document.write( "Question 932114: On the first day, I cycled 1/2 of the distance. On day 2, I cycled one half of the remaining distance. On day 3, I cycled three quarters of the remaining distance. On day 4, I cycled 10 miles.\r
\n" ); document.write( "\n" ); document.write( "On day 5 I cycled 2/3 of the remaining distance and on the final day I cycled the remaining 5 miles.
\n" ); document.write( "How far is it to my Grandmothers house?
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Algebra.Com's Answer #566066 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
On the first day, I cycled 1/2 of the distance.
\n" ); document.write( " On day 2, I cycled one half of the remaining distance.
\n" ); document.write( " On day 3, I cycled three quarters of the remaining distance.
\n" ); document.write( " On day 4, I cycled 10 miles.
\n" ); document.write( " On day 5 I cycled 2/3 of the remaining distance and
\n" ); document.write( " on the final day I cycled the remaining 5 miles.
\n" ); document.write( "How far is it to my Grandmothers house?
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\n" ); document.write( "Let d = distance to Grandma's house
\n" ); document.write( "End of day 1: \"1%2F2\"d remains
\n" ); document.write( "End of day 2: \"1%2F2\" * \"1%2F2\"d = \"1%2F4\"d remains
\n" ); document.write( "End of day 3: \"1%2F4\"d * \"1%2F4\" = \"1%2F16\"d remains
\n" ); document.write( "End of day 4: \"1%2F16\"d - 10 mi remain
\n" ); document.write( "End of day 5: \"1%2F3\"(\"1%2F16\"d - 10) = 5 (remained for the last day)
\n" ); document.write( "\"1%2F48\"d - \"10%2F3\" = 5
\n" ); document.write( "\"1%2F48\"d = 5 + \"10%2F3\"
\n" ); document.write( "\"1%2F48\"d = \"15%2F3\" + \"10%2F3\"
\n" ); document.write( "\"1%2F48\"d = \"25%2F3\"
\n" ); document.write( "multiply both sides by 48
\n" ); document.write( "d = 16(25)
\n" ); document.write( "d = 400 miles to grandma's house
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