SOLUTION: You drive at an average speed of 50mph to a discount shopping mall, spend two hours shopping, and then return at an average speed of 25mph. The entire trip take 8 hours. How far aw

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Question 1023723: You drive at an average speed of 50mph to a discount shopping mall, spend two hours shopping, and then return at an average speed of 25mph. The entire trip take 8 hours. How far away is the shopping mall?
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52771) About Me  (Show Source):
You can put this solution on YOUR website!
.
The sum of two numbers is 9 and the difference of their squares is 9. Find the numbers.?
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x + y = 9,        (1)
x%5E2+-+y%5E2 = 9.      (2)

Rewrite (2) in the form

(x+y)*(x-y) = 9,

then replace (x+y) by 9, due to (1). You will get

9*(x-y) = 9.

Next, reduce both sides by the factor 9. You will get

x - y = 1.        (3)

Now, you have thwo equations to solve,

x + y = 9,        (1')
x - y = 1.        (3')

Add them. You will get 2x = 9 + 10  --->  2x = 10  --->  x = 10%2F2 = 5.
Then from (1')  y = 9 - x = 9 - 5 = 4.

Answer.  x = 5, y = 4.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Time spent shopping is not travel time, so
+8+-+2+=+6+ hrs travel time
Let +d+ = distance to shopping mall
Let +t+ = time in hrs to get to shopping mall
+6+-+t+ = time in hrs to back from shopping mall
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Going to the mall:
(1) +d+=+50t+
Coming back from the mall:
(2) +d+=+25%2A%28+6+-+t+%29+
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By substitution:
(2) +50t+=+25%2A%28+6+-+t+%29+
(2) +50t+=+150+-+25t+
(2) +75t+=+150+
(2) +t+=+2+
Plug this result back into (1) or (2)
(1) +d+=+50t+
(1) +d+=+50%2A2+
(1) +d+=+100+
The shopping mall is 100 mi away
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check:
(2) +d+=+25%2A%28+6+-+t+%29+
(2) +d+=+25%2A%28+6+-2+%29+
(2) +d+=+25%2A4+
(2) +d+=+100+
OK