Speed, Distance and Time Basics

One Formula, Three Forms

Speed is the distance covered in one unit of time. Every question in this chapter starts from one relation:

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \qquad \text{Distance} = \text{Speed} \times \text{Time} \qquad \text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

• A scooter covers 135 km in 3 hours, so its speed is 135 ÷ 3 = 45 km/h.
• At 45 km/h for 2 hours 20 minutes it covers $45 \times \frac{7}{3}$ = 105 km, because 2 hours 20 minutes is $2\frac{1}{3} = \frac{7}{3}$ hours.
• 63 km at 45 km/h takes $\frac{63}{45} = \frac{7}{5}$ hours = 1 hour 24 minutes.

Keep the units matched. If the speed is in km/h, the time must be in hours and the distance in km. If the speed is in m/s, use metres and seconds. Most wrong answers in this chapter come from mixing the two.

Minutes into hours. Divide by 60. These come up often:

Going back, multiply the part of an hour by 60: 0.4 hour is 24 minutes, and 1.35 hours is 1 hour 21 minutes.

Speed from a time in minutes. A car covers 28 km in 35 minutes. Speed = $28 \div \frac{35}{60} = 28 \times \frac{60}{35}$ = 48 km/h. You can also think: 35 minutes for 28 km means 5 minutes for 4 km, so 60 minutes for 48 km.

Clock times. A lorry leaves at 9:40 am and covers 182 km at 52 km/h. Time = $\frac{182}{52} = 3\frac{1}{2}$ hours, so it arrives at 1:10 pm. Count across noon in two steps: 9:40 am to 12:40 pm is 3 hours, then 30 minutes more.

What the rest of the section covers. With one moving object, SSC asks about units, ratios of speed and time, average speed, late and early arrival, stoppages and round trips. When two things move at once (meeting, chasing, trains), relative speed takes over.

[SSC Tip] A decimal hour is not "hours.minutes". 2.4 hours is 2 hours 24 minutes, not 2 hours 40 minutes, and 3.75 hours is 3 hours 45 minutes. SSC builds a wrong option on exactly this slip.

Changing km/h into m/s and Back

One kilometre is 1000 m and one hour is 3600 s, so

$$1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{5}{18} \text{ m/s} \qquad 1 \text{ m/s} = \frac{18}{5} \text{ km/h} = 3.6 \text{ km/h}$$

• km/h to m/s: multiply by $\frac{5}{18}$. 72 km/h = $72 \times \frac{5}{18}$ = 20 m/s.
• m/s to km/h: multiply by $\frac{18}{5}$. 15 m/s = $15 \times \frac{18}{5}$ = 54 km/h.

Every multiple of 18 gives a whole number of m/s. These are worth knowing by heart:

Which way? The number in m/s is always the smaller one, a little over a quarter of the km/h number. If 54 km/h turns into something bigger than 54, you multiplied by the wrong fraction. When a question gives a length in metres and a speed in km/h (trains, police chases, races), change the speed to m/s before anything else.

Two quick ones.

• A sprinter runs 200 m in 25 s. That is 8 m/s, which is 8 × 3.6 = 28.8 km/h.
• A bus runs at 66 km/h for 15 s. $66 \times \frac{5}{18} = \frac{55}{3}$ m/s, and $\frac{55}{3} \times 15$ = 275 m.

Metres per minute. 1 km/h = $\frac{1000}{60} = \frac{50}{3}$ m per minute, so 12 km/h = 200 m per minute. Going back, 900 m in 4 minutes is 225 m per minute, which is 225 × 60 = 13,500 m an hour = 13.5 km/h.

Wheels. One full turn moves a wheel forward by its circumference, $\pi d$ (or $2\pi r$). A wheel of diameter 1.4 m moves $\frac{22}{7} \times 1.4$ = 4.4 m per turn. At 125 turns a minute that is 550 m per minute = 33,000 m an hour = 33 km/h. Using the radius where the formula needs the diameter halves the answer, and SSC keeps that half as an option.

