Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 16 Jul 2019, 13:50

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Car X is 40 miles west of Car Y. Both cars are traveling east, and Car

Author Message
TAGS:

### Hide Tags

Intern
Joined: 14 Jan 2013
Posts: 22
Location: India
Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

Updated on: 01 Jul 2019, 01:45
17
00:00

Difficulty:

15% (low)

Question Stats:

54% (02:28) correct 46% (02:27) wrong based on 104 sessions

### HideShow timer Statistics

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

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.
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

12 Jun 2013, 09:41
3
4
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?

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

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 1.5y-y=0.5y=y/2 miles per hour.

Time*Rate=Distance --> (8/3)*(y/2)=40 --> y=30 miles per hour.
_________________
Senior Manager
Joined: 13 May 2013
Posts: 414
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

15 Aug 2013, 11:07
5
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
##### General Discussion
Intern
Joined: 04 May 2013
Posts: 8
Schools: Yale '17
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

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.

Intern
Status: Student
Joined: 06 Oct 2013
Posts: 26
Location: Germany
Concentration: Operations, General Management
GMAT 1: 670 Q49 V35
GPA: 2.4
WE: Other (Consulting)
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

19 Jan 2014, 08:28
3
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.

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:

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

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

$$y\frac{m}{h} = \frac{80miles}{160min} = \frac{1miles}{2min} = \frac{60miles}{2hours} = 30\frac{miles}{hour}$$
_________________
Thank You = 1 Kudos
B.Sc., International Production Engineering and Management
M.Sc. mult., European Master in Management Candidate

_______________________________________________________

#1 Official GMAT Prep 1: 530 (Q41 V21), 10/10/13
#2 Manhattan GMAT CAT 1: 600 (Q43 V30), 12/17/13
#3 Manhattan GMAT CAT 2: 640 (Q43 V34), 01/13/14
#4 Manhattan GMAT CAT 3: 660 (Q45 V35), 01/16/14
#5 Manhattan GMAT CAT 4: 650 (Q45 V34), 01/18/14
#6 Manhattan GMAT CAT 5: 660 (Q42 V38), 01/21/14
#7 Official GMAT Prep 2: 640 (Q48 V30), 01/26/14
GMAT 670 Q49 V34 AWA5 IR6 - TOEFL ibt 110
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

19 Jan 2014, 10:08
2
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.

2 hours and 40 minutes = $$2\frac{2}{3}$$ hours= $$\frac{8}{3}$$ hours.
_________________
Intern
Status: Yes, I can and I will. 700+FTW
Joined: 30 Sep 2013
Posts: 20
Location: India
Concentration: Finance, Strategy
GMAT Date: 02-05-2014
GPA: 3.7
WE: Other (Retail)
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

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.

2 hours and 40 minutes = $$2\frac{2}{3}$$ hours= $$\frac{8}{3}$$ hours.

_________________
It takes time before all the things work together.
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

25 Jan 2014, 03:53
ROckHIsT wrote:
Do you have more such question in relative speed?

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

_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1787
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

30 Jul 2014, 20:28
1
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
Attachments

catc.png [ 5.8 KiB | Viewed 7920 times ]

_________________
Kindly press "+1 Kudos" to appreciate
CEO
Joined: 12 Sep 2015
Posts: 3847
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

29 Mar 2018, 07:44
Top Contributor
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 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
_________________
Test confidently with gmatprepnow.com
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6923
Location: United States (CA)
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

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
_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager
Joined: 04 Oct 2018
Posts: 167
Location: Viet Nam
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

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?

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
_________________
"It Always Seems Impossible Until It Is Done"
Intern
Joined: 14 Jan 2018
Posts: 18
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car  [#permalink]

### Show Tags

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 = 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.
_________________
If you have come so far, then GMAT is the only way out.
Re: Car X is 40 miles west of Car Y. Both cars are traveling east, and Car   [#permalink] 01 Jul 2019, 02:39
Display posts from previous: Sort by