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Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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Updated on: 01 Jul 2019, 01:45
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Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? A. 20 miles per hour B. 25 miles per hour C. 30 miles per hour D. 35 miles per hour E. 40 miles per hour Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time.
0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr.
However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason.
Thanks, Kiran
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Originally posted by kiranck007 on 12 Jun 2013, 09:24.
Last edited by Bunuel on 01 Jul 2019, 01:45, edited 3 times in total.
Renamed the topic, edited the question and the tags.




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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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12 Jun 2013, 09:41
kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? Answer: Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time.
0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr.
However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason.
Thanks, Kiran Say the rate of Car Y is y miles per hour and the rate of Car X is 1.5y miles per hour.. Since they are traveling in the same direction their relative speed is 1.5yy=0.5y=y/2 miles per hour. Time*Rate=Distance > (8/3)*(y/2)=40 > y=30 miles per hour.
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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15 Aug 2013, 11:07
Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going?
Basically, this problem is no different than if Y were stationary and it took x 2.66 hours to catch up to it. However, we need to consider that they are moving and that car X is traveling at a speed 1.5times y.
Speed of y = y Speed of x = 1.5y distance = 40
The relative speed of x compared to y is 1.5y  y = 0.5y
time = distance/speed 8/3 = 40/0.5y 4y=120 y=30
We use the relative speed in this formula (as opposed to the speed of x or y) because it's in the context of the time and distance x must cover to reach y. Thus, we use the relative speed of x compared to y. Once we obtain y, we can plug it into the actual speeds of x and y (1.5y and y respectively)
speed of y = 30




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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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19 Jan 2014, 08:08
Bunuel wrote: Time*Rate=Distance > 8/3*y/2=40 > y=30 miles per hour. How did we arrive to 8/3 hours.... as its given 2.40 hours which is 160 mins.... what have I mistaken out here, can someone point it out to me. Thanks in advance!