The Ratio Method: Same Distance, Same Time or Same Speed

Distance = Speed × Time links three things. Keep one of them fixed and the other two have a simple ratio:

Same distance, different speeds. If the speeds are 3 : 4, the times are 4 : 3. Suppose the slower one takes 20 minutes more. The gap is 4 − 3 = 1 part, so 1 part = 20 minutes and the times are 80 and 60 minutes. No formula, no equation.

"At a fraction of the usual speed". "Walking at $\frac{4}{5}$ of his usual speed, a man reaches 12 minutes late. What is his usual time?" Speeds usual : new = 5 : 4, so times usual : new = 4 : 5. The extra 1 part is 12 minutes, so the usual time is 4 × 12 = 48 minutes.

Faster works the same way. At $\frac{6}{5}$ of the usual speed, the times go 6 : 5. If he is then 10 minutes early, 1 part = 10 minutes and the usual time is 60 minutes.

Same time, different speeds. "Had Kunal cycled at 15 km/h instead of 12 km/h, he would have gone 9 km further in the same time. How far did he actually go?" The extra 3 km/h produced the extra 9 km, so he rode for 3 hours and covered 12 × 3 = 36 km.

Percentage changes. To cut a 3 hour trip to 2 hours 30 minutes, the times go 6 : 5, so the speeds must go 5 : 6, a rise of $\frac{1}{5}$ = 20%. The other way round, a 20% rise in speed (5 : 6) cuts the time from 6 parts to 5, which is $\frac{1}{6} = 16\frac{2}{3}\%$ less time. And if the speed falls by 20% (5 : 4), the time rises from 4 parts to 5, which is 25% more.

Speed ratio from distances and times. A covers 240 km in 4 hours and B covers 210 km in 5 hours. Their speeds are 60 and 42 km/h, so A : B = 10 : 7.

Time ratio from distance and speed ratios. The distances are 3 : 4 and the speeds 5 : 6. Time = distance ÷ speed, so the times are $\frac{3}{5} : \frac{4}{6}$. Multiply both by 30 to get 9 : 10.

Average Speed: Equal Distances and Equal Times

$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$$

This definition is the only safe one. The plain average of the speeds is right in just one special case.

Equal distances. Rekha cycles 72 km to her parents' town at 24 km/h and comes back at 36 km/h. Going takes 3 hours and returning takes 2 hours, so she covers 144 km in 5 hours. Her average speed is 28.8 km/h, not 30. She spends more time at the slower speed, so the average leans towards 24.

For two equal distances at speeds x and y, the same working always gives

$$\text{Average speed} = \frac{2xy}{x + y} = \frac{2 \times 24 \times 36}{24 + 36} = \frac{1728}{60} = 28.8 \text{ km/h}$$

Three equal distances. At speeds x, y and z the result is $\frac{3xyz}{xy + yz + zx}$, but there is a simpler route: pick a length for each part that every speed divides. With 15, 30 and 90 km/h, take each part as 90 km. The times are 6, 3 and 1 hours, so 270 km takes 10 hours and the average is 27 km/h. Formula check: $\frac{3 \times 15 \times 30 \times 90}{450 + 2700 + 1350} = \frac{121500}{4500} = 27$.

This "assume a distance" trick works for any number of equal parts. It also works when the question never gives the distance, because the answer does not depend on the distance you pick.

Equal times. A taxi runs 2 hours at 50 km/h and then 2 hours at 70 km/h. That is 100 + 140 = 240 km in 4 hours, an average of 60 km/h, which is the plain average. With equal times each speed gets the same weight, so $\frac{x + y}{2}$ is correct here and only here.

Quick checks on any answer.

• The average speed always lies between the lowest and the highest speed.
• With equal distances it is below the plain average of the speeds.
• If every speed is doubled, the average speed doubles too.

Why the harmonic form keeps appearing. A round trip, "goes to the office and comes back", "covers the first half and the second half", "three equal sides of a triangle": all of these are equal distances. Spot the phrase and reach for $\frac{2xy}{x + y}$ or for an assumed distance.

