Arithmetic - Time, Speed & Distance - Previous Year CAT/MBA Questions
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The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was
- (a)
14π
- (b)
18π
- (c)
16π
- (d)
12π
Answer: Option C
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Text Explanation :
Let the radius of A and B be a and b respectively. (a = 30 and b = 40)
Distance travelled by any wheel D = (Circumference C) × (Number of revolutions N)
∴ D ∝ R × N (∵ C ∝ Radius R)
∴ N ∝ 1/R
∴ Na/Nb = b/a = 40/30 = 4/3 ...(1)
It is given that Na = Nb + 5000 ...(2)
Solving (1) and (2), we get; Na = 20000 and Nb = 15000.
D = Ca × Na = 2πa × Na = vb × (3/4) [where vb is the velocity of B]
∴ 2π × (30 × 10-5) × 20000 = vb × (3/4)
∴ vb = 16π.
Hence, option (c).
Workspace:
A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?
- (a)
22
- (b)
20
- (c)
15
- (d)
23
Answer: Option C
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Text Explanation :
It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am
Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am
If they leave one by one every minute, the 45th motorcyclist would have left by 10:45 am to reach B at 11:00 am.
Thus, time taken by one motorcyclist to reach B from A = 15 minutes.
Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am
So, the last motorcyclist should have left A by 10:15 am
Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B
Hence, option (c).
Workspace:
John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?
Answer: 48
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Text Explanation :
Let the track length for John be 'a' and for Mary be 'b'
So, Distance travelled by John = 9a
Distance travelled by Mary = 5b
Now, Time taken by John = 9a/6 hours
Time taken by Mary = 5b/7.5 hours
We know that Time taken by John = Time taken by Mary
⇒ 9a/6 = 5b/7.5
⇒ a = 4b/9
Total track length = 325 meters
So, 4b/9 + b = 325 meters
⇒ 13b/9 = 325 meters
⇒ b = 225 meters
⇒ a = 100 meters
Mary jogs at 7.5 Kmph = 7.5 × 5/18
So, time taken = = 48 seconds.
Hence, 48.
Workspace:
Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at
- (a)
10 : 27 am
- (b)
10 : 25 am
- (c)
10 : 45 am
- (d)
10 : 18 am
Answer: Option A
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Text Explanation :
We are told that, by the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.
So, the speeds of A and B are in the ratio 60:40 or 3:2
Hence, the time they take to cover a particular distance will be in the ratio 2:3
We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.
That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12 + 12 + 6 = 30 minutes. (because 40% + 40% + 20% = 100%)
Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.
At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.
Time required to complete 60% of the track = 60% of 45 = 27 minutes.
Hence, B complete one single round at 10:27 AM.
Hence, option (a).
Workspace:
Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?
Answer: 15
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Text Explanation :
Let distance between X and Y be ‘d’ km and train T travels at speed ‘s’ kmph.
Therefore, speed of train S = s kmph
At 4 pm, T must have covered ‘s’ km.
Therefore, distance between the two
= (d – s) km
Time taken for the two to cover this distance = =
Distance covered by train T by now = s + =
By the given condition, =
Solving this, we get d = 15s
To travel from X to Y, train T takes 15s/s = 15 hours
Hence, 15.
Workspace:
Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is
Answer: 12
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Text Explanation :
Let distance between A and P be ‘x’ kms. Therefore, distance between P and B = (3x) kms
Let car 1 reaches point P in ‘t’ minutes. Therefore, car 2 takes (t + 60) minutes to reach P.
Speed of car 1 = x/t and speed of car 2 = 3x/(t + 60)
Hence, =
Solving this, we get
t = 12 minutes.
Hence, 12.
Workspace:
The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is
- (a)
4
- (b)
3
- (c)
6
- (d)
5
Answer: Option D
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Text Explanation :
Let the time taken by Narayan to cover 60 km be ‘t’ hours.
Therefore, Partha takes (t + 4) hours to cover 60 km.
Parth takes (t – 2) hours to cover half the distance.
