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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

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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.

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