[SSC Tip] Every average speed question carries the plain average as an option (30 km/h above). Ask first: equal distances or equal times? Only equal times allows $\frac{x + y}{2}$.

Average Speed: Unequal Parts and the Missing Speed

Given distances. A van goes 30 km at 45 km/h and then 50 km at 60 km/h. The times are $\frac{30}{45} = \frac{2}{3}$ hour and $\frac{50}{60} = \frac{5}{6}$ hour, which add up to $\frac{3}{2}$ hours. Average speed = $80 \div \frac{3}{2} = \frac{160}{3} = 53\frac{1}{3}$ km/h.

Parts given as fractions of the distance. "Half the journey at 30 km/h, one-third at 40 km/h and the rest at 20 km/h." Pick a total distance that makes every part easy, here 120 km:

• 60 km at 30 km/h: 2 hours
• 40 km at 40 km/h: 1 hour
• 20 km at 20 km/h: 1 hour

That is 120 km in 4 hours, so the average speed is 30 km/h.

Parts given as fractions of the time. "For $\frac{2}{5}$ of the time at 50 km/h and for the rest at 75 km/h." Now the speeds are weighted by time:

$$\frac{2}{5} \times 50 + \frac{3}{5} \times 75 = 20 + 45 = 65 \text{ km/h}$$

If you prefer numbers, take 5 hours: 100 + 225 = 325 km in 5 hours, again 65 km/h.

Weighting the speeds works only for fractions of the time. Doing it with the distance fractions of the 120 km question ($\frac{1}{2} \times 30 + \frac{1}{3} \times 40 + \frac{1}{6} \times 20 = 31\frac{2}{3}$) is wrong, and SSC uses that number as an option.

The missing speed. "A cyclist rides half of a route at 30 km/h. How fast must she ride the other half to average 40 km/h?" Take each half as 120 km. The first half takes 4 hours. The whole 240 km at 40 km/h should take 6 hours, so 2 hours are left for the second 120 km: 60 km/h. Check: $\frac{2 \times 30 \times 60}{30 + 60} = 40$.

When it cannot be done. Same cyclist, but now she wants an average of 60 km/h, twice her first speed. The whole 240 km at 60 km/h allows 4 hours, and she has already used all 4 on the first half. No speed is fast enough. Over two equal distances the average can never reach double the first speed.

Missing speed with the time given. A car must average 60 km/h for 5 hours, so it must cover 300 km. It does 3 hours at 50 km/h, which is 150 km. The last 2 hours must cover the other 150 km: 75 km/h.

Late and Early Arrival: Two Speeds, One Distance

The standard question: "Walking at 5 km/h, Tanu reaches school 7 minutes late. Walking at 6 km/h, she reaches 5 minutes early. How far is the school?"

Step 1: the time difference. One walk ends 7 minutes after the bell and the other 5 minutes before it, so the two walking times differ by 7 + 5 = 12 minutes = $\frac{1}{5}$ hour.

If both were late, you would subtract instead: late by 30 minutes and by 6 minutes means the walks differ by 24 minutes. Early by 20 and early by 5 means they differ by 15. Either way, turn the minutes into hours before you use km/h: leaving 12 minutes as 12 gives an answer 60 times too big.

Step 2, ratio method. Same distance, speeds 5 : 6, so the times are 6 : 5. The 1-part gap is 12 minutes, so the times are 72 and 60 minutes. Distance = 5 km/h for $\frac{72}{60}$ hours = 6 km.

Step 2, formula method. For speeds x and y (y the faster) and a time difference of t hours,

$$D = \frac{xy}{y - x} \times t = \frac{5 \times 6}{6 - 5} \times \frac{1}{5} = 6 \text{ km}$$

This comes straight from $\frac{D}{x} - \frac{D}{y} = t$. Use whichever you find faster; the ratio method also hands you both times.

The correct time. The 72 minute walk is 7 minutes late, so the correct time is 65 minutes. The 60 minute walk is 5 minutes early, which agrees. Checking with both speeds catches most slips.

