document.write( "Question 291540: A man covers a certain distance on scootr. Had he moved 3kmph faster, he would have taken 40min less,if he had moved 2kmph slower he would have taken 40 min more.The distance in (km)is \n" ); document.write( "
Algebra.Com's Answer #477365 by happyhaps(2)\"\" \"About 
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First of all, let us assume normal speed to be s and normal time be t.
\n" ); document.write( "Now the distance become st(Distance = speed x time).
\n" ); document.write( "According to the first statement, increased speed becomes s+3 and therefore time decreases by (t - 40/60) that is t - 2/3.
\n" ); document.write( "therefore the distance by new speed and time becomes (s + 3)(t - 2/3).
\n" ); document.write( "Now let us equate the first distance by the second one above since distance is same.
\n" ); document.write( "it becomes
\n" ); document.write( "st = (s+3)(t - 2/3)
\n" ); document.write( "Rearranging this becomes,
\n" ); document.write( "2s -9t = -6..........(I)\r
\n" ); document.write( "\n" ); document.write( "Now According to the second statement, decreased speed becomes s-2 and therefore increased time becomes t + 2/3
\n" ); document.write( "Again the distance by new speed and time becomes (s-2)(t + 2/3).
\n" ); document.write( "Now, again equate this distance by original distance st.
\n" ); document.write( "st = (s-2)(t + 2/3)
\n" ); document.write( "Rearranging this,
\n" ); document.write( "-2s + 6t = -4 ...... (II)\r
\n" ); document.write( "\n" ); document.write( "Now you have two linear equation (I) and (II)
\n" ); document.write( "Solve these to get values of s and t
\n" ); document.write( "which is s = 12 and t = 10/3\r
\n" ); document.write( "\n" ); document.write( "Now distance is st therefore 12 x 10/3 = 40 km \n" ); document.write( "
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