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car travelling on a straight road

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car travelling on a straight road

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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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.

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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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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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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.
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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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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.
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