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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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19 Jan 2014, 08:28
patternpandora wrote: Bunuel wrote: Time*Rate=Distance > 8/3*y/2=40 > y=30 miles per hour. How did we arrive to 8/3 hours.... as its given 2.40 hours which is 160 mins.... what have I mistaken out here, can someone point it out to me. Thanks in advance! You have to convert 160 minutes into hours because it is asked for mph! \(\frac{160}{60} = \frac{16}{6} = \frac{8}{3}\) You could also do that in the end: \((1.5y1y)\frac{m}{h} x 160min = 40miles\) \(0.5y\frac{m}{h} = \frac{40miles}{160min}\) \(y\frac{m}{h} = \frac{80miles}{160min} = \frac{1miles}{2min} = \frac{60miles}{2hours} = 30\frac{miles}{hour}\)
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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19 Jan 2014, 10:08
patternpandora wrote: Bunuel wrote: Time*Rate=Distance > 8/3*y/2=40 > y=30 miles per hour. How did we arrive to 8/3 hours.... as its given 2.40 hours which is 160 mins.... what have I mistaken out here, can someone point it out to me. Thanks in advance! 2 hours and 40 minutes = \(2\frac{2}{3}\) hours= \(\frac{8}{3}\) hours.
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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25 Jan 2014, 03:27
Do you have more such question in relative speed? Bunuel wrote: patternpandora wrote: Bunuel wrote: Time*Rate=Distance > 8/3*y/2=40 > y=30 miles per hour. How did we arrive to 8/3 hours.... as its given 2.40 hours which is 160 mins.... what have I mistaken out here, can someone point it out to me. Thanks in advance! 2 hours and 40 minutes = \(2\frac{2}{3}\) hours= \(\frac{8}{3}\) hours.
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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25 Jan 2014, 03:53
ROckHIsT wrote: Do you have more such question in relative speed?
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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30 Jul 2014, 20:28
Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time. 0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr. However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason. Thanks, Kiran[/size[/i][/color]] Refer diagram below \(160 Mins = \frac{160}{60} Hrs = \frac{8}{3}\) Time required by X to reach catch point = Time required by Y to reach catch point \(= \frac{8}{3}\) \(\frac{40+d}{\frac{150s}{100}} = \frac{d}{s}\) 3d = 2d + 80 d = 80 \(s = \frac{80}{\frac{8}{3}}\) s = 30
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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29 Mar 2018, 07:44
kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going?
Let's let Car X's original position be the initial starting point. So, when Car X is at the initial starting point, Car Y has already traveled 40 miles. My word equation involves the conditions when Car X catches up to Car Y. At that point, we can say: Car X's TOTAL distance traveled = Car Y's TOTAL distance traveled Car Y's total distance Let V = Car Y's speed (our goal is to find the value of V) From the time that Car X begins moving, Car Y drives for 2 2/3 hours (2 hours, 40 minutes). So Car Y's total distance = (time)(speed) = (2 2/3)(V) + 40 milesCar X's total distance We know that Car X is going 50% faster than Car Y. If Car Y's rate is V, then Car X's rate must be 1.5V We also know that Car X travels for 2 2/3 hours. So Car X's total distance = (time)(speed) = (2 2/3)(1.5V) miles Simplify: (2 2/3)(1.5V) = (8/3)(3/2) = 4V We're now ready to write our algebraic equation. Car X's total distance = Car Y's total distance4V = (2 2/3)(V) + 40 miles4/3V = 40 V = 40(3/4) V = 30
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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30 Mar 2018, 11:27
kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? We can let the rate of Car Y = r and the rate of Car X = 1.5r. Recall that 2 hours and 40 minutes = 2 2/3 hours = 8/3 hours. We can create the equation: 1.5r(8/3) = r(8/3) + 40 12r/3 = 8r/3 + 40 4r/3 = 40 4r = 120 r = 30
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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23 Dec 2018, 01:18
kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? Answer: Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time.
0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr.
However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason.
Thanks, Kiran Speed of Car Y = x => Speed of Car X = 1.5x (miles/h) Distance = 40 miles Catch up time = Distance between X&Y/ Different speed between X&Y => 40/(1.5x  x) = 2h40mins = 8/3 hours (40mins = 2/3 hour) => x = 30 miles/hr
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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01 Jul 2019, 02:39
kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? A. 20 miles per hour B. 25 miles per hour C. 30 miles per hour D. 35 miles per hour E. 40 miles per hour Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time.
0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr.
However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason.
Thanks, Kiran Using ratio approach to solve this problem It's given that speed of X is 50% more than speed of Y> Speed of X =3/2 * Speed of Y We can take speed of X=3k and Speed of Y=2K (I'm using K because I don't know the actual quantity) X will cover 40 miles in 2 hours 40 minutes which means that the difference between X and Y of 40 miles is covered by X in 2 hours 40 minutes Difference of X and Y covered by X (speed)= (40)/(8/3)= (40*3)/8 = 15 miles every hour> Equation 1 Difference of X and Y = 3k2k = K > Equation 2 Equating both of them we get k=15 miles per hour. Speed of Y= 2K = 2*15 mph= 30 miles per hour Speed of X= 3*15 mph= 45 miles per hour.
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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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28 Sep 2019, 04:37
speed of y ; a speed of x ; 1.5a given time ; 2 hrs 40 mins ; 160 mins ; 160/60 ; 8/3 hrs from given info we can say 1.5a*8/3 = 40+ 8*a/3 solve for a = 30 mph IMO C kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? A. 20 miles per hour B. 25 miles per hour C. 30 miles per hour D. 35 miles per hour E. 40 miles per hour Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time.
0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr.
However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason.
Thanks, Kiran



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Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car
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30 Oct 2019, 09:17
speed of y = s speed of x= 1.5s relative speed = .5s distance = 40 miles time ; 160 mins ; 160/60 ; 8/3 hrs 40=8/3 * .5s s= 30 mph IMO C kiranck007 wrote: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going? A. 20 miles per hour B. 25 miles per hour C. 30 miles per hour D. 35 miles per hour E. 40 miles per hour Let speed of Y = V , speed of X = 1.5 V, relative speed of X wrt Y (catch up speed of X) = 0.5V. Now catch up speed = (distance between X and Y)/time.
0.5 V = 40 miles / (8/3 hour), so V = 30 miles/hr.
However, what I don't understand is that the distance traveled by Car Y while Car X is catching up is not factored into the problem. Could someone clarify if my concern is valid, if not then reason.
Thanks, Kiran




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