Aptitude Time And DistancePage 3


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
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.


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
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.


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
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.


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
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.


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
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.


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
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.


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
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.


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
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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