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time speed dist

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time speed dist

by mitaliisrani » Mon Aug 16, 2010 9:41 pm
X and Y run between point A and point B which are 6 km apart. X starts at 10 a.m. from A, reaches B, and returns to A. Y starts at 10:30 a.m. from B, reaches A, and comes back to B. Their speeds are constant with Y's speed being twice that of X. While returning to their starting points they meet at a point which is exactly midway between A and B. When do they meet for the first time?

a) 10.33 1/2 am
b) 10.37 1/2 am
c) 10.33 am
d)10.33 2/3 am

OA A

Plz give approach
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by 4GMAT_Mumbai » Mon Aug 16, 2010 10:02 pm
Hi,

Let us see the approach ...

What is the distance ran by X before meeting Y at the mid point ? 6 + 3 = 9 km

What is the distance ran by Y before meeting X at the mid point ? 6 + 3 = 9 km

Time taken by X for running 9km = Time taken by Y for running 9km + 0.5 hours

If a km / h is the speed of X, 2a km / h is the speed of Y

So, (9 / a) = (9 / 2a) + 0.5

Solving this, one should be able to get 'a' to be 9 km / h

Coming to the 2nd part of the question,

By 10:30, X would have run 4.5 kms. The first meeting thus happens a little after 10:30 in the 1.5 km stretch closer to B.

Distance = 1.5 kms; Relative speed = 27 km / hr (as they are running in opposite directions) (9 km / h + 18 km / h)

Time = 1.5 / 27 = 3 / 54 hours

Converting into time,

Time = 3 * 60 / 54 = 10 / 3 minutes = 3 minutes and 20 seconds.

Mmm ... Not getting any of the answer choices ... Wondering what I am missing ... Help please !
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by mitaliisrani » Mon Aug 16, 2010 10:18 pm
4GMAT_Mumbai wrote:Hi,

Let us see the approach ...

What is the distance ran by X before meeting Y at the mid point ? 6 + 3 = 9 km

What is the distance ran by Y before meeting X at the mid point ? 6 + 3 = 9 km

Time taken by X for running 9km = Time taken by Y for running 9km + 0.5 hours

If a km / h is the speed of X, 2a km / h is the speed of Y

So, (9 / a) = (9 / 2a) + 0.5

Solving this, one should be able to get 'a' to be 9 km / h

Coming to the 2nd part of the question,

By 10:30, X would have run 4.5 kms. The first meeting thus happens a little after 10:30 in the 1.5 km stretch closer to B.

Distance = 1.5 kms; Relative speed = 27 km / hr (as they are running in opposite directions) (9 km / h + 18 km / h)

Time = 1.5 / 27 = 3 / 54 hours

Converting into time,

Time = 3 * 60 / 54 = 10 / 3 minutes = 3 minutes and 20 seconds.

Mmm ... Not getting any of the answer choices ... Wondering what I am missing ... Help please !
Hey dude your approach is absolutely correct..as you mentioned they meet after 10.30 ...and it takes them 3 minutes 20 seconds to meet..thus,,,..the y meet at 10.33 1/3am...which is option a
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by mitaliisrani » Mon Aug 16, 2010 10:35 pm
mitaliisrani wrote:
4GMAT_Mumbai wrote:Hi,

Let us see the approach ...

What is the distance ran by X before meeting Y at the mid point ? 6 + 3 = 9 km

What is the distance ran by Y before meeting X at the mid point ? 6 + 3 = 9 km

Time taken by X for running 9km = Time taken by Y for running 9km + 0.5 hours

If a km / h is the speed of X, 2a km / h is the speed of Y

So, (9 / a) = (9 / 2a) + 0.5

Solving this, one should be able to get 'a' to be 9 km / h

Coming to the 2nd part of the question,

By 10:30, X would have run 4.5 kms. The first meeting thus happens a little after 10:30 in the 1.5 km stretch closer to B.

Distance = 1.5 kms; Relative speed = 27 km / hr (as they are running in opposite directions) (9 km / h + 18 km / h)

Time = 1.5 / 27 = 3 / 54 hours

Converting into time,

Time = 3 * 60 / 54 = 10 / 3 minutes = 3 minutes and 20 seconds.

Mmm ... Not getting any of the answer choices ... Wondering what I am missing ... Help please !
Hey dude your approach is absolutely correct..as you mentioned they meet after 10.30 ...and it takes them 3 minutes 20 seconds to meet..thus,,,..the y meet at 10.33 1/3am...which is option a

Sorry an error in the first part 10/3 minutes is 3 1/3 (1/3rd is 20 seconds since 60 seconds is one minute)..you were correct all through out:)
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