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AFCAT Numerical Ability · Formula sheet

Speed, Time and Distance formulas

37 formulas and 46 common traps for AFCAT Numerical Ability Speed, Time and Distance, grouped by subtopic.

Full notes with worked examples

The Basic Formula, Units and Journeys in Parts

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Speed, distance, time and their units

Speed and the unit change

speed=distancetime,1 km/h=518 m/s\text{speed} = \dfrac{\text{distance}}{\text{time}}, \qquad 1 \text{ km/h} = \tfrac{5}{18} \text{ m/s}

Going and coming back at two speeds

Distance from a round trip

da+db=T  ⇒  d=abTa+b\dfrac{d}{a} + \dfrac{d}{b} = T \;\Rightarrow\; d = \dfrac{abT}{a + b}

One journey at two speeds, and rests on the way

A journey in two parts

xa+D−xb=T\dfrac{x}{a} + \dfrac{D - x}{b} = T

Common traps

Change the units before you divide

A 150 m train at 54 km/h: 150÷54150 \div 54 means nothing. Write 54 km/h as 15 m/s first, and the train passes a pole in 10 s.

Minutes are not decimal places

2 h 45 min is 2.75 h, not 2.45 h. Divide the minutes by 60.

The total time covers both legs

If the round trip takes 8 h, the 8 h is the ride out and the walk back together. Solving d15=8\tfrac{d}{15} = 8 uses the whole time on one leg.

Turn the minutes into hours

3 h 36 min is 3.6 h, not 3.36 h. Put the wrong figure into da+db=T\tfrac{d}{a} + \tfrac{d}{b} = T and every option is missed.

The two parts are x and what is left

Writing xa+xb=T\tfrac{x}{a} + \tfrac{x}{b} = T makes both parts the same length. The second part is D−xD - x.

No rest after the last stretch

A 5 km walk with a rest after every kilometre has 4 rests, not 5. The walker has arrived after the fifth.

Average Speed: Total Distance over Total Time

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Total distance over total time

Average speed

vˉ=d1+d2+…t1+t2+…\bar v = \dfrac{d_1 + d_2 + \dots}{t_1 + t_2 + \dots}

Speeds held for given times

Average over given times

vˉ=v1t1+v2t2t1+t2\bar v = \dfrac{v_1 t_1 + v_2 t_2}{t_1 + t_2}

Equal distances: there and back, or equal stretches

Average over two equal distances

vˉ=2aba+b\bar v = \dfrac{2ab}{a + b}

Fractions of the journey, and working back from an average

Average over fractions of the way

vˉ=1pa+qb(p+q=1)\bar v = \dfrac{1}{\tfrac pa + \tfrac qb} \quad (p + q = 1)

Stoppages and breaks

Minutes stopped per hour

stop per hour=v1−v2v1×60 min\text{stop per hour} = \dfrac{v_1 - v_2}{v_1} \times 60 \text{ min}

Common traps

Do not average the speeds

120 km at 60 km/h and 60 km at 20 km/h is 180 km in 5 h, 36 km/h. The mean of the two speeds, 40, counts the slow part as if it lasted no longer than the fast one.

Weight by time

The plain mean of the speeds is right only when the times are equal. 1 h at 35 km/h and 3 h at 55 km/h gives 50 km/h, not the mean 45, because the faster speed lasted three times as long.

The mean is too high

Out at 30 km/h and back at 70 km/h averages 42 km/h, not 50. The return is quick, so the slow leg fills most of the time.

Work out a changed speed first

'Out at 24 km/h and back 50% faster' means 36 km/h back. Find the second speed before using 2aba+b\tfrac{2ab}{a + b}: here 28.8 km/h.

Fractions of distance weight the times, not the speeds

A third of the way at 10 km/h and two thirds at 20 km/h is not 13×10+23×20≈16.7\tfrac13 \times 10 + \tfrac23 \times 20 \approx 16.7 km/h. Each part took 1 h, so the answer is 15 km/h.

Divide by the running speed

The 12 km lost each hour would have been run at 60 km/h, so it is 12 minutes. Dividing by the 48 km/h average gives 15 minutes, which is wrong.

Include the break, or leave it out, as asked

'Average speed including the break' puts the break in the time. 'Speed while driving' leaves it out. The options usually carry both.

Speed and Time in Inverse Proportion

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Speed ratios turn into time ratios

Same distance

t1t2=v2v1\dfrac{t_1}{t_2} = \dfrac{v_2}{v_1}

Through the distance: a new speed or a new time

Keeping to a schedule

required speed=distance lefttime left\text{required speed} = \dfrac{\text{distance left}}{\text{time left}}

A percent change in speed and the change in time

Speed and time move inversely

speed×ab  ⇒  time×ba\text{speed} \times \tfrac{a}{b} \;\Rightarrow\; \text{time} \times \tfrac{b}{a}

Common traps

Turn the ratio upside down

Speeds 4:54 : 5 do not give times 4:54 : 5. The faster car takes less time, so the times are 5:45 : 4.

