document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1207767>Question 1207767</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A 5-horsepower (hp) pump can empty a pool in 5 hours.A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool.After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool? </SPAN>\n" );
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document.write( "Answer: <font color=red>1.75 hours</font>\r<BR>\n" );
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document.write( "Explanation\r<BR>\n" );
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document.write( "The horsepower values are a distraction. Ignore them entirely.\r<BR>\n" );
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document.write( "Consider a 4000 gallon pool.<BR>\n" );
document.write( "The larger pump, which I'll call pump A, can empty the entire pool in 5 hours when working alone.<BR>\n" );
document.write( "A's unit rate is 4000/5 = 800 gallons per hour.<BR>\n" );
document.write( "Formula used: rate = (amountDone)/time\r<BR>\n" );
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document.write( "The smaller pump B can empty the entire pool in 8 hours when working alone.<BR>\n" );
document.write( "B's unit rate is 4000/8 = 500 gallons per hour.\r<BR>\n" );
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document.write( "When the two pumps work together, without either pump slowing down the other, their combined rate is 800+500 = 1300 gallons per hour.<BR>\n" );
document.write( "The two pumps work together for 2 hours. That drains 1300*2 = 2600 gallons and leaves 4000-2600 = 1400 gallons remaining.\r<BR>\n" );
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document.write( "Pump B breaks down at the 2 hour marker.<BR>\n" );
document.write( "Pump A now works alone to empty the remaining 1400 gallons of water.<BR>\n" );
document.write( "rate*time = amountDone<BR>\n" );
document.write( "time = amountDone/rate<BR>\n" );
document.write( "time = 1400/800<BR>\n" );
document.write( "time = 14/8<BR>\n" );
document.write( "time = (2*7)/(2*4)<BR>\n" );
document.write( "time = 7/4<BR>\n" );
document.write( "time = <font color=red>1.75 hours</font> is the amount of time needed for the larger pump to finish the job of emptying the pool.\r<BR>\n" );
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document.write( "The 4000 value mentioned isn't special. Feel free to change it to any other positive number you want. The final answer will still be the same regardless of the pool volume.<BR>\n" );
document.write( "I picked this value based on the LCM of 5 and 8, which is 40. Then I tacked on a few extra zeros to lead to a slightly more realistic pool volume. Even then 4000 is probably on the smaller side of things.\r<BR>\n" );
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document.write( "Another approach\r<BR>\n" );
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document.write( "one job = emptying the pool<BR>\n" );
document.write( "Pump A does one job in 5 hours when working alone<BR>\n" );
document.write( "A's unit rate is 1/5 of a job per hour<BR>\n" );
document.write( "Pump B does one job in 8 hours when working alone<BR>\n" );
document.write( "B's unit rate is 1/8 of a job per hour\r<BR>\n" );
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document.write( "Their combined unit rate is 1/8 + 1/5 = 5/40 + 8/40 = 13/40 of a job per hour.<BR>\n" );
document.write( "This assumes neither pump slows the other down.\r<BR>\n" );
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document.write( "In 2 hours the pumps work together to handle 2*13/40 = 13/20 of the job.<BR>\n" );
document.write( "1 - (13/20) = 20/20 - 13/20 = 7/20 of the job remains.\r<BR>\n" );
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document.write( "x = amount of extra time, in hours, pump A needs to work alone to finish the job<BR>\n" );
document.write( "This is after pump B stops working\r<BR>\n" );
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document.write( "rate*time = amountDone<BR>\n" );
document.write( "(1/5 of a job per hour)*(x hours) = 7/20 of a job remains<BR>\n" );
document.write( "(1/5)x = 7/20<BR>\n" );
document.write( "x = 5*7/20<BR>\n" );
document.write( "x = 5*7/(5*4)<BR>\n" );
document.write( "x = 7/4<BR>\n" );
document.write( "x = <font color=red>1.75 hours</font>\r<BR>\n" );
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document.write( "Extra info:<BR>\n" );
document.write( "1.75 hours = 60*1.75 = 105 minutes<BR>\n" );
document.write( "1.75 hours = 1 hour & 45 minutes  since 0.75 hr = 0.75*60 = 45 min<BR>\n" );
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