happens 4 times, not 5: the trip ends at the 40th kilometre and no rest is taken there.

Using the whole trip to find the catching up speed

To arrive on time after a slow start, divide the distance still left by the time still left. Dividing the whole distance by the whole time gives the planned average, not the speed needed now.

Relative Speed: Chasing and Meeting

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Closing a gap at the relative speed

Catching up

t = \frac{\text{gap}}{s_1 - s_2}

$s_1$ speed of the chaser
$s_2$ speed of the one ahead

How far each one runs before the catch

Distance run before the catch

d_{\text{ahead}} = \frac{s_2 \times \text{gap}}{s_1 - s_2} \qquad d_{\text{chaser}} = \frac{s_1 \times \text{gap}}{s_1 - s_2}

$s_1$ speed of the chaser
$s_2$ speed of the one ahead

A chase where the chaser starts later

Chase after a late start

t = \frac{s_2 \times \text{delay}}{s_1 - s_2}

$s_1$ speed of the chaser
$s_2$ speed of the one ahead
$t$ time from the chaser's start to the catch

Moving towards each other

Meeting head on

t = \frac{d}{s_1 + s_2} \qquad s_1 + s_2 = \frac{d}{t_1} \qquad s_1 - s_2 = \frac{d}{t_2}

$d$ distance between the two at the start
$t_1$ time to meet moving towards each other
$t_2$ time for the faster to catch the slower moving the same way

Common traps

Adding the speeds in a chase

When both run the same way, the gap closes at the difference of the speeds. The sum is only for movers coming towards each other.

Dividing metres by km/h

A 450 m gap closed at 3 km/h does not take 150 of anything. Change the gap to km (0.45) or the speed to metres a minute (50) before you divide.

Giving the chaser's distance when the other's is asked

The two distances differ by the starting gap, and both are usually among the options. Underline whose distance the question wants before you pick.

Measuring the catch point from the wrong start

When a question asks at what distance the thief is caught, the distance is counted from the chaser's starting point, which is the gap behind the thief's starting point.

Using the delay as the chase time

The delay only builds the lead. The chase itself lasts the lead divided by the relative speed, counted from the moment the chaser starts.

Adding the chase time to the wrong start

For a clock answer, add the chase time to the time the chaser set off, not to the time the one ahead set off.

Subtracting the speeds for movers coming towards each other

Head on, both close the gap, so the speeds add. The difference is only for one catching up with another going the same way.

Forgetting the early starter's lead

When one sets off an hour earlier, the distance still to close when the second starts is the full distance minus what the first covered in that hour.

Out and back: closing only the one-way distance

When one mover reaches the far end and turns back to meet the other, together they have covered twice the distance between the ends, not once.

Trains and Relative Speed

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A train passing a pole, a platform or a standing train

Train passing a fixed object

t = \frac{L_{\text{train}} + L_{\text{object}}}{s} \qquad s = \frac{p}{t_2 - t_1}

$L_{\text{object}}$ length of the platform, bridge or standing train (0 for a pole)
$s$ speed of the train in m/s
$p$ length of the platform

Trains passing moving trains or people

Two trains crossing

t_{\text{same}} = \frac{L_1 + L_2}{s_1 - s_2} \qquad t_{\text{opp}} = \frac{L_1 + L_2}{s_1 + s_2}

$L_1, L_2$ lengths of the two trains
$s_1, s_2$ speeds of the two trains, the first one the faster

Common traps

Leaving out the train's own length

To clear a platform the train covers the platform and its own length. Using the platform alone gives a time that is too short.

Adding a length for a pole or a standing man

A pole or a person has no length worth counting, so the distance is the train's length alone.

Using one train's length for two moving trains

Two trains have crossed only when their last coaches part, so the distance is the sum of both lengths.

Subtracting the speed of a walker coming towards the train

A walker coming towards the train meets it sooner, so the speeds add. Subtract only when the walker goes the same way as the train.

