- INTRODUCTION
- AVERAGE SPEED
- ACCELERATION AND DECELERATION
- RELATIVE SPEED
- APPLICATIONS OF RELATIVE SPEED
- RACES
- CLOCKS
- OTHER EXAMPLES
Time and Distance
Physical bodies can be either stationary or in motion. Motion occurs when a body of any shape or size changes its position with respect to any external stationary point. In motion, the body can either move at a constant speed or a variable speed which includes the case of acceleration and deceleration.
The mathematical model that describes motion has three variables namely speed, time and distance and the relationship is:
The units have to be consistent in this relationship. This formula is the source of the various formulae applied to the problems on the applications of time, speed and distance such as trains, boats and streams, clocks and races, circular motion and straight line motion.
Let’s formalize certain relationships, all of which stem from the formula.
- SPEED AND DISTANCE ARE DIRECTLY PROPORTIONAL WHEN TIME IS CONSTANT.
i.e. Speed 
Distance
For example, consider that two racers start running simultaneously from the same point in the same direction on a straight race course. The first racer runs at a speed of 10 kmph, while the second at a speed of 12 kmph. One hour later, a whistle is blown and both racers stop moving. Then, the ratio of the distance covered by the two racers is:
(Here, it is obvious that the time is constant; i.e. both racers run for 1 hour.)
Another Example: Two people, Anna and Bob, start walking simultaneously towards each other on a straight road, which is 10 km long. They meet each other 3 km from Anna’s starting point. Then, the ratio of their speeds will be:
(Again, the time of travel of both Anna and Bob is equal since they start walking simultaneously and stop when they meet.)
- DISTANCE AND TIME ARE DIRECTLY PROPORTIONAL WHEN SPEED IS CONSTANT.
i.e. Time
Distance
For example, consider that a man walks for 1 hour at a speed of 5 kmph and another man walks for 2 hours, also at 5 kmph. Then, the ratio of the distance covered by the first man to that covered by the second man is calculated as:
(Here, the speed of both men is equal to 5 kmph.)
- TIME IS INVERSELY PROPORTIONAL TO SPEED WHEN DISTANCE IS CONSTANT.
Example: Tarzan accidentally stepped on some thorns and could walk at only 4/5th his normal speed. Owing to this, he takes half an hour longer than usual to reach his destination. Then, his original time period can be calculated as follows:
(The distance covered by Tarzan in both cases is equal.)
Another Example: A girl cycles from her house to her school at 6 kmph and reaches there 10 minutes late. Had she gone at 7 kmph, she would have made it 2 minutes early. Then, the distance from her house to her school is calculated as:
Hence, Distance = 6 
84/60 = 8.4 km
(Again, the distance covered by the girl in both cases is equal to the distance between her house and school.)
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A car covers a distance of 100 km in 3 hours and when it returns it covers the same distance in 5 hrs. Find the ratio of the speed of the car in both directions.
Solution: Since the distance is constant, the speed is inversely proportional to the time taken.
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Ratio of the speeds = 5 : 3
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Two cyclists start moving towards each other from their houses. Their speeds are 10 km/hr and 11 km/hr respectively. Find the meeting point if the distance between their houses is 28 km.
Solution: Since both cyclists start at the same time and also meet at the same time, the time taken by both the cyclists is constant and, hence, the distance travelled by them is proportional to their speeds.
Ratio of the distances = Ratio of the speeds = 10 : 11
Distance covered by the first cyclist
Distance covered by the second cyclist
Hence, the two cyclists meet 40/3 km from the first cyclist’s house and 44/3 km from the second cyclist’s house. |
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Rajesh walks to and fro to a shopping mall. He spends 30 minutes shopping. If he walks at speed of 10 km an hour, he returns to home at 19.00 hours. If he walks at 15 km an hour, he returns to home at 18.30 hours. How fast must he walk in order to return home at 18.15 hours? [XAT 2009]
(1) 17 km/hour(2) 17.5 km/hour (3) 18 km/hour(4) 19 km/hour (5) None of the above
Solution: We know that Distance = Speed
With this we can say that time required in the first case will be 15x minutes and that in the second case will be 10x minutes.
Required speed = 15/0.75 = 20 km/hr Hence, option 5. |
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Siddhartha arrives at the office late everyday by half an hour. On a particular day he reduced his speed by 20% and, hence, arrived 45 minutes late. Now if he has to arrive on time, by what percent should he increase his speed?
Solution: Let Siddhartha’s daily speed and time be s and t respectively and the distance be d.
On reducing the speed by 20% he arrives 15 minutes later than his usual time.
