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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Problem related to time and distance [#permalink]
03 Jun 2013, 08:03

Bunuel wrote:

padmaranganathan wrote:

20. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day? (A) 100 (B) 120 (C) 140 (D) 150 (E) 160

Let \(t\) be the actual time and \(r\) be the actual rate.

"If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did" --> \((t+1)(r+5)-70=tr\) --> \(tr+5t+r+5-70=tr\) --> \(5t+r=65\);

"How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?" --> \((t+2)(r+10)-x=tr\) --> \(tr+10t+2r+20-x=tr\) --> \(2(5t+r)+20=x\) --> as from above \(5t+r=65\), then \(2(5t+r)+20=2*65+20=150=x\) --> so \(x=150\).

Answer: D.

OR another way:

70 miles of surplus in distance is composed of driving at 5 miles per hour faster for \(t\) hoursplusdriving for \(r+5\) miles per hour for additional 1 hour --> \(70=5t+(r+5)*1\) --> \(5t+r=65\);

With the same logic, surplus in distance generated by driving at 10 miles per hour faster for 2 hours longer will be composed of driving at 10 miles per hour faster for \(t\) hoursplusdriving for \(r+10\) miles per hour for additional 2 hour --> \(surplus=x=10t+(r+10)*2\) --> \(x=2(5t+r)+20\) --> as from above \(5t+r=65\), then \(x=2(5t+r)+20=150\).

Answer: D.

Note that the solutions proposed by dushver and dimitri92 are not correct (though correct answer was obtained). For this question we can not calculate neither \(t\) not \(r\) of the motorist.

Hope it helps.

Exactly, this is what I wanted to ask. Thanks again. Their solution means that they were traveling at speed S for t hours and only in the last hour they traveled by S+5 speed, which is wrong.

Re: If a motorist had driven 1 hour longer on a certain day and [#permalink]
10 Aug 2013, 11:15

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?

Re: If a motorist had driven 1 hour longer on a certain day and [#permalink]
10 Aug 2013, 11:32

1

This post received KUDOS

WholeLottaLove wrote:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?

(A) 100 (B) 120 (C) 140 (D) 150 (E) 160

the error is in highlited part: that should be: \(X = (r+10) * (t+2)\) (d is the actual distance)....X==>distance coveres when travelling with 10 miles more average speed and 2 hours more now you have to calculate X-d hope its clear _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: If a motorist had driven 1 hour longer on a certain day and [#permalink]
10 Aug 2013, 12:54

el1981 wrote:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100 (B) 120 (C) 140 (D) 150 (E) 160

.... used 3times s=vt and input 65=v+5t ,then got 150miles _________________

Re: If a motorist had driven 1 hour longer on a certain day and [#permalink]
13 Aug 2013, 10:33

blueseas wrote:

WholeLottaLove wrote:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

As you can see, I solved for d [(r+10) * (t+2)]then plugged in d for d+70 = (r+5) * (t+1). Can someone please explain why this is an incorrect approach?

(A) 100 (B) 120 (C) 140 (D) 150 (E) 160

the error is in highlited part: that should be: \(X = (r+10) * (t+2)\) (d is the actual distance)....X==>distance coveres when travelling with 10 miles more average speed and 2 hours more now you have to calculate X-d hope its clear

Thanks!

"How many more miles would he have covered than he actually did" doesn't refer to his actual distance but hypothetical miles. For example, if his actual distance (d) equaled 200 miles then his hypothetical distance (x) may = 50 miles which means his hypothetical total would be 200+50 = 250 miles.

We have two equations: d+70 = (r+5) * (t+1) d+70 = rt+r+5t+5 d+65 = rt+r+5t d = rt-r-5t-65

x = (r+10)*(t+2) x = rt+2r+10t+20 x - 2r - 10t - 20 = rt

Because we are looking for the difference between the hypothetical distance minus the actual distance we have to find x-d. x-d (rt+2r+10t+20) - (rt-r-5t-65) (rt+2r+10t+20) - rt+r+5t+65 3r +15t + 85

Re: If a motorist had driven 1 hour longer on a certain day and [#permalink]
19 Jan 2015, 08:13

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

If a motorist had driven 1 hour longer on a certain day and at a [#permalink]
24 Jan 2015, 14:18

I see thias as that, think is easier: 1) traveled 70 km in one additional hour, at an extra 5 m/h 2) so, in that un hour at that extra speed, advance 70 miles. s2 * t2 = d2 ------ s2 =(s1 + 5)-----s1=x ----t2=1; Therefore: (x+5) * 1 = 70 ---- x =65 3) now 65 plus 10 m/h, new speed is 75 m/h; then, in 2 additional hours, he adnvaced 2 *75 = 150 miles.

Re: If a motorist had driven 1 hour longer on a certain day and at an aver [#permalink]
24 Jan 2015, 17:58

Expert's post

Hi All,

This question can be solved by TESTing VALUES (as some of the explanations have noted). You can keep the values really small though and save some time on the calculations:

We're told that driving 1 extra hour AND at a speed that was 5 miles/hour faster (for the entire trip) would have increased TOTAL distance by 70 miles.

IF..... We originally drove for 1 hour at 60 miles/hour, then we would have traveled 60 miles.

Adding 1 extra hour and increasing speed by 5 miles/hour would give us..... 2 hours at 65 miles/hour, which gives us a total distance of 130 miles (which is 70 miles MORE than originally traveled).

So now we we've established the starting time, speed and distance, so we can answer the given question:

How many MORE miles would be traveled if the original time was increased by 2 hours AND the original speed was increased by 10 miles/hour?

1+2 = 3 hours 60 + 10 = 70 miles/hour 3 hours at 70 miles/hour = 210 miles

Since we originally traveled 60 miles, the 210 total miles is 150 miles MORE than originally traveled.

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...