So, to cover remaining distance i.e., 30 km, he takes (t + 4) – (t – 2) = 6 hours.
Hence, Partha's speed = 30/6 = 5 kmph
Hence, option (d).
Workspace:
On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is
Answer: 50
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Text Explanation :
If both cars travel towards east, they meet in 7 hours. Therefore the faster car starting from A covers the distance of separation of 350 km in 7 hours. Therefore,
Hence, 50.
Workspace:
Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is
Answer: 5
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Text Explanation :
The speeds of both the cars in the three legs of the journey are equal. Therefore the second car will reach point B as many minutes after the first car as it started after the first car.
For the first 50 km of the journey, the speed of the cars is 100 kmph. Therefore, the first car reaches a distance of 20 km
from point A after = hours
= 0.2 × 60 = 12 minutes.
When the first car reaches point B, the second car is 12 minutes away from point B. The speed of the car in the last 50 km distance is 25 kmph. Therefore the distance of the second car from point
Hence, 5.
Workspace:
Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is
- (a)
1 : 2
- (b)
2 : 7
- (c)
1 : 4
- (d)
2 : 9
Answer: Option C
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Text Explanation :
We have the following:
The cars 3 and 1 meet at point Q. Therefore, the car 3 covers 100 km in the time the car 1 covers 200 km. Therefore the ratio of the speeds of cars 3 and 1 i.e., C3 : C1 = 1 : 2.
The cars 2 and 3 meet at point P. Therefore, the car 2 covers 100 km in the time the car 3 covers 200 km. Therefore the ratio of the speeds of cars 3 and car 2 i.e., C3 : C2 = 2 : 1.
Combining the two statements, the ratio of the speeds of three cars i.e., C1: C2 : C3 = 4 : 1 : 2.
Hence, option (c).
Workspace:
A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is:
Answer: 20
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Text Explanation :
The speed in the second case is 5/4 times the speed in the first case.
Therefore, the time would be 4/5 times the time, i.e., 1/5 less.
This one fifth is 20 min. Therefore, the time taken in the first case is 100 min i.e. 100/60 = 5/3 hours.
The distance = 12 × 5/3 = 20 km.
Hence, 20.
Workspace:
A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is:
- (a)
√6 : √2
- (b)
√7 : 2
- (c)
2√5 : 3
- (d)
3 : 2
Answer: Option B
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Text Explanation :
Let the speed of the boat in still water and the speed of the river be x and y respectively.
And the distance between office and home be 'd'.
⇒ + =
⇒ =
⇒ 8(x2 – y2) = 4x2 – y2
⇒ x2/y2 = 7/4
⇒ x/y = √7/2
Hence, option (b).
Workspace:
In a 10 km race, A, B and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?
Answer: 1900
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Text Explanation :
In a 10 km race, A beats B by 1 km. So in the time A completes 10 km, B completes
9 km. Therefore ratio of their speeds is 10 : 9.
In a similar way, ratio of speeds of B and C is also 10 : 9
A : B : C
10 : 9
10 : 9
100 : 90 : 81
To get a common ratio for speeds of A, B, C we take LCM of 9 and 10 which is equal to 90 (∵ B is common to both ratios) and equate corresponding values of A and C is also taking B as 90
So if A runs 100m, C runs 81m.
So in a 10 km race, when A runs 10000 m C will run = 8100 m
∴ A will beat C by 10000 - 8100 or 1900 m.
Hence, 1900.
Workspace:
Arun drove from home to his hostel at 60 miles per hour. While returning home he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then the total distance traveled by him is
- (a)
55 miles
- (b)
60 miles
- (c)
65 miles
- (d)
70 miles
Answer: Option C
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Text Explanation :
Let ‘d’ miles be the distance for the onward journey from his home to his hostel.
Now time take for onward jouney = d/60.
For the return journey, time taken to cover half distance (i.e. d/2)
⇒ =
Now for the return journey, (after half the distance is covered) taking the bypass road will increase the distance by 5 miles i.e., the remaining distance of the journey is 'd/2 + 5' miles
Now, 'd/2 + 5' miles is covered at the speed of 50 miles per hour.
