BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT Prep - Maria's 40-mile trip

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Sun Dec 07, 2008 4:14 am

GMAT Prep - Maria's 40-mile trip

by jdawg » Mon Dec 08, 2008 11:27 pm
Looking for a "quick" method for this one:

During a 40-mile trip, Maria traveled at an average speed of x miles per hour for the first y miles of the trip and at an average speed of 1.25x miles per hour for the last 40-y miles of the trip. The time that Maria took to travel the 40 miles was what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip?

(1) x = 48

(2) y = 20

B

Thanks!
Top
Source: — Data Sufficiency |
Senior | Next Rank: 100 Posts
Posts: 93
Joined: Tue Sep 02, 2008 5:34 pm
Thanked: 7 times

Re: GMAT Prep - Maria's 40-mile trip

by hwiya320 » Tue Dec 09, 2008 7:06 am
jdawg wrote:Looking for a "quick" method for this one:

During a 40-mile trip, Maria traveled at an average speed of x miles per hour for the first y miles of the trip and at an average speed of 1.25x miles per hour for the last 40-y miles of the trip. The time that Maria took to travel the 40 miles was what percent of the time it would have taken her if she had traveled at an average speed of x miles per hour for the entire trip?

(1) x = 48

(2) y = 20

B
Thanks!
Ok, I had to stink about this one for a second.

You need to realize what you need to solve this equation and what the question is asking.

In simple word, question is asking, can you find out how long it took to travel with 48mph and 60mph vs. driving 48mph?

Let's look at it one by one,

1) X = 48. You have 48mph for this "Y" distance, and x*1.25, or 60mph for this 40 miles - "Y" miles.

Without knowing the distance of one, there is no way to tell the TIME it took to travel 40 miles.

I.E. if she travels 39 miles at 48mph and 1 mile at 60mph, that will be much longer than
traveling 1 mile at 48mph, and rest of 39 miles at 60mph right?

OK, so first statement is insufficient.

2) you have y = 20. So you know that half the time, she drove 48 mph, other half at 60mph. since they're proportionate, you can take the average of the two to get the speed.
(48+60)/2 = 54mph ONLY because the distances were identical.

Now you have 54mph/40 miles = time vs. 48mph/40miles = time, calculate the percentage.

so we have the time vs. time!

Remember, DS is not to solve the equation, but to realize what's necessary to calculate the problem.
Top
Master | Next Rank: 500 Posts
Posts: 279
Joined: Wed Sep 24, 2008 8:26 am
Location: Portland, OR
Thanked: 6 times

by pbanavara » Tue Dec 09, 2008 6:23 pm
Pretty quick method: Just know what is being asked for - Time. So quickly use the s=r*t formula , so t=s/r

so y/x + (40-y)/1.25x = 40/x

So you are asked what % is 40/x of y/x +(40-y)/1.25x. Let's call this z

So z = 40/x*1.25x/(40+.25y) = 40*1.25/40+.25y

So you just need to know y. Hence B is suff

- pradeep
Top
Senior | Next Rank: 100 Posts
Posts: 47
Joined: Wed Oct 08, 2008 11:29 am
Location: Toronto, Ontario
Thanked: 5 times

by anayeri » Tue Dec 09, 2008 6:27 pm
I used to arrogantly go about these questions thinking that I was good at word problems, but then I took 5 seconds to actually think and learn the rt=d equation, and it makes these questions 10 times easier.

Question asks what % of 40miles/X(miles/hr) is (Ymiles/Xmiles/hr)+(40-y miles)/1.25X(m/hr)?

So [(y/x) + (40-y)]/1.25x / (40/x).

This simplifies to: (0.25y+40)/(1.25*40), so the only unknown is y.

St 1) gives us x, but we don't care for x, cause all x's have been canceled out as we simplified the equation. INSUFF

St. 2) gives us y, which is exactly what we wanted/needed. SUFF.
Top
Master | Next Rank: 500 Posts
Posts: 266
Joined: Mon Oct 19, 2009 9:46 pm
Thanked: 8 times
GMAT Score:690

by pkw209 » Fri May 07, 2010 11:11 am
Great method. Thanks!
Top
Post new topic Post Reply
• Page 1 of 1