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If a motorist had driven 1 hour longer on a certain day and [#permalink]
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24 Mar 2008, 13:37
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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
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Re: PS: Motorist [#permalink]
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24 Mar 2008, 14:48
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D150
assuming t is the original travel time in hrs, and x is the original speed in miles per hour
from the first half  5t+(x+5)=70 => 5t+x=65 Now need to find out 10t+2(x+10) = 10t+2x+20 = 2(5t+x)+20 = 150



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Re: PS: Motorist [#permalink]
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24 Mar 2008, 19:32
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(t+1)(s+5) = d+70
expand to get ts+5t+s+5 = d+70, and recognize that ts = d (from standard equation speed = distance/time)
simplify down to 5t+s=65
now, (t+2)(s+10) = ts+10t+2s+20 = ts+2(5t+s)+20 = d+150
so your answer is 150



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Re: PS: Motorist [#permalink]
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24 Mar 2008, 19:41
el1981 wrote: i don't understand. how did you get 5t and x+5? 70 additional miles came from 2 sources  5 miles additional distance traveled for t hours (5t) + distance traveled in the additional hour. The later one is x+5 where in x is distance traveled in an hour with the original speed and 5 is the distance came from the additional speed (making total speed x+5 mph)



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Re: PS: Motorist [#permalink]
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25 Mar 2008, 13:37
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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 Interesting problem. first we have (r+5)(t+1)=d+70 > rt+r+5t+5=d+70. Also notice that rt=d thus we now have r+5t=65. Second equation. (r+10)(t+2)=d+x rt+2r+10t+20=d+x > plug in rt and 2(r+5t)=2(65) >130 > rt+150=rt+x > x=150. D



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Re: PS: Motorist [#permalink]
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25 Mar 2008, 19:19
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 D1=100, V1=50, T1=2 T2=3, V2=55 >D2= 165 T3=5, V3=65 > D3=325 D3D2=150 D
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Re: PS: Motorist [#permalink]
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25 Mar 2008, 23:47
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Hey I have a quick approach for this.
The guy covered 70 miles more in an hour by driving 5 miles/hour faster. Therefore his speed current is 70 Miles/hr and his original speed was 65 Miles /Hr (since he is traveling 5Miles/Hr faster). Therefore his new speed is 75Miles/Hr (65+10) and he will travel 150 (75x2) Miles more in 2 Hours.
Hence D.



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Re: PS: Motorist [#permalink]
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07 Jun 2010, 13:16
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my approach was similar too than the posts above ... but i'll share it anyway
avg rate of 5mph faster means that the motorist drove 5miles more in each hour. since he drove 5miles in the last (extra) hour too, so he drove 705=65 miles more in each of the hour earlier 65/5=13hrs this is the actual time he drove
avg rate of 10mph more means that he drove 10*13=130miles more in the first 13 hrs he covered 2*10=20miles more in the last 2 hours 130+20=150 more miles covered than he actually did
so D it is
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Re: Problem related to time and distance [#permalink]
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08 Jun 2010, 02:37
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 D.150, but I dont like my approach 1. (t+1)(v+5)=d+70 2. 5t+v =65 (taken from 1) 3. (t+2)(v+10)=[t+1 +1][v+5 + 5] =[ (t+1)(v+5)+ 5(t+1)+(v+5) +5]=d+70+80=d+150
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Re: Problem related to time and distance [#permalink]
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08 Jun 2010, 03:33
Hi sondenso, My approach was very similar to yours except for a minor change in the 3rd step;
"...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?" can be rephrased as;
(t+2)(v+10) = ? tv+10t+2v+20 = ? 10t+2v+20 = 2(5t+v)+20 = 2*65 (from 2) +20 = 150.