Both late. "At 4 km/h a man is 30 minutes late; at 5 km/h he is 6 minutes late." The gap is 24 minutes = $\frac{2}{5}$ hour, so

$$D = \frac{4 \times 5}{5 - 4} \times \frac{2}{5} = 8 \text{ km}$$

Check: 8 km takes 120 minutes at 4 km/h and 96 minutes at 5 km/h, so the correct time is 90 minutes, and 96 is indeed 6 minutes late.

The speed needed to be exactly on time. In the last case the correct time is 90 minutes = $\frac{3}{2}$ hours, so he must walk at $8 \div \frac{3}{2} = 5\frac{1}{3}$ km/h. This is not 4.5 km/h, the average of the two speeds.

The same idea with one speed. "At $\frac{4}{5}$ of the usual speed he is 12 minutes late" is the late-arrival question with only one trip given. It is solved by the ratio method in the Ratio Method part above.

Change of Speed: Saving Time, Making Up Delays and Faults

"If the speed were k km/h more, it would take t hours less." For a distance D at speed s,

$$\frac{D}{s} - \frac{D}{s + k} = t \Rightarrow s(s + k) = \frac{Dk}{t}$$

"A car travels 240 km. If its speed were 20 km/h more, the trip would take 1 hour less. Find its speed." So s(s + 20) = 240 × 20 = 4800. Since 60 × 80 = 4800, s = 60 km/h. Check: 240 ÷ 60 = 4 hours and 240 ÷ 80 = 3 hours.

In the exam you need not solve the quadratic. Either look for two numbers k apart whose product is $\frac{Dk}{t}$, or test the options: D ÷ option, minus D ÷ (option + k), should give t. "k km/h less, t hours more" is the same equation with s and s − k.

Making up a late start. "A bus starts 30 minutes late. By running 10 km/h faster than usual it covers 150 km and arrives on time. What is its usual speed?" Here t = $\frac{1}{2}$, so s(s + 10) = 150 × 10 × 2 = 3000, and s = 50 km/h (150 km takes 3 hours at 50 km/h and $2\frac{1}{2}$ hours at 60 km/h).

A fault on the way. "A car runs at 60 km/h for 1 hour. Then a tyre problem keeps it at $\frac{3}{4}$ of that speed, and it arrives 30 minutes late. How long is the trip?" After the fault, speeds go 4 : 3, so times on that stretch go 3 : 4. The extra 1 part is the 30 minutes of delay, so the stretch should have taken 3 × 30 = 90 minutes at 60 km/h, which is 90 km. Trip = 60 + 90 = 150 km.

[Tier 2] A fault, two cases. "After 45 km a train develops a fault and runs at $\frac{2}{3}$ of its speed, arriving 1 hour 30 minutes late. Had the fault come 30 km later, it would have been 1 hour 10 minutes late. Find the speed and the distance."

• At $\frac{2}{3}$ of the speed, the time on any stretch grows by $\frac{1}{2}$.
• The two cases differ only on one 30 km stretch, and moving the fault past it saves 90 − 70 = 20 minutes. So 20 minutes is $\frac{1}{2}$ of the normal time for 30 km, which is therefore 40 minutes. Speed = 30 km in 40 minutes = 45 km/h.
• In the first case the 90 minutes of delay is $\frac{1}{2}$ of the normal time after the fault, so that stretch takes 180 minutes, which is 135 km. Distance = 45 + 135 = 180 km.

Check: normally 180 ÷ 45 = 4 hours. With the fault, 1 hour + $\frac{135}{30}$ hours = 5.5 hours, which is 1 hour 30 minutes more.

[SSC Tip] In $s(s + k) = \frac{Dk}{t}$, t is in hours. "Saves 30 minutes" means t = $\frac{1}{2}$, so the right side doubles. Putting t = 30 is the usual slip, and it leads to a speed that does not match any option, or worse, matches a wrong one.

Stoppages and Rest Breaks

Speed with and without stoppages. "Without stoppages a bus runs at 60 km/h. With stoppages its average speed is 45 km/h. For how many minutes does it stop each hour?"