Equating (i) and (ii) we get, s
Hence, Siddhartha takes 1 hour daily to reach his
office. But even then he is 30 minutes late. Hence, if he has to reach
office in 30 minutes i.e. half of his current time then he should double
his speed (d = s Hence, he should increase his speed by 100% in order to reach office in time.
Alternatively, When distance is constant, then time and speed are inversely proportional. Hence, reducing speed by 20% means increasing time by 25%.
Thus, if Siddhartha was on time, he takes half an hour; whereas if he is late by 30 minutes, he takes 1 hour. So he needs to double his speed, or he has to increase his speed by 100%. |
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Travelling at 5/6th of his original speed, Manish is 5 minutes late in reaching his office from home. Find the original time he takes to travel to his office from home.
Solution: Let s be the original speed and t be the original time taken to reach office. Hence, d = s
Travelling at 5/6th of his original speed he takes 5 minutes more. Hence, d = (5/6)s
Equating (i) and (ii), we have s
Hence, he takes 25 minutes originally to reach his office.
Alternatively, Here, since distance is constant, speed is inversely proportional to time.
So if t is the original time and the new speed is 5/6 times the original speed, then the new time is 6/5 times of original time. Also, it is given that the new time is 5 minutes more than the original one.
Hence, (6/5)t = t + 5
Hence, t = 25 minutes |
The following conversion factors will be useful for solving problems on time and distance.
- 1 hour = 60 minutes = 3600 seconds
- 1 kilometre = 1000 meters
- 1 mile = 1.609 kilometres
- 8 km = 5 miles
- 1 yard = 3 feet
- 1 km/hr = 5/18 m/s
- 1 m/s = 18/5 km/hr
- 1 km/hr = 5/8 miles/hr
- 1 mile/hr = 22/15 feet/s
If an object travels a particular distance at different speeds during different time-intervals, then its average speed is calculated by dividing the total distance it covers by the total time it takes to cover the total distance.
Thus, if d1, d2 and d3 are the distances covered by an object in time intervals t1, t2 and t3 respectively, then the average speed is given by:
Corollary
- If the distance is constant, then the average speed is given by the harmonic mean of the individual speeds. If a and b are the respective individual speeds, then the average speed is given by:
Explanation:
Assume that a body travels a total distance of 2d km. It travels the first d km with a speed of a km/hr, and the remaining d km with a speed of b km/hr.
Then, the average speed s of the entire journey is given by:
To find the average speed when more than two different speeds are involved (and the distance is constant), use the formula:
where, n is the total number of individual speeds, and s1, s2, ... ,sn are the speeds.
- If the time is constant, then the average speed is given by the arithmetic mean of the given speeds. If a and b are the respective speeds, then the average speed is given by,
Explanation:
Assume that a body travels at a speed of a km/hr for time t and then travels at b km/hr again for time t. Then, the average speed Savg for the entire journey is given by:
To find the average speed when more than two different speeds are involved (and time is constant), use the formula:
where, n is the total number of individual speeds, and s1, s2, ... ,sn are the speeds.
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If Mike covers a distance of 300 km in three stretches of 100 km each with speeds 30 km/hr, 60 km/hr and 80 km/hr respectively, then what is the average speed of Mike throughout the journey?
Solution: Mike covers equal distances with different speeds each time. Let the speeds be a, b and c respectively. Hence, the average speed of the journey is the harmonic mean of these three speeds. Hence, the average speed is given by,
Hence, Savg = 48 km/hr |
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A man drives 150 kilometres to the seashore in 3 hours and 20 minutes. He returns from the shore to the starting point in 4 hours and 10 minutes. Let r be the average rate for the entire trip. Then the average rate for the trip going exceeds r, in kilometres per hour, by: [FMS 2010]
Solution:
r = 40 km/hr
Hence, option 1. |
In physics, acceleration is defined as the rate of change of velocity; meaning that a change in both magnitude and/or direction is an acceleration. However, in regards to solving problems in management exams, it is sufficient to consider acceleration as the rate of change of speed (i.e. magnitude only). In common speech, an increase in speed is termed acceleration; while a decrease in speed is termed deceleration. The unit of acceleration is length per squared second (m/s2).
The following two formulae come in handy while solving problems related to acceleration/ deceleration:
For example, if a car with an initial speed of 25 m/s is given an acceleration of 0.5 m/s2, then the time taken by the car to reach a speed of 40 m/s will be (40 – 25)/(0.5) = 30 seconds.
The word relative means with respect to or compared to one another. Relative speed refers to the speed of an object with respect to another which may be stationary or moving in the same or opposite direction. It should be noted here that in order for the speed to be relative, the frame of reference should be in either of the two objects. This can be better explained using a phenomenon we observe every day.
