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Timothy leaves home for school, riding his bicycle [#permalink]

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02 Sep 2012, 14:59

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Timothy leaves home for school, riding his bicycle at a rate of 9 miles per hour. Fifteen minutes 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 36 miles per hour, how far (in terms of miles) must she drive before she reaches Timothy?

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

I think is a 700 level problem but I tag it as 600/700, let me know. Either way I hope in an explanation Thanks

Sometimes these types of Combined Rate/Chase-Down questions can be really complicated, but this one is not so bad. You can actually solve it with some logic and a little bit of math (and avoid the longer calculation-heavy approach).

Based on what we're told in the prompt, we know that Timothy travels on his own at 9mph for 15 minutes. Since 15 minutes = 1/4 of an hour, Timothy traveled 2.25 miles before his Mom gets in the car to chase him down.

Since the Mom drives 36mph and Timothy is only going 9mph, she is traveling 4 TIMES FASTER than he is. In basic terms, this means that for every 1 mile Timothy travels, his Mom travels 4 miles.

So, if Timothy DID travel another 1 mile, he'd have traveled 2.25 + 1 = 3.25 miles while his Mom would have driven 4 miles (and gone past him). This means that his Mom caught him at some point between 2.25 miles and 3.25 miles. There's only one answer that 'fits'....

Re: Timothy leaves home for school, riding his bicycle [#permalink]

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02 Sep 2012, 19:41

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Timothy leaves home for school, riding his bicycle at a rate of 9 miles per hour. Fifteen minutes 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 36 miles per hour, how far (in terms of miles) must she drive before she reaches Timothy?

A) 1/3

B) 3

C) 4

D) 9

E) 12

Suppose Timothy's mom meets him after time 't'

Then distance traveled by his mom = distance traveled by him in (15mins) + distance traveled by him in time 't'

=> 36 * t = 9 *(15/60) + 9*t => t = (1/12) hrs

Distance traveled = 36 * t = 36 * (1/12) miles = 3miles

So, Answer will be B

And am not sure about the tag but i guess 600-700 is fine.

Timothy leaves home for school, riding his bicycle at a rate of 9 miles per hour. Fifteen minutes 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 36 miles per hour, how far (in terms of miles) must she drive before she reaches Timothy?

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

I think is a 700 level problem but I tag it as 600/700, let me know. Either way I hope in an explanation Thanks

Distance Timothy covers in 15 mins i.e. 1/4 hours = Rate*Time = 9/4 miles Time taken by mom to meet him = Distance/Relative Speed = (9/4)/(36 - 9) = 1/12 hours Distance traveled by mom in this time = 36 * (1/12) = 3 miles
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Re: Timothy leaves home for school, riding his bicycle [#permalink]

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23 Aug 2014, 19:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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In Time, Speed and Distance problems, it's a good idea to draw a rough diagram to visualize the given information.

Let the distance between the Home and the Point of Meeting be D miles. The question asks us to find the value of D.

As per the question, Timothy's mother left home 15 minutes after he did. But they both reached the Point of Meeting at the same time.

This means,

(Time taken by Timothy's mother to cover D miles) = (Time taken by Timothy to cover D miles) - (15 minutes)

Now, \(Time = \frac{Distance}{Speed}\)

So, the above equation becomes:

\(\frac{D}{36} = \frac{D}{9} - \frac{15}{60}\)

Upon solving this, we get, D = 3 miles

One important point to be careful of in Distance and Speed questions is: Use the same units for a particular quantity throughout the question. By this I mean, if the speed is expressed in miles per hour and travel time/ difference between time taken by 2 people is given in minutes, then you must remember to either convert the time from minutes to hours or convert the speed from miles per hours to miles per minutes. Throughout the question, you should use only one unit for time.

Re: Timothy leaves home for school, riding his bicycle [#permalink]

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13 May 2016, 03:39

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Timothy leaves home for school, riding his bicycle [#permalink]

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30 Jun 2016, 21:09

carcass wrote:

Timothy leaves home for school, riding his bicycle at a rate of 9 miles per hour. Fifteen minutes 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 36 miles per hour, how far (in terms of miles) must she drive before she reaches Timothy?

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

I think is a 700 level problem but I tag it as 600/700, let me know. Either way I hope in an explanation Thanks

In 15 mins, Timothy travels=9/4 miles. Now, let his mother takes x hours to reach him, traveling at 36mph. So, 36x=9x+9/4 x=1/12 hrs. Thus, the distance traveled by his mother to reach= 36*1/12=3 miles. Ans B

Re: Timothy leaves home for school, riding his bicycle [#permalink]

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04 Dec 2016, 20:35

carcass wrote:

Timothy leaves home for school, riding his bicycle at a rate of 9 miles per hour. Fifteen minutes 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 36 miles per hour, how far (in terms of miles) must she drive before she reaches Timothy?

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

I think is a 700 level problem but I tag it as 600/700, let me know. Either way I hope in an explanation Thanks

This is an 600-700 level question.

\(15minutes = \frac{15}{60}hour=\frac{1}{4}hour\)

When Timothy's mother starts driving, Timothy has traveled: \(\frac{1}{4}\times 9=\frac{9}{4}miles\)

Now, Timothy's speed is 9 miles/hour and Timothy's mother's speed is 36 miles/hour, hence Timothy's mother travels faster than Timothy \(36-9=27\) miles each hour.

At the first time, Timothy's mother is 9/4 miles far away from Timothy.

Hence, the time that Timothy's mother needs to catch up Timothy is: \(\frac{\frac{9}{4}}{36-9}=\frac{\frac{9}{4}}{27} =\frac{1}{12} hour\)

This means Timothy's mother must drive: \(\frac{1}{12}\times 36=3 miles\)

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