In every hour, the stops cost the bus 60 − 45 = 15 km of running. At its running speed of 60 km/h, 15 km takes $\frac{15}{60}$ hour = 15 minutes. In general, with x the speed without stops and y the speed with stops,

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

Ratio view: in one hour the bus covers 45 km instead of 60, so it moves for $\frac{45}{60} = \frac{3}{4}$ of the hour and stands still for the other $\frac{1}{4}$.

The reverse question. A train stops for 10 minutes in every hour and runs at 66 km/h when moving. It moves for 50 minutes an hour and covers $66 \times \frac{50}{60}$ = 55 km. So its speed including stoppages is 55 km/h, and a 165 km run takes 3 hours with the stops.

Running speed from the total time. A car covers 210 km in 4 hours, and its stops add up to 30 minutes. It was moving for $3\frac{1}{2}$ hours, so its running speed is 210 ÷ 3.5 = 60 km/h, while its average including the stops is 210 ÷ 4 = 52.5 km/h.

Rests on a walk. "A hiker walks 12 km at 4 km/h and rests for 10 minutes after every 3 km. How long does the walk take?" Walking time = 3 hours. She rests after the 3rd, 6th and 9th km, but not after the 12th, because the walk is over by then. That is 3 rests = 30 minutes, so the walk takes 3 hours 30 minutes.

Rule: n equal stages have n − 1 rests between them. A ladder of 4 stages has 3 gaps, not 4.

[SSC Tip] Divide the lost distance by the speed without stoppages. Dividing by the speed with stoppages ($\frac{15}{45} \times 60$ = 20 minutes) is the standard wrong option.

Round Trips, Two Modes and Distance from a Total Time

Going at one speed, coming back at another. "Ira cycles to the market at 12 km/h and walks back at 4 km/h. The round trip takes 2 hours. How far is the market?"

With the distance D, $\frac{D}{12} + \frac{D}{4} = 2$, so $\frac{D + 3D}{12} = 2$ and D = 6 km. In general, for a one-way distance D at x going and y returning, with total time T,

$$D = T \times \frac{xy}{x + y}$$

Ratio method: the speeds are 12 : 4 = 3 : 1, so the times are 1 : 3. The 2 hours split as $\frac{1}{2}$ hour and $1\frac{1}{2}$ hours, and D = 12 × $\frac{1}{2}$ = 6 km. The question asks for the one-way distance, so 12 km, the whole round trip, is the wrong answer here.

Walk one way, ride the other. "Walking to the station and riding back takes 5 hours 40 minutes. Walking both ways takes 8 hours 20 minutes. How long would riding both ways take?"

• Walking one way: 500 minutes ÷ 2 = 250 minutes = 4 hours 10 minutes.
• Riding one way: 340 − 250 = 90 minutes.
• Riding both ways: 180 minutes = 3 hours.

Short cut: ride both ways = 2 × (walk one way + ride back) − walk both ways = 680 − 500 = 180 minutes. Working in minutes avoids mixed fractions. It works the other way too: walk both ways = 2 × (walk + ride) − ride both ways.

Part of the way by each mode. "Mohan covers 50 km in 7 hours, partly on foot at 5 km/h and partly by cycle at 10 km/h. How far did he walk?" If all 7 hours were on the cycle he would cover 70 km, which is 20 km too much. Each hour on foot instead of on the cycle loses 5 km, so he walked for 4 hours: 20 km on foot, and 30 km by cycle in 3 hours. Check: 4 + 3 = 7 hours and 20 + 30 = 50 km.

Halves of a distance and a total time. "Half a journey by train at 60 km/h and the other half by car at 40 km/h takes 5 hours. How long is the journey?" Each half is $\frac{D}{2}$, so $\frac{D}{120} + \frac{D}{80} = 5$, which gives $\frac{2D + 3D}{240} = 5$ and D = 240 km. Check: 120 km by train takes 2 hours and 120 km by car takes 3 hours.

A stay in between. If the question adds a stay (say 30 minutes at the market), take it off the total time first and use the formula on the time that is left.