Boats and Streams

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Downstream and upstream speeds

Boats and streams

D = b + s \qquad U = b - s \qquad b = \frac{D + U}{2} \qquad s = \frac{D - U}{2}

$D$ downstream speed
$U$ upstream speed
$b$ speed of the boat in still water
$s$ speed of the stream

Boat and stream speeds from a ratio

Boat to stream from the two times

\frac{b}{s} = \frac{t_U + t_D}{t_U - t_D}

$t_U$ time upstream over a distance
$t_D$ time downstream over the same distance

Common traps

Forgetting to halve

The stream's speed is HALF the difference of the downstream and upstream speeds, and the boat's is half the sum. The difference itself is twice the stream's speed.

Giving the downstream speed as the boat's speed

The speed of the boat means its speed in still water. The downstream speed already includes the stream, so it is always more than the boat's speed.

Putting the time ratio straight in as the speed ratio

Over the same distance the longer time belongs to the slower, upstream, speed. Upstream and downstream times of 8 and 5 hours mean upstream and downstream speeds in the ratio 5 : 8.

Taking the speeds themselves as the unknowns

An equation such as 36 ÷ U + 20 ÷ D = 8 is not a straight-line equation in U and D. Take 1 ÷ U and 1 ÷ D as the unknowns and it is.

Stopping at the downstream and upstream speeds

These questions usually ask for the boat's speed or the stream's speed. After finding D and U, one more step remains: half their sum or half their difference.

Races

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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}

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

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}

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

Common traps

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.

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.

Circular Tracks

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The first meeting on a circular track

First meeting on a circular track

t_{\text{same}} = \frac{L}{s_1 - s_2} \qquad t_{\text{opp}} = \frac{L}{s_1 + s_2}

$L$ length of one lap
$s_1, s_2$ the two speeds, the first one the faster

Back at the starting point together

Together at the starting point

t = \text{LCM}\left(\frac{L}{s_1}, \frac{L}{s_2}\right)

$\frac{L}{s}$ a runner's time for one lap

Distinct meeting points on a circular track

Number of distinct meeting points

\text{same direction: } a - b \qquad \text{opposite directions: } a + b

$a : b$ the ratio of the speeds in lowest terms

Common traps

Using the difference of speeds for opposite directions

Running towards each other round the track, both close the lap, so the speeds add. The difference is only for runners going the same way.

Waiting for a full lap when one starts ahead

If one runner starts some way ahead on the track, the first overtaking closes only that start gap. A full lap is needed only when both start from the same point.

Counting a meeting after the race is over

In a race of fixed length, the meetings stop when the faster runner finishes. A meeting time later than his finishing time does not count.

Taking the LCM of the speeds

The LCM is taken of the lap TIMES. For speeds of 4 m/s and 6 m/s on a 120 m track the lap times are 30 s and 20 s, so they are at the start together after 60 s.

Giving the first meeting when the start is asked

Runners can meet anywhere on the track long before both are back at the starting point. Read whether the question asks for any meeting or a meeting at the start.

Using the speeds before reducing the ratio

Speeds of 12 m/s and 18 m/s give 5 meeting points running opposite ways, from the lowest ratio 2 : 3. Adding 12 and 18 to get 30 points is wrong.

Rounding up a part of a meeting

10.5 meetings in a period means 10 meetings. The eleventh comes after the period is over.

Time, Speed and Distance formulas

Speed, Time and Distance Basics

Distance equals speed times time

Changing km/h to m/s and back

Speed ratios when the time or the distance is fixed

Common traps

Average Speed

Average speed of a journey in parts

Common traps

Speed Changes and Equations

Speed and time on the same route

Two speeds over the same distance and the time between them

Halts, rests and the speed needed to arrive on time

Common traps

Relative Speed: Chasing and Meeting

Closing a gap at the relative speed

How far each one runs before the catch

A chase where the chaser starts later

Moving towards each other

Common traps

Trains and Relative Speed

A train passing a pole, a platform or a standing train

Trains passing moving trains or people

Common traps

Boats and Streams

Downstream and upstream speeds

Boat and stream speeds from a ratio

Common traps

Races

Beating by a distance or by a time

Head starts and dead heats

Common traps

Circular Tracks

The first meeting on a circular track

Back at the starting point together

Distinct meeting points on a circular track

Common traps

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