Shortcuts & Speed Methods

Units in One Step

A lot of marks in this chapter are lost on units, not on the method: a train length in metres with a speed in km/h, or 1 hour 12 minutes written as 1.12 hours. So this section starts with units and then goes topic by topic through ratios, average speed, trains, boats, races and tracks. Each method says when to use it, gives the quick working with one worked case, and ends with the places where it gives a wrong answer.

km/h and m/s without a pen

When to use it: the speed is in km/h but the lengths are in metres and the times in seconds (trains, races, tracks), or the other way round.

Fast way: the rule is × $\frac{5}{18}$ one way and × $\frac{18}{5}$, which is × 3.6, the other way. To do it in your head, count in 18s: every 18 km/h is 5 m/s.

• A multiple of 18 converts at once: 126 km/h is 7 × 18, so it is 7 × 5 = 35 m/s.
• A multiple of 9 gives a half: 153 km/h is 8.5 × 18, so it is 8.5 × 5 = 42.5 m/s.
• Going back, × 3.6 is × 4 minus a tenth of the result: 24 m/s gives 24 × 4 = 96, and 96 − 9.6 = 86.4 km/h.

Metres per minute

1 km/h is 1,000 m in 60 minutes, which is $\frac{50}{3}$ m a minute. So m/min × $\frac{3}{50}$ gives km/h. 250 m a minute is 250 × 3 ÷ 50 = 15 km/h.

Minutes into hours

Write the minutes over 60 and cancel before you divide. 9 minutes is $\frac{9}{60} = \frac{3}{20}$ hour, and 1 hour 12 minutes is $1\frac{12}{60}$ = 1.2 hours.

For example: a cyclist covers 2.7 km in 9 minutes. The speed is 2.7 ÷ $\frac{3}{20}$ = 2.7 × $\frac{20}{3}$ = 18 km/h, which is 5 m/s.

Where it fails:

• Multiplying by the wrong fraction. Do a quick check on the answer: people walk at 1 to 2 m/s and trains run at about 10 to 40 m/s. A train at 300 m/s or a walker at 18 m/s means the fraction is upside down.
• Converting the speed but not the time. 1 hour 12 minutes is 1.2 hours, not 1.12 hours.
• Mixing km and m, or hours and seconds, inside one sum. Pick one unit at the start and stay with it.

[SSC Tip] When the data is in metres and seconds but the answer is wanted in km/h, work the whole question in m/s and convert once at the end.

Same Distance: Speed and Time in Inverse Ratio

When to use it: the distance stays the same and the speed changes, or two people cover the same distance at different speeds.

Fast way: for a fixed distance, time goes the opposite way to speed. A speed ratio of a : b means a time ratio of b : a. Then one part does the rest.

For example: A and B cover the same road, with speeds in the ratio 7 : 9, and A takes 18 minutes more. The times are 9 : 7, so the gap of 2 parts is 18 minutes and 1 part is 9 minutes. A takes 81 minutes and B takes 63 minutes.

A fraction of the usual speed

When to use it: 'walking at $\frac{3}{4}$ of his usual speed, he is 20 minutes late'.

Fast way: at $\frac{a}{b}$ of the usual speed, the time becomes $\frac{b}{a}$ of the usual time, so the extra time is $\frac{b - a}{a}$ of the usual time. That gives

$$\text{Usual time} = \text{time lost} \times \frac{a}{b - a}$$

For $\frac{3}{4}$ of the speed this is 3 × the time lost; for $\frac{4}{5}$ it is 4 ×; for $\frac{2}{3}$ it is 2 ×. If the question gives a percentage, turn it into a fraction first: 25% slower is $\frac{3}{4}$, 20% faster is $\frac{6}{5}$, and half the speed is $\frac{1}{2}$, which doubles the time, so the time lost equals the usual time.

For example: walking at $\frac{3}{4}$ of his usual speed, Raju reaches school 14 minutes late. His usual time is 3 × 14 = 42 minutes. Check: the slow walk takes 42 × $\frac{4}{3}$ = 56 minutes, which is 14 more.

Faster than usual

At $\frac{a}{b}$ of the usual speed with a bigger than b, the time saved is $\frac{a - b}{a}$ of the usual time. 20% faster is $\frac{6}{5}$ of the speed, so the time saved is $\frac{1}{6}$ of the usual time. If that saving is 9 minutes, the usual time is 54 minutes.

