CAT 2002 — QA Question 31
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.
Answer & solution
Correct answer: 1.5 minutes
- A
4 minutes
- B
2.5 minutes
1.5 minutes
- D
Patient died before reaching the hospital
As per the description,
AB = 20 km
AG1 = BG3
2G1G2 = G2G3
∴ x = 2.5 km
y + 2y = 20 − 2x
∴ y = 5 km
∴ Time to cover A to G3 = = 20 min
∴ While coming back his speed is 60 km/hr.
∴ Time taken to cover the distance from G3 to A (i.e. 17.5 km) = 17.5 minutes
∴ Required time = 20 + 17.5 + 1 = 38.5 minutes
∴ The doctor will have 1.5 minutes to attend the patient.
Hence, option (c).