Use the distance left, not the whole route

After 110 of 280 km in 2 h, the bus needs 170 km in the 2 h left, which is 85 km/h. Dividing the whole 280 km by 2 h gives 140 km/h.

The two percentages differ

25% faster saves 20% of the time, not 25%. And 20% slower costs 25% more time, not 20%.

Late or Early: Finding the Usual Time and the Distance

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Slower than usual: the delay gives the usual time

The delay at a slower speed

delay=(qp−1)t\text{delay} = \left(\dfrac qp - 1\right) t

Faster than usual: the time saved, and a late start

The saving at a faster speed

saving=(1−ba)t\text{saving} = \left(1 - \dfrac ba\right) t

Two speeds and the gap between their times

The gap between two travel times

da−db=gap in hours\dfrac da - \dfrac db = \text{gap in hours}

Common traps

The fraction is of the usual time

At 45\tfrac45 of the speed the delay is a quarter of the usual time, not a fifth. The fifth is the drop in speed; the time grows by 54\tfrac54, which is a quarter more.

Answer what is asked

Some questions want the slower trip's time, not the usual time. With a usual time of 48 minutes and a 12 minute delay, the slow trip takes 60 minutes.

The saving is a fraction of the usual time

At 65\tfrac65 of the speed the saving is a sixth of the usual time, not a fifth. The time becomes 56\tfrac56 of usual.

A late start is a saving

'Started 20 minutes late but arrived on time' tells you the faster speed saved 20 minutes. It is not a delay to add.

Late and early add

8 minutes late at one speed and 8 minutes early at the other is a gap of 16 minutes, not 0. Subtract only when both are late or both early.

Minutes into hours before solving

With speeds in km/h, the gap must be in hours. Writing d3−d5=16\tfrac d3 - \tfrac d5 = 16 gives 120 km instead of 2 km.

Relative Speed: Meeting, Separating and Chasing

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Adding or subtracting the speeds

Relative speed

vrel=a+b  (opposite),vrel=a−b  (same way)v_{\text{rel}} = a + b \;(\text{opposite}), \qquad v_{\text{rel}} = a - b \;(\text{same way})

Meeting head on: when and where

Time and place of meeting

t=Da+b,dA=aa+bDt = \dfrac{D}{a + b}, \qquad d_A = \dfrac{a}{a + b} D

Meeting when one starts later

A late second start

t′=D−a t0a+bt' = \dfrac{D - a\,t_0}{a + b}

Chasing: catching up from behind

Catch-up time

t=head startvchaser−vleadert = \dfrac{\text{head start}}{v_{\text{chaser}} - v_{\text{leader}}}

A bird flying to and fro

The bird's distance

bird’s distance=vbird×Da+b\text{bird's distance} = v_{\text{bird}} \times \dfrac{D}{a + b}

Speed ratios from meeting and passing times

Times after meeting

vAvB=t2t1,tmeet=t1t2\dfrac{v_A}{v_B} = \sqrt{\dfrac{t_2}{t_1}}, \qquad t_{\text{meet}} = \sqrt{t_1 t_2}

Common traps

Moving apart still adds

Cars driving away from each other open the gap at the sum of their speeds, just as cars driving towards each other close it at the sum. Only the same direction subtracts.

The faster one covers more of the gap

At 50 km/h and 70 km/h the 360 km is shared as 150 and 210, in the ratio of the speeds, not 180 each.

Count from the right start

The 4 h after 8 am is from the second car's start. If the question counts from the first start, add the hour it ran alone.

The two runners cover different distances

The thief runs 450 m and the policeman 600 m: they differ by the head start. Read which one the question asks about.

A head start in time is not a distance

'Starts 1 h 30 min later' must be turned into a gap, 40×1.5=6040 \times 1.5 = 60 km, before dividing by the gain.

Do not follow the zigzags

Adding up each leg of the bird's flight is an endless sum. One multiplication, its speed times the meeting time, gives the total.

The quicker finish belongs to the faster train

With 16 h and 25 h after meeting, the train taking 16 h is the faster one, so the ratio is 5:45 : 4 and not 4:54 : 5.

Trains: Lengths, Platforms and Crossings

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Passing a pole or a standing man

Passing a point

t=Lvt = \dfrac{L}{v}

Crossing a platform or a bridge

Crossing a platform

t=L+Pvt = \dfrac{L + P}{v}

Length and speed from two crossings

Speed from two crossings

v=P2−P1t2−t1v = \dfrac{P_2 - P_1}{t_2 - t_1}

Passing someone who is moving

Passing a moving person

t=Lv∓ut = \dfrac{L}{v \mp u}

Two trains crossing or overtaking

Two trains

t=L1+L2v1±v2t = \dfrac{L_1 + L_2}{v_1 \pm v_2}

Common traps

Metres with metres per second

A 180 m train at 81 km/h: 180÷81180 \div 81 mixes metres with kilometres per hour. Change 81 km/h to 22.5 m/s first.

