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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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04 Dec 2015, 14:14
nycgirl212 wrote: Engr2012 wrote: You are absolutely correct to say that 't' is the total time taken when A and B work TOGETHER to finish the work. If you look at machinesaandbworkingtogethertaketminutestocomple147714.html#p1186258, it explains it really well and is as per your thinking. t is the total time taken by A and B together for the work, ta is the time taken by A to do the same work = t+64 and tb is the time taken by B to do the same work = t+25 Now the only thing that can speed up when 2 people do a job together is the rate at which the work is getting done. This rate = 1/t , 1/ta, 1/tb for rate when A/B work together, when A works alone and when B works alone respectively. Thus, per the rates: \(\frac{1}{t} = \frac{1}{ta}+ \frac{1}{tb}\) > \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > When you solve this equation, you get t = 40, > ta = t+64=104 and tb=t+25=65 > ta/tb = 104/65 = 8/5 or 8:5 . B is the correct answer. Hope this helps. Thanks, I totally follow the \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) but how do you get to t=40, you do (t+64)*(t+25) to get a common denominator? It just seems like a crazy quadratic.... Once you get, \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\), realize that 64 and 25 are perfect squares and as such the final answer must be a simple square root. \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > \(\frac{1}{t} = \frac{t+64+t+25}{(t+64)*(t+25)}\) > (t+64)*(t+25) =t*(t+64+t+25) >\(t^2+89t+64*25 = 2t^2+89t\) > \(t^2 = 64*25\)> \(t = 8*5 = 40\)



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Machines A and B, working together, take t minutes to comple [#permalink]
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04 Dec 2015, 14:20
Engr2012 wrote: nycgirl212 wrote: Engr2012 wrote: You are absolutely correct to say that 't' is the total time taken when A and B work TOGETHER to finish the work. If you look at machinesaandbworkingtogethertaketminutestocomple147714.html#p1186258, it explains it really well and is as per your thinking. t is the total time taken by A and B together for the work, ta is the time taken by A to do the same work = t+64 and tb is the time taken by B to do the same work = t+25 Now the only thing that can speed up when 2 people do a job together is the rate at which the work is getting done. This rate = 1/t , 1/ta, 1/tb for rate when A/B work together, when A works alone and when B works alone respectively. Thus, per the rates: \(\frac{1}{t} = \frac{1}{ta}+ \frac{1}{tb}\) > \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > When you solve this equation, you get t = 40, > ta = t+64=104 and tb=t+25=65 > ta/tb = 104/65 = 8/5 or 8:5 . B is the correct answer. Hope this helps. Thanks, I totally follow the \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) but how do you get to t=40, you do (t+64)*(t+25) to get a common denominator? It just seems like a crazy quadratic.... Once you get, \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\), realize that 64 and 25 are perfect squares and as such the final answer must be a simple square root. \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > \(\frac{1}{t} = \frac{t+64+t+25}{(t+64)*(t+25)}\) > (t+64)*(t+25) =t*(t+64+t+25) >\(t^2+89t+64*25 = 2t^2+89t\) > \(t^2 = 64*25\)> \(t = 8*5 = 40\) how do you get from \(2t^2+89t\) to \(t^2 = 64*25\)



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Machines A and B, working together, take t minutes to comple [#permalink]
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04 Dec 2015, 17:40
nycgirl212 wrote: Engr2012 wrote: Once you get, \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\), realize that 64 and 25 are perfect squares and as such the final answer must be a simple square root.
\(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > \(\frac{1}{t} = \frac{t+64+t+25}{(t+64)*(t+25)}\)
> (t+64)*(t+25) =t*(t+64+t+25) >\(t^2+89t+64*25 = 2t^2+89t\) > \(t^2 = 64*25\)> \(t = 8*5 = 40\)
how do you get from \(2t^2+89t\) to \(t^2 = 64*25\) You are missing the fact that \(t^2+89t+64*25 = 2t^2+89t\) is 1 equation entirely. You are considering that you get \(2t^2+89t\) from \(t^2+89t+64*25\) This is not true. You need to look at the complete equation, \(t^2+89t+64*25 = 2t^2+89t\) > cancelling \(89t\) from both sides and transferring \(t^2\) from LHS to RHS you get, \(64*25=t^2\) > \(t^2=64*25\)>\(t=8*5 = 40\) Hope this helps.