Where it fails:

• Keeping the ratio the same way round. Faster means less time, so 7 : 9 in speed is 9 : 7 in time.
• Taking the time lost as the new time. Being 14 minutes late does not mean the walk now takes 14 minutes.
• Treating the percent change in speed as the percent change in time. 25% faster saves 20% of the time, and 20% slower adds 25%.
• Using it when the distance changes too. Then go back to distance = speed × time.

Average Speed in One Line

When to use it: a trip is split into parts at different speeds and the average speed for the whole trip is asked.

The rule behind all of it: average speed = total distance ÷ total time. The shortcuts below are this rule with the arithmetic done in advance.

Equal distances at two speeds

$$\text{Average} = \frac{2xy}{x + y}$$

For example: a car goes to a town at 35 km/h and comes back at 63 km/h. The average is $\frac{2 \times 35 \times 63}{98}$ = 45 km/h, not the 49 you get by adding and halving.

Equal distances at three speeds

$$\text{Average} = \frac{3xyz}{xy + yz + zx}$$

In the exam it is quicker to take each part as the LCM of the speeds. At 8, 10 and 40 km/h, take each part as 40 km. The times are 5, 4 and 1 hours, so 120 km take 10 hours and the average is 12 km/h.

Equal times

If the trip is split into equal times rather than equal distances, the plain average is right. 3 hours at 44 km/h and 3 hours at 56 km/h give 50 km/h.

The missing speed

If the average for a round trip is A and one way is done at x, the other way must be done at

$$y = \frac{Ax}{2x - A}$$

For example: a man averages 24 km/h for a round trip and goes out at 20 km/h. He must come back at $\frac{24 \times 20}{40 - 24}$ = 30 km/h.

Where it fails:

• Adding and halving when the distances are equal. That is right only for equal times.
• Using $\frac{2xy}{x + y}$ when the parts are of different lengths, such as 'the first 40% at one speed and the rest at another'. Go back to total distance ÷ total time, with a handy total such as 100 km.
• Forgetting that the missing-speed formula needs 2x bigger than A. Going out at 20 km/h, you can never average 40 km/h for the round trip, however fast you come back.

Late and Early at Two Speeds, and the Speed-Change Quadratic

Late at one speed, early at another

When to use it: 'at x km/h he is m minutes late; at y km/h he is n minutes early. Find the distance.'

Fast way:

$$D = \frac{\text{product of speeds} \times \text{time difference}}{\text{difference of speeds}}$$

The time difference is m + n when one trip is late and the other early, and the gap between m and n when both are late (or both early). Put it in hours.

For example: walking at 4 km/h, Meena is 10 minutes late; at 6 km/h she is 10 minutes early. The time difference is 20 minutes = $\frac{1}{3}$ hour, so D = $\frac{4 \times 6 \times \frac{1}{3}}{2}$ = 4 km. Check: 4 km takes 60 minutes at 4 km/h and 40 minutes at 6 km/h, so the right time is 50 minutes.

Both late: at 30 km/h a bus is 40 minutes late, and at 40 km/h it is 10 minutes late. The difference is 30 minutes = $\frac{1}{2}$ hour, and D = $\frac{30 \times 40 \times \frac{1}{2}}{10}$ = 60 km.

'If the speed were k km/h more, it would take t hours less'

Fast way: with usual speed x, $\frac{D}{x} - \frac{D}{x + k} = t$, which tidies up to

$$x(x + k) = \frac{Dk}{t}$$

So you want two numbers k apart whose product is $\frac{Dk}{t}$. Usually one of the options gives them at once.

For example: a 420 km trip would take 1 hour less if the speed were 10 km/h more. x(x + 10) = 4,200, and 60 × 70 = 4,200, so the speed is 60 km/h. Check: 7 hours at 60 km/h and 6 hours at 70 km/h.

If the speed is k km/h less and the trip takes t hours more, use x(x − k) = $\frac{Dk}{t}$.

Where it fails:

• Leaving the time in minutes. With speeds in km/h, the time difference must be in hours.
• Adding the minutes when both trips are late, or subtracting when one is late and the other early.
• Taking the negative root of the quadratic. A speed is always positive.
• Answering with the wrong number of the pair. In x(x + k) the smaller number is the usual speed; in x(x − k) the larger one is.