Suppose we are travelling in a train (so, the frame of reference is inside this train) and we observe another train coming towards us from the opposite direction on a parallel track. The speed of the second train will seem much faster than what it actually is. On the other hand, if the second train were moving at the same speed, in the same direction and on a parallel track, then it will appear to be stationary when seen from the first train. So what we observe is actually the speed of the second train ‘relative’ to our own speed (which is equal to the speed of the first train).
- TRAVELLING IN OPPOSITE DIRECTIONS
When two objects are moving in opposite directions, towards each other or away from each other (as shown in the following two figures) on a straight line with speeds u and v, then their relative speed = u + v
Now, if there is no end point (i.e. some point reaching which the vehicles would change their directions and move in the opposite one), then in the case of the first diagram, the two vehicles will first meet at some point between their initial positions, and will then proceed to move away from one another. In the case of the second diagram, the two vehicles will never meet, and will keep moving apart.
However, if there is an end point, then (in case of the first diagram) the two vehicles will first meet each other somewhere in-between and will proceed to move apart. So far, the relative speed remains u + v. The faster of the two vehicles will then reach its end point and reverse its direction. Now, both vehicles will move in the same direction and hence the relative speed becomes |u – v| (see TRAVELLING IN THE SAME DIRECTION). This continues until the slower vehicle also reaches its end point at which point the two vehicles again move in opposite directions; so the relative speed is again u + v. This pattern of alternating the relative speed between u + v and |u – v| every time one of the vehicles reaches its end point keeps continuing. (Refer to the following diagram; here the car is assumed to be faster than the truck.)
IMPORTANT:
- If both objects reach their end points simultaneously, then the relative speed will remain as it was (i.e. it won’t change from |u – v| to u + v or vice-versa.)
- Most questions on this sub-topic ask about the distance covered by
the vehicles at various meetings. At the first meeting, the distance covered
by the two vehicles will be the distance between the two end points (say
d). The individual distances covered by the car and the truck will be
in the ratio v : u (since speed
distance).
From the diagram, it is clear that before the second meeting, both the car and the truck have individually covered distance d, and have together covered another d. Since distance d was already covered in the first meeting, a distance of 2d was covered between the two meetings. Hence, total distance covered for the second meeting by both vehicles together was (d + 2d = 3d).
Similarly, for each subsequent meeting, a further distance of 2d will be covered as compared to the distance covered in the previous meeting. So, for (say) the nth meeting, both vehicles together will cover a distance of (d + 2(n – 1)d).
IMPORTANT:
- This scenario may not always hold true. For example, it is possible that the truck is so slow that the car meets the truck for the first time, reaches the end point, meets the truck again (second time) on its reverse journey and reaches the other end point before the truck reaches its end point for the first time. This will obviously lead to a different set of calculations than that given above. It is advised that when the student encounters these types of problems, he/she visualizes it before trying to solve it.
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Two boats, traveling 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? [CAT 2004]
Solution: The boats together travel 10 + 5 = 15 km in 60 minutes.
Hence, option 3. |
- TRAVELLING IN THE SAME DIRECTION
When two objects are moving in the same direction on a straight line at speeds u and v (as shown in the following figure), where the faster one is either drawing closer to the slower object or moving away from the slower object, then the relative speed of one object with respect to the other = |u – v|.
In the above figure, if the frame of reference is inside the car, then according to the frame, the car is stationary and the relative speed of the truck with respect to the car is |u – v|. If u > v, then the truck appears to approach the seemingly stationary car with this speed; and if u < v, then the truck appears to move backward away from the car with the same relative speed.
If the frame of reference is inside the truck, then according to the frame, the truck is stationary while the car appears to move with a speed of |u – v|. If v > u, then the car appears to move forward, away from the truck with this speed; and if v < u, then the car appears to move backward, toward the truck with the same speed.
Thus, the relative speed will always be positive; only the direction of motion varies.
Now, let us consider the possibility of end points for both vehicles, and that the two vehicles start moving from the same initial point. The scenario will be similar to that of the earlier one (i.e. TRAVELLING IN OPPOSITE DIRECTIONS). Here too, the relative speed will alternate from |u – v| to u + v every time one of the vehicles reaches its end point. (Refer to the diagram; again, the car is assumed to be faster than the truck.)
As apparent from the above diagram, the distance covered by the two vehicles between any two consecutive meetings will be 2d. Hence, for (say) the nth meeting, both vehicles together will cover a distance of (2n)d.