Solved Examples

Question 1

A cyclist covers 750 m in 2 minutes 30 seconds. What is his speed in km/h? (a) 18 km/h (b) 5 km/h (c) 15 km/h (d) 22.5 km/h

Answer. 2 minutes 30 seconds is 150 seconds, so his speed is 750 ÷ 150 = 5 m/s. To get km/h, multiply by $\frac{18}{5}$: 5 × 3.6 = (a) 18 km/h. 5 km/h is the m/s number with the wrong unit stuck on it.

Question 2

A car runs at 65 km/h for 3 hours 36 minutes. How far does it go? (a) 218.4 km (b) 195 km (c) 234 km (d) 260 km

Answer. 36 minutes is $\frac{36}{60} = \frac{3}{5}$ = 0.6 hour, so the time is 3.6 hours, not 3.36. Distance = 65 × 3.6 = (c) 234 km.

Question 3

A bus leaves a town at 7:50 am and travels 189 km at a steady 54 km/h. At what time does it reach? (a) 11:40 am (b) 10:50 am (c) 11:05 am (d) 11:20 am

Answer. Time = 189 ÷ 54 = 3.5 hours, which is 3 hours 30 minutes (not 3 hours 50 minutes). From 7:50 am, 3 hours takes us to 10:50 am, and 30 more minutes gives (d) 11:20 am.

Question 4

A wheel of radius 28 cm makes 375 full turns every minute. What is its speed in km/h? (Take $\pi = \frac{22}{7}$.) (a) 19.8 km/h (b) 39.6 km/h (c) 11 km/h (d) 79.2 km/h

Answer. One turn moves the wheel by its circumference: $2 \times \frac{22}{7} \times 28$ = 176 cm. In a minute it goes 375 × 176 = 66,000 cm = 660 m. In an hour that is 660 × 60 = 39,600 m, which is (b) 39.6 km/h. Using $\pi r$ instead of $2\pi r$ halves the answer to 19.8 km/h.

Question 5

Walking at $\frac{8}{11}$ of his usual speed, Raghav reaches his office 9 minutes late. How long does he usually take? (a) 33 minutes (b) 72 minutes (c) 24 minutes (d) 27 minutes

Answer. Same distance, so time goes the opposite way to speed. Speeds usual : now = 11 : 8, so times usual : now = 8 : 11. He is 3 parts slower, and those 3 parts are the 9 minutes he is late, so 1 part = 3 minutes. His usual time is 8 × 3 = (c) 24 minutes.

Question 6

The speeds of A and B are in the ratio 5 : 8. To cover the same distance, A takes 36 minutes more than B. How long does B take? (a) 1 hour (b) 1 hour 36 minutes (c) 12 minutes (d) 1 hour 20 minutes

Answer. For the same distance the times are in the inverse ratio, 8 : 5. A takes 3 parts more than B, and those 3 parts are 36 minutes, so 1 part is 12 minutes. B takes 5 parts = 60 minutes = (a) 1 hour, and A takes 96 minutes.

Question 7

Had Harsh driven at 52 km/h instead of 40 km/h, he would have covered 36 km more in the same time. How far did he actually drive? (a) 156 km (b) 120 km (c) 90 km (d) 144 km

Answer. Same time, so the extra distance comes only from the extra speed. 12 km/h more gave 36 km more, so he drove for 36 ÷ 12 = 3 hours. At 40 km/h that is 40 × 3 = (b) 120 km. 156 km is what he would have covered at 52 km/h, which the question did not ask.

Question 8

A cyclist covers 66 km in 3 hours and a runner covers 39 km in 3 hours 15 minutes. What is the ratio of the cyclist's speed to the runner's speed? (a) 22 : 13 (b) 13 : 22 (c) 6 : 11 (d) 11 : 6

Answer. Find each speed first. Cyclist: 66 ÷ 3 = 22 km/h. Runner: 3 hours 15 minutes is $3\frac{1}{4} = \frac{13}{4}$ hours, so 39 ÷ $\frac{13}{4}$ = 39 × $\frac{4}{13}$ = 12 km/h. Ratio = 22 : 12 = (d) 11 : 6.