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Machines A and B, working together, take t minutes to comple [#permalink]
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07 Dec 2015, 14:38
Engr2012 wrote: nycgirl212 wrote: Engr2012 wrote: Once you get, \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\), realize that 64 and 25 are perfect squares and as such the final answer must be a simple square root.
\(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > \(\frac{1}{t} = \frac{t+64+t+25}{(t+64)*(t+25)}\)
> (t+64)*(t+25) =t*(t+64+t+25) >\(t^2+89t+64*25 = 2t^2+89t\) > \(t^2 = 64*25\)> \(t = 8*5 = 40\)
how do you get from \(2t^2+89t\) to \(t^2 = 64*25\) You are missing the fact that \(t^2+89t+64*25 = 2t^2+89t\) is 1 equation entirely. You are considering that you get \(2t^2+89t\) from \(t^2+89t+64*25\) This is not true. You need to look at the complete equation, \(t^2+89t+64*25 = 2t^2+89t\) > cancelling \(89t\) from both sides and transferring \(t^2\) from LHS to RHS you get, \(64*25=t^2\) > \(t^2=64*25\)>\(t=8*5 = 40\) Hope this helps. got it! that helps a lot, thank you!



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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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14 Oct 2016, 06:51
GMATPASSION wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 rate of A: 1/(t+64) rate of B: 1/(t+25) rate together: 1/t 1/t = 1/(t+64) +1/(t+25) 1/t = t+25+t+64/(t+64)(t+25) cross multiply: t(2t+89) = (t+64)(t+25) extend t^2 + 89t = t^2 + 25t+64t +64*25 we can either leave 64*25 or use the technique doubling and halving: 64*25 = 32*50 = 16*100 = 1600. now, we can get rid of one t^2, 89t and we have: t^2 = 1600 well, that's 40^2. we can also see that 64 is 8^2 and 25 is 5^2. thus, t must be 8*5 = 40. to get back to the problem.. t=40. time for A: 40+64 = 104 time for B: 40+25 = 65 rate: 104/65 A, C, D  out right away. between B and E there is no way from 104 to get 64 and from 65 to get 25... answer must be B alternatively, divide both 104 and 65 by 13 and get 8/5



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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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07 Nov 2017, 14:48
VeritasPrepKarishma wrote: skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 It's great if you can use the method of elimination to arrive at the final answer on the actual test. For practice questions, ensure that you understand how to solve it logically too so that if the options do not allow for elimination, you still know how to solve it. Why does machine A take 64 mins extra while working alone? Because machine B is not working. So machine B used to do this work in t mins when both were working together. Machine A now takes 64 mins for the work that machine B used to do in t mins. Ratio of time taken by A:B = 64:t Similarly, why is machine B taking 25 mins extra? Because machine A used to do this work in t mins when both were working together. Working alone, machine B finishes this portion of the work in 25 mins. Ratio of time taken by A:B = t:25 64/t = t/25 t = 40 Required ratio = 40/25 = 8/5 hi mam one simple question ratio of time taken by A : B = 64 : t how this is that it takes 'A' 64 minutes extra to complete some portion of the work formerly done by 'B' in 't' minutes...? isn't it such that it takes "A + B" to complete the work in "t" minutes....? thanks in advance, mam



