Time Speed and Distance Questions for CAT


Question 1: I walk to a town at \(3_2^1\) kmph, rest there for 45 minutes and ride back at \(7_2^1\) kmph. Find the distance to the town, if the total time spent by me is 6 hrs 37 min.
[1] 14 km
[2] 7km
[3] 5 km
[4] 8 km
Option:

Let the distance be D.

Therefore, \(\begin{array}{l}\frac{D}{{3.5}} + \frac{{45}}{{60}} + \frac{D}{{7.5}} = 6\frac{{37}}{{60}}\\ \Rightarrow D = 14\;cm\end{array}\)

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Question 2: A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 seconds and 10 seconds respectively. The length of the train (in metres) is
[1] 45
[2] 54
[3] 50
[4] 72
Option:

Let the length of train be x km and its speed be y km/hr.

Then, \(\frac{x}{{y - 2}} = \frac{9}{{60 \times 60}}{\rm{ and }}\frac{x}{{y - 4}} = \frac{{10}}{{60 \times 60}}\)

=> 9y – 3600x = 18 and 10y -3600x =40

=> y- 400x = 2 and y – 360x =4

Therefore, 40x = 2 or \(x\) m

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Question 3: A man reaches his office 30 min late, if he walks from his home at 3 km per hour and reaches 40 min early if he walks at 4 km per hour. How far is his office from his house?
[1] 7km
[2] 14 km
[3] 5 km
[4] 3 km
Option:

Time gained = 30 + 40 = 70 min =\(\frac{{70}}{{60}}\)hrs.

Let the distance be x km.

Therefore, \(\frac{x}{3} - \frac{x}{4} = 70 \times \frac{1}{{60}} \Rightarrow x = 14\) km

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Question 4: Raju, walking at the rate of 6 kmph, covers a certain distance in three hours. In how much time will Raju cover this distance running at the speed of 18 kmph?
[1] 1 hour
[2] 3 hours
[3] 60 hours
[4] 22 hours
Option:

Let the distance be X.

=>Distance = Speed x Time taken = 6 x 3 = 18 km.

Now, speed = 18 km/hr.

=>Time taken = Distance/Speed = 18/ 18 = 1 hour.

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Question 5: Two cyclists start together to travel to a certain destination, one at the rate of 4 kmph and the other at the rate of 5 kmph. Find the distance if the former arrives half an hour after the latter.
[1] 2 km
[2] 10m
[3] 10000m
[4] 1 km
Option:

Let the distance be X km. Now Time= distance/ speed.

=>Time for the first rider = X/14.

Similarly, time for the second cyclist = X/5.

Given that: X/14 – X/5 =1/2

=> X= 10 km = 10000 m

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Question 6: Two trains 121 m and 99 m in length respectively are running in opposite directions, one at the rate of 40 kmph and the other at the rate of 32 kmph. How long will they take to be completely clear of each other from the moment they meet?
[1] 110 sec
[2] 99 sec
[3] 88 sec
[4] 11 sec
Option:

Time taken to cover\( = \left( {121 + 99} \right) = 220\)

at the rate of (32 + 40), i.e. 72 kmph = \(\frac{{220}}{{\left( {72 \times \frac{5}{{18}}} \right)}}\)sec =11 sec.

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Question 7: A jeep travels a distance of 100 km at a uniform speed. If the speed of the jeep is 5 kmph more, then it takes 1 hour less to cover the same distance. The original speed of the jeep is
[1] 20 kmph
[2] 25 kmph
[3] 30 kmph
[4] 50 kmph
Option:

Let the original speed of the jeep be x kmph.

\( \Rightarrow \frac{{100}}{x} - \frac{{100}}{{x + 5}} = 1\)

Solving this, we get x = 20 kmph.

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Question 8: Two athletes cover the same distance at the rate of 10 and 15 kmph respectively. Find the distance travelled when one takes 15 minutes longer than the other.
[1] 7.5 m
[2] 750 km
[3] 7.5 km
[4] 15 km
Option:

Let the distance be D km.