Sp time taken to cover ' + 5' miles is (d/2 + 5)/50 =
Total time for return journey = + =
Now as time taken for return journey is half an hour more than onward journey.
⇒ =
Hence, d = 30
Total distance covered (for onward and return journey) = d + d/2 + d/2 + 5
= 30 + 30/2 + 30/2 + 5 = 65 miles
Hence, option (c).
Workspace:
A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 pm at a point which is 168 km away from A. What is the distance, in km, between A and B?
- (a)
364
- (b)
378
- (c)
380
- (d)
388
Answer: Option B
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Text Explanation :
Motorbike travels 168 kms from 1 pm till 3:40 pm i.e. in 2 hours 40 mins = hours.
⇒ Speed of motorbike = = 63 kmph
∴ Speed of Car = 2 × Speed of Motorbike = 2 × 63 = 126 km/hr
Now as the car travels for 1 hour 40 minutes.
Distance travelled by the car = 126 × or 210 km
Distance between A and B = Distance travelled by Motorbike + Distance travelled by Car
= 168 + 210 = 378 km
Hence, option (b).
Workspace:
Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to
- (a)
6:15 am
- (b)
6:30 am
- (c)
6:45 am
- (d)
7:00 am
- (e)
7:15 am
Answer: Option B
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Text Explanation :
The angles of the triangle formed by A, B and C tell us that ABC is a right-angle triangle, with right-angle at vertex C, 30° at vertex A and 60° at vertex B.
Since AB = 500 km, in 30°-60°-90° triangle ABC, we get,
AC = 250 km and BC = 250 km
The train, travels at 50 km/hr. It will travel from B to C (i.e. 250 km) in 5 hours. Since it leaves at 8:00 a.m., it will reach C at 1:00 p.m.
Now, Rahim must be at C latest by 12:45 p.m. (15 minutes before the train)
Travelling at 70 km/hr, he will take approximately 6.2 hours to travel from A to C. Therefore, he must leave at least by
12.75 – 6.2 = 6.55 hours after mid-night.
This is a little after 6:30 a.m. If he leaves by 6:45 a.m., he will not make it to point C 15 minutes before the train arrives.
Hence, option (b).
Workspace:
Answer the next 2 questions based on the information given below.
Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes the schedule of an airline operating non-stop flights between A and B. All the times indicated are local and on the same day.
Assume that planes cruise at the same speed in both directions. However, the effective speed is influenced by a steady wind blowing from east to west at 50 km per hour.
What is the time difference between A and B?
- (a)
1 hour and 30 minutes
- (b)
2 hours
- (c)
2 hours and 30 minutes
- (d)
1 hour
- (e)
Cannot be determined
Answer: Option D
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Text Explanation :
Let the speed of the plane be x kmph.
Then the speed from B to A is (x – 50) kmph and that from A to B is (x + 50) kmph.
Note that the plane travels from B to A, halts for 1 hour and travels back to B, all in 12 hrs.
∴ 3000/(x – 50) + 1 + 3000/(x + 50) = 12
Solving this we get x = 550. [you can check options from the 2nd question of this set to figure out the value of x.]
Speed of plane = 550 kmph
Now, the plane takes 3000/500 = 6 hrs to travel from B to A.
It reaches A when the time at B is 8:00 am + 6 hrs = 2:00 p.m.
When plane reaches A, time at B is 2 p.m. while time at A is 3 p.m.
∴ The time difference between A and B is 1 hour.
Hence, option (d).
Workspace:
What is the plane’s cruising speed in km per hour?
- (a)
700
- (b)
550
- (c)
600
- (d)
500
- (e)
Cannot be determined
Answer: Option B
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Text Explanation :
As calculated in the first question, the cruising speed of the plane is 550 kmph.
Hence, option (b).
Workspace:
Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?
- (a)
3
- (b)
3.5
- (c)
4
- (d)
4.5
- (e)
5
Answer: Option C
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Text Explanation :
Arun has travelled 60 km when Barun starts.
