PSA_LEVEL-UP_QNR46. Sumit walks from his hostel to his college CMEK at a speed of 10 km/h usually; he gets there late by 10 minutes. The very next day he increases his speed by 5 km; he finds that he is still 2 minutes late. If his college starts at 9:30 am, at what time should he start from his hostel to reach college exactly on time (he prefers to stick to his usual walking speed)? What is the distance between his hostel and college?
a. 9:06 am, 4 km
b. 9:05 am, 4 km
c. 9:00 am, 6 km
d. 9:06 am, 8 km
Relation between speed, time and distance is given as
Distance ‘d’ is constant here. (Hostel to College distance is fixed).
Sumit’s initial speed given is 10 km/h. Next day, his speed is 15 km/h.
Thus, speed becomes 3/2 times.
If speed ‘s’ becomes 3/2 times, time will become 2/3 times. [so that product of new speed and new time results in the same distance ‘d’].
If earlier time was t, the new time will be (2/3)* t. You can observe that ‘t’ reduces by one-third (1/3).
It is given in the question that Sumit was initially 10 minutes late; after increasing his speed he is 2 minutes late. This means net time taken is reduced by 8 minutes.
Therefore, one third of ‘t’ is nothing but 8 minutes.
(1/3)*t = 8
=> t= 24 minutes.
Sumit takes 24 minutes to reach his college at his usual speed of 10 km/h.
He should start for his college exactly at 9:06 am.
Distance between his hostel and college = (24/60) * 10 = 4 km. Option a.