Direct proportionality between time and distance (when the speed is constant) time µ distance
A car moves for 2 hours at a speed of 25 kmph and another car moves for 3 hours at the same speed. Find the ratio of distances covered by the two cars.
Since, the speed is constant, we can directly conclude that time µ distance.
Since, the times of travel are 2 and 3 hours respectively, the ratio of distances covered is also 2/3
Direct Proportionality between speed and distance (when the time is constant) speed µ distance
A body travels at S1 kmph for the first 2 hours and then travels at S2 kmph for the next two hours. Here two motions of one body are being described and between these two motions the time is constant hence speed will be proportional to the distance travelled
Two cars start simultaneously from A and B respectively towards each other with speeds of S1 kmph and S2 kmph. They meet at a point C…. Here again, the speed is directly proportional to the distance since two motions are described where the time of both the motions is the same, that is, it is evident here that the first and the second car travel for the same time.
In such a case the following ratios will be valid:
Q. A car travels at 30 km/h for the first 2 hours of a journey and then travels at 40 km/h for the next 2 hours of the journey. Find the ratio of the distances travelled at the two speeds.
Since time is constant between the two motions described, we can use the proportionality between speed and distance
Q. Two cars leave simultaneously from points A and B on a straight line towards each other. The distance between A and B is 100 km. They meet at a point 40 km from A. Find the ratio of their speeds
Since time is the same for both the motions described, we have ratio of speed = ratio of distance.
SA / SB = 40/60 = 2/3
Q. Two cars move simultaneously from points A and B owards each other. The speeds of the two cars are 20 m/s and 25 m/s respectively. Find the meeting point if d(AB) = 900 km
For the bodies to meet, the time of travel is constant (since the two cars have moved simultaneously).
Hence, speed ratio = distance ratio
4/5 = distance ratio
Hence, the meeting point will be 400 km from A and 500 km from B
Inverse proportionality between speed and time (when the distance is constant)
A body travels at S1 kmph for the first half of he journey and then travels at S2 kmph for the second half of the journey. Here two motions of one body are being described and between these two motions the distance travelled is constant. Hence he speed will be inversely proportional to the time travelled for.
Two cars start simultaneously from A and B respectively towards each other. They meet at a point C and reach their respective destinations B and A in t1 and t2 hours respectively… Here again, the speed is inversely proportional to the time since two motions are described where the distance of both the motions is the same, that is, it is evident here that the first and the second car travel for the distance, viz., AB.
In such a case, the following ratio will be valid
S1S2=t2t1 or S1∗t1=S2∗t2=S3∗t3=K
Since Distance = speed * time, When one of either speed or time is changed, the other quantity is also changed to maintain the same distance.
|Speed is increased by %||Time is reduced by %|
Q. A train meets with an accident and moves at 3/4 its original speed. Due to this, it is 20 minutes late. Find the original time for the journey beyond the point of accident.
Solving mathematically S1 /S2 = t2 /(t2 + 32)
5/7.5 = t2/(t2 + 32)
5t2 + 160 = 7.5 t2
t2 = 160/2.5 = 64 minutes
Hence, the distance is given by 7.5 × 64/60 = 8 km
Q. A train meets with an accident and moves at 3/4 its original speed. Due to this, it is 20 minutes late. Find the original time for the journey beyond the point of acciden
we get that a 25% reduction in speed leads to a 33.33% increase in time.
But, 33.33% increase in time is equal to 20 minutes increase in time.
Hence, total time (original) = 60 minutes
Time taken by a train of length X metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover X metres.
Time taken by a train of length X metres to pass an object of width Y meters is equal to the time taken by the train to cover X + Y metres.
Suppose two trains or two bodies are moving in the same direction at A m / s and B m/s, where A > B, then their relatives speed = (A – B) m / s.
Suppose two trains or two bodies are moving in opposite directions at A m / s and B m/s, then their relative speed is = (A + B) m/s.
If two trains of length a metres and b metres are moving in opposite directions at u m / s and v m/s, then time taken by the trains to cross each other = (a + b)/(u+v) sec.
If two trains of length a metres and b metres are moving in the same direction at u m / s and v m / s, ; then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec.
If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then (A’s speed) : (B’s speed) = b1/2:a1/2
If speed of boat in still water is X km/hr and speed of stream is Y km/hr then speed of boat downstream is (X+Y) km/hr.
