dwurster Posted December 13, 2007 Posted December 13, 2007 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? I think the aim is to solve this equation: 1/A + 1/B + 1/C = 1/X Therefore we have three further equations, haven't we?! ;) 1/A + 1/B = 1/6 1/B + 1/C = 1/10 1/C + 1/A = 1/7,5 Is there any possibility to solve this problem without using and transforming these three equations? (it takes a lot of time and I get some distorted numbers :( ) I were very happy if anybody could help me?!? Thanks a lot... Quote
krusta80 Posted December 13, 2007 Posted December 13, 2007 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? I think the aim is to solve this equation: 1/A + 1/B + 1/C = 1/X Therefore we have three further equations, haven't we?! ;) 1/A + 1/B = 1/6 1/B + 1/C = 1/10 1/C + 1/A = 1/7,5 Is there any possibility to solve this problem without using and transforming these three equations? (it takes a lot of time and I get some distorted numbers :( ) I were very happy if anybody could help me?!? Thanks a lot... Work = 1 job Rate = Rate_A + Rate_B + Rate_C Time = ? So, we need to find the sum of the individual rates. This can be done by finding the value of each rate. Fortunately, we have three equations for three unknowns...we just need to set them up using work equations: (Rate_A + Rate_B)*6 = 1 (Rate_C + Rate_B)*10 = 1 (Rate_A + Rate_C)*7.5 = 1 Matrix time! Rate_A Rate_B Rate_C | RHS 1 1 0 | 1/6 0 1 1 | 1/10 1 0 1 | 1/7.5 Rate_A Rate_B Rate_C | RHS 0 0 1 | (1/6 - 1/10 - 1/7.5)/-2 0 1 0 | 1/10 + 1/2*(1/6 - 1/10 - 1/7.5) 1 0 0 | 1/7.5 + 1/2*(1/6 - 1/10 - 1/7.5) Now we sum down the RHS values to get our new rate: 1/20+1/15-1/12 + 1/10 +1/12-1/20-1/15+1/7.5+1/12-1/20-1/15 = 1/20+1/12+1/15 = (3+5+4)/60 = 12/60 = 1/5 1 = 1/5*t t = 5 days Quote
krusta80 Posted December 13, 2007 Posted December 13, 2007 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? I think the aim is to solve this equation: 1/A + 1/B + 1/C = 1/X Therefore we have three further equations, haven't we?! ;) 1/A + 1/B = 1/6 1/B + 1/C = 1/10 1/C + 1/A = 1/7,5 Is there any possibility to solve this problem without using and transforming these three equations? (it takes a lot of time and I get some distorted numbers :( ) I were very happy if anybody could help me?!? Thanks a lot... Usually with stuff like this, the choices help make the problem easier. What choices were given? Quote
dwurster Posted December 13, 2007 Author Posted December 13, 2007 Thanks for your quick help!! Unfortunately I haven't any answer choices for this task, sorry. Otherwise I obviously would have posted them... Quote
yowas Posted December 13, 2007 Posted December 13, 2007 This was a long problem. The only way I know to do it is using the approach you started. Plug and chug. I got 5 days. Quote
iisan Posted December 14, 2007 Posted December 14, 2007 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? I think the aim is to solve this equation: 1/A + 1/B + 1/C = 1/X Therefore we have three further equations, haven't we?! ;) 1/A + 1/B = 1/6 1/B + 1/C = 1/10 1/C + 1/A = 1/7,5 Is there any possibility to solve this problem without using and transforming these three equations? (it takes a lot of time and I get some distorted numbers :( ) I were very happy if anybody could help me?!? Thanks a lot... DW: Just add all the three equations 2(1/A + 1/B + 1/C) = (5+3+4)/30 = 12/30 1/A + 1/B + 1/C = 1/5 Ans is 5. Quote
dwurster Posted December 14, 2007 Author Posted December 14, 2007 very nice! here we go... ;) (solving it with equations / a matrix would have been very time consuming) thx! Quote
