shankar.ashwin wrote:A leaves home at 9: 30 a.m. daily to reach his office at 10:30 a.m. He travels by his bike at a speed of 36 kmph. One a particular day, he left home at the usual time but his bike broke down after travelling some distance. After waiting for 15 minutes he walked to the office at a speed of 18 kmph and reached the office at 11 a.m. Find the distance for which A had to walk?
A)27
B) 9
C)18
D) 22.5
E) None of these
IMO
B. Distance walked is 9 Kms
Time = 1 hr (10:30 - 9:30)
Speed = 36 Km/Hr (given, the speed with bike)
Therefore,
total distance = 36 Kms
Now, he drives the bike for some distance. Let the
time on bike = T1
Therefore, his
distance traveled on bike = Speed x Time =
36 x T1
He walks the remaining distance.
Let's assume that the
time to complete the walk = T2
Therefore,
distance he walked = Speed x Time =
18 x T2
Total time he took when his bike broke was 1:30 Hrs (given 10:30am - 9am)
He waits for 15 mins
Therefore, the
net time for bike ride (T1) + time for walking (T2) = 1:30 Hrs - 15 mins = 1Hr 15 mins =
5/4 Hrs
So, we know the total distance is 36 km
and
putting all the values together, we get
Total distance = 36T1 + 18T2 = 36
and we know
T1 + T2 = 5/4
Solving the equations, we would get T2 = 1/2 and therefore,
distance walked = 1/2 * 18 = [spoiler]9 Kms = answer choice B[/spoiler]