If speed of boat in still water is X km/hr and speed of stream is Y km/hr then speed of boat in going upstream [against flow] is (X- Y) km/hr.
If speed of boat downstream is a km/hr and speed of boat upstream is b km/hr.
Then speed of boat in still water = 1/2(a+b)
Then speed of stream = 1/2(a-b)
Q. A train crosses a pole in 8 seconds. If the length of the train is 200 metres, find the speed of the train.
In this case, it is evident that the situation is one of the train crossing a stationary object without length.
Thus, ST = 200/8 = 25 m/s so 25 × 18/5 = 90 kmph.
Q. A train crosses a man travelling in another train in he opposite direction in 8 seconds. However, the train requires 25 seconds to cross the same man if the trains are travelling in the same direction. If the length of the first train is 200 metres and that of the train in which the man is sitting is 160 metres, find the speed of the first train.
Here, the student should understand that the situation is one of the train crossing a moving object without length. Thus the length of the man’s train is useless or redundant data.
(Speed of first train + Speed of second train) * time taken to cross when moving in opposite direction = length of first train
(Speed of first train – Speed of second train) * time taken to cross when moving in same direction = length of first train
Equating RHS we finally get Speed of first train as 59.4 kmph / 16.5 mps
Acceleration is defined as the rate of change of speed. Acceleration can be positive (speed increases) or negative (speed decreases also known as deceleration)
The unit of acceleration is speed per unit time (e.g. m/s2)
For instance, if a body has an initial speed of 5 m/s and a deceleration of 0.1 m/s2 it will take 50 seconds to come to rest
Final speed = Initial speed + Acceleration × Time
Q. Water flows into a cylindrical beaker at a constant rate. The base area of the beaker is 24 cm2. The water level rises by 10 cm every second. How quickly will the water level rise in a beaker with a base area of 30 cm2.
The flow of water in the beaker is 24 cm2 × 10 cm/s = 240 cm3/s.
If the base area is 30 cm2 then the rate of water level rise will be 240/30 = 8 cm/s
Q. A 2 kilowatt heater can boil a given amount of water in 10 minutes. How long will it take for a less powerful heater of 1.2 kilowatts to boil the same amount of water?
The heating required to boil the amount of water is 2 × 10 = 20 kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 20/1.2 minutes = 16.66 minutes.
Q. A 2 kilowatt heater can boil a given amount of water in 10 minutes. How long will it take for a less powerful heater of 1.2 kilowatts to boil double the amount of water?
When the water is doubled, the heating required is also doubled. Hence, heating required = 40 kilowatt minutes. At the rate of 1.2 kilowatt, this heat will be generated in 40/1.2 = 33.33 minutes
Problems on clocks are based on the movement of the minute hand and that of the hour hand as well as on the relative movement between the two. It is best to solve problems on clocks by considering a clock to be a circular track having a circumference of 60 km and each kilometre being represented by one minute on the dial of the clock. Then, we can look at the minute hand as a runner running at the speed of 60 kmph while we can also look at the hour hand as a runner running at an average speed of 5 kmph.
Since, the minute hand and the hour hand are both moving in the same direction, the relative speed of the minute hand with respect to the hour hand is 55 kmph, that is, for every hour elapsed, the minute hand goes 55 km (minute) more than the hour hand.
Number of right angles formed by a clock:
A clock makes 2 right angles between any 2 hours. Thus, for instance, there are 2 right angles formed between 12 to 1 or between 1 and 2 or between 2 and 3 or between 3 and 4 and so on
However, contrary to expectations, the clock does not make 48 but 46 right angles in a day. This happens because whenever the clock passes between the time period 2–4 or between the time period 8–10 there are not 4 but only 3 right angles.
This happens because the second right angle between 2–3 (or 8–9) and the first right angle between 3–4 (or 9–10) are one and the same, occurring at 3 or 9.
Exactly the same situation holds true for the formation of straight lines. There are 2 straight lines in every hour. However, the second straight line between 5–6 (or 11–12) and the first straight line between 6–7 (or 12–1) coincide with each other and are represented by the straight line formed at 6 (or 12)
Q. At what time between 2–3 p.m. is the first right angle in that time formed by the hands of the clock?
At 2 p.m. the minute hand can be visualised as being 10 kilometres behind the hour hand. (considering the clock dial to be a race track of circumference 60 km such that each minute represents a kilometre).
