GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Nov 2019, 16:47 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # An ambulance, an hour after starting, meets with an accident which....

Author Message
TAGS:

### Hide Tags

Intern  B
Joined: 05 Dec 2017
Posts: 20
GPA: 3.35
An ambulance, an hour after starting, meets with an accident which....  [#permalink]

### Show Tags

1
6 00:00

Difficulty:   95% (hard)

Question Stats: 23% (03:12) correct 77% (03:07) wrong based on 43 sessions

### HideShow timer Statistics

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
Math Expert V
Joined: 02 Aug 2009
Posts: 8150
An ambulance, an hour after starting, meets with an accident which....  [#permalink]

### Show Tags

1
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
_________________
Manager  B
Joined: 01 Jan 2016
Posts: 53
GPA: 3.75
WE: Engineering (Energy and Utilities)
Re: An ambulance, an hour after starting, meets with an accident which....  [#permalink]

### Show Tags

3
1
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
_________________

If you like it, give kudos. Thanks The price of success is hard work, dedication to the job at hand, and the determination that whether we win or lose, we have applied the best of ourselves to the task at hand.
~Vince Lombardi
Manager  S
Joined: 09 Nov 2015
Posts: 144
Re: An ambulance, an hour after starting, meets with an accident which....  [#permalink]

### Show Tags

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.
Intern  Joined: 19 May 2019
Posts: 3
An ambulance, an hour after starting, meets with an accident which....  [#permalink]

### Show Tags

A man goes to the fair with his and dog. Unfortunately man misses his son which he realizes 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to son and comes back to the man to show him the direction of his son. He keeps moving to and fro at 60 m/min between son and father, till the man meets the son. What is the distance traveled by the dog in the direction of the son? ans 1000 m

Posted from my mobile device An ambulance, an hour after starting, meets with an accident which....   [#permalink] 08 Aug 2019, 08:13
Display posts from previous: Sort by

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