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If a motorist had driven 1 hour longer on a certain day and

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Re: If a motorist had driven 1 hour longer on a certain day and [#permalink] New post 01 Jul 2012, 03:50
1)
(V+5)(T+1)-VT=70
(V+10)(T+2)-VT=X
============
2)
5T+V=65
10T+2V=X-20
2*65=X-20 =>X=150
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Re: Problem related to time and distance [#permalink] New post 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 hours plus driving 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 hours plus driving 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.
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Re: If a motorist had driven 1 hour longer on a certain day and [#permalink] New post 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?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

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

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

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
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Re: If a motorist had driven 1 hour longer on a certain day and [#permalink] New post 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
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Re: If a motorist had driven 1 hour longer on a certain day and [#permalink] New post 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?

d=r*t
d+70 = (r+5) * (t+1)

d = (r+10) * (t+2)

(r+10) * (t+2) + 70 = (r+5) * (t+1)
rt+2r+10t+20 +70 = rt + r + 5t + 5
r + 10t + 90 = 5t + 5

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

But from here...I am stuck!
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Re: If a motorist had driven 1 hour longer on a certain day and [#permalink] New post 16 Oct 2013, 18:08
orignal speed = s ;

if you increase s by 5 m/hr, distance travelled is 70 miles in that extra 1 hour, therefore :

==> (s+5)*1=70 ; s=65

Now, in 2 more hours with an increased 10m/hr over original speed, distance traveled :

d=(65+10)*2 ; d=150. Ans (D)
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Re: If a motorist had driven 1 hour longer on a certain day and   [#permalink] 16 Oct 2013, 18:08
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