SOLUTION: at the beinning of a walk jack roberto and juanita are 5.5 miles apart. If they leave at the same time and walk in the same direction, roberto over takes juanita in 11 hours. If th

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Question 661236: at the beinning of a walk jack roberto and juanita are 5.5 miles apart. If they leave at the same time and walk in the same direction, roberto over takes juanita in 11 hours. If they walk toward eachother, they meet in 1 hour. what are their speeds?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
It is possible to tell from the problem who walks faster
Walking in the same direction, roberto over takes juanita.
If Roberto walked slower than her, he would never catch her
so he is the faster walker
Let +r+ = his speed in mi/hr
Let +j+ = her speed in mi/hr
-------------------
I can think of Juanita as standing still and Roberto as
approaching at the difference of their speeds
(1) +5.5+=+%28+r+-+j+%29%2A11+
-------------------
Walking toward each other, think of Juanita
as standing still and add their speeds
(2) +5.5+=+%28+r+%2B+j+%29%2A1+
---------------------
(1) +5.5+=+11r+-+11j+
and
(2) +5.5+=+r+%2B+j+
(2) +j+=+5.5+-+r+
substitute (2) into (1)
(1) +5.5+=+11r+-+11%2A%28+5.5+-+r+%29+
(1) +5.5+=+11r+-+60.5+%2B+11r+
(1) +22r+=+60.5+%2B+5.5+
(1) +22r+=+66+
(1) +r+=+3+
and, since
(2) +j+=+5.5+-+r+
(2) +j+=+5.5+-+3+
(2) +j+=+2.5+
3 mi/hr is his speed in mi/hr
2.5 mi/hr is her speed in mi/hr
check:
In 11 hrs, she walks +d+=+2.5%2A11+
+d+=+27.5+ mi
In 11 hrs, he walks +d+=+3%2A11+=+33+
+33+-+27.5+=+5.5+, so he makes up
the 5.5 mi difference
OK