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While driving from Aville to Btown, Harriet drove at a constant spee
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19 Feb 2015, 09:55
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73% (03:01) correct 27% (02:39) wrong based on 343 sessions
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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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19 Feb 2015, 11:00
X = Time from A to B Y = Time from B to A X+Y=5 hrs total travel time 115X=135Y since the distances are the same
X+Y=5 => 135X+135Y=675 115X=135Y => 115X135Y=0 Combined you get 250X=675 hrs X=2.7 hrs * 60 mins/hr= 162



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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19 Feb 2015, 11:27
Its D. Here it goes: Distance will be same both sides... hence total distance will D+D= 2D Total time given+ 5hrs Speed per side also given.. 115km/ph and 135 km/ph Speed= Distanct/Time D/s1 + D/s2= total time (d/115) + (d/135) = 5 solving it D comes to 621 km Hence each side is D/2 : 621/2 = 310.5 km Time for first journey is 310.5/115 which is 162 mins Thanks
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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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19 Feb 2015, 12:49
Celestial09 wrote: Its D. Here it goes: Distance will be same both sides... hence total distance will D+D= 2D Total time given+ 5hrs Speed per side also given.. 115km/ph and 135 km/ph Speed= Distanct/Time D/s1 + D/s2= total time (d/115) + (d/135) = 5 solving it D comes to 621 km Hence each side is D/2 : 621/2 = 310.5 km Time for first journey is 310.5/115 which is 162 mins Thanks
Kudos please if my solution is correct! Hello, I have struggles coming from (D/115) + (D/135) = 5 to D = 621 could you explain the calculation steps ?



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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19 Feb 2015, 13:06
Hello! Thanks for addressing your doubt. here it goes: (d/115) + (d/135) = 5 kindly note.. its addition of two fractions... so I kindly take 5 common from both denominators... hence you will be left with d/23 + d/27 and that 5 will cross multiply to other side of equation.. resulting d/23 + d/27 = 25 then kindly solve it ... 27d +23 d = 25*23*27 d will come down to 310.5 which i particularly took as total distance 310.5*2 = 621 km hope it makes sense. thanks LaxAvenger wrote: Celestial09 wrote: Its D. Here it goes: Distance will be same both sides... hence total distance will D+D= 2D Total time given+ 5hrs Speed per side also given.. 115km/ph and 135 km/ph Speed= Distanct/Time D/s1 + D/s2= total time (d/115) + (d/135) = 5 solving it D comes to 621 km Hence each side is D/2 : 621/2 = 310.5 km Time for first journey is 310.5/115 which is 162 mins Thanks
Kudos please if my solution is correct! Hello, I have struggles coming from (D/115) + (D/135) = 5 to D = 621 could you explain the calculation steps ?



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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22 Feb 2015, 12:37



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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11 Mar 2015, 21:07
Answer = D = 162 Let distance between A & B = d Setting up the time equation \(\frac{d}{115} + \frac{d}{135}= 5\) \(d = \frac{23 * 27}{2}\) (Need not have to solve) Time required in minutes from A to B \(= \frac{23*27}{2*115} * 60 = 162\)
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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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23 Mar 2015, 08:48
Distance from A to B is the same as distance from B to A. Hence
115x=135y x/y=135/115 => x/y=27/23 so x= 27/50 of the total time (50=27+23). total time x+y=5h= 300minutes,
x= 300m*(27/50)= 162m
Answer. D



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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19 Jul 2015, 11:00
Hi everyone,
I don't see why we divide the distance 625 by 250 and not 115, while Harriet drives from A to B at 115mph... Does not make sense to me.
But at the same time, dividing 625 by 115 gives us a time of 5+ hours, which is also impossible because the whole trip A > B > A takes 5 hours. Can someone please give some insights?
Many thanks in advance



