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.
Hence \( \frac{t_A}{t_B} = \frac{d_A}{d_B} \)
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:
\( \frac{S_1}{S_2} = \frac{D_1}{D_2} \)
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
\( \frac{S_1}{S_2} = \frac{D_1}{D_2} \)
= 3/4
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.
S_{A} / S_{B} = 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
\( \frac{S_1}{S_2} = \frac{t_2}{t_1} \) or \( S_1*t_1 = S_2*t_2 = S_3*t_3 = K \)
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 % |
---|---|
10 | 9.09 |
20 | 16.66 |
25 | 20 |
33.33 | 25 |
50 | 33.33 |
60 | 37.5 |
75 | 42.85 |
100 | 50 |
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 S_{1} /S_{2} = t_{2} /(t_{2} + 32)
5/7.5 = t_{2}/(t_{2} + 32)
5t_{2 + 160 = 7.5 t2}
t_{2} = 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) = \( b^{1/2} : a^{1/2} \)
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, S_{T} = 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/s^{2})
For instance, if a body has an initial speed of 5 m/s and a deceleration of 0.1 m/s^{2} 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 cm^{2}. 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 cm^{2}.
The flow of water in the beaker is 24 cm^{2} × 10 cm/s = 240 cm^{3}/s.
If the base area is 30 cm^{2} 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.
The relative speed of two bodies moving around a circle in the same direction is taken as S1 – S2. Also, when two bodies are moving around a circle in the opposite direction, the speed of the two bodies is taken to be S1 + S2
Three or more bodies start moving simultaneously from the same point on the circumference of the circle, in the same direction around the circle. They will first meet again in the LCM of the times that the fastest runner takes in totally overlapping each of the slower runners.
For instance, if A, B, C and D start clockwise from a point X on the circle such that A is the fastest runner hen we can define T_{AB} as the time in which A completely overlaps B, T_{AC} as the time in which A completely overlaps C and T_{AD} as the time in which A completely overlaps D. Then the LCM of T_{AB} , T_{AC} and T_{AD will be the time in which A, B, C and D will be together again for the first time}.
if A, B and C start from a point X on the circle such that T_{A} , T_{B} and T_{C} are the times in which A, B and C respectively cover one complete round around the circle, then they will all meet together at the starting point in the LCM of T_{A} , T_{B} and T_{C}.
Ans .
34.28
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
Ans .
48
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.
Ans .
7.5
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
Ans .
283.33
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
Ans .
96
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) = b^ 1/2 : a^1/2. so we get Hence, the Bombay Mail would travel at 96 kmph.
Ans .
12
From the figure above we see that Shyam would have walked a distance of 4 + 4 + 4 = 12 km. (G to P1, P1 to G and G to P2).
Ans .
279.166
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
Ans .
7.5
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).
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.
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 seconds
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.
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.
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.
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.
Ans .
None of these
[73.5 × 136]/38.
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.
Ans .
48 m
At 3/4th speed, extra time = 1/3 rd of time = 16 minutes. Normal time = 48 minutes.
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
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.
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
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 circle). Thus, the cyclist’s speed would be 628/2 = 314 meters/minute.
Ans .
60 kmph
The average speed would be given by: 120 * 1* 40 * 3 / 4 = = 60 kmph
Ans .
1000 kmph
By increasing the speed by 33.33%, it would be able to reduce the time taken for travel by 25%. But since this is just able to overcome a time delay of 30 minutes, 30 minutes must be equivalent to 25% of the time originally taken. Hence, the original time must have been 2 hours and the original speed would be 750 kmph. Hence, the new speed would be 1000 kmph.
Ans .
3 kmph
Solve through options using trial and error. For usual speed 3 kmph we have: Normal time Æ 2/3 hours = 40 minutes. At 4 kmph the time would be 2/4 hrs, this gives us a distance of 10 minutes. Hence option (c) is correct
Ans .
25 kmph
The required speed s would be satisfying the equation: 300/s – 300/(s + 2) = 5 Solving for s from the options it is clear that s = 25
Ans .
48 kmph
Solve this question using the values given in the options. Option (d) can be seen to fit the situation given by the problem as it gives us the following chain of thought: If the average speed of the faster train is 48 kmph, the average speed of the slower train would be 32 kmph. In this case, the time taken by the faster train (192/48 = 4 hours) is 2 hours lesser than the time taken by the slower train (192/32 = 6 hours). This satisfies the condition given in the problem and hence option (d) is correct
Ans .
400 km
At 40 kmph, Harsh would cover (200/60) × 40 km. = 400/3 km. = 133.33 km. This represents the distance by which Vijay would be ahead of Harsh, when Vijay reaches the endpoint means in essence that Vijay must have travelled for 133.33/20 hours Æ 6.66 hours Hence, the distance is 60 × 6.66 = 400 km
Ans .