\( \Rightarrow \frac{D}{{10}} - \frac{D}{{15}} = \frac{{15}}{{60}} \Rightarrow \frac{D}{{30}} = \frac{1}{4} \Rightarrow D = 7.5km\)

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Question 9: A motorcyclist covers 4 successive 4 km stretches at speeds of 20 kmph, 30 kmph, 40 kmph, and 50 kmph respectively. Find the average speed over the total distance.
[1] 40.2 kmph
[2] 31.2 kmph
[3] 50.3 kmph
[4] 36 kmph
Option:

Average speed =\(\frac{{Total{\rm{ distance covered }}}}{{{\rm{ time taken }}}} = \frac{{16}}{{\frac{4}{{20}} + \frac{4}{{30}} + \frac{4}{{40}} + \frac{4}{{50}}}}\)

\( = \frac{{16 \times 600}}{{120 + 80 + 60 + 48}} = \frac{{9600}}{{308}} = 31.17\) kmph

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Question 10: Ram and Shyam travel the same distance at the speeds of 10 kmph and 15 kmph respectively. If Ram takes 30 min longer than Shyam, then the distance travelled is
[1] 15 km
[2] 2 km
[3] 10 km
[4] 30 km
Option:

. By a previously explained logic, \(x/10 - x/15 = 1/2\)

\( \Rightarrow X = 15km\)

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Question 11: A vendor takes 10 hours to walk to a certain place and cycle back. He would have gained three hours if he had covered both routes on cycle. How long would he have taken to walk both ways?
[1] 6.5 hrs
[2] 26 hrs
[3] 33 hrs
[4] 13 hrs
Option:

One way walk +one way cycling = 10 hrs. Now multiply both sides by 2.

We get 2-way walk + 2-way cycling = 20 hrs ....(1)

Also given: 2-way cycling = 7 hrs ....(2)

Subtract (2) from (1) to get: 2-way walk = 20 - 7 = 13 hrs.

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Question 12: If Sita walks at 5 kmph, she misses her train by 10 minutes. If she walks at 7 kmph, she reaches the station 10 minutes early. How much distance does she walk to the station?
[1] 5.8 km
[2] 35.6 km
[3] 10.6 km
[4] 92 km
Option:

Let the distance be D.

\(\frac{D}{5} - \frac{D}{7} = \frac{{10 + 10}}{{60}}\)

\( \Rightarrow D = \frac{{35}}{6} = 5.8km\)

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Question 13: Rahul can cover the distance to his school in 15 minutes. He covers 2/3rd of it at the rate of 5 kmph and the remaining at the rate of 10 kmph. The total distance to the school is
[1] 15 km
[2] 1.5 km
[3] 3 km
[4] 6km
Option:

Let the distance to his school be x km.

According to the given conditions,

\(\frac{{2x}}{3} \times \frac{1}{5} + \frac{x}{3} \times \frac{1}{{10}} = \frac{{15}}{{60}}\)

\( \Rightarrow \frac{{4x + x}}{{30}} = \frac{1}{4} \Rightarrow x = \frac{{30}}{{4 \times 5}} = \frac{3}{2} = 1.5km\)

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Question 14: A friend is spotted by Lalloo at a distance of 200 m. When Lalloo starts to approach him, the friend also starts moving in the same direction as Lalloo. If the speed of his friend is 15 kmph, and that of Lalloo is 20 kmph, then how far will the friend have to walk before Lalloo meets him?
[1] 600 m
[2] 0.6 m
[3] 6km
[4] 900 m
Option:

Lalloo is unfortunate that the friend is moving away from him.

(Because the friend moves in same direction as Lalloo).

relative speed= 20- 15= 5,kmph. distance= 200 m.

Thus, Lalloo will meet his friend when he gains 200 m over him.