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Sukriti and Saloni are athletes. Sukriti covers a distance of 1 km in 5 minutes and 50 seconds, while Saloni covers the same distance in 6 minutes and 4 seconds. If both of them start together and run at uniform speed, by what distance will Sukriti win a 5 km mini marathon: [IIFT 2009]
(1) 150 m(2) 200 m (3)175 m(4) 225 m
Solution: Sukriti runs 1 km in 5 minutes and 50 seconds = 350 seconds. Saloni runs 1 km in 6 minutes and 4 seconds = 364 seconds.
Hence, option 2. |
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A thief escaped from a prison with a speed 20 km/hr and after 2 hours the police followed the thief with a speed 30 km/hr. When will the police catch the thief?
Solution: In two hours, the thief would have covered 40 km. Thus, the relative distance between the thief and the police is 40 km. Thus,time taken by the police to catch the thief
Alternatively, Let the speed of the thief be st and that of the police be sp. Also, let the time duration of the entire chase (i.e. time from when the police started chasing to when the thief was caught) be t. Since the distance covered by both parties is the same, the ratio of the speeds will be inversely proportional to the ratio of the time periods. Thus,
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REMEMBER:
- In Example 6, the first method used the concept of relative speed. Here, the frame of reference was assumed to be either the thief or the police. In the alternative method, the frame of reference was assumed to be some external stationary point and the speeds of each object was considered individually. The importance of selecting the appropriate frame of reference is clear from the above example; it was much more convenient solving it using the former method.
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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. [CAT 2005]
Question 1: At what time do Ram and Shyam first meet each other? (1) 10:00 a.m.(2) 10:10 a.m. (3) 10:20 a.m.(4) 10:30 a.m.
Solution 1: 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 Ram meets Shyam (2.5 Hence, option 2.
Question 2: At what time does Shyam overtake Ram? (1) 10:20 a.m.(2) 10:30 a.m. (3) 10:40 a.m.(4) 10:50 a.m.
Solution 2: 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) 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 2. |
The concept of relative speed can be used to solve problems based on trains. When two trains are moving in the same direction or in opposite directions, the total distance required to be travelled before they cross each other completely is equal to the sum of the lengths of the two trains. This is because when we say cross, we mean that the end of one train must pass the end of the other train, in case of trains moving in opposite directions. If the trains are moving in the same direction, then in order for them to cross each other, the end of the faster train must pass the start of the slower one. This distance is covered at the relative speed of the two trains.
Let S1 be the speed of a train of length d1, S2 be the speed of another train of length d2, Sobj be the speed of a moving object of negligible length, and let t be the time taken for crossing.
If the train of length d1 crosses a stationary object of negligible length, then the Time-Speed-Distance formula can be modified to:
If the train of length d1 crosses the moving object (at Sobj) of negligible length, then the Time-Speed-Distance formula can be modified to:
If the train of length d1 crosses the train of length d2 (meaning S1 >; S2), then the Time-Speed-Distance formula can be modified to:
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A train crosses a man travelling in another train in the opposite direction in 10 seconds. But the train requires 30 seconds to cross the same man if the train were travelling in the same direction. If the length of the first train is 180 meters and that of the other train in which the man is sitting is 120 meters, then find the speed of the first train.
Solution: Here the length of the man’s train is redundant. Let a be the speed of the first train and b be the speed of the man, which is the speed of his train.
Thus, when the two trains are running in opposite directions, their relative speed would be the sum of their speeds. Hence,
a + b = 180/10 = 18 m/s … (i)
Similarly, when the two trains are moving in the same direction, their relative speed would be the difference of their speeds. Hence,
a – b = 180/30 = 6 m/s… (ii)
Solving (i) and (ii) simultaneously, we get, a = 12 m/s
Hence, speed of the first train = 12 m/s |
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The Ghaziabad-Hapur-Meerut EMU and the Meerut-Hapur-Ghaziabad EMU start at the same time from Ghaziabad and Meerut and proceed towards each other at 16 km/hr and 21 km/hr, respectively. When they meet, it is found that one train has travelled 60 km more than the other. The distance between two stations is: [IIFT 2007]
(1) 445 km(2) 444 km (3) 440 km(4) 450 km
Solution: Let the distance covered by the slower train (i.e. the 16 km/hr one) be x km. Then, the distance covered by the faster train will be (x + 60) km. Since the two trains start simultaneously and both stop when they meet each other, the time of travel of both trains is equal. Hence, Distance
Hence, option 2. |
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Two super-fast trains, which travel at speeds of 120 km/hr and 150 km/hr, left the Beijing Railway Station at exactly the same time, and were headed in the same direction. Thirty minutes later, another super-fast train left Beijing Station in the same direction as the previous two. It crossed the faster train an hour and a half after it crossed the slower one. Find the third train’s speed. (Note: Assume all trains to be of negligible length.)