8 km/h
In 6 minutes, the car goes ahead by 0.6 km. Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph, since, the pedestrian is walking at 2 kmph, hence, the net speed is 8 kmph.
Ans .
330/23 m/s
Ans .
120 km/h
direct formula based questions
Ans .
3600
In four hours, the train will travel 180 km (180,000 metres). The number of poles would be 180,000/50 = 3600.
Ans .
12 km
You can solve this question using the options. Option (a) fits the given situation best as if we take the distance as 12 km he would have taken 1 hour to go by car and 4 hours to come back walking —a total of 5 hours as given in the problem.
Ans .
80 km/h
To reduce the time of the journey by 25%, he should increase his speed by 33.33% or 1/3rd . Thus,required speed = 80 kmph.
Ans .
600 km
If the car does half the journey @ 30 kmph and the other half at 40 kmph it’s average speed can be estimated using weighted averages. Since, the distance traveled in each part of the journey is equal, the ratio of time for which the car would travel would be inverse to the ratio of speeds. Since, the speed ratio is 3:4, the time ratio for the two halves of the journey would be 4:3. The average speed of the car would be: (30 × 4 + 40 × 3)/7 = 240/7 kmph. It is further known that the car traveled for 17.5 hours (which is also equal to 35/2 hours). Thus, total distance = average speed × total time = (240 × 35)/(2 × 7) = 120 × 5 = 600 km
Ans .
24 km
d/6 + d/4 = 10 Æ d = 24 km
Ans .
10 km
By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minute drop in his time for travel—since he goes from being 10 minutes late to only 4 minutes late.) Hence, his time originally must have been 30 minutes. Hence, the required distance is 20 kmph × 0.5 hours = 10 km
Ans .
51.18 km/h
The distance covered in the various phases of his travel would be: 10 km + 25 km + 50 km + 60 km. Thus the total distance covered = 145 km in 2 hours 50 minutes so 145 km in 2.8333 hours so 51.18 kmph
Ans .
106 km
Total distance/Total time = 1590/15 = 106 kmph
Ans .
40 min/h
Since the train travels at 60 kmph, it’s speed per minute is 1 km per minute. Hence, if it’s speed with stoppages is 40 kmph, it will travel 40 minutes per hour.
Ans .
10
Since Chandu is moving at a speed of 10 m/s and he has to cover 360 km or 360000 meters, the time taken would be given by 360000/10 seconds = 36000 seconds = 36000/60 minutes = 600 minutes = 10 hours.
Ans .
5 kmph
30/(15 + x) + 30/(15 – x) = 4 hrs 30 minutes. At x = 5, the equation is satisfied.
Ans .
58 m
When Ajay does 600 metres, Vijay does 540 m. When Vijay does 500 metres, Anjay does 475 m Thus, Ajay : Vijay : Anjay = 600 : 540 : 513. Thus, Ajay would beat Anjay by (87 × 2/3) = 58 m in a 400 m race
Ans .
3 km/h
Look for the solution by thinking of the factors of 91. It can be seen that 91/13 + 91/7 = 7 + 13 = 20 hours. This means that the speed of the boat in still water is 10 kmph and the speed of the water flow would be 3 kmph. Option (a) is correct
Ans .
2 kmph
14/(5 + x) + 9/(5 – x) = 5 x = 2, fits this equation
Ans .
2/11 hours past 2 p.m.
If we consider the clock to be a circle with circumference 60 km, the speed of the Minute hand = 60 kmph, while the speed of the hour hand = 5 kmph. The relative speed = 55 kmph. At 2 PM, the distance between the two would be seen as 10 km. This would get covered in 10/55 = 2/11 hours. Option (a) is correct.
Ans .
11
The hands would be together once in each hour. However, the 12 noon time would be counted in both 11 to 12 and 12 to 1. Hence, the no. of times = 12 – 1 = 11.
Ans .
4p cm
(1/4) × 2pr = 4p (Since r = 8 cm).
Ans .
0.5 (x + y)
The speed of the boat in still water is the average of the upstream and downstream speeds. (x + y)/2
Ans .
22 km/h
10/(x – 2) + 10/(x + 2) = 55/60 = 11/12 hours. x = 22 fits the expression.
Ans .
8 km/h
The given situations are satisfied with the speed of the boat as 8 kmph and the speed of the stream as 3 kmph. Option (c) is correct.
Ans .
19
Rate in still water = (16 + 22)/2 = 19 kmph
Ans .
80 m
The length of the train would be given by: 36 × 5/18 × 8 = 80 meters.
Ans .
2.88 km
Upstream speed = 4.8 kmph Downstream speed = 7.2 kmph. d/4.8 + d/7.2 = 1 Solving we get d = 2.88 km.
Ans .
