SOLUTION: john drove home at 40 mph. average speed on return is 24mph. return trip took 1/2 hour longer because of traffic. how for did she travel to appointment

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Question 606222: john drove home at 40 mph. average speed on return is 24mph. return trip took 1/2 hour longer because of traffic. how for did she travel to appointment
Found 2 solutions by mathie123, josmiceli:
Answer by mathie123(224) About Me  (Show Source):
You can put this solution on YOUR website!
For these types of questions, I find it useful to make a table that compares the way there vs the way back (speed, distance and time).
We also need to recall that speed=distance/time
Okay, so for the way there we drove at 24 mph. We do not know the distance so I will call it d. and similarily we do not know the time. I will call the time y.
For the way back we drove at 40 mph a distance of d and a time of y-0.5.
Using our speed equation we know that distance=speed*time (just rearranging it).
Now we can make equations using what we have:
For the trip there:
24%2Ay=d
and the way back:
40%28y-0.5%29=d

Now I will let you do the rest. You have two equations with two unknowns (d and y). We just need to find d.

Try it out and if you are still not able to get the answer let me know:)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+d+=+r%2At+
On 1st trip,
+r%5B1%5D+=+40+ mi/hr
(1) +d+=+40t%5B1%5D+
-----------
On 2nd trip,
+r%5B2%5D+=+24+ mi/hr
+t%5B2%5D+=+t%5B1%5D+%2B+.5+
(2) +d+=+24%2A%28+t%5B1%5D+%2B+.5+%29+
-----------
By substitution:
+40t%5B1%5D+=+24t%5B1%5D+%2B+12+
+16t%5B1%5D+=+12+
+t%5B1%5D+=+.75+
and, since
(1) +d+=+40t%5B1%5D+
(1) +d+=+40%2A.75+
(1) +d+=+30+
The appointment is 30 mi away
check:
(2) +d+=+24%2A%28+t%5B1%5D+%2B+.5+%29+
(2) +d+=+24%2A%28+.75+%2B+.5+%29+
(2) +d+=+24%2A1.25+%7D%7D%0D%0A%282%29+%7B%7B%7B+d+=+30+
OK