Solution: The speeds of the two trains are 120 km/hr and 150 km/hr respectively. Hence, in 30 minutes they will each travel 60 km and 75 km respectively. Hence, in 30 minutes, the relative distance between the third train and the slower train will be 60 km; and that between the faster train and the third train will be 75 km.
Hence, if the speed of the third train is u km/hr and the time taken by the third train to cross the slower train is x hours, then we have,
Also,
Since the third train is faster than both the other trains, its speed must be 180 km/hr. |
The basic equation of time, speed and distance is also useful to solve problems of boats and streams.
Downstream motion of a boat is its motion in the same direction as the flow of the stream. Upstream motion of a boat is its motion in the opposite direction to the flow of the stream.
If the speed of the boat in still water is Sb and the speed of the stream or river is Ss then,
Effective speed of the boat when it is moving downstream, Sd = Sb + Ss
Effective speed of the boat when it is moving upstream, Su = Sb – Ss
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If the speed of the boat in still water is 4.5 km/hr and speed of the stream is 2.5 km/hr, find the total time required by the boat to travel down 28 km and up 12 km.
Solution: Sd = Sb + Ss = 4.5 + 2.5 = 7 km/hr
Su= Sb – Ss = 4.5 – 2.5 = 2 km/hr
Time for downward travel = 28/7 = 4 hr Time for upward travel = 12/2 = 6 hr Total time required = 4 + 6 = 10 hr |
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A boat covers 18 km downstream in 4 hr and 18 km upstream in 12 hr. Find the speed of the boat in still water and speed of the stream.
Solution:
= 3 km/hr Speed of stream Ss
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A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. The speed of the boat in still water is: [IIFT 2008]
(1) 3 km/hour(2) 4 km/hour (3) 8 km/hour(4) None of the above
Solution: Let the time taken to go of 30 km upstream be 3x hours. Then the time taken to go 40 km upstream is 4x hours. Let the time taken to go 44 km downstream be 4y hours. Then the time taken to go 55 km downstream is 5y hours. Then we have: 3x + 4y = 10 …(i) 4x + 5y = 13 …(ii) Solving for x and y, we have, x = 2 and y = 1
Hence, option 3. |
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A cyclist drove one kilometer, with the wind in his back, in three minutes and drove the same way back, against the wind in four minutes. If we assume that the cyclist always puts constant force on the pedals, how much time would it take him to drive one kilometer without wind? [SNAP 2008]
Solution: Time taken by cyclist in the direction of the wind is 3 minutes. Let the cyclist’s speed be v km/min. Let the speed of the wind be u km/min.
We have (v – u) Thus, 24u = 1
Hence, option 2. |
In circular motion, instead of a straight track, we talk about a circular track where two objects can run in the same direction or in opposite directions with different speeds.
In the case of objects running in the same direction, the faster object overtakes the slower one. Whenever the faster object comes in contact with the slower object, this is known as overlapping or lapping of the slower object by the faster object.
The relative speed of two objects moving around a circle in the same direction, starting at the same point is taken as (x – y), where x and y are the speeds of the faster and slower objects respectively.
So, the time taken by the two objects, starting from the same initial point, to meet each other for the first time if they are running on a circular track in the same direction
This is because, when moving in the same direction, the distance by which the faster object would be ahead of the slower one, at the time of their first meeting, will be equal to the circumference of the track. Hence, their relative distance will be the circumference of the track.
If more than two objects start moving simultaneously around a circular track from the same point, in the same direction, then they will meet again for the first time at a time which is given by the LCM of the times that the fastest runner takes in totally overlapping (defined earlier) each of the slower runners. For example, if A, B and C start moving in a clockwise direction from the same initial point on the circle where C is the fastest runner, and if we define TAC as the time in which C completely overtakes A and TBC as the time in which C completely overtakes B, then the LCM of TAC and TBC will be the time in which A, B and C will meet again for the first time.
The relative speed of two objects moving around a circle in opposite directions, starting from the same point is taken as (x + y), where x and y are the speeds of the two objects.
Hence, the time taken by them to meet each other for the first time:
Moreover, if two (or more) objects start moving simultaneously around a circular track in the same or opposite directions, starting from the same initial point, then they will meet again for the first time at the starting point at a time which is given by the LCM of the times taken by each object to complete one round.
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Amit and Sunil start running together on a circular track in opposite directions with speeds 30 m/s and 50m/s. The length of the circular track is 600 m. When will they meet for the first time and when will they meet at the starting point for the first time?