=> time required = distance / speed = 0.2/5 = 1/25 hrs.

=>Distance travelled by the friend in 1/25 hrs. (when Lalloo catches up him)

=> Time x Speed = 1/25 x 15 = 3/5 km = 600 m.

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Question 15: A distance is covered at a certain speed in a certain time. If the double of this distance is covered in four times the time, then what is the ratio of the two speeds?
[1] 1.5 : 0.7
[2] 1 : 1.9
[3] 4 : 2
[4] 6 : 1
Option:

Case I : Distance D Speed \({S_1}\) Time \(D/{S_1}\)

Case II : Distance 2D Speed \({S_2}\) Time \(4\left( {D/{S_1}} \right)\)

=> Speed for case II = \({S_2}\) = Distance/Time = \(2D/\left( {4D/{S_1}} \right) = {S_1}/2\)

Hence, speed for case I : speed for case II. = \({S_1}:{S_1}/2 = 1:1/2 = 2:1\)

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Question 16: Raghunandan completes 9/30th of his total journey by ship, 5/18th by a horse carriage and the remaining 10 km on foot. The length of the total journey is
[1] 450/19 km
[2] 24.62 km
[3] 27.2 km
[4] 92 km
Option:

Let the total length of journey, performed by Raghunandan be x km.

According to the given condition, \(x - \frac{9}{{30}}x - \frac{5}{{18}}x = 10\)

\( \Rightarrow \frac{{90x - 27x - 25x}}{{90}} = 10 \Rightarrow 38x = 10 \times 90\)

Or \(x = \frac{{10 \times 90}}{{38}} = \frac{{450}}{{19}}km\)

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Question 17: Walking at 4/7th of his usual speed, Ramu gets late by 15 minutes. Find the time he would have taken walking at his usual speed.
[1] 25 min
[2] 20 min
[3] 30 min
[4] 24 min
Option:

Speed = 4/7th of the usual speed.

=>Time required = 7/4th of the usual time.

=> (7/4th of the usual time)- (usual time required) = 15/60 hours.

=> usual time = 1/3 hours = 20 min.

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Question 18: If I take as much time in running 15 m as a car takes in covering 36 m, then find the distance I can cover in the time taken by the car to travel 1 km.
[1] 416.6 m
[2] 41.6 m
[3] 4.16 km
[4] 41.6 km
Option:

36m = 15 m.

\(1000m \equiv \frac{{15}}{{36}} \times 1000 = 416.6m\)

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Question 19: Ramesh runs 22.4 kmph. How many metres does he run in five minutes?
[1] 1866m
[2] 18.66 km
[3] 2.866 km
[4] 1.432 km
Option:

Speed = 22.4 km/hr = 22.4 x 5/18 m/s = 6.22 m/s.

=>In 5 min (300 sec), distance covered = 6.22 x 300 = 1866 m.

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Question 20: A handcart has to cover a distance of 120 km in 15 hours. If it covers half the distance in 4/7th of the time, then what speed should it maintain in order to cover the remaining journey in the scheduled time?
[1] 9.33 kmph
[2] 4.6 kmph
[3] 3.1 kmph
[4] None of these
Option:

Time left = 1 - 4/7 = 3/7th of the total scheduled time.

3/7 x 15 = 45/7 hrs; Distance left = 60 km.

Required speed = 60/(45/7) = 9.33 km/hr.

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Question 21: The time in which a train 180 m long, travelling at the speed of 30 kmph, will pass a signal pole is
[1] 21.6 min
[2] 0.216 min
[3] 21.6 sec
[4] None of these
Option:

Distance to be covered = 180 m.

Speed = 30 kmph. =>Time = 0.18/30 hrs = 21.6 sec.

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Question 22: The time taken by a 90 m long train, running at the speed of 18 kmph to cross a bridge 270 m long, is
[1] 70 sec
[2] 146 sec
[3] 72 sec
[4] None of these
Option:

Distance to be covered = 90 + 270 = 360 m. Speed = 18 km/hr.

