car travelling on a straight road

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vittalgmat: Legendary Member; Posts: 621; Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members; Thanked: 33 times

by vittalgmat » Mon May 04, 2009 11:33 pm

While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y ?

(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour.

(2) 3 minutes ago car X was 1/2 mile ahead of car Y.

[spoiler]This is an old gmat paper Q. The OA is D, but I have trouble justifying how stmt 2 is sufficient. If the car was 1/2 mile ahead of Y 3 mins ago, how can I find out how many mins later it will be 2 miles ahead.

IMO, stmt 2 tells us the instantaneous position of the car X at a point 3 mins ago. Only if I know the constant speed, can I answer the Q.
right ??[/spoiler]

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Source: Beat The GMAT — Data Sufficiency | Mon May 04, 2009 11:33 pm

bluementor: Master | Next Rank: 500 Posts; Posts: 418; Joined: Wed Jun 11, 2008 5:29 am; Thanked: 65 times

Re: car travelling on a straight road

by bluementor » Tue May 05, 2009 12:15 am

vittalgmat wrote:While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y ?

(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour.

(2) 3 minutes ago car X was 1/2 mile ahead of car Y.

[spoiler]This is an old gmat paper Q. The OA is D, but I have trouble justifying how stmt 2 is sufficient. If the car was 1/2 mile ahead of Y 3 mins ago, how can I find out how many mins later it will be 2 miles ahead.

IMO, stmt 2 tells us the instantaneous position of the car X at a point 3 mins ago. Only if I know the constant speed, can I answer the Q.
right ??[/spoiler]

No, you don't really need to know the actual speeds (if they are known to be constant). Think about it like this:

3 minutes ago, X is ahead of Y by 0.5 miles.
Now, X is ahead of Y by 1 mile.

So, in 3 minutes, X has managed to move away further from Y by 0.5 miles. This is the important fact in statement 2.

Given this info, we can calculate that in the next 3 minutes, X will gain a 0.5 mile lead (so it will be 1.5 miles ahead of Y) and 3 minutes after that, X will gain another 0.5 mile lead (so it will be 2 miles ahead of Y).

So X (and Y) need to travel for a further 6 minutes before X is ahead of Y by 2 miles.

hope this helps.

-BM-

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vittalgmat: Legendary Member; Posts: 621; Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members; Thanked: 33 times

by vittalgmat » Tue May 05, 2009 10:35 am

Thanks BM. I had discounted/forgotten to consider that NOW car X was ahead of Y by 1 mile.. So in 3 mins, the car has travelled 0.5 miles, so sufficient.

thanks

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clawhammer: Master | Next Rank: 500 Posts; Posts: 132; Joined: Sun Aug 15, 2010 10:08 amFollowed by: 1 member; Thanked: 3 times

by clawhammer » Sat Nov 20, 2010 8:36 am

Can anyone please describe the steps of solving this problem with information provided in (A) - as if it were a PS question.

I tried: X travels @ 50mph, Y travels @ 40mph
let x be the distance traveled by Y when X is 2 miles ahead. so X will travel x+2 miles.
so according to my logic, the time in which X is 2 miles ahead of Y = x/40 = (x+2)/50

but how to calculate the extra time needed by x to be 2 miles ahead? (i can feel im getting something wrong as X is already 1 mile ahead)

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sat Nov 20, 2010 8:48 am

clawhammer wrote:Can anyone please describe the steps of solving this problem with information provided in (A) - as if it were a PS question.

I tried: X travels @ 50mph, Y travels @ 40mph
let x be the distance traveled by Y when X is 2 miles ahead. so X will travel x+2 miles.
so according to my logic, the time in which X is 2 miles ahead of Y = x/40 = (x+2)/50

but how to calculate the extra time needed by x to be 2 miles ahead? (i can feel im getting something wrong as X is already 1 mile ahead)

Don't make unnecessary complications when solving a problem.
You're looking for the time after which X will be 2 mile ahead of Y.
As X is already 1 mile ahead of Y, look for the time after which X will be another 1 mile ahead of Y.

As they are going in the same direction, their relative speed = (50 - 40) mph = 10 mph. This means X travels 10 mile more than Y in 1 hour (= 60 minutes).

Therefore X will be 1 mile ahead of Y in 6 minutes.

Rahul Lakhani
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clawhammer: Master | Next Rank: 500 Posts; Posts: 132; Joined: Sun Aug 15, 2010 10:08 amFollowed by: 1 member; Thanked: 3 times

by clawhammer » Sat Nov 20, 2010 9:10 am

Thanks Rahul.

I was just trying to understand if it were possible to make an algebraic equation out of this. usually when in doubt, I find it easier to translate the problem to an equation. Since the general logic is to come up with a time required for X or Y and then to make an equation based on the logic that time required for both cars will be same.

I now tried, x/40 = (x+1)/50 and solve for x, but forgot x is the distance not time, so there's one additional step. time required is x+1/50 or 5/50 or 1/10 hr = 6 mins.

Seems your approach is much better. Is there something similar if the cars are traveling at opposite directions?

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sat Nov 20, 2010 9:25 am

clawhammer wrote:Usually when in doubt, I find it easier to translate the problem to an equation.

That's a nice methodical approach.
But do not blindly follow the question and form equation. Try to understand the question properly. This saves a lot time! Blind following may result in an equation like (x - y + 10)/2 = (y + 5)... Which is an weird way of saying x = 3y!

clawhammer wrote:Is there something similar if the cars are traveling at opposite directions?

If they are traveling in opposite direction, their relative speed will be the sum of their individual speeds.

Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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