Abhishek009 Posted February 3, 2011 Posted February 3, 2011 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? I were very happy if anybody could help me?!? Thanks a lot... Lets take the total work to be 30 units... A + B can do the total work in 6 days so in one day they can do (30 / 6) 5 units B + C can do the total work in 10 days so in one day they can do (30 / 10) 3 units A + C can do the total work in 7.5 days so in one day they can do (30 / 7.5) 4 units Now add those 2 ( A + B + C ) in one day does ( 5 + 3 +4 )12 units Now A + B + C produces 6 units .... So they can complete the total work in 30 / 6 = 5 days... Solved the problem without using equations and fractions... Quote
jevul Posted March 15, 2011 Posted March 15, 2011 1/A + 1/B = 1/6 --------> 1/A = 1/6 - 1/B >>> (1) 1/B + 1/C = 1/10 -------> 1/B = 1/10 - 1/C >>> (2) 1/C + 1/A = 1/7.5 -------> 1/C = 1/7.5 - 1/A >>> (3) 1/A + 1/B + 1/C = 1/X SUNSTITUTE equation (1), (2) & (3) (1/6 - 1/B) + (1/10 - 1/C) + (1/7.5 - 1/A) = 1/X [1/6 + 1/10 + 1/7.5] - [1/A + 1/B + 1/C] = 1/X |_______________| |_____________| 2/5 - 1/X = 1/X 2/5 = 2/X X = 5 days Quote
nbgb Posted July 8, 2012 Posted July 8, 2012 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? I think the aim is to solve this equation: 1/A + 1/B + 1/C = 1/X Therefore we have three further equations, haven't we?! ;) 1/A + 1/B = 1/6 1/B + 1/C = 1/10 1/C + 1/A = 1/7,5 Is there any possibility to solve this problem without using and transforming these three equations? (it takes a lot of time and I get some distorted numbers :( ) I were very happy if anybody could help me?!? Thanks a lot... Ideal {6*(A + B) - w, 10*(B + C) - w, (15*(A + C))/2 - w, (A + B + C)*d - w}={-5*w + d*w, 30*C - w, 15*B - w, 10*A - w} {{A -> 0, B -> 0, C -> 0, w -> 0}, {B -> (2*A)/3, C -> A/3, d -> 5, w -> 10*A}} d=5. Quote
nbgb Posted July 8, 2012 Posted July 8, 2012 {6*(A + B) - w, 10*(B + C) - w, (15*(A + C))/2 - w, (A + B + C)*d - w} {-5*w + d*w, 30*C - w, 15*B - w, 10*A - w} {{A -> 0, B -> 0, C -> 0, w -> 0}, {B -> (2*A)/3, C -> A/3, d -> 5, w -> 10*A}} d=5. Quote
nbgb Posted July 8, 2012 Posted July 8, 2012 idel {6*(A + B) - w, 10*(B + C) - w, (15*(A + C))/2 - w, (A + B + C)*d - w} ={-5*w + d*w, 30*C - w, 15*B - w, 10*A - w} {{A -> 0, B -> 0, C -> 0, w -> 0}, {B -> (2*A)/3, C -> A/3, d -> 5, w -> 10*A}} d=5. Quote
Brent Hanneson Posted August 11, 2012 Posted August 11, 2012 I'm trying to solve the following work problem: Working together, Andy and Bob can do a job in 6 days. Bob and Cindy can do the same job in 10 days. Cindy and Andy can do it in 7.5 days. How long will it take if they all work together? To avoid a lot of fractions, you can apply a number of "work units" to the entire job such that we're dealing with nice numbers. Let's say that the entire job requires 30 work units (since 6, 10 and 7.5 all divide nicely into 30) Let A = number of work units Andy can complete in 1 day Let B = number of work units Bob can complete in 1 day Let C = number of work units Cindy can complete in 1 day Andy and Bob can do a job in 6 days: So, it takes them 6 days to complete 30 work units. That means they can complete 5 work units per day . In other words, A+B=5 Bob and Cindy can do the same job in 10 days: So, it takes them 10 days to complete 30 work units. That means they can complete 3 work units per day . In other words, B+C=3 Cindy and Andy can do it in 7.5 days: So, it takes them 7.5 days to complete 30 work units. That means they can complete 4 work units per day . In other words, A+C=4 So, we have A+B=5 B+C=3 A+C=4 Add them all together, to get 2A + 2B + 2C = 12 Divide both sides by 2 to get: A+B+C=6 In other words, all 3 people can complete a total of 6 work units each day. If the total job is 30 work units, it will take them 5 days to complete the job Cheers, Brent Quote
pranava Posted August 21, 2012 Posted August 21, 2012 LCM of 6 days, 10 days and 7.5 days is 30. With that A+B=5 C+B=3 and C+A=4 2(A+B+C)=12, therefore A+B+C=6. All of them together can complete in 6 days. Quote
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