Tuesday, 1 November 2011

Arithmetic-16 (CAT-2004)

If a man cycles at 10 km/hr, he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?
(1) 11 km/hr
(2) 12 km/hr
(3) 13 km/hr
(4) 14 km/hr

Solution:

Let the total distance be d.
If a man cycles at 10 km/hr, he arrives at a certain place at 1 p.m
Speed = distance/time => time = t1= distance/speed = d/10 -----(1)
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m
t2= distance/speed = d/15                           ----- (2)
In the second case he reaches 2 hours early, which means,
t1 – t2 = 2 => d/10 – d/15 = 2 => d((15-10)/(10*15)) = 2 => d/30 = 2 => d=60 km
At what speed must he cycle to get there at noon ie., 12’O clock
In this case let his speed be x and time taken be t3.
He has to reach 1 hour earlier when compared to the first case.
Hence, t1- t3 = 1 hour => d/10 – d/x = 1 => 60/10 – 60/x = 1
=> 60/x = 6-1 = 5 => x = 60/5 = 12
Hence his speed must be 12 kmph in order to reach at noon