Also, the first right angle between 2–3 is formed when the minute hand is 15 kilometres ahead of the hour hand.
Thus, the minute hand has to cover 25 kilometres over the hour hand. If the minute hand covers 55 km over the hour hand in 60 mins. Then it takes 5/11 of an hour which is = 27 (3/11) minutes to cover 25 km.
Hence, the required answer is: 2 : 27 : 16.36 seconds.
Solved Question Papers
The Sinhagad Express left Pune at noon sharp. Two hours later, the Deccan Queen started from Pune in the same direction. The Deccan Queen overtook the Sinhagad Express at 8 p.m. Find the average speed of the two trains over this journey if the sum of their average speeds is 70 km/h.
The ratio of time for the travel is 4:3 (Sinhagad to Deccan Queen). Hence, the ratio of speeds would be 3:4. Since, the sum of their average speeds is 70 kmph, their respective speeds would
be 30 and 40 kmph respectively. Use alligation to get the answer as 34.28 kmph
Walking at 3/4 of his normal speed, Abhishek is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is
When speed goes down to three fourth (i.e. 75%) time will go up to 4/3rd
(or 133.33%) of the original time. Since, the extra time required is 16 minutes, it should be equated to 1/3
rd of the normal time. Hence, the usual time required will be 48 minutes.
Ram and Bharat travel the same distance at the rate of 6 km per hour and 10 km per hour respectively. If Ram takes 30 minutes longer than Bharat, the distance travelled by each is
Since, the ratio of speeds is 3:5, the ratio of times would be 5:3. The difference in the times would be 2 (if looked at in the 5:3 ratio context.) Further, since Ram takes 30 minutes longer, 2
corresponds to 30. Hence, using unitary method, 5 will correspond to 75 and 3 will correspond to
45 minutes. Hence at 10 kmph, Bharat would travel 7.5 km
Two trains for Mumbai leave Delhi at 6 : 00 a.m. and 6 : 45 am and travel at 100 kmph and 136 kmph respectively. How many kilometres from Delhi will the two trains be together?
The train that leaves at 6 am would be 75 km ahead of the other train when it starts. Also, the relative speed being 36 kmph, the distance from Mumbai would be:
(75/36) × 136 = 283.33 km
Lonavala and Khandala are two stations 600 km apart. A train starts from Lonavala and moves towards Khandala at the rate of 25 km/h. After two hours, another train starts from Khandala at the rate of 35 km/h. How far from Lonavala will they will cross each other?
When the train from Khandala starts off, the train from Lonavala will already have covered 50 kms. Hence, 550 km at a relative speed of 60 kmph will take 550/60 hrs. From this, you can get
the answer as: 50 + (550/60) * 25 = 279.166 km
Walking at 3/4 of his normal speed, a man takes 2(1/2) hours more than the normal time. Find the normal time.
When his speed becomes 3/4th , his time would increase by 1/3rd
. Thus, the normal time = 7.5 hrs. (since increased time = 2.5 hrs).
Alok walks to a viewpoint and returns to the starting point by his car and thus takes a total time of 6 hours 45 minutes. He would have gained 2 hours by driving both ways. How long would it have taken for him to walk both ways?
8 h 45 min
7 h 45 min
5 h 30 min
6 h 45 min
Ans .8 h 45 min
Since he gains 2 hours by driving both ways (instead of walking one way) the time taken for driving would be 2 hours less than the time taken for walking. Hence, he stands to lose another
two hours by walking both ways. Hence his total time should be 8 hrs 45 minutes.
Sambhu beats Kalu by 30 metres or 10 seconds. How much time was taken by Sambhu to complete a race 1200 meters.
6 min 30 s
3 min 15 s
12 min 10 s
2 min 5 s
Ans .6 min 30 s
Kalu’s speed = 3 m/s.
For 1200 m, Kalu would take 400 seconds and Sambhu would take 10 seconds less. Hence, 390
Ram and Shyam run a race of 2000 m. First, Ram gives Shyam a start of 200 m and beats him by 30 s. Next, Ram gives Shyam a start of 3 min and is beaten by 1000 metres. Find the time in minutes in which Ram and Shyam can run the race separately.
Ans .4, 5
When Ram runs 2000 m, Shyam runs (1800 – 30s)
When Ram runs 1000 m, Shyam runs (2000 – 180s).