Time = 0.36/18 = 0.02 hrs = 0.02 x 60 x 60 sec = 72 sec.

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Question 23: Find the time taken by two trains, one 180 m long and the other 270 m long, to cross each other, if they are running at speeds of 46 kmph and 54 kmph respectively. Consider both possible cases of motion.
[1] 202.5, 16.2 sec
[2] 160, 100 sec
[3] 108.45, 15.6 sec
[4] 204.5, 14.8 sec
Option:

Case I: Motion in same direction

=>Relative speed = 54 — 46 = 8 km/hr.

Distance to be covered = 180 + 270 = 450 m.

=>Time = 0.450/8 = 0.056 hrs = 202.5 sec.

Case II: Motion in opposite direction.

=> Relative Speed = 54 + 46=100 km/hr.

Distance to be covered = 180 + 270= 450 m.

=>Time = 0.450/100 = 0.00045 hrs = 16.2 sec.

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Question 24: A person going from Pondicherry to Ootacamond travels 120 km by steamer, 450 km by rail and 60 km by horse transit. The journey occupies 13 hours 30 minutes, and the speed of the train is three times that of the horse-transit and \(1_2^1\) times that of the steamer. Find the speed of the train.
[1] 20 kmph
[2] 60 kmph
[3] 10 kmph
[4] 50 kmph
Option:

If speed of horse is X, then speed of steamer = 2X and

speed of train = 3X => \(\frac{{120}}{{2x}} + \frac{{450}}{{3x}} + \frac{{60}}{x} = 13.5\)

=> X = 20 km/hr => 3X = 60 km/hr.

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Question 25: A railway passenger counts the telegraph posts as he passes them. If they are 50 metres apart and the train is going at 48 kmph, how many posts will he pass per minute?
[1] 16
[2] 20
[3] 24
[4] 10
Option:

48 km/hr = 48 x 5/18 m/sec = \(\frac{{48 \times 5 \times 60}}{{18}}\)m/min. = 800 m/min.

:. Number of posts = 800/50 = 16.

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Question 26: A and B start on a journey at the same time. B travels at 4/7th of A's rate, and arrives 3 hours 15 minutes after him. How long did each take to complete the whole journey?
[1] \(\frac{{91}}{{12}}\)hours, \(\frac{{13}}{3}\) hours
[2] \(\frac{{89}}{{12}}\) hours,\(\frac{{13}}{5}\) hours
[3] \(\frac{{91}}{{10}}\) hours,\(\frac{{13}}{3}\) hours
[4] \(\frac{{91}}{{12}}\) hours,\(\frac{{13}}{7}\) hours
Option:

A's speed : B's speed = 7: 4.

A’s time : B's time = 4 : 7 = 4X, 7X.

\( \Rightarrow 7X - 4X = 3\frac{{15}}{{60}} \Rightarrow X = \frac{{13}}{{12}}\)

\( \Rightarrow 7X = \frac{{91}}{{12}}{\rm{ hrs}},{\rm{ }}4X = \frac{{13}}{3}\) hrs.

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Question 27: A ship 156 km from the shore springs a leak which admits 2.5 metric tons of water In \(6_2^1\) minutes. A quantity of 68 metric tons would suffice to sink it, but the pumps can throw out 15 metric tons in an hour. The average rate of sailing so that it just reaches the shore as it begins to sink should be
[1] 18 kmph
[2] 60 kmph
[3] 15 kmph
[4] 10 kmph
Option:

In one minute, amount flowing in = \(\frac{{15}}{{39}}\) MT.

In one minute, amount thrown out = \(\frac{{15}}{{60}} = \frac{1}{4}\)MT.

Effective rate of filling in one hour= \(\left( {\frac{{15}}{{39}} - \frac{1}{4}} \right)MT = \frac{{21}}{{56}}\) MT/Min.

