document.write( "Question 1206959: Hi
\n" ); document.write( "Water is leaking fron a tank at a constant rate . At 230pm the tank is 3/5 full .At 325pm the tank was 2/7 full. At what time will the tank be empty.
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Algebra.Com's Answer #844720 by math_tutor2020(3816)\"\" \"About 
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\n" ); document.write( "Answer: 4:15 PM\r
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\n" ); document.write( "\n" ); document.write( "Explanation\r
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\n" ); document.write( "\n" ); document.write( "x = number of minutes after 2:30 PM
\n" ); document.write( "y = fraction representing how full the tank is.
\n" ); document.write( "y = 0 means the tank is completely empty.
\n" ); document.write( "y = 1 means the tank is completely full.\r
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\n" ); document.write( "\n" ); document.write( "The water is leaking at a constant rate. Which will mean we can use a linear equation. \r
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\n" ); document.write( "\n" ); document.write( "We have the two points on this line which are (0,3/5) and (55, 2/7)
\n" ); document.write( "The 55 refers to 55 minutes after 2:30 PM to arrive at 3:25 PM.\r
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\n" ); document.write( "\n" ); document.write( "Note: The jump from 2:30 to 3:30 is 1 hour, aka 60 minutes, so we subtract off 5 minutes to go from 2:30 to arrive at 3:25.\r
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\n" ); document.write( "\n" ); document.write( "I'll skip a few steps but the equation of the line through those points mentioned is y = (-1/175)x + 3/5\r
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\n" ); document.write( "\n" ); document.write( "We want to know when the tank will be empty, so we want to determine x when y = 0.\r
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\n" ); document.write( "\n" ); document.write( "y = (-1/175)x + 3/5
\n" ); document.write( "0 = (-1/175)x + 3/5
\n" ); document.write( "(1/175)x = 3/5
\n" ); document.write( "x = 175*(3/5)
\n" ); document.write( "x = 105
\n" ); document.write( "It takes 105 minutes to go from 3/5 full to completely empty.\r
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\n" ); document.write( "\n" ); document.write( "105 min = 60 min + 45 min
\n" ); document.write( "105 min = 1 hr + 45 min\r
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\n" ); document.write( "\n" ); document.write( "Now add this duration to 2:30 to see where we end up.
\n" ); document.write( "From 2:30 to 3:30 is 1 hour.
\n" ); document.write( "From 3:30 to 4:00 is 30 minutes.
\n" ); document.write( "From 4:00 to 4:15 is another 15 minutes (30+15 = 45 minutes total)\r
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\n" ); document.write( "\n" ); document.write( "Or...
\n" ); document.write( "The jump from 2:30 PM to 4:30 PM is 2 hours
\n" ); document.write( "Rewind the clock 15 minutes to go from 4:30 PM to 4:15 PM
\n" ); document.write( "Rewinding those 15 minutes means the 2 hour duration is adjusted to 1 hour+45 minutes. \r
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\n" ); document.write( "\n" ); document.write( "Therefore, the time gap from 2:30 PM to 4:15 PM is 1 hour+45 minutes aka 105 minutes.\r
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\n" ); document.write( "\n" ); document.write( "Tutor @mananth made a mistake.
\n" ); document.write( "The tank doesn't start off 100% full, so we don't need to drain 100% of the contents.
\n" ); document.write( "Instead we need to drain 3/5 = 0.6 = 60% of the contents.\r
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\n" ); document.write( "\n" ); document.write( "If you solved (11/35)/55 = 1/x then you'll get x = 175
\n" ); document.write( "While this is a correct solution to this equation, the equation itself is set up wrong.
\n" ); document.write( "The \"1\" on the right hand side should be 3/5 to represent the fractional amount we need to drain.
\n" ); document.write( "The \"1\" on the right hand side represents 100% full. \r
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\n" ); document.write( "\n" ); document.write( "If you solved (11/35)/55 = (3/5)/x, then you'll get x = 105 which then leads to 4:15 PM as mentioned above.
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