Bunuel wrote:
On a 20 km tunnel connecting two cities A and B there are three gutters. The distance between gutter 1 and 2 is half the distance between gutter 2 and 3. The distance from city A to its nearest gutter, gutter 1 is equal to the distance of city B from gutter 3. On a particular day the hospital in city A receives information that an accident has happened at the third gutter. The victim can be saved only if an operation is started within 40 minutes. An ambulance started from city A at 30 km/hr and crossed the first gutter after 5 minutes. If the driver had doubled the speed after that, what is the maximum amount of time the doctor would get to attend the patient at the hospital? Assume 1 minute is elapsed for taking the patient into and out of the ambulance.
(A) 4 minutes
(B) 2.5 minutes
(C) 1.5 minutes
(D) 1.5 minutes Same Though
(E) Patient died before reaching the hospital
Explanation:
let that the distance of the gutter 1 and A to be x
The gutter 1 and 2 to be y
From the question, we will get that
x+y+2y+x = 20
2x + 3y = 20.
Since we know that the speed x is 30 km/hr and so x = 30*5/60 = 2.5 km ; correspondingly the value of y will be 5km.
Now we know that the ambulance doubles its speed to 60km/hr or we can write 1 km per minute. So the time taken for the remaining journey = 15*2 + 2.5 = 32.5.
Since, it takes 1 min to load and unload the patient so total time will be = 5 + 32.5 + 1 = 38.5 minutes.
So, doctor gets = 40 - 38.5 = 1.5 min to attend the patient.
IMO-C or D