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by vkb16 » Sun Oct 11, 2009 4:09 am
Stacy and heather are 20 miles apart and walk towards each other on the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather walked when the two meet?
7
8
9
10
11

OA B

how do u solve the sum with the sum of rates method (i.e here the sum of rates is 11)
My procedure was.. In 24 min S walks 24/60*6= 2.4 MILES

thus, the two will meet in (20-2.4)/(5+6) hours = 19.6/11 hrs

in this time, H must have walked (19.6/11)*5 = 8.9 miles...

Where did I go wrong??
thanks
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Re: rate problem-mgmat

by dmateer25 » Sun Oct 11, 2009 4:25 am
vkb16 wrote:Stacy and heather are 20 miles apart and walk towards each other on the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from her original destination has Heather walked when the two meet?
7
8
9
10
11

OA B

how do u solve the sum with the sum of rates method (i.e here the sum of rates is 11)
My procedure was.. In 24 min S walks 24/60*6= 2.4 MILES

thus, the two will meet in (20-2.4)/(5+6) hours = 19.6/11 hrs

in this time, H must have walked (19.6/11)*5 = 8.9 miles...

Where did I go wrong??
thanks
You did everything right except your subtraction was incorrect.

(20-2.4) = 17.6

(17.6/11) * 5 = 8
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by vkb16 » Wed Oct 14, 2009 8:21 pm
damn the silly errors!
thanks for pointing it out though!
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