A bullock cart has to cover a distance of 80 km in 10 hours. If it covers half of the journey in (3/5)th time, what should be its speed to cover the remaining distance in the time left ?
a)5 km/hr | b)10 km/hr |
c)15 km/hr | d)18 km/hr |
Total Distance to cover in 10 hours = 80km
if it covers 40 km in 3/5 th of time i.e. 40 km in 6 hours
therefore remaining time = 10 - 6 = 4 hours
remaining distance = 40 km
therefore speed required = (40/4) km/hr = 10 km/hr
A man, on tour, travels first 160 km at 65 km/hr and the next 160 km at 80 km/hr. the average speed for the first 320 km of the tour, is :
a)71.11 km/hr | b)71.31 km/hr |
c)71.21 km/hr | d)71.41 km/hr |
Case Ist: Distance = 160 km., Speed = 64 km/hr
Case IInd: Distance = 160 km., Speed = 80 km/hr
Therefore average speed for 320 km of tour = (2 x 64 x 80)/(64 + 80) = (2 x 64 x 80)/144
= 71.11 km/hr
By walking at ¾ of his usual speed, a man reaches his office 25 minutes later than usual. His usual time is :
a)60 minutes | b)70 minutes |
c)75 minutes | d)80 minutes |
Let initial time taken = t hr
time taken when travelled at speed 3/4 of usual = (4/3) t hours
therefore as given (4/3) t - t = 25
thus t/3 = 25 => t = 75 min
A certain distance is covered at a certain speed. If half of this distance is covered in triple the time, the ratio of the two speeds is :
a)3 : 1 | b)2 : 1 |
c)6 : 1 | d)1 : 1 |
Let a km be covered in b hours
Then the speed of object in first case = a/b km/hr
As half of this distance is covered in triple the time
Then the speed of object in second case = (a/2)/3b = (a/6b) km/hr
Therefore, Ratio od speed = (a/b) : (a/6b) = 1/1 : 1/6
= 6 : 1
A student walks from his house at 5 km/hr and reaches his school 10 minutes late. If his speed had been 6 km/hr he would have reached 15 minutes early. The distance of his school from his house is :
a)12.5 km | b)25 km |
c)37.5 km | d)50 km |
Let the distance of school from house be x km.
The difference of the time taken = 25 min
Therefore, Time in first case be T1 = x/5
Time in second case be T2 = x/6
therefore, T1 - T2 = 25/60
=> x/5 - x/6 = 25/60
=> (6x - 5x)/30 = 25/60
=> x = 25/2
=> x = 12.5
A boat goes downstream in half the time it takes to go upstream then the ratio between the speed of the boat in still water to that of stream is :
a)3 : 1 | b)1 : 2 |
c)1 : 3 | d)2 : 1 |
Let the speed of the boat in still water be x km/hr and that of stream be y km/hr. Also let k be the distance travelled, then
k/(x + y) = 1/2 [(k/(x - y)]
or 2x - 2y = x + y
x = 3y
=> x : y = 3 : 1
Two towns A and B are 250 km apart. A bus starts from A and B at 6 AM at a speed of 40 km/hr. At the same time another bus starts from B to A at a speed of 60 km/hr. The time of their meeting is :
a)8 : 30 AM | b)8 AM |
c)9 AM | d)9 : 15 AM |
Let the buses meet x hours after 6 AM
So, the distance covered by the two buses is 250 km
Therefore 40x + 60x = 250
=> x = 250/100 = 2.5 hours
= 2 hours 30 min
thus they will meet at 8:30 AM
A man standing on a railway platform observes that a train going in one direction takes 4 seconds to pass him. Another train of some length going in the opposite direction takes 5 seconds to pass him. The time taken (in seconds) by the two trains will be :
a)32/9 | b)33/7 |
c)40/9 | d)49/9 |
Let the length of each train be l metres
=> Speed of first train = (l/4) m/s.
and Speed of second train = (l/5) m/s.
As both trains are moving in opposite direction
=> Relative Speed = [(l/4) + (l/5)] m/s. = 9l/20 m/s.
Time taken to cross each other = time taken to covel 2l metres at speed (9l/20) m/s.
= (2l/9l/20) s = (20 x 2)/9 = 40/9 sec.
A police car is ordered to chase a speeding car that is 5 km ahead. The car is travelling at an average speed of 80 km/hr and the police car pursues it at an average speed of 100 km/hr. how long does it takes for the police car to overtake the other car ?
a)17 minutes | b)19 minutes |
c)13 minutes | d)15 minutes |
Distance travelled by thief car in one hour = 80 km
Distance travelled by police car in one hour = 100 km
So police travels extra 20 km in 1 hour
So to overtake thief, police car has to travel 5 km extra
Therefore, time = 5/20 = 1/4 hour = 15 mins
a)9 km/hr | b)8 km/hr |
c)4 km/hr | d)3 km/hr |
Two trains starts runnings at the same time from two stations to 210 km apart and going in opposite directions cross each other at a distance of 100 km from one of the station. The ratio of their speed is :
a)11 : 9 | b)10 : 11 |
c)11 : 10 | d)9 :11 |
Let their relative speed be x km/hr and y km/hr respectively.
Then the time taken by I train to cover 110 km = time taken by train II to cover 100 km
Thus, x/110 = y/100 or x/y = 110/100
therefore x : y = 11 : 10
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