Then: 2000/1000 = 1800 – 30s / 2000 – 180s
Solving , we get s = 6.66 m/s
Thus, Shyam’s speed = 400 m/minute and he would take 5 minutes to cover the distance. Option (b) fits.
A starts from a point that is on the circumference of a circle, moves 600 metre in the North direction and then again moves 800 metre East and reaches a point diametrically opposite the starting point. Find the diameter of the circle?
Ans .1000 m
The diameter of the circle would be given by the hypotenuse of the right triangle with legs 600 and
800 respectively. Hence, the required diameter = 1000 meters.
X and Y are two stations 600 km apart. A train starts from X and moves towards Y at the rate of 25 km/h. Another train starts from Y at the rate of 35 km/h. How far from X they will cross each other?
Ans .250 km
The distance would get divided in the ratio of speeds (since time is constant). Thus, the distance
ratio would be 5 : 7 and required distance = 5/12 × 600 = 250 km.
Two trains A and B start from station X to Y, Y to X respectively. After passing each other, they take 12 hours and 3 hours to reach Y and X respectively. If train A is moving at the speed of 48 km/h, the speed of train B is
Ans .96 km/h
The time taken before their meeting would be given by t
2 = 12 × 3 = 36 Æ t = 6 hours. This means
that their ratio of speeds is 1:2. Since train A is traveling slower, the speed of train B would be
double the speed of train A. Required answer = 48 × 2 = 96.
Two trains for Patna leave Delhi at 6 a.m. and 6:45 a.m. and travel at 98 kmph and 136 kmph respectively. How many kilometres from Delhi will the two trains meet?
None of these
Ans .None of these
[73.5 × 136]/38.
A and B travel the same distance at the rate of 8 kilometre and 10 kilometre an hour respectively. If A takes 30 minutes longer than B, the distance travelled by B is
Ans .20 km
Solve using options. The value in option (d) fits the situation as 20/8 – 20/10 = 2.5 – 2 = 0.5 hours = 30 minutes.
Walking at 3/4 of his usual speed, a man is 16 minutes late for his office. The usual time taken by him to cover that distance is
Ans .48 m
At 3/4th speed, extra time = 1/3
rd of time = 16 minutes. Normal time = 48 minutes.
Two cars started simultaneously toward each other from town A and B, that are 480 km apart. It took the first car travelling from A to B 8 hours to cover the distance and the second car travelling from B to A 12 hours. Determine at what distance from A the two cars meet.
Ans .288 km
The speed of the first car would be 60 kmph while the speed of the second car would be 40 kmph.
The relative speed of the two cars would be 100 kmph. To cover 480 km they would take 480/100
= 4.8 hours Æ In 4.8 hours, the car traveling from A to B would have traveled 4.8 × 60 = 288 kms
A car travels 1/3 of the distance on a straight road with a velocity of 10 km/h, the next 1/3 with a velocity of 20 km/h and the last 1/3 with a velocity of 60 km/h. What is the average velocity of the car for the whole journey?
Ans .18 km/h
Assume a distance of 60 km in each stretch. Get the average speed by the formula. Total distance/
Total time = 180/10 = 18 kmph.
A person travelled a distance of 200 kilometre between two cities by a car covering the first quarter of the journey at a constant speed of 40 km/h and he remaining three quarters at a constant speed of x km/h. If the average speed of the person for the entire journey was 53.33 km/h what is the value of x?
Ans .60 km/h
The total time taken by the motorist would be 200/53.333 = 200 × 3/160 = 3.75 hours = 3 hours 45 minutes. In the first half of the journey the motorist covers 1/4th the distance @ 40kmph. This
means that he takes 50/40 = 1.25 hours = 1 hour 15 minutes in covering the first 50 kms. This also
means that he covers the remaining distance of 150 km in 2 hours 30 minutes so a speed of 60
kmph. Hence, option (b) is correct
A cyclist moving on a circular track of radius 100 metres completes one revolution in 2 minutes. What is the average speed of cyclist (approximately)?
Ans .314 m/minute
The length of the circular track would be equal to the circumference of the circle. In 2 minutes
thus, the cyclist covers 3.14 × 200 = 628 meters (using the formula for the circumference of a
Thus, the cyclist’s speed would be 628/2 = 314 meters/minute.