GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Feb 2019, 10:01

### 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

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Valentine's day SALE is on! 25% off.

February 18, 2019

February 18, 2019

10:00 PM PST

11:00 PM PST

We don’t care what your relationship status this year - we love you just the way you are. AND we want you to crush the GMAT!

# Aaron travels from town x to town y and then back from town y to town

Author Message
TAGS:

### Hide Tags

Manager
Joined: 15 Feb 2018
Posts: 243
Aaron travels from town x to town y and then back from town y to town  [#permalink]

### Show Tags

10 Oct 2018, 12:35
1
00:00

Difficulty:

45% (medium)

Question Stats:

69% (01:43) correct 31% (02:37) wrong based on 79 sessions

### HideShow timer Statistics

Aaron travels from town X to town Y and then back from town Y to town X, taking different routes in each direction. If his speed when travelling from town X to town Y is 36 miles per hour, and his speed when travelling in the opposite direction is 48 miles per hour, what is his average speed for the entire journey?

1) The length of the return trip is 24% of the entire distance traveled.

2) The length of the return trip is 100 miles.

*I noted difficulty as 600-700. Maybe it is 700. Please let me know what you think the difficulty should be.
Director
Joined: 19 Oct 2013
Posts: 508
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: Aaron travels from town x to town y and then back from town y to town  [#permalink]

### Show Tags

10 Oct 2018, 14:11
1
philipssonicare wrote:
Aaron travels from town X to town Y and then back from town Y to town X, taking different routes in each direction. If his speed when travelling from town X to town Y is 36 miles per hour, and his speed when travelling in the opposite direction is 48 miles per hour, what is his average speed for the entire journey?

1) The length of the return trip is 24% of the entire distance traveled.

2) The length of the return trip is 100 miles.

*I noted difficulty as 600-700. Maybe it is 700. Please let me know what you think the difficulty should be.

D1 = T1 * V1
T1 = D1/V1

Avg speed = (Total distance)/(Total time)

Statement one tells us that D1 = 0.24Dtotal

D1 + D2 = Dtotal

0.24Dtotal + D2 = Dtotal

D2 = 0.76Dtotal.

T1 = 0.24Dtotal/36

T2 = 0.76Dtotal/48

Avg speed = Dtotal/ (T1+T2)

1 equation 1 variable we can find the answer

A is sufficient

Statement 2

gives us D2 but that is not enough to find anything.
Insufficient

Posted from my mobile device
Manager
Joined: 15 Feb 2018
Posts: 243
Re: Aaron travels from town x to town y and then back from town y to town  [#permalink]

### Show Tags

11 Oct 2018, 15:40
3
Nice solution Salsanousi.
I've elaborated below so that others can see "1 equation 1 variable we can find the answer" (as I certainly didn't when I started).

Average speed = $$\frac{Total Distance}{Total Time}$$
D1T1V1 and D2T2V2 6 variables
D1T136mph and D2T248mph 4 unknown variables (speed taken from the original condition)

1)0.76D, T1, 36mph and 0.24D, T2, 48mph
T=$$\frac{D}{V}$$

T1=$$\frac{0.76D}{36}$$

T2=$$\frac{0.24D}{48}$$

Now we have the respective Ts. Want to find the respective Ds
D1=36·$$\frac{0.76D}{36}$$=36·0.76D

D2=48·$$\frac{0.24D}{48}$$=48·0.24D

Average speed = $$\frac{Total Distance}{Total Time}$$

Average speed = $$\frac{36·0.76D+48·0.24D}{\frac{0.76D}{36}+\frac{0.24D}{48}}$$

Multiply through by 100 to get rid of the decimals

=$$\frac{3600·76D+4800·24D}{\frac{76D}{3600}+\frac{24D}{4800}}$$

Simplify the denominator

=$$\frac{\frac{D(3600·76+4800·24)}{1}}{\frac{47D}{1800}}$$

=$$\frac{1800D(3600·76+4800·24)}{47D}$$

Cancel D from the numerator and denominator

=$$\frac{1800(3600·76+4800·24)}{47}$$

Shows that average speed is a number, we just don't need to work out what that number is.
Sufficient

2) D1T136mph and D2100miles48mph 3 unknown variables
We can't solve for 3 unknown variables with one equation

Anyone that reads this answer, kudos please! It took me over half an hour with all of the formatting!
Director
Joined: 19 Oct 2013
Posts: 508
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Aaron travels from town x to town y and then back from town y to town  [#permalink]

### Show Tags

Updated on: 11 Oct 2018, 16:00
philipssonicare

By the way in your elaborated solution you added some extra steps that can be avoided.

For avg speed it will be = Dtotal/(0.24Dtotal/36 + 0.76Dtotal/48)

the only variable in here is Dtotal which cancels out.

What you did was to multiply V1 * T1 and also V2 * T2 to get Dtotal. This extra step can be avoided. I hope it makes sense and is clear

I apologize for my formatting as I solved the question using the mobile, I will edit it later from the computer.

Posted from my mobile device

Originally posted by Salsanousi on 11 Oct 2018, 15:54.
Last edited by Salsanousi on 11 Oct 2018, 16:00, edited 1 time in total.
Manager
Joined: 15 Feb 2018
Posts: 243
Re: Aaron travels from town x to town y and then back from town y to town  [#permalink]

### Show Tags

11 Oct 2018, 15:58
I was simply going out of my way to try to make it as clear as possible
Thanks once again for your input
Re: Aaron travels from town x to town y and then back from town y to town   [#permalink] 11 Oct 2018, 15:58
Display posts from previous: Sort by