document.write( "Question 1177332: A kayak can travel 24 miles downstream in 2 hours, while it would take 12 hours to make the same trip upstream. Find the speed of the kayak in still water, as well as the speed of the current. Let k represent the speed of the kayak in still water, and let c represent the speed of the current. \n" ); document.write( "

Algebra.Com's Answer #853739 by ikleyn(53646) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "A kayak can travel 24 miles downstream in 2 hours, while it would take 12 hours to make the same trip upstream.
\n" ); document.write( "Find the speed of the kayak in still water, as well as the speed of the current.
\n" ); document.write( "Let k represent the speed of the kayak in still water, and let c represent the speed of the current.
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document.write( "Let k be the rate of the kayak in still water (in miles per hour)\r\n" );
document.write( "and c be the rate of the current (in the same units).\r\n" );
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document.write( "Then the effective rate of the kayak downstream is k + c\r\n" );
document.write( "and  the effective rate of the kayak   upstream is k - c.\r\n" );
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document.write( "From the problem, the effective rate of the kayak downstream is the distance of 24 miles \r\n" );
document.write( "divided by the time of 2 hours   = 12 mph.\r\n" );
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document.write( "                  The effective rate of the plane upstream is the distance of 24 miles \r\n" );
document.write( "divided by the time of 12 hours   = 2 mph.\r\n" );
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document.write( "So, we have two equations to find 'k' and 'c'\r\n" );
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document.write( "    k + c = 12,    (1)\r\n" );
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document.write( "    k - c =  2.    (2)\r\n" );
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document.write( "To solve, add equations (1) and (2).  The terms 'c' and '-c' will cancel each other, and you will get\r\n" );
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document.write( "    2k = 12 + 2 = 14  --->   k = 14/2 = 7.\r\n" );
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document.write( "Now from equation (1)\r\n" );
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document.write( "     v = 12 - u = 12 - 7 = 5.\r\n" );
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document.write( "ANSWER.  The rate of the kayak in still water is 7 mph.  The rate of the current is 5 mph km/h.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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\n" ); document.write( "\n" ); document.write( "This solution produces the same answer as in the post by @mananth, but has an advantage
\n" ); document.write( "that it does not contain excessive calculations that the solution by @mananth has.\r
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\n" ); document.write( "\n" ); document.write( "We, the tutors, write here our solutions not only to get certain numerical answer.
\n" ); document.write( "We write to teach - and, in particular, to teach solving in a right style.\r
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\n" ); document.write( "\n" ); document.write( "This style solving presented in my post, is straightforward with no logical loops.
\n" ); document.write( "The solution presented in the post by @mananth has two logical loops.\r
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document.write( "    One loop in the @mananth post is writing\r\n" );
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document.write( "        24/(k+c) = 2   --->   2k + 2c = 24  --->  /2  --->  k+c = 12,\r\n" );
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document.write( "    while in my solution I simply write for the effective rate \r\n" );
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document.write( "        k + c = 24/2 = 12.\r\n" );
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document.write( "    Second loop in the @mananth post is writing\r\n" );
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document.write( "                       24/(x-y) = 12   --->   12x - 12c = 24  --->  /12  --->  k-c = 2,\r\n" );
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document.write( "    while in my solution I simply write for the effective rate upstream\r\n" );
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document.write( "        u - v = 24/12 = 2.\r\n" );
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\n" ); document.write( "\n" ); document.write( "It is why I presented my solution here and why I think it is better than the solution by @mananth:
\n" ); document.write( "- because it teaches students to present their arguments in a straightforward way, without logical zigzags.\r
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\n" ); document.write( "\n" ); document.write( "@mananth repeats his construction of solution with no change for all similar problems on floating
\n" ); document.write( "with and against the current simply because his COMPUTER CODE is written this way.
\n" ); document.write( "But this way is not pedagogically optimal - in opposite, it is pedagogically imperfect.\r
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