Solution: Time taken to meet each other for the first time = Circumference/Relative speed = 600/80 = 7.5 s Time taken by Amit to complete a round = 600/30 = 20 s Time taken by Sunil to complete a round = 600/50 = 12 s Time when they meet at the starting point for the first time = LCM (20, 12) = 60 s |
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Two runners are running on a circular track of radius 14 metres. When the two runners start running simultaneously in the same direction, they meet each other every 22 seconds. When they start running simultaneously in opposite directions, they meet each other every 10 seconds. Now, the two runners run on a straight 100 metre track in the same direction. If the runners run at the same speed as before, then how much head-start (in metres) must the faster runner give the slower one, so that they both reach the finish line together?
Solution: Since the radius of the circular track is 14 metres,its circumference
Let the speeds of the faster and slower runners be SF and SS respectively.
Now, when they run in the same direction, they meet every 22 seconds. Hence,
When they run in opposite directions, they meet every 10 seconds. Hence,
Solving equation (i) and (ii) simultaneously, we get,
SF = 12.8/2 = 6.4
Now, we move on to the second part of the question. Two runners, with speeds 6.4 m/s and 2.4 m/s, run in the same direction on a 100 m track. Let x be the head-start that the faster runner gives the slower one. Once the slower runner is ahead of the faster one by x metres, the time for which the two runners run before reaching the finish will be the same. Hence,
Hence, the faster runner gives the slower one a head-start of 62.5 metres. |
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Two aeroplanes were flying in a circle. When they moved in opposite directions, they met each other every 3 hours; and when they moved in the same direction, they met each other every 15 hours. They realized that when moving in opposite directions, the distance between the two planes reduced from 95 km to 10 km every 12 minutes. Determine the speed of the faster aeroplane (in km/hr).
Solution: Let the speeds of the two planes be S1 and S2 respectively. It is given that when the planes moved in opposite directions, the distance between the planes reduced from 95 to 10 km in 12 minutes. So, in a time period of 12 minutes, the planes covered a relative distance of 95 – 10 = 85 km with a relative speed of S1 + S2
When the aeroplanes move in opposite directions, they meet every 3 hours. Hence,
Hence, Circumference of the path = 3
When the aeroplanes move in the same direction, they meet every 15 hours. Hence,
From (i) and (ii), we get,
S1 = (425 + 85)/2 = 510/2 = 255 km/hr
Hence, speed of the faster aeroplane is 255 km/hr. |
A contest of speed between contestants is called a race. If all the contestants reach the finishing line at the same time, then the race is called a dead heat. The winners distance is equal to the length of the race.
A beats B by x meters implies that in a race of L meters B is x meters behind A, who is at the finishing line; which means that when A covers L meters, B has covered (L – x) meters in the same time.
A gives B a start of x meters means that in a race of L meters, A starts the race only when B has covered x meters, which implies that if both A and B run at the same speeds, then when A covers L meters, B would have covered (L + x) meters.
A beats B by t seconds implies that when A and B start together from the starting point, A reaches the finishing point t seconds before B finishes.
A gives B a start of t seconds implies that A starts the race t seconds after B starts from the starting point.
A beats B by x meters or t seconds means B runs x meters in t seconds.
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In a 1200 m race, A can beat B by 120 m and in a 1000 m race, B can beat C by 50 m. Find the distance by which A beats C in a 800 m race.
Solution: When A covers 1200 m, B covers 1080 m. So, when A covers 800 m, B will cover 720 m. Similarly, when B covers 1000 m, C covers 950 m. So, when B covers 720 m, C will cover 684 m. Hence, when A covers 800 m, C will cover 684 m. So, A will beat C by 116 m in a race of 800 m. |
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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? [CAT 2004]
(1) Karan and Arjun reach the finishing line simultaneously. (2) Arjun beats Karan by 1 metre. (3) Arjun beats Karan by 11 metres. (4) Karan beats Arjun by 1 metre.
Solution: In the first race when Karan runs 100 m, Arjun runs 90 m.
In the second race, when Karan runs 110 m,
Hence, option 4. |
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Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race? [CAT 2001]
(1) 36 meters(2) 48 meters (3) 60 meters(4) 72 meters
Solution: Let the length of track is d meters.
Solving we get, d = 48 m Hence, option 2.
Alternatively, When A finished, the gap between B and C was 6 m. After B ran another 12 m, the gap is increased by another 2 m.
