GRE Time/Speed/Distance Problems - Tips and Tricks you must know with examples
1.
Speed: The rate at which anything covers a particular distance is called its speed ⇒ Speed = Distance Travelled / Time Taken
Generally, speed is expressed in the following units: miles/hr, km/hr, m/sec, m/min, etc.
1. A policeman goes after a thief who is 176 m ahead of him. When and where will the policeman catch the thief when they run at the rates of 11440 and 10560 meters per hour respectively?
Sol. Time to \(\frac{catch }{ overtake} = \frac{lead distance }{ difference of speeds} = \frac{176 }{ (11440 – 10560) }= \frac{176 }{ 880} = \frac{1}{5} hours = 12\) minutes ⇒ The time required to overtake the thief = 12 min.
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b) The distance from the starting point \(= \frac{11440 \times 12}{60}\) kms = 2288 meters
2. If A goes from X to Y at U km/hr and comes back from Y to X at V km/hr, then Average speed during the whole journey \(= \frac{2UV }{ (U + V)}\) km/hr.
2. If I walk at the rate of 4 kms an hour, I reach my destination 30 min too late; If I walk at the rate of 5 kms an hour I reach 30 minutes too soon. How far is my destination ?
Sol. Let time taken be T hrs for the distance to be covered at the normal speed (neither fast nor slow). Then we have 4 (T + 0.5) = 5 (T – 0.5) {Note : 0.5 here is 30 min} ⇒ T = 4.5 hours ⇒ Distance \(= 4 (T + 0.5) = 4 \times 5 = 20\) km.
3. If a man changes his speed in the ratio m : n then the ratio of times taken becomes n : m.
3. A man rows 18 kms down a river in 4 hours with the stream and returns in 12 hours; find his speed and also the velocity of the stream.
Sol. Speed with the stream \(= \frac{18}{4} = 4.5\) kms an hour. ⇒ Speed against the stream \(= \frac{18}{12} = 1.5\) kms an hour. ⇒ Speed of the stream \(= \frac{1}{2} (4.5 – 1.5) = 1.5\) kms an hour and his speed = 4.5 – 1.5 = 3 kms an hour
4. If three men cover the same distance with speeds in the ratio a : b : c, the times taken by these three will be resp. in the ratio \(\frac{1}{a} : \frac{1}{b} : \frac{1}{c}\).
4. A, B and C can walk at the rates of 3, 4, 5 kms an hour. They start from X at 1, 2, 3 o’clock respectively; when B catches up with A, B sends him back with a message to C; when will C get the message ?
Sol. In one hour A covers 3 kms Now (B – A) = 1 kmph hence B catches up A after 3 hours i.e. at 5 o’ clock. Now upto 5 o’ clock A has covered 12 kms and C has covered 10 kms. Hence Distance between A & C = 2 kms and their relative speed (3 + 5) = 8 kmph. To cover 2 kms at 8 kmph, Time \(=\frac{ 2}{8}\) hours = 15 min. Hence C gets the message at 15 minutes past 5 o’clock.
Relative Speed:When two objects travel in the same direction, relative speed = difference of speedsWhen two objects travel in opposite directions, relative speed = sum of speeds5. When two objects travel in the same direction, relative speed = difference of speeds time to catch / overtake = lead distance / difference of speeds
5. A student walks to school at the rate of 2.5 kms an hour and reaches 6 minutes too late. Next day he increases his speed by 2 kms an hour and then reaches there 10 minutes too soon. Find the distance of the school from his home.
Sol. Let t be the usual time We have \(⇒ 2.5 \times (t + \frac{1}{10}) = 4.5 (t – \frac{1}{6})\), or \(t = \frac{1}{2}\) hours. Hence distance \(= 2.5 (\frac{1}{2} + \frac{1}{10}) = 2.5 \times \frac{6 }{10} = 1.5\) km.
6. When two objects travel in the Opposite directions, relative speed = sum of speeds time to meet = lead distance / sum of speeds.
6. A man can row in still water a distance of 4 kms in 20 minutes and 4 kms with the current in 16 min. How long will it take him to row the same distance against the current ?
Sol. \(X = \frac{4}{(\frac{20}{60})} = 12\), \(X + Y = 4 \times \frac{60 }{16} = 15\) or \(Y = 3\). or Time \(= \frac{4 }{ (X – Y)} = 4 \times \frac{60 }{9} = \frac{80}{3}\) minutes
7. If the speed of a boat (or man) in still water be \(X \frac{km}{hr}\), and the speed of the stream (or current) be \(Y \frac{km}{hr}\), then
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a) Speed of boat with the stream (or Downstream or D/S) \(= \frac{(X + Y) km}{hr}\)
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b) Speed of boat against the stream (or upstream or U/S) \(= \frac{(X – Y) km}{hr}\)
We have \(X = \frac{[(X + Y) + (X – Y)] }{ 2}\) and \(Y = \frac{[(X + Y) – (X – Y)] }{ 2} ⇒\) Boat’s speed in still water \(= \frac{[Speed downstream + Speed upstream] }{ 2}\) Speed of current \(= \frac{[Speed downstream – Speed upstream] }{ 2}\)
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