[SSC Tip] With options, skip the algebra. Divide D by an option and by (option + k), and see whether the two times differ by t. Each try takes a few seconds.

Stoppages and Two-Mode Journeys

Stoppages

When to use it: 'without stoppages a bus runs at x km/h; with stoppages it averages y km/h. For how many minutes does it stop per hour?'

Fast way:

$$\text{Stop time per hour} = \frac{x - y}{x} \times 60 \text{ minutes}$$

Divide by the faster speed, the one without stops.

For example: 80 km/h without stops and 68 km/h with stops. Stop time = $\frac{12}{80}$ × 60 = 9 minutes per hour. Why it works: in each hour the bus covers 68 km, which at 80 km/h needs only 51 minutes of running. For the other 9 minutes it stands still.

For the stop time on a whole trip, multiply by the number of hours the trip takes with stops, or find (time with stops) − (time without stops) directly.

Walk one way, ride back

When to use it: two of these three times are given and the third is asked: walk one way and ride back, ride both ways, walk both ways.

Fast way: call one-way walking w and one-way riding r. Then 'walk and ride' = w + r, 'ride both' = 2r and 'walk both' = 2w. So

Walk both ways = 2 × (walk one way and ride back) − (ride both ways)

For example: walking one way and riding back takes 3 hours 50 minutes, and riding both ways takes 2 hours 20 minutes. Walking both ways takes 7 hours 40 minutes − 2 hours 20 minutes = 5 hours 20 minutes. Check: riding one way is 1 hour 10 minutes, so walking one way is 2 hours 40 minutes.

Run it the other way when walking both ways and riding both ways are given: walk one way and ride back is the average of the two.

Where it fails:

• Dividing by the slower speed in the stoppage formula. Here $\frac{12}{68}$ × 60 gives about 10.6 minutes, a common wrong option.
• Giving the stop time per hour when the stop time for the whole trip is asked.
• Mixing hours and minutes in two-mode questions. Turn everything into minutes first.

Relative Speed, the Meeting Point and the After-Meeting Rule

Relative speed in one line

When to use it: two things move at once and the gap between them matters.

Fast way: moving in opposite directions (towards each other or apart), the gap changes at the sum of the speeds. Moving the same way, it changes at the difference. Then time = gap ÷ relative speed.

Where they meet

Fast way: starting together from the two ends of a road D long, they meet after $\frac{D}{a + b}$, and the distances they cover are in the ratio of their speeds, a : b. So the meeting point is $D \times \frac{a}{a + b}$ from A's start.

For example: towns P and Q are 351 km apart. A car leaves P at 42 km/h and, at the same moment, another leaves Q at 36 km/h. They meet after 351 ÷ 78 = 4.5 hours, and the car from P has done 351 × $\frac{42}{78}$ = 189 km.

One starts later: let the early one run alone first, then share the rest. Two cities are 360 km apart. A car leaves the first at 7 a.m. at 36 km/h, and a car leaves the second at 9:30 a.m. at 54 km/h. By 9:30 the first car has done 90 km, so 270 km are left at 90 km/h: 3 hours. They meet at 12:30 p.m., 90 + 108 = 198 km from the first city.

After meeting: the square-root rule

When to use it: two people start at the same moment from A and B towards each other, and after they meet they take $t_1$ and $t_2$ more hours to reach the far ends.

Fast way:

$$\frac{S_1}{S_2} = \sqrt{\frac{t_2}{t_1}}$$

and they met $\sqrt{t_1 t_2}$ hours after starting.

For example: after meeting, Anil takes 3 hours 36 minutes more and Sunil takes 2 hours 30 minutes more. In hours these are 3.6 and 2.5, so Anil : Sunil = $\sqrt{2.5 : 3.6}$ = $\sqrt{25 : 36}$ = 5 : 6, and they met after $\sqrt{3.6 \times 2.5}$ = 3 hours. Check: in 3 hours they cover 15 and 18 parts of the road; then Anil needs 18 ÷ 5 = 3.6 hours and Sunil 15 ÷ 6 = 2.5 hours.