Barun overtakes Arun in 60/(40 – 30) = 6 hrs
In this time, Barun travels 6 × 40 = 240 km from the starting point.
Kiranmala overtakes Arun at the same point.
Kiranmala takes 240/60 = 4 hrs to reach there.
Arun takes 240/30 = 8 hrs to reach there.
∴ Kiranmala starts 8 – 4 = 4 hrs after Arun.
Hence, option (c).
Workspace:
Answer the next 2 questions based on the information given below.
Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.
At what time do Ram and Shyam first meet each other?
- (a)
10:00 a.m.
- (b)
10:10 a.m.
- (c)
10:20 a.m.
- (d)
10:30 a.m.
Answer: Option B
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Text Explanation :
The distance between points A and B is 5 km.
Ram starts at 9 a.m. from A at a speed of 5 km/hr. So, he reaches B at 10:00 a.m.
When Ram reaches B (i.e. at 10.00 a.m., or 15 minutes after Shyam started from A), Shyam (running at a speed of 10 km/hr) is 15/60 × 10 = 2.5 km away from A.
Ram meets Shyam = 10 minutes after 10:00 a.m., i.e. at 10:10 a.m.
Hence, option (b).
Workspace:
At what time does Shyam overtake Ram?
- (a)
10:20 a.m.
- (b)
10:30 a.m.
- (c)
10:40 a.m.
- (d)
10:50 a.m..
Answer: Option B
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Text Explanation :
Ram reaches B at 10.00 a.m. while Shyam reaches B at 10:15 a.m.
At 10:15 a.m., Ram is (15/60) × 5 = 1.25 km away from B.
Shyam overtakes Ram 1.25/(10 – 5) = 0.25 hrs = 15 minutes after 10:15 a.m., i.e. at 10:30 a.m.
Hence, option (b).
Workspace:
Karan and Arjun run a 100 metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?
- (a)
Karan and Arjun reach the finishing line simultaneously.
- (b)
Arjun beats Karan by 1 metre.
- (c)
Arjun beats Karan by 11 metres.
- (d)
Karan beats Arjun by 1 metre.
Answer: Option D
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Text Explanation :
In the first race when Karan runs 100 m, Arjun runs 90 m.
∴ Ratio of Karan’s and Arjun’s speed = 10 : 9
In the second race, when Karan runs 110 m, Arjun runs = 99 m
∴ Karan beats Arjun by 1 m.
Hence, option (d).
Workspace:
If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?
- (a)
11 km/hr
- (b)
12 km/hr
- (c)
13 km/hr
- (d)
14 km/hr
Answer: Option B
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Text Explanation :
Let x hrs be the time required at 15kmph to reach the place at 11:00 am
∴ Distance that the man has to cycle = 15x km
Now, = x + 2
∴ 15x = 10x + 20
∴ 5x = 20
∴ x = 4 hrs
∴ Distance to be travelled = 60 km
∴ To reach the place at noon, the man should cycle for x + 1 = 5 hrs.
Let the new speed be y.
∴ 5 =
∴ y = 12 km/hr
Hence, option (b).
Workspace:
Two boats, travelling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide?
- (a)
- (b)
- (c)
- (d)
Answer: Option C
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Text Explanation :
The distance between the boats one minute before they collide will be same as the total distance they travel in last one minute.
The boats together travel 10 + 5 = 15 km in 60 minutes.
∴ In one minute they can travel km.
∴ They are km part, one minute before they collide.
Hence, option (c).
Workspace:
A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?
- (a)
4
- (b)
8
- (c)
16
- (d)
32
Answer: Option C
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Text Explanation :
Let πr = x.
∴ The circumference of the circular path = 2x.
∴ The sprinter covers 2x metres in one round. The time taken to complete each round is as under:
Thus, we see that the ratio of time taken for the second round to that for the 1st round = 16.
Similarly, ratio of time taken for the third round to that for the 2nd round = 16 and so on.
Hence, option (c).
Workspace:
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