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A hare and a turtle decided to race each other, the race beginning at an oak tree and finishing at a pine tree. The turtle’s speed was only half of that of the hare. However, the hare got caught cheating and was forced to hop back to the oak tree and start again, after it had already covered 2/3rd of the total distance. The turtle, slow but steady and honest, won the race by 8 minutes. How long did the hare take to complete the race?
Solution: Let the total distance from the oak tree to the pine tree be d kilometres. Let the hare’s speed be x km/min; so the turtle’s speed = x/2 km/min.
Hence, the hare ran 2d/3 kilometres, got caught cheating, then ran back another 2d/3 km to the oak tree; then finally ran d km from the oak to the pine tree.
Time taken by the turtle to complete the race
Since the turtle beat the hare by 8 minutes, hence (Time taken by the hare to complete the race) – (Time taken by the turtle to complete the race) = 8
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A clock dial has 60 divisions called minute spaces. It is the distance covered by the minute hand in one minute. So, the minute hand covers 60 minute spaces in an hour. The hour hand covers only 5 minute spaces in an hour.
Hence, the minute hand moves faster than the hour hand. In 1 hour, the minute hand moves 55 minute spaces more than the hour hand.
Hence, the time taken by the minute hand to gain x minute spaces over the hour hand= (60/55)x = (12/11)x minutes
- Right angles formed in a clock: A right angle is said to be
formed when an angle of 90
is formed between the hour and minute hands (i.e. when there are exactly 15
minute spaces between the two hands). In each hour, two right angles are
always formed. However, it should be noted that during the interval from 2
o’clock to 4 o’clock, one right angle is shared between the hours 2 o’clock to
3 o’clock and 3 o’clock to 4 o’clock (i.e. the right angle formed at 3
o’clock). The same logic applies to the interval from 8 o’clock to 10 o’clock
(the right angle formed at 9 o’clock is shared between the hours 8-9 and
9-10). Hence, in a 12-hour period, the clock will only make 22 right angles,
NOT 24.
- Straight lines formed in a clock: A straight line is said
to be formed when an angle of 0
or 180
is formed between the hour and minute hands (i.e. there are 0 or 30 minute
spaces between the two hands). Similar to the right angles, two straight lines
are formed in each hour and the straight lines formed at 6 o’clock and 12
o’clock are both shared between two different hourly intervals (the straight
line formed at 6 o’clock is shared between the intervals 5-6 and 6-7, while
that formed at 12 o’clock is shared between the intervals 11-12 and
12-1).
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At what time between 5 o’clock and 6 o’clock do the minute and hour hands of a clock coincide?
Solution: At 5 o’clock, the minute hand is at 12 and the hour hand is at 5. They are 25 minute spaces apart. Thus, if, say, the hour hand moves x minute spaces ahead, then in order to overlap, the minute hand must move (25 + x) minute spaces ahead. That is, the minute hand has to gain 25 minute spaces over the hour hand. The time taken by the minute hand to gain 25 minute
spaces over the hour hand = (12/11)
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REMEMBER:
- Every hour, the minute hand and hour hand coincide or overlap each other once. But, between 11 a.m. to 1 p.m. and between 11 p.m. to 1 a.m., the hands overlap only once, that is, at 12 p.m. and 12 a.m. respectively. Hence, the hour and minute hands overlap 22 times in 24 hours.
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When are the minute and hour hand at 180( with respect to each other between 1 p.m. and 2 p.m.?
Solution: At 1 p.m., the minute hand is at 12 and the hour hand is at 1.
They are 5 minute spaces apart. In order for the hour and minute hand to be at 180( with respect to each other, they must be 30 minute spaces apart. So, after 1 p.m., the minute hand has to gain (5 + 30) minute spaces over the hour hand.
Time taken by the minute hand to gain 35 minute
spaces over the hour hand = (12/11)
Hence, the hour hand and minute hand will be at 180( with respect to each other between 1 p.m. and 2 p.m. at 1:38:11 p.m. [ |
0.18 minutes 
11 seconds].
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A man on his way to dinner shortly after 6:00 p.m.
observes that the hands of his watch form an angle of 110 [FMS 2010]
Solution: When the hour hand and minute hand form an angle of
110
1. The minute hand is behind the hour hand. The speed of the minute hand is 6 Initial distance between the hour and the minute
hands at 6:00 pm is 180
2. The minute hand is ahead of the hour hand.
Hence option 2. |
If a clock shows the time as 1:05 when the correct time is 1:00, then the clock has gained 5 minutes. If a clock shows the time as 2:18 when the correct time is 2:25, then the watch has lost 7 minutes.
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A clock is currently showing the right time as 12 p.m. If, after this, it starts gaining 10 minutes every hour, what time will it show at 10 p.m.?
