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Distance Time problem

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Intern
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Distance Time problem [#permalink]

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New post 22 Feb 2009, 06:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Timothy leaves home for school, riding his bicycle at a rate of 9mph. 15mins after he leaves, his mother sees Timothy's math homework lying on his bed and immediately leaves home to bring it to him. If his mother drives at 36mph, how far (in terms of miles) must she drive before she reaches Timothy?

A. 1/3
B. 3
C. 4
D. 9
E. 12

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Re: Distance Time problem [#permalink]

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New post 22 Feb 2009, 07:28
Hi Vemuri,


Let D miles be the distance travelled after which Timothy's mother catches him.

Since Timothy started 15 min. earlier than her mother, he has a lead of 2.25 miles. (Since, distance = 9 * (15/60) = 2.25 miles).

Now, we know that Timothy travelled for 15 min. more than his mother did. If we subtract this extra time (which Timothy used) from the total time taken by him to travel distance D, we will get the time required by Timothy to travel distance (D-2.25) miles. This time will be the same as that required by her mother to travel.

Just find their respective times and equate them. You will get the answer.


Remaining distance for Timothy to travel = (D - 2.25) miles

So, time taken by Timothy to travel the above distance = {(D-2.25)/9} hrs. ---> Equation 1

Now time taken by her mother to travel distance D with speed 36 mph = (D/36) hrs. ---> Equation 2


By equating 1 & 2,
{(D-2.25)/9} = (D/36)

we get,
D = 3 miles i.e., option B.


Hope that helps.


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Technext
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Re: Distance Time problem [#permalink]

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New post 22 Feb 2009, 07:55
Vemuri wrote:
Timothy leaves home for school, riding his bicycle at a rate of 9mph. 15mins after he leaves, his mother sees Timothy's math homework lying on his bed and immediately leaves home to bring it to him. If his mother drives at 36mph, how far (in terms of miles) must she drive before she reaches Timothy?

A. 1/3
B. 3
C. 4
D. 9
E. 12


D is the distance

Time traveled by T = D/9
Time traveled by Mom = D/36

As per question,

D/9 = D/36 + 15/60
4D = D + 9
3D = 9
D = 3

Option B

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Re: Distance Time problem [#permalink]

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New post 22 Feb 2009, 08:44
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Different approach: Relative velocity.

Relative velocity of mother vs Tim = 36 - 9 = 27mph.
Distance to catch up = 9/4
Time to catch up = (9/4)/27

Distance traveled by Mom in above time = 36(9/4)/27 = (36 * 9)/(27 * 4) = 3 => B

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Re: Distance Time problem [#permalink]

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New post 22 Feb 2009, 10:48
ConkergMat wrote:
Different approach: Relative velocity.

Relative velocity of mother vs Tim = 36 - 9 = 27mph.
Distance to catch up = 9/4
Time to catch up = (9/4)/27

Distance traveled by Mom in above time = 36(9/4)/27 = (36 * 9)/(27 * 4) = 3 => B

Nice approach ConkergMat.

Here goes my +1.
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+++ Believe me, it doesn't take much of an effort to underline SC questions. Just try it out. +++
+++ Please tell me why other options are wrong. +++

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Re: Distance Time problem   [#permalink] 22 Feb 2009, 10:48
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