Wednesday, 1 August 2018

Question of the day


1. A man takes 6 hrs 30 min in walking to a certain place and riding back. He would have gained 2 hrs 10 min by riding both ways. How long would he take to walk both ways?
(a) 8 hr 30 min 
(b) 8 hr 40 min 
(c) 8 hr 50 min
(d) 9 hr 
(e) None of these

2. A Train which travels at the uniform rate of 10 m a second leaves Madras for Arconum at 7 a.m. At what distance from Madras will it meet a train which leaves Arconum for Madras at 7.20 a.m., and travels one-third faster than the former does, the distance from Madras to Arconum being 68 km?
(a) 40 
(b) 38 
(c) 36
(d) 34 
(e) None of these

3. A, B and C can walk at the rates of 3, 4 and 5 km an hour respectively. They start from Poona at 1, 2, 3 o’clock respectively. When B catches A, B sends him back with a message to C. When will C get the message?
(a) 5.15 
(b) 5.20 
(c) 5.30
(d) 6.00 
(e) None of these

4. Two men start together to walk a certain distance, one at  3 (3/4)  km an hour and the other at 3 km an hour. The former arrives half an hour before the latter. Find the distance.
(a) 7 km 
(b) 7.5 km 
(c) 8 km
(d) 8.5 km 
(e) None of these

3 comments:

  1. 1. Ans.(b)
    Sol. Walking + Riding = 6 hrs 30 min ……(1)
    2 Riding = 6 hrs 30 min – 2 hrs 10 min
    = 4 hrs 20 min ……(2)
    Solving the above two relations (equations);
    2 × (1) – (2) gives
    2 walking = 13 hrs – 4 hrs 20 min = 8 hrs 40 minutes

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  2. 2. Ans.(c)
    Sol. S1 = 10 * 18/5 = 36 km/hr
    S2 = 36 + 36/3 = 48 km/hr
    Difference in time = T1 – T2 = 7 am – 7.20 am = 1/3 hr
    Distance of meeting point from madras

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  3. 3. Ans.(a)
    Sol. Speed Starting time

    A takes a lead of 3 km from B.
    Relative speed of A and B = 4 – 3 = 1 km/hr
    B catches A after 3/1 = 3 hrs, i.e., at 2 + 3 = 5 o’clock.
    A returns at 5 o’clock and from a distance of 3 × 4 = 12 km from Poona.
    In the meantime C covers a distance of 5 × 2 = 10 km from Poona.
    Thus, A and C are 12 – 10 = 2 km apart at 5 O’clock.
    Relative speed of A and C = 3 + 5 = 8 km/hr.
    Thus, they meet after 2/8 = 1/4 hr = 15 min.
    Thus, C will get the message at 5.15 o’clock.

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