My Math Forum Problem with Direct and Inverse Variation

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 August 6th, 2012, 05:20 AM #1 Member   Joined: Jul 2012 Posts: 32 Thanks: 1 Problem with Direct and Inverse Variation A and B can do a job in 12 days. After working for two days. they are assisted by C, who works at the same rate as A. The work takes 6 1/4 (Six, 1 by 4th) days more to finish. In how many days will B alone do the work? Please help Thanks from Dkp23
 August 6th, 2012, 09:51 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 Re: Problem with Direct and Inverse Variation Giving you the solution may be nice (to you!) but will not really HELP you. Here's an example, where the "job" is picking 840 apples, A picking @ 20 per minute, B picking @ 15 per minute: After 2 minutes (2 days in your case), apples left to be picked = 840 - 2(20)(15) = 770 C now arrives, picking @ 20 (same as A), so combined speed = 20 + 15 + 20 = 55 apples per minute. Time to complete job = 770 / 55 = 14 minutes. Now, using that as a "guide", let's see YOUR stuff
 August 7th, 2012, 07:41 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 Re: Problem with Direct and Inverse Variation Using my example, problem can be worded this way (yours too!): A and B travel distance k miles at combined speed a+b in 24 hours. If, after 2 hours, A doubles speed, then remainder can be travelled in 14 hours. If B travelled alone, how long would it take him to travel the k miles? And you'd be able to "see" clearly what's "happening": Code: At combined speeds, distance k miles is travelled in 24 hours: @(a + b)mph ------------------------------k miles---------------------------------> 24 hours So: k / (a + b) = 24 ; simplify to get: a = (k - 24b) / 24 [1] After 2 hours, A doubles his speed; remaining distance travelled in 14 hours: @(a + b)mph @(2a + b)mph ----2(a + b) miles---->2 hours---------------k - 2(a + b) miles-------> 14 hours So: (k - 2(a + b)) / (2a + b) = 14 ; simplify to get: 30a + 16b = k [2] Substitute [1] in [2]: 30[(k - 24b) / 24] + 16b = k ; simplify to get: b = k / 56 So B's speed is k/56, therefore takes 56 hours if alone. Yours would be: A and B travel distance k miles at combined speed a+b in 12 hours. If, after 2 hours, A doubles speed, then remainder can be travelled in 25/4 hours. If B travelled alone, how long would it take him to travel the k miles? OK? By the way, you could use distance 1 (k = 1) and get 1/speed (1/56 in my example). And that's usually how these are solved; like: in 1 hour, A and B combined do 1/12 of job. BUT I simply hate to use that way when trying to show someone for first time.
 August 8th, 2012, 05:11 PM #4 Member   Joined: Jul 2012 Posts: 32 Thanks: 1 Re: Problem with Direct and Inverse Variation Thank You!!
 August 8th, 2012, 06:50 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Problem with Direct and Inverse Variation I would turn the statement "A and B can do a job in 12 days" into the equation: $12A+12B=1$ After A and B work for two days, 5/6 of the job is left to do. Then we are told. "they are assisted by C, who works at the same rate as A" which means C = A. And the statement "The work takes 6 1/4 days more to finish" gives us: $(2A+B)\frac{25}{4}=\frac{5}{6}\:\therefore\:30A+15 B=2$ Solving the two equations gives: $A=C=\frac{1}{20},\,B=\frac{1}{30}$ Therefore, it would take B 30 days to complete the job working alone.

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# a & b can do a job in 12 days.after working for two days,they are assisted by C,who works at the same rate as. A.The work takes 25/4 days

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