Where it fails:

• Using the difference of speeds for people coming towards each other, or the sum for people going the same way.
• Forgetting the late start. The one who starts later has travelled for less time.
• Turning the square-root rule upside down. The one who takes longer after the meeting is the slower one.
• Leaving out the square root. 25 : 36 comes straight from the times; the speeds are in its square root, 5 : 6.
• Using the rule when the two did not start at the same moment.

Chases with a Head Start

When to use it: one person (a policeman, a second car) chases another who is ahead on the same road.

Fast way:

$$\text{Time to catch} = \frac{\text{head start}}{\text{faster speed} - \text{slower speed}}$$

If the head start is given as time, turn it into distance first: head start = slower speed × head-start time.

The distance the chaser runs before the catch is

$$\text{head start} \times \frac{\text{chaser's speed}}{\text{difference of speeds}}$$

For example (head start in metres): a thief is 300 m ahead of a policeman. The thief runs at 10 km/h and the policeman at 13 km/h. The gap closes at 3 km/h, which is $\frac{5}{6}$ m/s, so the catch takes 300 ÷ $\frac{5}{6}$ = 360 seconds = 6 minutes. The policeman runs 300 × $\frac{13}{3}$ = 1,300 m, and the thief runs 1,000 m.

Another one (head start in hours): a lorry leaves a depot at 7 a.m. at 36 km/h. A car leaves the same depot at 9 a.m. at 60 km/h on the same road. By 9 a.m. the lorry is 72 km ahead, and the gap closes at 24 km/h, so the car catches it 3 hours later, at 12 noon, 180 km from the depot.

The gap just before the catch

How far apart are they 5 minutes before the catch? Relative speed × 5 minutes, whatever the head start was. At a relative speed of 6 km/h that is 6 × $\frac{5}{60}$ = 0.5 km.

Where it fails:

• Using the sum of the speeds. In a chase both move the same way, so it is always the difference.
• Giving the thief's distance when the policeman's is asked, or the other way round. The chaser always runs the head start more.
• Counting from the wrong start. For the lorry and the car, the 3 hours run from 9 a.m., not from 7 a.m.
• Keeping km/h with a gap in metres. Change the relative speed to m/s (or the gap to km) first.

Trains: Add the Lengths, Use Relative Speed

The one rule: the time a train takes to pass something is

$$\text{Time} = \frac{\text{total length to clear}}{\text{relative speed}}$$

For example: a 210 m train runs at 63 km/h, which is 17.5 m/s. It passes a pole in 210 ÷ 17.5 = 12 s and a 140 m bridge in 350 ÷ 17.5 = 20 s. It passes a man walking the same way at 9 km/h at a relative 54 km/h = 15 m/s, in 14 s.

Two trains

Trains of 135 m and 165 m run at 45 km/h and 27 km/h. In opposite directions the relative speed is 72 km/h = 20 m/s, so they clear 300 m in 15 s. In the same direction it is 18 km/h = 5 m/s, so the faster one needs 300 ÷ 5 = 60 s.

Length and speed from two crossings

A goods train passes a pole in 20 s and a 250 m platform in 30 s. The extra 10 s are spent on the platform alone, so the speed is 250 ÷ 10 = 25 m/s (90 km/h) and the length is 25 × 20 = 500 m.

[Tier 2] Speed ratio from crossing times

Two trains pass a standing man in $t_1$ and $t_2$ seconds, and pass each other, running in opposite directions, in $t_3$ seconds. Then

$$S_1 : S_2 = (t_3 - t_2) : (t_1 - t_3)$$

With 21 s, 15 s and 17 s the ratio is 2 : 4 = 1 : 2.

Where it fails:

• Adding a platform for a pole, or forgetting it for a bridge.
• Giving a man, a pole or a signal a length. They have none.
• Using the sum for trains going the same way. Same direction means the difference.
• Keeping km/h with metres and seconds. Change to m/s first.

Boats and Streams: Half the Sum, Half the Difference

When to use it: any boat, swimmer or rower in a river.

Fast way: with the boat's speed in still water b and the stream's speed s:

• downstream speed d = b + s, and upstream speed u = b − s
• b = $\frac{d + u}{2}$ (half the sum) and s = $\frac{d - u}{2}$ (half the difference)

For example: a boat covers 66 km downstream in 3 hours and 50 km upstream in 5 hours. So d = 22 and u = 10, which makes the boat's own speed 16 km/h and the stream's 6 km/h.