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While driving from Aville to Btown, Harriet drove at a constant spee
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04 Sep 2016, 13:17
5hr = 300min. If harriet spend equal hrs on each leg she will spend 150min on each. Since speed AB is less than speed BA and distance on each leg is the same, time spent on AB is more than 150min, which mean we can eliminate ans. A, B and C.
Now let plug in ans. D or E and verify which one give same distance on each leg.
D. t= 162min * leg AB > d = 115.162/60 = 18630/60 * leg BA > d = 135*138/60 = 18630/60
so the correct ans. is D



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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13 Nov 2016, 10:35
Rates and Distance: Q: Find t on the way in? When you are given the same distance on the way in and back. ex: d and you are given the rates for back and forth. ex: 115 and 135 and you are given the total time. ex: 5
step1: organize the formulas like below. Use 5t for the time on the way back = very important.
rt=d 115 (t) = d 135 (5t) = d
step2: Make the two equations equal and solve for t
115(t) = 135 (5t) t= 27/10 hours
step3: check the convention of time that the question wants you to answer and convert if needed
In this question, you are asked for minutes, so you need to do one more thing before jumping into the answers
27/10 hours * 60minutes = 162 minutes



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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13 Nov 2016, 10:49
Distance taken = speed*time Let Harriet took x hrs to go from A to B and y hrs to come from B to A. Distance=115*x=135*y x+y=5 (total time=5 hrs.) we need to find x. y=5x Substituting in Distance equation. 115*x=135(5x) Solve to get x=2.7hrs 2.7*60=162mins. Answer=D.
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While driving from Aville to Btown, Harriet drove at a constant spee
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13 Nov 2016, 12:54
Bunuel wrote: While driving from Aville to Btown, Harriet drove at a constant speed of 115 kilometers per hour. Upon arriving in Btown, Harriet immediately turned and drove back to Aville at a constant speed of 135 kilometers per hour. If the entire trip took 5 hours, how many minutes did it take Harriet to drive from Aville to Btown?
A. 138 B. 148 C. 150 D. 162 E. 168
Kudos for a correct solution. average speed for round trip≈125 kph 125*5=625 round trip miles 625/2=312.5 one way miles 312.5/115=2.72 hours 2.72*60=163.2 minutes closest to 162 D



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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22 Nov 2016, 02:53
Bunuel wrote: While driving from Aville to Btown, Harriet drove at a constant speed of 115 kilometers per hour. Upon arriving in Btown, Harriet immediately turned and drove back to Aville at a constant speed of 135 kilometers per hour. If the entire trip took 5 hours, how many minutes did it take Harriet to drive from Aville to Btown?
A. 138 B. 148 C. 150 D. 162 E. 168
Kudos for a correct solution. distance travelled is the same in each direction thus time is inversely proportional to speed , total time = 5 , let first leg time t and 2nd leg time = 5t 115/135 = 5t / t thus , t in hours = 5*27/50 , in minutes = 300*27/50 = 6*27 = 162..........D



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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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02 Feb 2018, 23:32
I solved in following way; Rate of driving from Aville to Btown is 115 kmph Rate of driving from Btown to Aville is 135 kmph Therefore, the ratio of rate is 115:135 Since the distance is same, the ratio of time would be reciprocal of ratio of rate.So, ratio of time would be \(135:115=27:23; 27x:23x\) Total time taken is 5 hours=300 minutes \(27x+23x=50x\) Given, \(50x=300\) \(x=6\) So, time taken to drive from Aville to Btown would be \(27x=27*6=162\)
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Re: While driving from Aville to Btown, Harriet drove at a constant spee
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03 Feb 2018, 07:27
Bunuel wrote: While driving from Aville to Btown, Harriet drove at a constant speed of 115 kilometers per hour. Upon arriving in Btown, Harriet immediately turned and drove back to Aville at a constant speed of 135 kilometers per hour. If the entire trip took 5 hours, how many minutes did it take Harriet to drive from Aville to Btown?
A. 138 B. 148 C. 150 D. 162 E. 168
Kudos for a correct solution. Let the time from A to B be t1 and time from B to A be t2. \(t1 + t2 = 5\) \(115*t1 = 135*t2\) > \(t2 = \frac{23}{27}t1\) \(t1 = 2.7 Hours.\)
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