12 km/h
Vijay takes 9 hours to return upstream after going for 6 hours downstream. Solve using options. Option (d) fits as we get Downstream speed = 18 kmph Æ distance = 18 × 6 =108 km Also, upstream speed = 12 kmph Æ distance = 12 × 9 = 108 km
Ans .
1.4 km/h
Upstream speed = 3.2 kmph Downstream speed = 6 kmph. Thus, speed of stream = 1.4 kmph.
Ans .
12 m
Speed of running of the train = 1.25 km/hr. With stoppage, an effective speed of 60 kmph means that the time of travel per hour would be 60/1.25 = 48 minutes. Thus, the train stops for 12 minutes per hour
Ans .
210 km
When the second train leaves Muzaffarpur, the first train would have already traveled 30 km. Now, after 9 AM, the relative speed of the two trains would be 10 kmph (i.e. the rate at which the faster train would catch the slower train). Since the faster train has to catch up a relative distance of 30 km in order for the trains to meet, it would take 30/10 = 3 hours to catch up. Distance from Muzaffarpur = 70 × 3 = 210 km
Ans .
0.1 km
20 × (5/18) × 18 = 100 m = 0.1 km.
Ans .
12
Distance to be covered = 120 meters. Speed = 10m/s Æ Time required = 120/10 =12 seconds.
Ans .
20
When Shyam does 500, Vinay does 375. Since Vinay has a start of 140 m, it means that Vinay only needs to cover 360 m to reach the destination. When Vinay does 360, Shyam would cover 480 m and lose by 20 m. (Since the ratio of their speeds is 3 : 4)
Ans .
10
Solve using options 1.33 m/s fits perfectly. When A scores 60 points B scores 45, and C scores 40. Thus, when B scores 90, C would score 80. So, B can give C 10 points in 90
Ans .
1.33 m/s
Solve using options 1.33 m/s fits perfectly.
Ans .
20 m
Speed of Vinay = 5 m/s, Speed of Ajay = 4m/s. In a hundred meter race, Vinay would take 20 seconds to complete and in this time Ajay would only cover 80 meters. Thus, Vinay beats Ajay by 20 meters in a hundred meter race.
Ans .
30 km/h
(Sf – Ss) × 60 = 200 Where Sf and Ss are speeds of the faster and slower train respectively Æ Sf – Ss = 3.33 Also, (Sf + Ss) × 10 = 200. Æ Sf + Ss = 20. Solving we get Ss = 8.33 m/s = 8.33 × 18/5 = 30 kmph.
Ans .
4 h
Since A to C is double the distance of A to B, it is evident that the time taken for A to C and back would be double the time taken from A to B and back (i.e. double of 6.5 hours = 13 hours). Since going from A to C takes 9 hours, coming back from C to A would take 4 hours (Since 9 + 4 = 13).
Ans .
16.25 km/h
15 km upstream in 80 minutes Æ 15/1.33 = 11.25 kmph. (upstream speed of the boat). Thus, still water speed of the boat = 11.25 + 5 = 16.25 kmph
Ans .
None of these
8/(12 – 2) = 8/10 = 0.8 hours
Ans .
200 metres
(20 × 5/18) × 36 = Lt so Lt = 200 m.
Ans .
6 km/h
Upstream speed = 40/8 = 5 kmph. Downstream speed = 49/7 = 7 kmph. Speed in still water = average of upstream and downstream speed = 6 kmph.
Ans .
1.6875 km
x/9 + x/3 = 3/4 Æ 4x/9 = 3/4 Æ x = 27/16 kms = 1.6875 kms.
Ans .
4 kmph
Solve through options. For option (c) at 4 kmph, the boat would take exactly 4 hours to cover the distance.
Ans .
4 km/h
In order to solve this, you first need to think of the speed of the river flow (if the speed of the boat in still water is 3 kmph). If we take the speed of the river flow as s, we get downstream speed as 3 + s and upstream speed as 3 – s. 10/(3 – s) – 10/(3 + s) = 8 hours Æ s = 2kmph. Note: It is obvious that since the difference between the downstream time and the upstream time is 8 hours, the upstream and downstream speeds would both be factors of 10. The only value of s such that both 3 + s and 3 – s are factors of 10 is s = 2. If the boat needs to reach 10 km downstream in 100 minutes (1.66 hours) it means: 10/1.66 = 6 kmph is the downstream speed. Since, the speed of the stream is 2 kmph, the required speed of the boat = 4 kmph
Ans .
75.6 km/h
7 × St = Lt 25 × St = Lt + 378 Solving,St = 21 m/sec. = 21 × 18/5 = 75.6 kmph.
Ans .
9 km/h, 3 km/h
From the situation described in the first condition itself we can see that the speed of coming back has to be double the speed of going downstream. Checking the options, only option (a) fits this condition i.e. Downstream speed = 2 × Upstream speed. Hence, option (a) is correct.