document.write( "Question 113764This question is from textbook Algebra Structure and method
\n" ); document.write( ": At 8:00 a.m the Smith's left a campground,driving at 48mi/h. At 8:20 a.m the Garcia's left the same campground and followed the same route, driving at 60 mi/h. At what time did they overtake the Smith's?
\n" ); document.write( "Here is what I have but I don't know if it is right
\n" ); document.write( "60t=48(t+1/3)
\n" ); document.write( "60t=48t+16
\n" ); document.write( "60t-48t=48t-48t+16
\n" ); document.write( "12t=16
\n" ); document.write( "12t/12=16/12
\n" ); document.write( "t=14/12or 11/3
\n" ); document.write( "They overtook the Smiths at 1:20
\n" ); document.write( "Is this right?
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Algebra.Com's Answer #82773 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
The Smiths and Garcias both travel the same distance
\n" ); document.write( "\"D%5Bs%5D+=+D%5Bg%5D\" I'll just call the distance \"D\"
\n" ); document.write( "To find their elapsed time, start a stopwatch when the
\n" ); document.write( "Garcias leave at 8:20. Stop the stopwatch when they meet.
\n" ); document.write( "Call that elapsed time \"t\"hrs. But the Smiths have already been
\n" ); document.write( "driving for 20 min., so their time is \"t+%2B+1%2F3\"hrs
\n" ); document.write( "\"D+=+r%2At\"
\n" ); document.write( "\"D%5Bs%5D+=+48%28t+%2B+1%2F3%29\"
\n" ); document.write( "\"D%5Bg%5D+=+60t\"
\n" ); document.write( "\"48%28t+%2B+1%2F3%29+=+60t\"
\n" ); document.write( "\"48t+%2B+16+=+60t\"
\n" ); document.write( "\"12t+=+16\"
\n" ); document.write( "\"t+=+4%2F3\"hrs or,
\n" ); document.write( "\"t+=+80+min\"
\n" ); document.write( "This is the Garcias elapsed time
\n" ); document.write( "8:20 + 1 hr 20 min = 9:40 answer
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