Time till it just begins to sink = \(\frac{{68}}{{21/156}} = 505\) minutes.

Speed required = (156/505) = 0.3 km/min = 0.3 x 60 km/hr = 18 km/hr.

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Question 28: A train 90 m long is travelling at the rate of 40 kmph. A man is running at a speed of 5 kmph in a direction opposite to the motion of the train. The time taken by the train to cross the man is
[1] 7.2 sec
[2] 92 sec
[3] 2 sec
[4] 66 sec
Option:

Distance to be covered = 90 m. Relative speed = 40 + 5 = 45 km/hr.

Time = 0.09/(45) = 7.2 sec.

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Question 29: A train travelling at 25 kmph leaves Delhi at 9 am and another train travelling at 35 kmph starts at 2 pm in the same direction. How many km from Delhi are they together?
[1] \(437_2^1\) km
[2] 43.7 km
[3] 35 km
[4] 12.5 km
Option:

The first train has a start of 25 x 5 km and

the second train gains (35 - 25) or 10 km per hour.

=> The second train will gain 25 x 5 km in 25 x 5 /10 or \(12_2^1\) hrs.

=> The required distance from Delhi = \(12_2^1 \times 35{\rm{ km }} = 437_2^1km\).

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Question 30: Walking at 3/4 of his usual rate, a man is \(1_2^1\) hours too late. Find his usual time.
[1] \(4_2^1\) hours
[2] ( \(1_2^1\) x 3) hours
[3] (3 + ) \(1_2^1\)) hours
[4] All of these
Option:

Since the man walks at 3/4 of his usual rate, the time that he takes is 4/3 of his usual time.

=> 4/3 of usual time= usual time + \(1_2^1\)hours.

=> 1/3 of usual time = \(1_2^1\) hours.

Usual time = (\(1_2^1\) x 3) hours = ( 3 + \(1_2^1\) ) hours = \(4_2^1\) hours.

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Question 31: 1 have to be at a certain place at a certain time and find that I shall be 40 minutes late, if I walk at the rate of 4 kmph and 40 min early if I walk at the rate of 5 kmph. What is the usual time required by me to reach my destination?
[1] 7 hrs
[2] 14 hrs
[3] 6 hrs
[4] 3.5 hrs
Option:

Let t be the usual time.

:. Distance covered D \( = 4 \times \left( {t + \frac{2}{3}{\rm{ }}} \right) = 5 \times \left( {t - \frac{2}{3}{\rm{ }}} \right)\)

\( \Rightarrow t = 6\)

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Question 32: Two men A and B walk from P to Q, a distance of 21 km, at 3 and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. The distance from P to R is
[1] 12 km
[2] 18 km
[3] 10 km
[4] 24 km
Option:

When B meets A at R, B has walked the distance PQ + QR and A the distance PR.

That is, both of them have together walked twice the distance from P to Q, i.e. 42 km.

Now, the speed of A and B are in the ratio 3 : 4, and they have walked 42 km.

Hence, the distance PR travelled by A = 3/7 of 42 km = 18 km.

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Question 33: A person can row at the rate of 3 kmph in still water. If the river is running at 1.2 kmph, then it takes him 2 hours to row to a place and back. How far is the place?
[1] 5 km
[2] 2.5 km
[3] 6 km
[4] 8 km
Option:

X = 3 kmph, Y = 1.2 kmph. Since time taken = Distance/Speed = 2 hrs.

=> D/4.2 + D/1.8 = 2 =>(1.8 D + 4.2 D)(4.2 x 1.8).

=> D = 2.52 km.

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Question 34: The current of a stream runs at 5 kmph. A motor launch goes 25 km upstream and back again to the starting point in 275/2 minutes. The speed of the motor launch in the absence of any current is
[1] 22.8 kmph
[2] 10.1 kmph
[3] 31.6 kmph
[4] 6 kmph
Option:

Let X = Speed of motor launch.

