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.

Hope it helps!
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9*1/4=9/4 miles Timothy ran before his mom started

36-9=27m/h is a rate to achieve Timothy

9/4:27=1/12 h is a time mom will run

36*1/12=3 miles mom need to run

B

Hi All,

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'....

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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This is a simple relative speed question. Here is a conceptual post on relative speed: http://www.veritasprep.com/blog/2012/07 ... elatively/

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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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.

Hope this helped!

Japinder
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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

36t=9(t+1/4)
t=1/12 hour
1/12*36=3 miles

3) Tim’s rate = 9miles/hr
Fifteen minutes later, mom leaves at rate of 36 miles/hr. Distance mom drives to reach Tim?
D R T
Tim 9 x
Mom 36 x-1/4

Distance is the same because mom is "catching up" to Tim. Set distances equal to each other.
• 9x = 36(x-1/4)
• 9x = 36x – 9 → 9 = 27x → x = 1/3 hours

Mom's time is (x- 1/4) or (1/3 - 1/4) = 1/12 hours.

Distance = (36 miles/hour)(1/12 hours) = 3 miles.

Answer: B

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:
\(\frac{1}{12}\times 36=3 miles\)

The answer is B
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T covers 9/4 miles before his mom starts.

So the time is 9/4* 1/27= 1/12 hr.(27 is the relative speed)

Distance Timothy's mom has to cover = 36 * 1/12 = 3 miles.

Hence Ans:B

So, Timothy has a 15 minute advantage over his mother, which translates into being 9/4 miles ahead, since he's riding his bycicle at 9 mph.

His mother leaves their home at 36 mph, as it is said. This means she covers 9 miles per 15 minutes.

Let T be the amount of periods of 15 minutes:

\((9/4)*T + 9/4 = (36/4)*T\)

\(T = 1/3\)

So, Timothy will be caught 5 minutes after his mother has left their home.

5 minutes are 1/12 of 1 hour, so his mother will catch him after having driven for 3 miles.

We can let the time, in hours, of Tim’s mother = t, and thus Tim’s time = t + 1/4. Using the formula distance = rate x time, we see that Tim’s distance is 9(t + 1/4) = 9t + 9/4 and his mother’s distance is 36t, and thus:

9t + 9/4 = 36t

9/4 = 27t

9 = 108t

9/108 = t

1/12 = t

Thus, Tim’s mother drives 36 x 1/12 = 3 miles before she reaches him.

Answer: B
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T pedals 9/4 miles in 1/4 hour
M gains 27 mph on T
(9/4)/27=1/12 hour for M to catch T
(1/12 hour)(36 mph)=3 miles
B

Chase has a headstart of 9.(1/4)=9/4 miles

Let x represent the time it takes his mum to catch up to him.

36.x=9.x+9/4

x=1/12 hour

at 36 mph, his mum will catch up with him at 36/12=3 miles.

I think this is a low 600's question.

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