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Med Level Rate Question

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Med Level Rate Question

by seal4913 » Fri Mar 23, 2012 2:49 pm
Westville and Eastown are 55 miles apart. Derek sets off from Westville at 1pm travelling towards Eastown at an average speed of 60 miles per hour. Emily sets of from Eastown at 1:10pm and drives towards Westville at an average speed of 40 miles per hour. At what time will they pass each other on the road between Westville and Eastown?

A) 1:27pm
B) 1:30pm
C) 1:33pm
D) 1:37pm
E) 1:43pm

Here's my method:

[spoiler] 60 x t + (1/6) = 60t + 10
40 x t = 40t
Total: = 55

So 60t + 10 + 40t = 55; 100t + 10 =55; 100t = 45; t = 45/100 = 9/20 so i convert that to hours by times it by 3 to get 27/60 so it takes 27 minutes. Therefore the later person "t" started at 1:10pm so I had 27 minutes to that and get D 1:37pm[/spoiler]
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by killer1387 » Fri Mar 23, 2012 4:23 pm
at 1:10 pm Emily starts.
in 10 min. derek covers=10miles

remaining distance=55-10=45 miles

now v=60+40
d=45
t=45*60/100=27 min

hence they meet at 1:37 pm

OPTION D
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by rijul007 » Sat Mar 24, 2012 12:07 am
At 1:10,
Derek has travelled = 10/60 * 60 = 10 miles

Distance between Emily and Derek = 55-10 = 45

The speed at which they are getting close to each other = 40+60 =100 miles/hr
Time taken = 45/100 hrs = 9/20 hrs = 9/20 * 60 = 27 mins

They pass each other at = 1:37 PM

Option D
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by tennisstar » Sun Mar 25, 2012 2:49 pm
Hi,

Can you please explain your method in detail? How did you set up the first few equations?

60 x t + (1/6) = 60t + 10
40 x t = 40t
Total: = 55

It will help many of us.

Thx much,

seal4913 wrote:Westville and Eastown are 55 miles apart. Derek sets off from Westville at 1pm travelling towards Eastown at an average speed of 60 miles per hour. Emily sets of from Eastown at 1:10pm and drives towards Westville at an average speed of 40 miles per hour. At what time will they pass each other on the road between Westville and Eastown?

A) 1:27pm
B) 1:30pm
C) 1:33pm
D) 1:37pm
E) 1:43pm

Here's my method:

[spoiler] 60 x t + (1/6) = 60t + 10
40 x t = 40t
Total: = 55

So 60t + 10 + 40t = 55; 100t + 10 =55; 100t = 45; t = 45/100 = 9/20 so i convert that to hours by times it by 3 to get 27/60 so it takes 27 minutes. Therefore the later person "t" started at 1:10pm so I had 27 minutes to that and get D 1:37pm[/spoiler]
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by seal4913 » Mon Mar 26, 2012 3:11 am
tennisstar wrote:Hi,

Can you please explain your method in detail? How did you set up the first few equations?

60 x t + (1/6) = 60t + 10
40 x t = 40t
Total: = 55

It will help many of us.

Thx much,
So for rate the equation is Rate x Time = Work or Distance. We know D's rate is 60mph since he starts first it's T(the unknown) plus his headstart. 10 minutes is the same as 1/6 as 10 min/60mins is 1/6.

So his work/distance would be 60 x t + 1/6 which equals 60t + 10.

Do the same for Emily, rate x time = work is 40 x t = 40t.

This equals thier total distance 55 miles.

So since they are travling towards each other you add it. So 60t + 10 + 40t = 55. Slove for t and get 9/20 which is minutes so I times it by 3 to get it in an hour so it's 27 mins over 1hour. So the time is 27 minutes. I know Emily started at 1:10pm and i add 27 mins and get 1:37pm.

Let me know if that helps

So we know
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by doclkk » Mon Mar 26, 2012 3:42 pm
My method - is slow but it's really simple to think about.

Derek is going 60 miles an hour or 1 Mile per minute.

After 10 minutes, he has already traveled 10 miles.

At 1:10, Emily leaves, she's going 2/3 miles per minute. At this point they are 45 miles from each other. When are they 0 miles from each other?

So combine the two rates.Together they are traveling 5/3 miles per minute or 5 miles every 3 minutes. There are 45 miles left. So what's 45 / 5 = 9.

They have to do 5 miles every 3 minutes 9 times.

so that's 27 minutes. 27 minutes + 1:10 = 1:37.

Answer is D.
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