Time, Speed and Distance for CUET General Aptitude

Q: How do you convert km/h to m/s and back?

Multiply km/h by 5/18 to get m/s, and multiply m/s by 18/5 to get km/h. The factor comes from 1 km = 1,000 m and 1 hour = 3,600 s, so 1 km/h is 1000/3600 = 5/18 m/s.

Q: How do boats and streams questions work?

With the boat's still-water speed b and the stream's speed s, downstream speed is b + s and upstream speed is b - s. So the still-water speed is half their sum, and the stream speed is half their difference.

About 10 min to learn · 71 questions

Every time, speed and distance question in CUET's General Aptitude Test is a version of one relation: speed equals distance divided by time. What catches students is rarely the formula but a slip in units or reading. This lesson covers unit conversion, average speed, late-and-early problems, relative speed and chases, meeting points and circular tracks, trains crossing poles, platforms and each other, and boats and streams, with worked examples and traps.

Syllabus area: General Mental Ability, Numerical Ability

On this page · 9 sections

01The core idea
02Speed basics
03Relative speed
04Trains
05Boats and streams
06Exam shortcuts
07Common mistakes
08Quick revision
09Check your understanding

The core idea

Speed is distance divided by time, and every question in this chapter rearranges that one relation. Before using it, put all three quantities in matching units: km/h with kilometres and hours, m/s with metres and seconds. Most wrong answers here come from mixed units, not hard mathematics.

Two proportion facts save time. For a fixed distance, speed and time are inversely proportional: speeds in $3 : 5$ mean times in $5 : 3$ . For a fixed time, distance is directly proportional to speed.

Speed basics

What is the basic formula linking time, speed and distance?

Speed = distance ÷ time, which rearranges to distance = speed × time and time = distance ÷ speed.

Speed, distance and time

\text{Speed} = \frac{\text{distance}}{\text{time}} \qquad \text{Distance} = \text{speed} \times \text{time}

How do you convert km/h to m/s and back?

Multiply km/h by $\tfrac{5}{18}$ to get m/s, and multiply m/s by $\tfrac{18}{5}$ to get km/h. The factor comes from 1 km = 1,000 m and 1 hour = 3,600 s, so 1 km/h is $\tfrac{1000}{3600} = \tfrac{5}{18}$ m/s.

km/h	18	36	54	72	90	108
m/s	5	10	15	20	25	30

Example 1 · Converting units

A metro train runs at 81 km/h. How far does it go in 40 seconds?

Speed: $81 \times \tfrac{5}{18} = 22.5$ m/s.
Distance: $22.5 \times 40 = 900$ m.

Answer

900 m

Watch out

Divide minutes by 60: 2 hours 24 minutes is $2 + \tfrac{24}{60} = 2.4$ hours, but 2 hours 15 minutes is 2.25 hours, not 2.15. And 1 m/s is 3.6 km/h, so the km/h figure is always the bigger one.

How do you find average speed?

Average speed is total distance divided by total time. It is almost never the plain average of the speeds, because more time is spent at the slower speed. For the same distance at speeds $x$ and $y$ , the average is $\tfrac{2xy}{x + y}$ ; only for the same time at each speed is it $\tfrac{x + y}{2}$ .

Average speed, equal distances

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

Example 2 · There and back

Sunita drives to her office at 30 km/h and returns along the same road at 45 km/h. Find her average speed.

Equal distances, so use $\tfrac{2xy}{x + y}$ .
$\tfrac{2 \times 30 \times 45}{30 + 45} = \tfrac{2700}{75} = 36$ km/h.

Answer

36 km/h

The tempting 37.5 km/h is the plain average and is wrong.

Exam shortcut

For three equal distances, take each part as the LCM of the speeds, find each time, and divide total distance by total time.
90 km at 45 km/h and 70 km at 35 km/h take 2 hours each, so the average is $160 \div 4 = 40$ km/h. It equals the plain mean only because the times happen to be equal. Always check.
A target average fixes the total time allowed. Subtract the time already used to find the speed needed for the rest.

How do you solve "late and early" problems?

Find the difference between the two travel times, then write distance over one speed minus distance over the other equals that difference. Late by $p$ minutes and early by $q$ minutes lie on opposite sides of the right time, so the difference is $p + q$ ; late by both means $p - q$ .

Example 3 · Late and early

At 6 km/h Farhan reaches the bus stop 8 minutes late; at 8 km/h he is 7 minutes early. How far is the stop?

Time difference: $8 + 7 = 15$ minutes, which is $\tfrac{1}{4}$ hour.
$\tfrac{d}{6} - \tfrac{d}{8} = \tfrac{1}{4}$ , so $\tfrac{d}{24} = \tfrac{1}{4}$ .
$d = 6$ .

Answer

6 km

Exam shortcut

At a fraction of the usual speed, say $\tfrac{4}{5}$ , the time becomes $\tfrac{5}{4}$ of the usual, so the extra time is $\tfrac{1}{4}$ of the usual time.

Relative speed

What is relative speed and when do you add or subtract speeds?

Relative speed is how fast the gap between two moving objects changes. In opposite directions, towards or away from each other, add the speeds. In the same direction, subtract the smaller from the larger.

Situation	Relative speed	Time taken
Moving towards each other	$a + b$	gap ÷ $(a + b)$
Moving apart from one point	$a + b$	separation ÷ $(a + b)$
One chasing the other	$a - b$	head start ÷ $(a - b)$

Example 4 · A late start

A tractor leaves a farm at 6:00 a.m. at 16 km/h. A jeep leaves the same farm at 7:15 a.m. at 36 km/h on the same road. When does the jeep catch the tractor?