Solution: The clock gains 10 minutes in one hour. Hence, after
10 hours, the clock must have gained 10
Thus, the time on the clock at 10 p.m. will be 11:40 p.m. |
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A clock is showing the right time at 12 p.m. After this it started gaining 15 minutes per hour. So, when this clock shows 5 pm, what is the actual time?
Solution: Let the actual number of hours passed be x. Time gained in one hour = 15 minutes = 1/4 hour So, time gained in x hours = (1/4)x hours So hours passed as shown by the clock = x + (1/4)x = (5/4)x … (i)
Also, the clock is showing 5 p.m. Thus, hours passed as shown by clock = 5 … (ii) Equating equations (i) and (ii), we get, (5/4)x = 5 Thus, x = 4 hours The actual time has increased by 4 hours after 12 p.m.
Hence, the right time is 4:00 p.m. |
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Sangeeta and Swati bought two wristwatches from Jamshedpur Electronics at 11.40 a.m. IST. After purchasing they found that when 60 minutes elapses on a correct clock (IST), Sangeeta’s wristwatch registers 62 minutes whereas Swati’s wristwatch registers 56 minutes. Later in the day Sangeeta’s wristwatch reads 10 p.m., then the time on Swati’s wristwatch is: [XAT 2009]
(1) 8:40 p.m.(2) 9:00 p.m. (3) 9:20 p.m.(4) 9:40 p.m. (5) Cannot be calculated
Solution: It was 11.40 a.m. when Sangeeta and Swati bought the watches. According to Sangeeta’s watch, it was now 10 p.m.; that is, her watch has shown time elapsing by 10 hours, 20 minutes (= 620 minutes).
It is given that when the actual time elapses by 60 minutes, Sangeeta’s watch shows time elapsing by 62 minutes. Hence, Actual time Time on Sangeeta’s watch 60 minutes --------------> 62 minutes ? <-------------- 620 minutes
Hence, the actual time elapsed by (620 It is given that when the actual time elapses by 60 minutes, Swati’s watch shows time elapsing by 56 minutes. Hence,
Actual time Time on Swati’s watch 60 minutes -------------> 56 minutes 600 minutes -------------> ?
Hence, the time on Swati’s watch elapsed by (600
Hence, the time on her watch is 11.40 a.m. + (9 hours, 20 minutes) = 9.00 p.m. Hence, option 2. |
Speed of minute hand = 6
per min [360
in 60 min]
Speed of hour hand = 1/2
per min [30
in 60 min]
Since both hands move in the same direction,
Relative speed of minute hand with respect
to hour hand = 6 
1/2 = 5½
per min
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At what time between 3 p.m. and 4 p.m. will the minute hand coincide with the hour hand?
Solution: At 3 o’clock, the hour hand is ahead of the minute
hand by 90
It takes 16 minutes and 21.81 seconds after 3 p.m. for the hour hand to meet the minute hand, i.e. at 3:16:21.81 p.m. |
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At what time between 5 o’clock and 6 o’clock will the hands of a watch be at right angles?
Solution: At 5 o’clock, the angle between the minute hand and
the hour hand is 150 degrees. The minute hand has to gain 150
This means that the hands of the watch will at right angles for the first time at 5 : 10 : 54.54 p.m. The minute and hour hands will also form a right angle when the minute hand gains 150 + 90 = 240 degrees over the hour hand.
This means that the hands of the watch will at right angles for the second time at 5:43:38.18 p.m. |
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The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the graph below
[CAT 2001]
Question 1: Manasa makes the 200 km trip from Mumbai to Pune at a steady speed of 60 km/hour. What is the amount of petrol consumed for the journey?
(1) 12.5 litres(2) 13.33 litres (3) 16 litres(4) 19.75 litres
Solution 1: Time for the trip = 200/60 = 10/3 hours
Hence, option 2.
Question 2: Manasa would like to minimize the fuel consumption for the trip by driving at the appropriate speed. How should she change the speed?
(1) Increase the speed (2) Decrease the speed (3) Maintain the speed at 60 km/hour (4) Cannot be determined
Solution 2: If the speed is 40 km/hour, then fuel consumption is given by, (200/40)
Hence, option 2. |
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Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top of the escalator after having taken 25 steps. While Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up? [CAT 2001]
(1) 40(2) 50 (3) 60(4) 80
Solution: Assume Shyama takes 3 steps and Vyom takes 2 steps in 6 seconds and let the escalator moves up by x steps per second.
Vyom took 20
Similarly, in Vyom’s case, the height of the stairway = (20 + 60x) steps
Hence, option 2. |