Round trips

For a trip of D km each way, the total time is

$$T = \frac{D}{b + s} + \frac{D}{b - s} = \frac{2Db}{b^2 - s^2}$$

Turned round, it gives the distance: $D = \frac{T(b^2 - s^2)}{2b}$.

For example: a boat goes at 11 km/h in still water, the stream runs at 3 km/h, and the round trip takes 5 hours 30 minutes. D = $\frac{5.5 \times 112}{22}$ = 28 km. Check: 28 ÷ 14 = 2 hours down and 28 ÷ 8 = 3.5 hours up.

Ratio of the times

If going up takes k times as long as coming down the same distance, then $\frac{b + s}{b - s} = k$, which gives b : s = (k + 1) : (k − 1). Upstream twice as long means b : s = 3 : 1, so a boat that does 21 km/h in still water is on a 7 km/h stream.

Where it fails:

• Taking b as the average speed for a round trip. The average for going and coming back is $\frac{du}{b}$, which is always less than b. For the boat above (d = 14, u = 8) it is $\frac{112}{11}$, about 10.2 km/h, not 11.
• Mixing up the two speeds. Downstream is always the faster one, and the stream is slower than the boat, or the boat could not go upstream at all.
• Reading 'speed of the man' or 'rate in still water' as d. It means b.

Races: Turn Leads into Speed Ratios

When to use it: 'A beats B by x m', 'by t seconds', 'gives a start of h m', 'dead heat'.

Fast way: a race result is a snapshot. When A finishes L m, B has run L − x m in the same time, so

$$S_A : S_B = L : (L - x)$$

For example: in a 200 m race A beats B by 16 m. A : B = 200 : 184 = 25 : 23.

Beats by x metres or t seconds

Fast way: 'by x m or t s' means B runs those last x m in t s. So B's speed is $\frac{x}{t}$, B's time is $L \div \frac{x}{t}$, and A's time is t seconds less.

For example: in a 750 m race, A beats B by 30 m or 5 s. B runs at 6 m/s and takes 125 s, so A takes 120 s.

Turning a lead in seconds into metres

Multiply the seconds by the loser's speed. In a 900 m race A finishes in 2 minutes 50 seconds and beats B by 10 s. B takes 180 s, so B runs at 5 m/s, and the lead is 10 × 5 = 50 m.

Head start and dead heat

'A gives B a start of h m' means B runs only L − h m. For a dead heat (both finish together), L : (L − h) must equal A : B. If A runs at $\frac{8}{5}$ of B's speed and gives a 75 m start, $\frac{L}{L - 75} = \frac{8}{5}$, so L = 200 m.

Three runners

In a 200 m race A beats B by 20 m, and in another 200 m race B beats C by 10 m. When A runs 200 m, B runs 180 m, and C runs 180 × $\frac{190}{200}$ = 171 m. So A beats C by 29 m, not 30.

Where it fails:

• Adding the leads in a three-runner question.
• Converting seconds with the winner's speed. The extra seconds belong to the loser, so use his speed.
• Mixing up a start and a lead. A start shortens the loser's race; a lead is what he still has to run when the winner finishes.

Circular Tracks: First Meeting and Meeting at the Start

When to use it: two or three runners start together from one point on a circular track.

Fast way:

• First meeting anywhere, same direction: $\frac{\text{track}}{a - b}$, because the faster one must gain one full lap.
• First meeting anywhere, opposite directions: $\frac{\text{track}}{a + b}$, because together they must cover one lap.
• First meeting at the starting point: the LCM of the lap times, whichever way they run.

For example: a 420 m track, runners at 5 m/s and 2 m/s. The lap times are 84 s and 210 s.

• Same direction: 420 ÷ 3 = 140 s.
• Opposite directions: 420 ÷ 7 = 60 s.
• At the start: LCM(84, 210) = 420 s, which is 7 minutes.

With three runners, all three are first together at the start after the LCM of the three lap times. To find when they first meet anywhere while running the same way, take the LCM of the pairwise first-meeting times.

Lap times that are fractions

If the lap times come out as fractions, change them to seconds, or use LCM of fractions = $\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$. On a 2 km track at 15 km/h and 25 km/h, the lap times are 8 minutes and $\frac{24}{5}$ minutes. LCM($\frac{40}{5}$, $\frac{24}{5}$) = $\frac{120}{5}$ = 24 minutes. In opposite directions they first meet after 2 ÷ 40 hours = 3 minutes.

