Jamil1992Mehedi wrote:
An ambulance, an hour after starting, meets with an accident which detains it for a 30 minutes, after which it proceeds at \(\frac{3}{4}\) of its former rate and arrives \(3\frac{1}{2}\) hours late. Had the accident happened \(90\) km farther along the line, it would have arrived only 3 hours late. The Approximate length of the trip in miles was.
A. 370 miles
B. 373 miles
C. 376 miles
D. 379 miles
E. 382 miles
Interesting Question!
So the difference in time taken by the ambulance to reach it's destination in the 2 cases (1/2 hr) is the time taken to travel 90 km at the different speeds.
Let the original speed=R1 and the reduced speed be =R2= 3/4 R1
In case 1,
the ambulance traveled the 90 km at R2 i.e (3/4 R1)
so, t1=d/s = 90 / R2 = 90 / (3/4)R1
In case 2,
Ambulance traveled the 90 km at speed R1
so t2=d/s = 90/R1
From question we know, t1-t2=0.5 hours
so, 90/(3/4)R1 - 90/R1=0.5
Solving,
R1= 60 km/hr; R2= 45 km/hrIn case 1,
Before the accident:
The ambulance traveled an hour before the accident. So distance traveled at 60 km/hr= 60 km
After the accident:
let the distance traveled after accident be d'
due to the accident, the ambulance took 3 extra hours to travel this distance [total delay(3.5) -wait time (0.5hrs)=3hrs]
so, d'/45=d'/60+3
solving, d'=540 km
Total distance=60+540=600 km= 600 * 0.62=373 km approx =Option B
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