Add the train's own length

A 200 m train crossing a 300 m platform must cover 500 m. Using 300 m stops the clock while most of the train is still on the platform.

The man on the platform is a pole

Passing a man who stands on the platform takes the train's length only. Adding the platform to that crossing double counts it.

The man in the train

A man in one train watching the other go past sees only the other train's length go by. Adding both trains' lengths is for when the trains clear each other completely.

Both lengths, and subtract for the same direction

Clearing another train covers L1+L2L_1 + L_2, not one length. And overtaking on a parallel track closes at the difference of the speeds, which is why it takes so much longer.

Boats and Streams: Downstream, Upstream and Still Water

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Downstream and upstream speeds

With and against the stream

vdown=b+s,vup=b−sv_{\text{down}} = b + s, \qquad v_{\text{up}} = b - s

Still water and stream from the two speeds

Half the sum, half the difference

b=d+u2,s=d−u2b = \dfrac{d + u}{2}, \qquad s = \dfrac{d - u}{2}

Equal times and ratios of time

Equal times

d1b+s=d2b−s\dfrac{d_1}{b + s} = \dfrac{d_2}{b - s}

Out and back in a total time

A round trip on a river

db+s+db−s=T\dfrac{d}{b + s} + \dfrac{d}{b - s} = T

Two trips, two equations

Two trips

p1x+q1y=T1,p2x+q2y=T2,x=1u,  y=1dp_1 x + q_1 y = T_1, \quad p_2 x + q_2 y = T_2, \quad x = \tfrac1u,\; y = \tfrac1d

Boats meeting, chasing and crossing the river

Two boats closing

vclose=(b1+s)+(b2−s)=b1+b2v_{\text{close}} = (b_1 + s) + (b_2 - s) = b_1 + b_2

Common traps

Still water is not downstream

'Speed of the boat' in these questions means its speed in still water, bb. A boat that goes 20 km/h downstream on a 3 km/h stream has b=17b = 17, not 20.

Half the difference

With 15 km/h down and 7 km/h up, the stream is 4 km/h, not 8. The difference 8 counts the stream twice, once added and once taken away.

Twice the time is half the speed

'Takes twice as long upstream' makes the downstream speed double the upstream speed. Writing b−s=2(b+s)b - s = 2(b + s) turns the stream backwards.

The round trip average is below b

At b=6b = 6 and s=2s = 2 the round trip averages 5135\tfrac13 km/h, not 6. The slow upstream leg lasts longer than the quick one.

Solve for the reciprocals

Taking uu and dd as the unknowns leaves 12u+36d=6\tfrac{12}{u} + \tfrac{36}{d} = 6, which is not linear. With x=1ux = \tfrac1u and y=1dy = \tfrac1d it is.

The stream cancels for boats that meet

One boat gains the stream and the other loses it, so the meeting time depends only on the boats' own speeds. Adding the stream to both counts it twice.

Races: Beating, Head Starts and Dead Heats

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Beating by a distance or by a time

Speed ratio from a margin

vAvB=DD−x\dfrac{v_A}{v_B} = \dfrac{D}{D - x}

Head starts and dead heats

Start for a dead heat

start=D(1−vBvA)\text{start} = D\left(1 - \dfrac{v_B}{v_A}\right)

Three runners: chaining the results

C's distance when A finishes

(D−x)(D−y)D\dfrac{(D - x)(D - y)}{D}

Common traps

Beaten by a distance, or by a time

'Beats by 10 s' is a gap in time: B's time is 10 s more. It is not 10 m to subtract from the race.

Count the start once

With an 80 m start in an 800 m race, B has 720 m to run. When A wins by 30 m, B has run 690 m, not 770.

Distance margins do not add

Beating B by 25 m and B beating C by 50 m does not make 75 m: while A runs the race, B runs only 225 m, so C falls behind B by less than 50 m. The answer is 70 m.

Circular Tracks: Meetings, Laps and the Starting Point

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The first meeting: opposite ways or the same way

First meeting on a track

topp=La+b,tsame=La−bt_{\text{opp}} = \dfrac{L}{a + b}, \qquad t_{\text{same}} = \dfrac{L}{a - b}

Counting meetings and laps

Meetings in a time T

n=T vrelL    (whole meetings only)n = \dfrac{T\,v_{\text{rel}}}{L} \;\;(\text{whole meetings only})

Back at the start together

Together at the start

Tstart=LCM(t1,t2,t3)T_{\text{start}} = \text{LCM}(t_1, t_2, t_3)

Common traps

The same way needs a whole lap

Two runners going the same way from one point next meet when the faster has gained a full lap, not half a lap. Half a lap ahead is the moment they are furthest apart.

Whole meetings only

If the time allows 3.6 laps of gain, A has lapped B 3 times. The fourth lapping has not happened yet.

Meeting is not meeting at the start

At 6 m/s and 4 m/s on a 300 m track, opposite runners first meet after 30 s, but they are both at the start only after the LCM of their lap times, 50 s and 75 s, which is 150 s.

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