[Tier 2] How many meeting points

Write the speeds in their lowest ratio a : b. Going the same way, they meet at a − b different points on the track; going opposite ways, at a + b points. At 5 : 2 that is 3 points the same way and 7 points the opposite way.

Where it fails:

• Giving the LCM when the first meeting anywhere is asked. They usually meet several times before they are both back at the start.
• Using the sum for the same direction, or the difference for opposite directions.
• Taking the LCM of the speeds instead of the lap times.
• Mixing km/h with metres. Change the units once, at the start.

Option Elimination

When to use it: before the full working, or when the clock is running down.

Fast way: run these quick checks on the four options.

• Average speed over equal distances lies between the two speeds and nearer the slower one. The plain average is almost always a trap option.
• Crossing time for a platform, a bridge or another train must be more than the time to pass a pole.
• Same direction is slower. Overtaking takes longer than crossing an oncoming train.
• Boats: downstream > still water > upstream, and the stream is slower than the boat.
• Meeting point is nearer the slower one's start.
• Chases: the chaser runs farther than the head start.
• Races: in a three-runner question, A beats C by less than the sum of the two leads.
• Circular tracks: the first meeting at the start is a whole number of laps for every runner.

For example: a car goes out at 28 km/h and returns at 42 km/h. Its average speed is (a) 35 km/h (b) 33.6 km/h (c) 36.4 km/h (d) 31.5 km/h.

(a) is the plain average and (c) is nearer the faster speed, so both go. Between (b) and (d), work it out: $\frac{2 \times 28 \times 42}{70}$ = 33.6, so the answer is (b).

Another one: a 225 m train passes a pole in 9 s. How long does it take to pass a 250 m platform? (a) 10 s (b) 19 s (c) 9 s (d) 6 s.

(c) and (d) are not longer than the pole time, so they go. (a) is 250 ÷ 25, the platform alone. The speed is 25 m/s and the length to clear is 475 m, so the time is 19 s: (b).

Where it fails:

• When two options pass every check, you still have to work it out.
• The 'nearer the slower' check is for equal distances. If much more distance is done at the faster speed, the average can sit nearer the faster one.

[SSC Tip] Trap options come from a short list: the plain average of the speeds, the time without the platform, the sum of the speeds in a chase, a ratio the wrong way round, the leads added in a three-runner race, and the LCM given for a first meeting anywhere. If your answer is one of these, go over that step again before you mark it.

Solved Examples

The answers below use the quick methods from this section, with a rough time for each.

Question 1

A train covers 2.1 km in 1 minute 24 seconds. What is its speed in km/h?

Answer. 1 minute 24 seconds is 84 s, so the speed is 2,100 ÷ 84 = 25 m/s. Multiply by 3.6: 90 km/h.

Time check: about 10 seconds.

Question 2

Walking at $\frac{6}{7}$ of his usual speed, a man reaches his office 8 minutes late. What is his usual time for the walk?

Answer. At $\frac{6}{7}$ of the speed the time becomes $\frac{7}{6}$ of the usual time, so the extra $\frac{1}{6}$ of the usual time is 8 minutes. Usual time = 6 × 8 = 48 minutes.

Time check: about 8 seconds.

Question 3

A and B cover the same distance in 5 hours and 6 hours. A is 9 km/h faster than B. How long is the distance?

Answer. Times 5 : 6 means speeds 6 : 5. The gap of 1 part is 9 km/h, so A goes at 54 km/h and B at 45 km/h. Distance = 54 × 5 = 270 km.

Time check: about 15 seconds.

Question 4

A man drives up a hill road at 18 km/h and comes back down the same road at 27 km/h. What is his average speed for the whole trip?

Answer. Equal distances, so $\frac{2 \times 18 \times 27}{45}$ = 21.6 km/h. The plain average, 22.5 km/h, is the trap.

Time check: about 10 seconds.

Question 5

A cyclist covers three equal stretches of a road at 6 km/h, 10 km/h and 15 km/h. What is his average speed for the whole road?