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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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08 Nov 2017, 04:53
gmatcracker2017 wrote: VeritasPrepKarishma wrote: skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 It's great if you can use the method of elimination to arrive at the final answer on the actual test. For practice questions, ensure that you understand how to solve it logically too so that if the options do not allow for elimination, you still know how to solve it. Why does machine A take 64 mins extra while working alone? Because machine B is not working. So machine B used to do this work in t mins when both were working together. Machine A now takes 64 mins for the work that machine B used to do in t mins. Ratio of time taken by A:B = 64:t Similarly, why is machine B taking 25 mins extra? Because machine A used to do this work in t mins when both were working together. Working alone, machine B finishes this portion of the work in 25 mins. Ratio of time taken by A:B = t:25 64/t = t/25 t = 40 Required ratio = 40/25 = 8/5 hi mam one simple question ratio of time taken by A : B = 64 : t how this is that it takes 'A' 64 minutes extra to complete some portion of the work formerly done by 'B' in 't' minutes...? isn't it such that it takes "A + B" to complete the work in "t" minutes....? thanks in advance, mam Check this post: https://gmatclub.com/forum/ifdevworks ... l#p1957822I have explained this concept in detail. Let me know if something is still unclear.
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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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08 Nov 2017, 06:08
gmatcracker2017 wrote: VeritasPrepKarishma wrote: skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 It's great if you can use the method of elimination to arrive at the final answer on the actual test. For practice questions, ensure that you understand how to solve it logically too so that if the options do not allow for elimination, you still know how to solve it. Why does machine A take 64 mins extra while working alone? Because machine B is not working. So machine B used to do this work in t mins when both were working together. Machine A now takes 64 mins for the work that machine B used to do in t mins. Ratio of time taken by A:B = 64:t Similarly, why is machine B taking 25 mins extra? Because machine A used to do this work in t mins when both were working together. Working alone, machine B finishes this portion of the work in 25 mins. Ratio of time taken by A:B = t:25 64/t = t/25 t = 40 Required ratio = 40/25 = 8/5 hi mam one simple question ratio of time taken by A : B = 64 : t how this is that it takes 'A' 64 minutes extra to complete some portion of the work formerly done by 'B' in 't' minutes...? isn't it such that it takes "A + B" to complete the work in "t" minutes....? thanks in advance, mam hi mam I have read the explanation you have provided in another post, and perhaps, there is nothing more to explain, thanks. but my little doubt still centers on a simple concept: "whatever work B did in 't' minutes is now done by A in 64 minutes". yes, A has to expend some extra minutes on the work previously done by B, but how much time B actually needs to do that extra job..? when A and B work together, they need 't' minutes to complete the job.. if supposed that, when A and B work together, they need 5 minutes to finish the job, and A alone needs 8 minutes to complete the job. Here, A has to spend some extra 3 minute on completion the job. So far so good. So, according to you, time taken by A : time taken by B = 3 : 5 here it is assumed that, to do the extra job done by A, B needs 5 minutes, but according to the fact, given, 5minute is required by A and B when they work together Sorry to knock you again, mam



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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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09 Dec 2017, 05:55
VeritasPrepKarishma wrote: skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 It's great if you can use the method of elimination to arrive at the final answer on the actual test. For practice questions, ensure that you understand how to solve it logically too so that if the options do not allow for elimination, you still know how to solve it. Why does machine A take 64 mins extra while working alone? Because machine B is not working. So machine B used to do this work in t mins when both were working together. Machine A now takes 64 mins for the work that machine B used to do in t mins. Ratio of time taken by A:B = 64:t Similarly, why is machine B taking 25 mins extra? Because machine A used to do this work in t mins when both were working together. Working alone, machine B finishes this portion of the work in 25 mins. Ratio of time taken by A:B = t:25 64/t = t/25 t = 40 Required ratio = 40/25 = 8/5 Why do you say: "So machine B used to do this work in t mins when both were working together."? Is this true? Machines A and B used to do this work in t mins not machine B.



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Re: Machines A and B, working together, take t minutes to comple [#permalink]
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10 Dec 2017, 23:19
gmatcracker2017 wrote: hi mam I have read the explanation you have provided in another post, and perhaps, there is nothing more to explain, thanks. but my little doubt still centers on a simple concept: "whatever work B did in 't' minutes is now done by A in 64 minutes". yes, A has to expend some extra minutes on the work previously done by B, but how much time B actually needs to do that extra job..? when A and B work together, they need 't' minutes to complete the job.. if supposed that, when A and B work together, they need 5 minutes to finish the job, and A alone needs 8 minutes to complete the job. Here, A has to spend some extra 3 minute on completion the job. So far so good. So, according to you, time taken by A : time taken by B = 3 : 5 here it is assumed that, to do the extra job done by A, B needs 5 minutes, but according to the fact, given, 5minute is required by A and B when they work together Sorry to knock you again, mam Ok, think of the job as assembling toys. Say we have 45 toys to assemble. A is assembling toys and B is assembling toys. Together they start at 12 noon and are done in 2 ( = t) hours i.e. at 2 pm. So in 2 hrs, say A assembled 20 toys and B assembled 25 toys. They got done by 2 pm since they were working together. Now think  Say only A started working at 12 o clock. By 2 o clock, A assembled 20 toys (since that is A's speed). But we still have 25 toys left. Why? Because B did not work. Otherwise B would have assembled these 25 toys by 2 pm. Now A works alone till 4:30 pm to assemble these 25 toys i.e. for another 2.5 hrs. So can we say that whatever work was done by B in 2 (= t) hrs was done by A in the additional time? If B took 2 hrs to assemble 25 toys and A took 2.5 hrs to do the same work, So their ratio of time take which is 2:2.5 (= 4:5) is also their ratio of speeds since they did the same work in this time? Does this help?
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