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Machine A can finish of a work in 4 hours, while В ca
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13 Feb 2011, 13:18
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Machine A can finish of a work in 4 hours, while В can finish of the work in 2 hours, and С can finish of the work in 8 hours. How many hours will it take for A, B, and С together to finish the work? (A) (B) (C) (D) (E)
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Re: Machine A can finish of a work in 4 hours, while В ca
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13 Feb 2011, 13:25



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Re: Machine A can finish of a work in 4 hours, while В ca
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13 Feb 2011, 13:55
banksy wrote: Machine A can finish of a work in 4 hours, while В can finish of the work in 2 hours, and С can finish of the work in 8 hours. How many hours will it take for A, B, and С together to finish the work? (A) (B) (C) (D) (E) General formula for multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously. For example if: Time needed for A to complete the job is A hours; Time needed for B to complete the job is B hours; Time needed for C to complete the job is C hours; ... Time needed for N to complete the job is N hours; Then: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}\), where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously. Check this for more: wordtranslationsrateswork104208.html#p812628For our original question it'll be: 1/4+1/2+1/8=1/T > 7/8=1/T > T=8/7. Answer: 8/7 hours. Please provide answer choices for PS questions.
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Re: Machine A can finish of a work in 4 hours, while В ca
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14 Feb 2011, 15:11
Thank you=)
But I still got a question. Is it ok to use this formula if we know that machines finish different parts of work(A1/4 of all work, B1/6...etc)...I thought we can use it only if we know that machines A,B,C finish whole work with different rates.



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Re: Machine A can finish of a work in 4 hours, while В ca
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14 Feb 2011, 15:33
banksy wrote: Thank you=)
But I still got a question. Is it ok to use this formula if we know that machines finish different parts of work(A1/4 of all work, B1/6...etc)...I thought we can use it only if we know that machines A,B,C finish whole work with different rates. If we are given that A can do 1/4 of the job in say 4 hours and B can do 1/6 of the same job in 2 hours then A can do the whole job in 4*4=16 hours and B can do the whole job in 6*2=12 hours so 1/16+1/12=1/T. Or as the formula for multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\) basically represents the sum of the rates of these entities (as rate is reciprocal of time and we can add the rates of the entities to get combined rate) then the rate of A would be rate=job/time=(1/4)/4=1/16 and the rate of B would be rate=job/time=(1/6)/2=1/12 > 1/16+1/12=1/T. Check the link in my previous post for more.
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Re: Machine A can finish of a work in 4 hours, while В ca
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15 Feb 2011, 03:39
Bunuel, can you please explain me something. machine A in one hour make 1/4 of the work machine B in one hour make 1/2 of the work machine C in one hour make 1/8 of the work so in the first hour of work the machines together made 7/8 of the work. Means  we need to finish 1/8 of the work. why it will be wrong to say that we need one hour + 1/8 hour to finish the work with all the 3 machines?
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Re: Machine A can finish of a work in 4 hours, while В ca
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15 Feb 2011, 05:26
144144 wrote: Bunuel, can you please explain me something. machine A in one hour make 1/4 of the work machine B in one hour make 1/2 of the work machine C in one hour make 1/8 of the work so in the first hour of work the machines together made 7/8 of the work. Means  we need to finish 1/8 of the work. why it will be wrong to say that we need one hour + 1/8 hour to finish the work with all the 3 machines? Everything is correct till the red part: yes, in 1 hour three machines together can do 7/8 of the job and 1/8 of the job is to be done, which means that 7 parts out of 8 are done and 1 part is to be done, so 1/7 th of what is already done is to be done. Three machines will need 1/7 th of an hour more to finish the whole job > 1 hour +1/7 hours = 8/7 hours, the same result as above. Of course this can be done much simpler: 7/8 job/hour > whole job in 1/(7/8)=8/7 hours (time is reciprocal of rate).
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Machine A can finish of a work in 4 hours, while В ca
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15 Feb 2011, 22:27
Ye, i got it. thanks for the explanation. have a great day. +1
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Re: Machine A can finish of a work in 4 hours, while В ca
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12 May 2018, 02:17
banksy wrote: Machine A can finish of a work in 4 hours, while В can finish of the work in 2 hours, and С can finish of the work in 8 hours. How many hours will it take for A, B, and С together to finish the work? (A) (B) (C) (D) (E) "R" stands for rate
R(A) = 1/4
R(B) = 1/2
R(C) = 1/8
Combined rate = 1/4+1/2+1/8 = 7/8
Time = 8/7 hrs
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