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An ambulance, an hour after starting, meets with an accident which....

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An ambulance, an hour after starting, meets with an accident which....  [#permalink]

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New post 05 Feb 2018, 07:30
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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
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An ambulance, an hour after starting, meets with an accident which....  [#permalink]

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New post 05 Feb 2018, 07:57
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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


This is NOT a GMAT question..
requires intensive calculations and by changing km to miles in answer, it is made more complicated..
cannot be from any reliable source..

But if you want to do it from LOGIC point of view..
normal time be x for remaining distance, but time taken = \(\frac{4x}{3}\)..
so difference in time = \((1+\frac{1}{2}+\frac{4x}{3})-(1+x)=\frac{7}{2}....x=9\)...
now take second case of going further 90km which lessens the difference by 1/2...
if it loses 90 km in 1/2 hr..
total difference was 3 + 1/2 out of which 1/2 was for repair .. so difference in travelling is 3+1/2-1/2=3 hrs..
it will lose 3 hrs in 90*6=540 km..
..

now it travelled for 10 hr hours, which is 1 hr + 540 km...
so 540 in 9hr... so 60 km per hour..
so total distance = 540+60=600km..
change into miles.. approx 372.8( ofcourse I used a calculator for it)..

B
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Re: An ambulance, an hour after starting, meets with an accident which....  [#permalink]

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New post 05 Feb 2018, 09:04
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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/hr

In 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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Re: An ambulance, an hour after starting, meets with an accident which....  [#permalink]

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New post 09 Feb 2018, 00:56
Let the normal speed of the ambulance be 's', the time it would have taken had it not met with the accident and been able to travel the entire distance at its normal speed be 't' and the distance it travels be 'd'. Then, t=(d/s).
Now, had the accident occurred 90kms further along the route the ambulance could have traveled 90kms more at its normal speed and 0.5 hrs could thus have been saved. So we can conclude that the difference between the times for the ambulance to cover 90kms at 's' km/hr and 90 kms at (0.75)s km/hr is 0.5 hrs. So, 90/(0.75)s - 90/s = 0.5. So, 's'=60 km/h.
Now, as per the first scenario, the ambulance has the accident 1 hr after starting, there is an half hour delay and it travels at three-fourth of its normal speed the rest of the trip and thus takes 3.5 hrs more to reach its destination. So before the accident it travels 60 kms in 1 hr and for the rest of the trip (d-60)kms it travels at 45 km/hr. We can thus formulate the following equation summing up the different time segments of the trip:
1hr + 0.5hr (delay) + {(d-60 kms)/45}hrs = (t + 3.5)hrs. But t=d/s=d/60
Therefore, (d-60)/45=t+2. D=600 kms= 372.8 miles. Ans B.
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Re: An ambulance, an hour after starting, meets with an accident which.... &nbs [#permalink] 09 Feb 2018, 00:56
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