Question 9

A man drives from P to Q at 42 km/h and comes back from Q to P at 70 km/h. What is his average speed for the whole trip? (a) 52.5 km/h (b) 56 km/h (c) 54 km/h (d) 50 km/h

Answer. The two legs are the same distance, so use $\frac{2xy}{x + y} = \frac{2 \times 42 \times 70}{42 + 70} = \frac{5880}{112}$ = (a) 52.5 km/h. Or assume PQ = 210 km: 5 hours going and 3 hours back, so 420 km in 8 hours, again 52.5 km/h. 56 km/h is the plain average of 42 and 70, which would be right only if he drove for equal times.

Question 10

A person covers three equal distances at 14 km/h, 21 km/h and 42 km/h. What is the average speed for the whole distance? (a) $25\frac{2}{3}$ km/h (b) 24 km/h (c) 20 km/h (d) 21 km/h

Answer. Take each distance as 42 km, because 14, 21 and 42 all divide 42. The times are 3, 2 and 1 hours, so 126 km takes 6 hours. Average speed = 126 ÷ 6 = (d) 21 km/h.

Question 11

A man covers $\frac{1}{4}$ of a journey at 18 km/h, $\frac{1}{2}$ of it at 45 km/h and the rest at 30 km/h. What is his average speed for the journey? (a) 31 km/h (b) 30 km/h (c) 34.5 km/h (d) 32 km/h

Answer. Pick a length that makes every part easy: 180 km. Then 45 km at 18 km/h takes $2\frac{1}{2}$ hours, 90 km at 45 km/h takes 2 hours, and the last 45 km at 30 km/h takes $1\frac{1}{2}$ hours. That is 180 km in 6 hours, so the average speed is (b) 30 km/h.

Question 12

A woman drives the first half of a journey at 40 km/h. At what speed must she drive the second half so that her average speed for the whole journey is 60 km/h? (a) 120 km/h (b) 80 km/h (c) 100 km/h (d) 90 km/h

Answer. Take each half as 120 km. The first half takes 120 ÷ 40 = 3 hours. At an average of 60 km/h the whole 240 km should take 4 hours, so only 1 hour is left for the second 120 km. She needs (a) 120 km/h. 80 km/h is the "equal times" answer (40 and 80 average to 60), but here the halves are equal distances, and $\frac{2 \times 40 \times 80}{120}$ is only $53\frac{1}{3}$.

Question 13

Walking at 4 km/h, a student reaches school 9 minutes late. Walking at 5 km/h, she reaches 6 minutes early. How far is the school from her home? (a) 1 km (b) 4 km (c) 5 km (d) 6 km

Answer. One walk is late and the other early, so the two walking times differ by 9 + 6 = 15 minutes. The speeds are 4 : 5, so the times are 5 : 4, and the 1-part gap is 15 minutes. So the slow walk takes 5 × 15 = 75 minutes, and the distance is 4 × $\frac{75}{60}$ = (c) 5 km. Check: at 5 km/h, 5 km takes 60 minutes; the correct time is 66 minutes, so 75 is 9 late and 60 is 6 early.

Question 14

A car reaches its destination 25 minutes late when it runs at 36 km/h, and 5 minutes late when it runs at 45 km/h. How far is the destination? (a) 90 km (b) 45 km (c) 72 km (d) 60 km

Answer. Both trips are late, so the times differ by 25 − 5 = 20 minutes. Speeds 36 : 45 = 4 : 5 give times 5 : 4, so 1 part = 20 minutes and the trip at 36 km/h takes 100 minutes. Distance = 36 × $\frac{100}{60}$ = (d) 60 km. (Formula: $\frac{36 \times 45}{9} \times \frac{20}{60} = 60$.)