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Re: Problem related to time and distance [#permalink]
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08 Jun 2010, 07:54
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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" > \((t+1)(r+5)70=tr\) > \(tr+5t+r+570=tr\) > \(5t+r=65\); "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?" > \((t+2)(r+10)x=tr\) > \(tr+10t+2r+20x=tr\) > \(2(5t+r)+20=x\) > 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.
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Re: motorist  work problem [#permalink]
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26 Aug 2011, 11:16
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Let original speed be s, distance be d, and time t d=st Also, d+70=(s+5)(t+1) d+70=st+s+5t+5 st+70=st+s+5t+5 s+5t=65 And, d1=(s+10)(t+2) d1=st+2s+10t+20 so, d1d=2s+10t+20 =2(s+5t)+20 =2(65)+20 =130+20=150 OA D.
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Re: motorist  work problem [#permalink]
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26 Aug 2011, 11:30
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The motorist covers 70 miles in an hour when he drives 5 miles/hr faster => He is currently driving at 65 miles/hr If he drives 10 miles/hr faster for 2 more hours he would cover 75*2 = 150 miles (D) it is.
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Re: motorist  work problem [#permalink]
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04 Oct 2011, 12:14
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Lets take x as time and y as avg speed If he has driven 1 hr longer and 5 km/hr faster the distance covered Is 70 (X+1)*(Y+5) – XY = 70 XY+Y+5+5X – XY = 70 Y+5X = 65 Scenario 2 (X+2)(Y+10)XY = D 2x+10Y+20 = D 2(X+5Y)+20 = D 2*65 +20 = D D = 150
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Re: motorist  work problem [#permalink]
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04 Oct 2011, 13:27
GyanOne wrote: The motorist covers 70 miles in an hour when he drives 5 miles/hr faster => He is currently driving at 65 miles/hr
If he drives 10 miles/hr faster for 2 more hours he would cover 75*2 = 150 miles
(D) it is. despite the correct answer, the logic here is flawed, isn't it ? Quote: The motorist covers 70 miles in an hour when he drives 5 miles/hr faster M drives less than 70miles in one hour. Because M drove more than the original distance in the time before the extra hour starts. So that delta + another hour's worth of driving (at the elevated speed) is 70miles.



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Re: motorist  work problem [#permalink]
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06 Oct 2011, 04:47
igemonster wrote: despite the correct answer, the logic here is flawed, isn't it ? Quote: The motorist covers 70 miles in an hour when he drives 5 miles/hr faster M drives less than 70miles in one hour. Because M drove more than the original distance in the time before the extra hour starts. So that delta + another hour's worth of driving (at the elevated speed) is 70miles. Yes, 70 miles is the sum of extra distance covered by increasing speed + extra distance covered in an hour. But remember, this is a PS question. You do have the correct answer in the options and there is only one correct option. I will just assume any case and whatever I get will be my answer. Say the original speed was 60 miles/hr. The speed increased to 65. So the motorist would have traveled 5 miles + 65 miles (= 70) extra. (He travels for only one hour initially.) If he instead increases speed by 10 miles/hr, his speed becomes 70. In the initial 1 hr, he will travel an extra 10 miles and then in additional 2 hrs he will travel an extra 70*2 = 140 miles. So he will travel an extra 150 miles. I could have just as well assumed the original speed to be 55 miles/hr. If the speed increases to 60 and extra distance covered is 70, it means that the motorist travels for 2 hrs initially. 70 = 5 + 5 + 60 If instead, the speed increases by 10 miles/hr, the speed becomes 65. Extra distance covered = 10+10+2*65 = 150 In any case, the answer has to be the same since it is a PS question. I could also have assumed the original speed to be 65 miles/hr (as done above) If the speed increases to 70 and I cover 70 miles extra, it means I didn't travel at all before. If the speed instead becomes 75, I travel 2*75 = 150 miles extra.
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Re: motorist  work problem [#permalink]
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15 Oct 2011, 23:31
We can make use of the below 3 equations (s+5)(t+1) = 70+d (s+10)(t+2) = x and st=d
On solving we will get x=150 (D)



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Re: motorist  work problem [#permalink]
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16 Oct 2011, 09:58
Solution : sspeed. ttime. ddistance.
Two equations will be formed. As per the conditions given we`ll get,
(s+5)(t+1)  st = 70.....(1) on simplifying this,we get s + 5t = 65....(a)
Another one will be, (s+10)(t+2)  st = d on simplifying this we`ll get,
2s + 10t +20 = d...(b) substituting (a) in (b)
we get 65*2 + 20 = 150.



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Re: motorist  work problem [#permalink]
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18 Oct 2011, 20:11
Let original speed be x
Extra distance travelled in 1 hr: 70 = (x + 5)*1 or x+5 = 70  (1)
Now, question asks us to find: d = (x+10)*2 d = (x+5+5)*2
From (1), d = (70+5)*2 = 150. Hence (D).



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Re: motorist  work problem [#permalink]
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10 Mar 2012, 11:55
tm=(1+t) dm=70+d rm=(r+5)
Being r,t and d the actual rate, time and distance.
If you realize that the motorist has actually covered a distance of 70 miles in 1 hour, going at (r+5)speed, it is much easier...
70= (r+5)*1 then r=65mph
If the new rate is 10 mph more then it is r+10=75mph
Therefore in two hours the distance covered at this speed is 150miles




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