By 7:15 the tractor is $16 \times 1\tfrac{1}{4} = 20$ km ahead.
The jeep gains $36 - 16 = 20$ km an hour, so it needs 1 hour.

Answer

8:15 a.m.

Meetings and tracks

Two people starting together from the ends of a road meet at a point that divides it in the ratio of their speeds. With speeds $3 : 2$ , the faster has covered $\tfrac{3}{5}$ of the road.
If after meeting they need $t_1$ and $t_2$ more hours, their speeds are in the ratio $\sqrt{t_2} : \sqrt{t_1}$ .
On a circular track, runners in the same direction first meet after track ÷ $(a - b)$ , in opposite directions after track ÷ $(a + b)$ . To meet again at the start, take the LCM of the lap times.

Trains

How do you solve train problems?

A train has length, so while crossing something it covers its own length plus the length of whatever it crosses. Then use the right speed: its own for a still object, the sum for something coming the other way, the difference for something going the same way.

A train shown twice beside a platform: first with its front at the start of the platform, then with its rear at the far end. The front moves through the platform's length plus the train's own length. — To clear a platform, the front of the train travels the platform's length plus its own length.

The train crosses	Distance covered
A pole, signal or standing person	The train's length
A platform, bridge or tunnel	Train length + platform length
Another train	The sum of the two lengths

Example 5 · Overtaking

A 250 m train at 63 km/h overtakes a 200 m train running at 45 km/h on a parallel track. How long does overtaking take?

Same direction: $63 - 45 = 18$ km/h, which is 5 m/s.
Distance: $250 + 200 = 450$ m.
Time: $450 \div 5 = 90$ s.

Answer

90 seconds

Example 6 · Pole and tunnel

A train passes a pole in 15 s and a 300 m tunnel in 27 s. Find its length.

The extra 12 s are spent covering the tunnel's 300 m, so the speed is $300 \div 12 = 25$ m/s.
Length: $25 \times 15 = 375$ m.

Answer

375 m

Exam shortcut

The two-crossings pattern is common: the difference in times is the time to cover the difference in distances. And a passenger sitting in another train is a point moving at that train's speed, so only the first train's length is covered.

Boats and streams

How do boats and streams questions work?

With the boat's still-water speed $b$ and the stream's speed $s$ , downstream speed is $b + s$ and upstream speed is $b - s$ . So the still-water speed is half their sum, and the stream speed is half their difference.

Boats and streams

b = \frac{\text{down} + \text{up}}{2} \qquad s = \frac{\text{down} - \text{up}}{2}

Example 7 · Still water and stream

A boat goes 42 km downstream in 3 hours and 20 km upstream in 2 hours. Find the speed in still water and the stream's speed.

Downstream: $42 \div 3 = 14$ km/h. Upstream: $20 \div 2 = 10$ km/h.
Still water: $(14 + 10) \div 2 = 12$ km/h. Stream: $(14 - 10) \div 2 = 2$ km/h.

Answer

12 km/h and 2 km/h

Exam shortcut

For a round trip, find the upstream and downstream times separately and add: $\tfrac{d}{b - s} + \tfrac{d}{b + s}$ = total time.
An object floating in the river is at rest relative to the water, so a rower who goes away from it for 15 minutes and turns back at the same effort takes 15 minutes to return, whatever the stream.

Exam shortcuts

Know the conversions. 18 km/h is 5 m/s, and multiples follow.
Equal distances. Average speed $\tfrac{2xy}{x + y}$ .
Late and early. Add the minutes for opposite sides, subtract for the same side.
Trains. Add the lengths of everything crossed.
Sense checks. An average speed lies between the speeds, nearer the slower one for equal distances. A still-water speed lies between upstream and downstream.

Common mistakes

The plain mean of speeds for equal distances.
Minutes as decimals. 1 hour 15 minutes is 1.25 hours.
Forgetting to convert km/h before multiplying by seconds.
Leaving out a length when a train crosses a platform or another train.
The wrong direction rule. Same direction subtracts; opposite adds.
Late and early as $p - q$ .
Ignoring the head start when one traveller starts later.
Not halving in boats and streams.
Answering the wrong quantity. The slower train, the length rather than the speed, the distance from a particular town.

Quick revision

Basics

Speed = distance ÷ time
km/h $\times \tfrac{5}{18}$ = m/s
Fixed distance: speed and time inversely proportional

Average speed

Total distance ÷ total time
Equal distances: $\tfrac{2xy}{x + y}$
Equal times: $\tfrac{x + y}{2}$

Relative speed

Opposite: $a + b$ . Same direction: $a - b$
Chase: head start ÷ $(a - b)$
Meeting point divides the road in the speed ratio

Trains and boats

Pole: train length. Platform: train + platform
Two trains: both lengths, relative speed
$b = \tfrac{\text{down} + \text{up}}{2}$ , $s = \tfrac{\text{down} - \text{up}}{2}$

Ratios of speeds and times come from ratio and proportion, average speed is a cousin of the weighted average in averages, and rates carry over to time and work.

Next · 08Quick revision

Check your understanding

71 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Speed, conversion and average speedQ1 of 71Q1

Speed, conversion and average speed

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Speed, conversion and average speed

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