Y = Speed of current = 5 kmph. Distance = 25 km.

Time required = 275/(2 x 60) hrs = 25/(X + 5) +25/(X — 5).

\( \Rightarrow 2.3 = \frac{{25\left( {X - 5} \right) + 25\left( {X + 5} \right)}}{{\left( {X + 5} \right)\left( {X - 5} \right)}} = \)>\(2.3 = 50{\rm{X}}/\left( {{X^2} - 25} \right)\)

\( \Rightarrow 2.3{X^2} - 50X - 57.2 = 0\)

Apply the quadratic formula to get.

X= speed of launch = (50 ± 55)/4.6.

Neglect the negative sign => X= 105/4.6 = 22.82 kmph.

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Question 35: A man can row with the stream at the rate of 20 kmph and against the stream at 5 kmph. The man's rate in still water is
[1] 7.5 kmph
[2] 12.5 kmph
[3] 17.5 kmph
[4] 15.5 kmph
Option:

X + Y = 20, X - Y = 5 => X = 12.5 kmph.

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Question 36: A boat goes 50 km upstream in 10 hours and a distance of 40 km downstream in 8 hours. The speed of the boat in still water is
[1] 0 kmph
[2] 10 kmph
[3] 7.2 kmph
[4] 5 kmph
Option:

Speed upstream = 50/10 = 5 kmph.

Speed downstream = 40/8 = 5 kmph.

Since both these speeds are equal, it means that the water is still.

=>Required answer = 5 kmph

Two cyclists start from the same place to ride in the same direction. A starts at noon at 8 kmph while B starts at 2 pm at the rate of 10 kmph.

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Question 37: How far will A have ridden before he is overtaken by B?
[1] 76 km
[2] 80 km
[3] 84 km
[4] 75 km
Option:

By the time B starts, A has already covered 16 km.

Difference between B's speed and A's speed is 2 km/hr.

So, total time for B to catch up with A will be \(\frac{{16}}{2}\) i.e.. 8 hrs.

:. At 10 pm, B will overtake A.

Distance covered by A before he is overtaken by B = 10 x 8 = 80 km.

i.e. A and B will be 5 km apart at \(\frac{{11}}{2}\) hrs after A's start and \(\frac{{21}}{2}\) hrs after B's start, i.e. at 7.30 pm and 12.30 am next day.

Hence required answers are 80 km, 7.30 pm and 12.30 am next day.

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Question 38: At what times A and B will be 5 km apart?
[1] 7.30 pm on the same day and 1.30 am on the next day.
[2] 7.30 pm on the same day and 12.30 am on the next day.
[3] 8.30 pm on the same day and 1.30 am on the next day.
[4] 8.30 pm on the same day and 12.30 am on the next day.
Option:

By the time B starts, A has already covered 16 km.

Difference between B's speed and A's speed is 2 km/hr.

So, total time for B to catch up with A will be \(\frac{{16}}{2}\) i.e.. 8 hrs.

:. At 10 pm, B will overtake A.

Distance covered by A before he is overtaken by B = 10 x 8 = 80 km.

i.e. A and B will be 5 km apart at \(\frac{{11}}{2}\) hrs after A's start and \(\frac{{21}}{2}\) hrs after B's start, i.e. at 7.30 pm and 12.30 am next day.

Hence required answers are 80 km, 7.30 pm and 12.30 am next day.

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Question 39: A train runs at 45 kmph. How far does it go in 6 seconds?
[1] 45 m
[2] 60 m
[3] 75 m
[4] 55 m
Option:

\(45 \times \frac{5}{{18}} \times 6 = 75m\)

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Question 40: If the speed of train is 92.4 kmph, then how many metres does it cover in 20 minutes?
[1] 30800 m
[2] 32800 m
[3] 38200 m
[4] 35500 m
Option:

In 60 minutes, the train covers 92400 m.

:. in 20 minutes, the train covers 30800 m.

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