Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself:Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is
[2019]
Explanation
The minimum time travelled by Car 1 = 6 hours.
Minimum time travelled by Car 2 = 5 hours
Now, the time taken can be of any value -> (6,5) (7,6) ...... (1000,999)
However, since we want to calculate the highest possible percentage by which speed of car 2 could exceed car 1, we consider the lowest time taken - 6 hours and 5 hours, respectively.
Let the distance between the start and finish be 30 Kms (LCM of 6 and 5)
So, Car 1 travels at 5 Kmph and Car 2 travels at 6 Kmph
Percentage increase in speed by Car 2 = x 100 = x 100 = 20%
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself: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
[2019]
Explanation
Circumference of A and B are in the ratio 3 : 4
So, Ratio of Distance travelled in one revolution by A and B = 3: 4
Since they travel the same distance,
Ratio of number of revolutions of A and B = 4 : 3 ...(1)
We know that each wheel of A requires 5000 more revolutions than B
So, the Ratio of number of revolutions of A and B = (n + 5000) : n ...(2)
So, comparing (1) and (2)
Number of revolutions of A and B are 20000 and 15000 respectively
So, we know Bike B does 15000 revolutions in 45 minutes
Distance travelled = 2 x π x r x Number of revolutions
Speed of Bike B = mph
Speed of Bike B = 2 x π x 2/5 x 5 x 4 Kmph
Speed of Bike B = 16 π Kmph
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself: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?
[2019]
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.
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself: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
[2019]
Explanation
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.
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself:One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport 1/3 of his total journey time, while Bimal took each mode of transport 1/3 of the total distance. The percentage by which Bimal's travel time exceeds Amal's travel time is nearest to
[2019]
Explanation
Let the distance be = 60 Kms.
So, Bimal travels a distance of 20Kms in each mode and Amal travels 10, 20 and 30 Kms respectively in 1 hour each
Time taken by Bimal = Time taken to travel 20 kms in 10 Kmph + Time taken to travel 20 kms in 20 Kmph + Time taken to travel 20 kms in 30 Kmph
Time taken by Bimal = 2 + 1 + hours = 3 + hours
Extra time taken by Bimal = hour
Percentage increase in time = x 100 = x 100 = 22
So, Percentage increase in time = 22
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself: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:
[TITA 2018]
Correct Answer : 12
Explanation
Let us take the distance AP = x and BP = 3x
It’s also given that SpeedCar-2 = 1/2 SpeedCar-1
S1 = S, S2 = S/2
Car 2 takes one hour more than Car 1
T2 = t minutes T2 = (t + 60) minutes
We know that Time = Distance/Speed
TCar-1 = t TCar-2 = 6t
We know, TCar-2 takes (t + 60) minutes
6t = t + 60
5t = 60
t = 12 minutes
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself: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
(TITA 2018)
Correct Answer : 50
Explanation
Given that 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.
They meet after one hour or their relative velocity is (a + b)km/hr and their relative distance is 350kms
a + b = 350/1 which is equal to 350 kms/hr
If they both move towards east, then they meet in 7 hrs.
We have to find the difference between their speeds, in km per hour. B is catching upon A, so once again their relative distance will be 350kms.
Relative speed = b – a , B travels faster so that it can meet a.
⇒ b - a = 50 km/hr
We know that b + a = 350 km/hr
By this we can find the values of a and b but they have asked for only the difference between their speeds which is b – a = 50km/hr
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Question for CAT Previous Year Questions: Speed, Time & Distance
Try yourself: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
(TITA 2018)
Correct Answer : 5 km
Explanation
Given that 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 Car 1 is 20km away from A and it travels at 100kmph and the time taken is 20/100 = 1/5 hr
So car 2 is 12 minutes behind car 1.
We have to find the distance in km, between car 2 and B when car 1 reaches B.
If car 1 reaches B, car 2 will take 12 minutes to reach b.
Distance between car 2 and B is 25 x (1/5) = 5 kms
The distance, in km, between car 2 and B when car 1 reaches B is 5 kms.
FAQs on Speed, Time & Distance CAT Previous Year Questions with Answer PDF
1. How to solve speed, time, and distance problems in CAT?
Ans. To solve speed, time, and distance problems in CAT, follow these steps:
1. Understand the problem: Read the question carefully and identify what is given and what needs to be found.
2. Determine the formula: Use the formula Distance = Speed × Time to determine the relationship between the given quantities.
3. Substitute values: Substitute the given values into the formula.
4. Solve for the unknown: Solve the equation to find the unknown quantity.
5. Check the answer: Verify if the solution makes sense in the context of the problem.
2. What are the key concepts to understand in speed, time, and distance problems for CAT?
Ans. The key concepts to understand in speed, time, and distance problems for CAT are:
1. Speed: It represents how fast an object is moving and is usually measured in units like km/hr or m/s.
2. Time: It represents the duration or period during which an object moves and is usually measured in units like hours or seconds.
3. Distance: It represents the total length covered by an object in motion and is usually measured in units like kilometers or meters.
4. Relationship: The relationship between speed, time, and distance is given by the formula Distance = Speed × Time.
3. Can you provide an example of a speed, time, and distance problem in CAT?
Ans. Sure! Here's an example of a speed, time, and distance problem in CAT:
A train travels a distance of 300 km in 5 hours. What is its average speed in km/hr?
Solution:
Given: Distance = 300 km, Time = 5 hours
Using the formula Distance = Speed × Time, we can rearrange it to find Speed = Distance / Time.
Substituting the given values, we have Speed = 300 km / 5 hours = 60 km/hr.
Therefore, the average speed of the train is 60 km/hr.
4. Is it important to convert units in speed, time, and distance problems for CAT?
Ans. Yes, it is important to convert units in speed, time, and distance problems for CAT to ensure consistency and accuracy in calculations. If the given values are in different units, convert them to a common unit before solving the problem. For example, if the speed is given in km/hr and the time is given in minutes, convert the time to hours by dividing it by 60. This will help in applying the formula correctly and obtaining the desired result.
5. How can I improve my problem-solving speed for speed, time, and distance problems in CAT?
Ans. To improve problem-solving speed for speed, time, and distance problems in CAT, you can:
1. Practice regularly: Solve a variety of speed, time, and distance problems to become familiar with different types of questions and their solutions.
2. Learn shortcuts: Memorize common formulas and techniques to solve problems quickly, such as the formula Distance = Speed × Time.
3. Analyze previous year questions: Analyze previous year CAT questions related to speed, time, and distance to understand the patterns and strategies used in solving them.
4. Use approximation: When the answer choices are far apart, approximate values to quickly eliminate incorrect options and narrow down the possible answers.
5. Time yourself: Set a time limit while practicing and try to solve problems within that time frame. This will help you improve your speed and efficiency during the actual exam.