Answer. Take each stretch as 30 km, the LCM of the speeds. The times are 5, 3 and 2 hours, so 90 km take 10 hours and the average is 9 km/h.

Time check: about 15 seconds.

Question 6

Cycling at 14 km/h, Priya reaches the station 9 minutes late. At 21 km/h she is 6 minutes early. How far is the station?

Answer. One late and one early, so the time difference is 9 + 6 = 15 minutes = $\frac{1}{4}$ hour. D = $\frac{14 \times 21 \times \frac{1}{4}}{7}$ = 10.5 km. Check: 45 minutes at 14 km/h and 30 minutes at 21 km/h, so the right time is 36 minutes.

Time check: about 15 seconds.

Question 7

A car covers 300 km at a steady speed. If the speed were 5 km/h more, the trip would take 2 hours less. Find the speed.

Answer. x(x + 5) = $\frac{300 \times 5}{2}$ = 750, and 25 × 30 = 750. So the speed is 25 km/h. Check: 12 hours at 25 km/h and 10 hours at 30 km/h.

Time check: about 20 seconds.

Question 8

Without stoppages a train runs at 96 km/h; with stoppages it averages 84 km/h. For how many minutes does it stop in an hour?

Answer. $\frac{96 - 84}{96}$ × 60 = 7.5 minutes. Dividing by 84 instead gives about 8.6 minutes, which is wrong.

Time check: about 10 seconds.

Question 9

Two trains start at the same time from X and Y, which are 448 km apart, and run towards each other at 52 km/h and 60 km/h. How far from X do they meet?

Answer. They close the gap at 112 km/h, so they meet after 4 hours. The train from X has run 52 × 4 = 208 km.

Time check: about 10 seconds.

Question 10

Two cars start at the same time from A and B towards each other. After they pass, the car from A takes 2 hours 40 minutes to reach B, and the car from B takes 6 hours to reach A. If the car from A runs at 60 km/h, how fast is the other car?

Answer. In hours the two times are $\frac{8}{3}$ and 6. Car from A : car from B = $\sqrt{6 : \frac{8}{3}}$ = $\sqrt{9 : 4}$ = 3 : 2. So the other car runs at 60 × $\frac{2}{3}$ = 40 km/h. It makes sense: the car from B needs more time after the meeting, so it is the slower one.

Time check: about 20 seconds.

Question 11

A policeman spots a thief 175 m away. The thief runs at 6 km/h and the policeman chases him at 8 km/h. How far will the policeman run before he catches the thief?

Answer. Policeman's distance = 175 × $\frac{8}{8 - 6}$ = 700 m. The gap closes at 2 km/h = $\frac{5}{9}$ m/s, so the catch takes 315 seconds, which is 5 minutes 15 seconds.

Time check: about 15 seconds.

Question 12

Two trains, 245 m and 115 m long, run on parallel tracks in the same direction at 68 km/h and 50 km/h. How long does the faster one take to pass the slower one completely?

Answer. Length to clear = 360 m. Relative speed = 18 km/h = 5 m/s. Time = 360 ÷ 5 = 72 seconds.

Time check: about 12 seconds.

Question 13

A boat goes 52 km downstream in 4 hours and 30 km upstream in 5 hours. Find the speed of the stream.

Answer. d = 13 km/h and u = 6 km/h. The stream is half the difference: 3.5 km/h. The boat does half the sum, 9.5 km/h, in still water.

Time check: about 12 seconds.

Question 14

In a 400 m race, A beats B by 32 m or 4 seconds. How long does A take to finish the race?

Answer. B runs the last 32 m in 4 s, so B's speed is 8 m/s and B takes 400 ÷ 8 = 50 s. A takes 4 s less: 46 seconds.

Time check: about 12 seconds.

Question 15

Two cyclists start together from the same point on a circular track 1.5 km long and ride in opposite directions at 20 km/h and 30 km/h. After how many minutes will they first be together at the starting point?

Answer. Lap times: 1.5 ÷ 20 hours = 4.5 minutes and 1.5 ÷ 30 hours = 3 minutes. LCM($\frac{9}{2}$, $\frac{6}{2}$) = $\frac{18}{2}$ = 9 minutes. They meet somewhere on the track much sooner, after 1.5 ÷ 50 hours = 1.8 minutes, so 1 minute 48 seconds is the trap.

Time check: about 25 seconds.