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Speed - Distance Query

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Speed - Distance Query

by MI3 » Tue May 31, 2011 1:44 am
Q. Deb normally drives to work in 45 minutes at an average speed of 40 miles per hour. This week, however, she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20% . If Deb averages between 12 and 16 miles per hour when biking, how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives?
A. 135
B. 105
C. 95
D. 75
E. 45

Answer I got was D, am I correct?

Thanks,
M
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by Frankenstein » Tue May 31, 2011 1:48 am
Hi,
Even I got D

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by sivaelectric » Tue May 31, 2011 2:31 am
Did you exact 75 or somewhere close to that because I am not getting it. Can you please explain how you solved.
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by Frankenstein » Tue May 31, 2011 2:54 am
sivaelectric wrote:Did you exact 75 or somewhere close to that because I am not getting it. Can you please explain how you solved.
Hi,
Distance she travels is 40.(45/60) = 30 miles
Her bike distance is 30(1-20/100) = 24 miles
In order to ensure that she arrives at work at the same time as when she arrives, we have to consider the worst possible scenario that her average speed is 12mph.
So time taken =24/12 = 2hours =120 mins
So, difference = 120-45 = 75 mins

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by sivaelectric » Tue May 31, 2011 2:57 am
Hey frankenstein, whats the significance of
If Deb averages between 12 and 16 miles per hour when biking
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by Frankenstein » Tue May 31, 2011 3:05 am
sivaelectric wrote:Hey frankenstein, whats the significance of
If Deb averages between 12 and 16 miles per hour when biking
Hi,
When she travels @12mph, she will reach in 24/12 = 2hours (that is the longest time she can take)
When she travels @16mph, she will reach in 24/16 = 1.5hours (that is the shortest time she can take)
In this ques we are asked to find the time by which she needs to start to ensure that she arrives at time.So, we have to consider the longest time as she cannot be slower than this.

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by bubbliiiiiiii » Tue May 31, 2011 3:09 am
When driving:
avg. speed = 40 mph
Total Time = 45 mins = .75 hrs
Total Distance = speed x time = 40 x .75 = 30 miles

When biking:
distance = 80% of driving distance = 24 miles.
Average speed = 12-16 mph

Case (i) at 12 mph
time = distance/speed = 24/12 = 2 hrs = 120 mins

[icase (ii) at 16 mph[/i]
time = distance/speed = 24/16 = 1.5 hrs = 90 mins

Time taken when she drives is 45 mins. Thus, to cover up the additional time she has to start early from home. The additional time can be obtained by substracting 45 from both the cases we get,

Case (i): 120-45 = 75 mins

Case (ii): 90-45 = 45 mins

So the extra time needed should be between 45 and 75 mins.

Can you please help me understand what am I doing wrong here?
Regards,

Pranay
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by sivaelectric » Tue May 31, 2011 3:10 am
Oh god :( bad on my part for. careless reading. Thanks Frankenstein. :)
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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