Aptitude Time And DistancePage 3
17. | | The ratio of speeds of A and B is 3 : 2. B started running from P to wards East at 8:00 am. After one hour, A started running from P in the same direction. When will A meet B? | | (a)10:00 am | | (b)11:00 am | | (c)12:00 am | | (d)cannot be determined |
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Answer is: BLet the speed of A be 3X kmph, speed of B is 2X kmph.
in one hour B covers a distance of 2X km.
As, at 9:00 am A and B are traveling in the same direction the relative speed
= (3X - 2X) = X kmph.
To catch B, A has to gain 2X km.
It takes 2 hours to gain 2X km.
hence, at 9:00 am + 2 hours = 11:00 am, they meet.
18. | | The ratio of speeds of Tarun and Varun is 4 : 3. Tarun starts to chase Varun who is at a distance of 10 km from Tarun and is running away from him. In order to catch Varun, how many kilometers should Tarun cover? | | (a)30 km | | (b)40 km | | (c)50 km | | (d)cannot be determined |
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Answer is: BLet the speed of Tarun be 4X kmph.
The speed of Varun is 3X kmph.
If Tarun covers a distance of 4X km, he gains X km over Varun;
to gain X km, he shall cover 4X km.
Distance between Tarun and Varun is 10 km.
in order to gain 10 km over Varun , Tarun has to cover 40 km.
19. | | Two cyclists A and B are cycling around a circular track of 150 m length with respective speeds of 10 m/sec and 15 m/sec. What is the time taken by them to meet for the first time right at their starting point? | | (a)60 seconds | | (b)30 seconds | | (c)20 seconds | | (d)cannot be determined |
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Answer is: BThe time taken by A to complete one round = 150/10 = 15 seconds.
The time taken by B to complete one round = 150/15 = 10 seconds.
hence, After a time which is the LCM of (15,10) seconds, they will meet right at the starting point i.e. after 30 seconds.
20. | | Three cyclists A, B and C with respective speeds of 8 m/sec , 10 m/sec and 12 m/sec are cycling around a circular track of length 120 m. After how many seconds will they meet for the first time right at their starting point? | | (a)60 seconds | | (b)120 seconds | | (c)240 seconds | | (d)30 seconds |
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Answer is: AThe times taken by the three cyclists to complete one round each are
A = 120/8 sec
B = 120/10 sec
C = 120/12 sec.
i.e. 15 sec, 12 sec, 10 sec respectively.
Hence, After a time which is the LCM of (15, 12, 10) sec, they will meet right at the starting point for the first time after 60 seconds.
21. | | A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is: | | (a)100 kmph | | (b)110 kmph | | (c)120 kmph | | (d)130 kmph |
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Answer is: CLet speed of the car be X kmph.
Then, speed of the train = (150/100)X = (3/2) kmph
75/X - 75/(3/2)X = 125/(10 x 60)
75/X - 50/X = 5/24
X = (25 x 24)/5 = 120 kmph.
22. | | A man has to walk a distance of 16 km from his house to his friend's house. He walks at a speed of 8 kmph and after every 30 minutes he takes rest for 10 minutes. How much time does he take to reach his friend's house? | | (a) 2 hours 10 minutes | | (b)2 hours 30 minutes | | (c)2 hours 40 minutes | | (d)2 hours 20 minutes |
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Answer is: BHis entire walking time = 16/8 i.e. 2 hours. So, he travels in four thirty minutes laps
with a rest of 10 min. between every two successive laps.
So, there are three such resting times of 10 min in between.
So, the total time taken = walking time + resting time
= 2 hours + 3 (10 min.) = 2 hours 30 minutes.
23. | | Mr. Raaz started half an hour later than usual for the market place. But by increasing her speed to (3/2) times his usual speed he reached 10 minutes earlier then usual. What is his usual time for this journey? | | (a)2 hours | | (b)1 hour | | (c)45 minutes | | (d)1 hour 15 minutes |
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Answer is: ALet Mr. Raaz's usual speed be 's' (in kmph) and let the time of his journey be 't' (in hours).
So distance traveled = d = st {t - (2/3) x (3/2) s}
so, 2t = 3t - 2
t = 2 hours.
24. | | A train T1, left station S1 for station S2 at 11:00 am at speed of 72 kmph. Another train T2 left station S2 for S1 at 10:00 am at a speed of 108 kmph. At what time would the trains be 24 km apart for the first time, the distance between station S1 and S2 is 240 km. | | (a)11:48 am | | (b)12:00 noon | | (c)11:24 am | | (d)11:36 am |
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Answer is: DLet the 2 trains be 24 km apart after t hours from the train T , starts from S1
72t + 108(t + 1) = (240 - 24)
180t = 108
t = 108/180 hour, which is 36 min.
At 11:36 am T1 and T2 were 24 km apart.
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