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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?

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 miles


Car 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 distance
4V = (2 2/3)(V) + 40 miles
4/3V = 40
V = 40(3/4)
V = 30

Cheers,
Brent
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Bunuel
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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Bunuel
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:

\((1.5y-1y)\frac{m}{h} x 160min = 40miles\)

\(0.5y\frac{m}{h} = \frac{40miles}{160min}\)

\(y\frac{m}{h} = \frac{80miles}{160min} = \frac{1miles}{2min} = \frac{60miles}{2hours} = 30\frac{miles}{hour}\)
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patternpandora
Bunuel
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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Do you have more such question in relative speed?


Bunuel
patternpandora
Bunuel
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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Do you have more such question in relative speed?


Theory on Distance/Rate Problems: distance-speed-time-word-problems-made-easy-87481.html

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
All PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64
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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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kiranck007
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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kiranck007
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:
30 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
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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kiranck007
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 = 3k-2k = 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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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
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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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
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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GMATPrepNow
kiranck007
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 miles


Car 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 distance
4V = (2 2/3)(V) + 40 miles
4/3V = 40
V = 40(3/4)
V = 30

Cheers,
Brent


hi GMATPrepNow Brent :)

what is the reason that you attribute 40 miles to Y Car ? Car X still have to do more than 40 miles :) I ask this question because i see other solutions in this thread, attributing 40 miles to car X as well :?

Or maybe it doesn`t matter :grin:

thanks :)
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GMATPrepNow


hi GMATPrepNow Brent :)

what is the reason that you attribute 40 miles to Y Car ? Car X still have to do more than 40 miles :) I ask this question because i see other solutions in this thread, attributing 40 miles to car X as well :?

Or maybe it doesn`t matter :grin:

thanks :)

All that really matters is that we start with a 40-mile gap between the two cars.

However, since Car X is 40 miles west of Car Y, and since the cars are travelling east, we see that Car Y is originally 40 miles ahead of Car X.
So, when the two cars are at the same position, Car X will have travelled 40 miles further than Car Y.
To account for this (and create an equation) we need to add 40 miles to Car Y's travel distance.

Does that help?

Cheers,
Brent
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kiranck007
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

The questions uses relative speed. Both cars are going in the same direction (East) with X 40 miles behind Y. The ratio of their speeds is 3:2 so the difference between them is the speed of X relative to Y. This speed is what will decide the time X takes to overtake Y
Distance between X and Y = 40 miles
Time taken by X to overtake Y = 2 hr 40 mins = 8/3 hrs
Speed of X relative to Y = 40/(8/3) = 15 mph

3 : 2 Diff on ratio scale = 1 which is actually 15. Hence the multiplier is 15 and their speeds are 45 mph (of X) and 30 mph (of Y).

Answer (C)


Check this video to know how to recognise relative speed questions: https://youtu.be/wrYxeZ2WsEM
Check these videos to learn how to use ratios to do these calculations orally:
https://youtu.be/5ODENGG5dvc
https://youtu.be/7ASEIvxYPCM
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Decided to plugin the answer choices for this one.
Starting from the middle, say speed of car Y is 30mph, so the speed of car X will be 45mph.

D = 40 miles
S = 45mph - 30mph = 15mph (relative speed)
T = 40/15 hours = 8/3 hours

Now 8/3 hours * 60 = 160 minutes = 120 minutes (2 hours) + 40 mins
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