Question 15

A train covers 450 km. If its speed were 15 km/h more, it would take 1 hour less for the journey. What is its speed? (a) 90 km/h (b) 75 km/h (c) 60 km/h (d) 50 km/h

Answer. Test the options. At 75 km/h the trip takes 450 ÷ 75 = 6 hours; at 75 + 15 = 90 km/h it takes 450 ÷ 90 = 5 hours. That saves exactly 1 hour, so the speed is (b) 75 km/h. (By algebra, s(s + 15) = 450 × 15 = 6750, and 75 × 90 = 6750.)

Question 16

A train leaves 20 minutes late. To reach its destination 160 km away on time, it runs 16 km/h faster than its usual speed. What is its usual speed? (a) 96 km/h (b) 64 km/h (c) 80 km/h (d) 72 km/h

Answer. The faster run must save 20 minutes = $\frac{1}{3}$ hour. Try 80 km/h: the usual time is 160 ÷ 80 = 2 hours, and at 96 km/h it is 160 ÷ 96 = $1\frac{2}{3}$ hours. The saving is $\frac{1}{3}$ hour, so the usual speed is (c) 80 km/h. Putting t = 20 instead of $\frac{1}{3}$ into the formula is the slip that throws everything off.

Question 17

A train runs at 72 km/h. After 1 hour an engine fault brings its speed down to $\frac{2}{3}$ of that, and it reaches its destination 1 hour late. How long is the journey? (a) 216 km (b) 144 km (c) 288 km (d) 180 km

Answer. After the fault the speed goes 3 : 2, so the time on the rest of the route goes 2 : 3. The extra 1 part is the 1 hour of delay, so the rest of the route normally takes 2 hours, which is 72 × 2 = 144 km. Add the first hour's 72 km: (a) 216 km. Check: normally 3 hours; with the fault, 1 hour + 144 ÷ 48 = 4 hours.

Question 18

Without stoppages a bus runs at 75 km/h, and with stoppages its average speed is 60 km/h. For how many minutes does the bus stop in each hour? (a) 15 minutes (b) 10 minutes (c) 20 minutes (d) 12 minutes

Answer. Each hour the bus loses 75 − 60 = 15 km because of the stops. At its running speed of 75 km/h, 15 km takes $\frac{15}{75}$ hour = $\frac{1}{5}$ hour = (d) 12 minutes. Dividing the 15 km by 60 km/h, the speed with stoppages, gives 15 minutes, which is the usual wrong option.

Question 19

Ankit walks to the station and rides back in 3 hours 20 minutes. If he rides both ways, he takes 2 hours 10 minutes. How long would he take to walk both ways? (a) 5 hours 30 minutes (b) 4 hours 30 minutes (c) 2 hours 15 minutes (d) 4 hours 10 minutes

Answer. Riding both ways is 130 minutes, so riding one way is 65 minutes. Walk + ride = 200 minutes, so walking one way is 200 − 65 = 135 minutes. Walking both ways = 270 minutes = (b) 4 hours 30 minutes. Short cut: 2 × 200 − 130 = 270.

Question 20

A man covers 38 km in 5 hours, partly on foot at 4 km/h and partly by cycle at 10 km/h. How far does he walk? (a) 30 km (b) 12 km (c) 8 km (d) 10 km

Answer. Suppose all 5 hours were on the cycle: 50 km, which is 12 km too much. Each hour spent walking instead of cycling loses 10 − 4 = 6 km, so he walked for 12 ÷ 6 = 2 hours, which is 4 × 2 = (c) 8 km. Check: 3 hours by cycle is 30 km, and 8 + 30 = 38 km.

Question 21

A man cycles to a lake at 15 km/h and comes back by the same road at 10 km/h. The round trip takes 2 hours 30 minutes. How far is the lake? (a) 15 km (b) 30 km (c) 12.5 km (d) 18 km

Answer. Speeds 15 : 10 = 3 : 2, so the times are 2 : 3. The 150 minutes split as 60 and 90 minutes. Going takes 1 hour at 15 km/h, so the lake is (a) 15 km away. Check: coming back, 15 km at 10 km/h is $1\frac{1}{2}$ hours, and 1 + $1\frac{1}{2}$ = $2\frac{1}{2}$ hours.