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A biker travels a certain distance. If he had travelled 4

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rxs0005
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A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 16:16
A biker travels a certain distance. If he had travelled 4 mph faster he would have taken 30 min less. If he had travelled 3 mph slower he would have taken 30 min more. Find the original speed.



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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 18:43
I am afraid thats not the correct answer
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 19:16
Let original rate be R and time taken be T. Original distance traveled is RT

If he travels 4 mph faster, he covers it 1/2 an hr less than the original time, which means

(R-4)(t-1/2) = RT (or) 4t - 1/2R = 2 ----- 1

If he travels 3 mph slower, it takes him 1/2 hr more than the original time
(R-3)(t+1/2) = RT (or) -3t + 1/2R = 3/2 ----- 2

solving 1 & 2, I get t = 7/2 hrs and R = 24 mph

what's the OA?
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 20:01
12 mph.

Let x be the distance and s be original speed.

From 1 : x/s - x/(s+4) = 1/2

From 2: x/(s-3) - x/s = 1/2

Subtract 1 from 2 and solve for s = 12 mph
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 21:33
24

Let r=rate, t=time, and x=distance

rt=x

(r+4)(t-.5)=x=rt

(r-3)(t+.5)=x=rt

multiply both equations out you get:

rt+4t-.5r-2=rt

rt-3t+.5r-1.5=rt

cancel rt from both sides of each equation you get:

4t-.5r+2=0

-3t+.5r-1.5=0

multiply first equation by 3 and the 2nd equation by 4 you get:

12t-1.5r-6=0

-12t+2r-6=0

subtract equation 2 from equation 1 you get:

.5r-12=0 or

.5r=12 therefore:

r=24
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 22:40
24

Let r=rate, t=time, and x=distance

rt=x

(r+4)(t-.5)=x=rt

(r-3)(t+.5)=x=rt

multiply both equations out you get:

rt+4t-.5r-2=rt

rt-3t+.5r-1.5=rt

cancel rt from both sides of each equation you get:

4t-.5r+2=0

-3t+.5r-1.5=0

multiply first equation by 3 and the 2nd equation by 4 you get:

12t-1.5r-6=0

-12t+2r-6=0

subtract equation 2 from equation 1 you get:

.5r-12=0 or

.5r=12 therefore:

r=24


Good explanation...I got 24 as well.
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   05 Jan 2005, 22:40
rxs0005 wrote:
A biker travels a certain distance. If he had travelled 4 mph faster he would have taken 30 min less. If he had travelled 3 mph slower he would have taken 30 min more. Find the original speed.


original speed = 24mph
distance = 84 miles
original time = 3 hours 30 minuets
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   06 Jan 2005, 06:08
24 is the correct answer
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vprabhala
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   06 Jan 2005, 07:36
Late.. but I get 24 ...too
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Re: A biker travels a certain distance. If he had travelled 4 [#permalink] New post   06 Jan 2005, 09:17
banerjeea_98 wrote:
12 mph.

Let x be the distance and s be original speed.

From 1 : x/s - x/(s+4) = 1/2

From 2: x/(s-3) - x/s = 1/2

Subtract 1 from 2 and solve for s = 12 mph


Ooops Calculation mistake, gotta be careful :( ....solve above and we get 24mph and not 12mph



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Re: A biker travels a certain distance. If he had travelled 4 [#permalink]
06 Jan 2005, 09:17
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