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A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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20 Mar 2013, 18:13
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A machine puts c caps on bottles in m minutes. How many hours will it take to put caps on b bottles? A. 60bm/c B. bm/60c C. bc/60m D. 60b/cm E. b/60cm guys this is how i got to solve the above problem;i found at the rate at which the machine produces a bottle in the said minutes hence c/m.time eqauls work done over rate;in that case,b/c/60m=60bm/c...following my algorithm,i got option A for an answer yet the problem holds B for an answer.what could possibly go wrong?
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Last edited by Bunuel on 21 Mar 2013, 03:25, edited 1 time in total.
Renamed the topic and edited the question.



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Re: A machine puts c caps on bottles [#permalink]
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20 Mar 2013, 18:44
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The machine puts c caps in m minutes or c caps in m/60 hours. Rate = c/m/60 = 60c/m. To put b caps the machine will take: Time = Work/Rate T = b/60c/m = bm/60c Answer is B. In your algorithm, by mistake you converted minutes into hours by multiplying minutes with 60, you should have divided minutes by 60.
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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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21 Mar 2013, 04:28
chiccufrazer1 wrote: A machine puts c caps on bottles in m minutes. How many hours will it take to put caps on b bottles?
A. 60bm/c B. bm/60c C. bc/60m D. 60b/cm E. b/60cm
guys this is how i got to solve the above problem;i found at the rate at which the machine produces a bottle in the said minutes hence c/m.time eqauls work done over rate;in that case,b/c/60m=60bm/c...following my algorithm,i got option A for an answer yet the problem holds B for an answer.what could possibly go wrong? A machine puts c caps on bottles in m minutes > the rate of the machine is (rate) = (job)/(time) = c/m cap/minute. The time needed to put caps on b bottles (time) = (job)/(rate) = b/(c/m) minutes = bm/c minutes = bm/(60c) hours. Answer: B.
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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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21 Mar 2013, 09:42
Fantastic The answer is B. Word problem. We need to get to find out the unit (how much in 1 type) to get to the answer!
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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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01 Feb 2014, 09:16
chiccufrazer1 wrote: A machine puts c caps on bottles in m minutes. How many hours will it take to put caps on b bottles?
A. 60bm/c B. bm/60c C. bc/60m D. 60b/cm E. b/60cm
guys this is how i got to solve the above problem;i found at the rate at which the machine produces a bottle in the said minutes hence c/m.time eqauls work done over rate;in that case,b/c/60m=60bm/c...following my algorithm,i got option A for an answer yet the problem holds B for an answer.what could possibly go wrong? A machine puts c caps in m minutes per bottle. How many hours will it take to put caps on b bottles? I think this wording is more clear Cheers J



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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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29 Mar 2014, 00:48
jlgdr wrote: chiccufrazer1 wrote: A machine puts c caps on bottles in m minutes. How many hours will it take to put caps on b bottles?
A. 60bm/c B. bm/60c C. bc/60m D. 60b/cm E. b/60cm
guys this is how i got to solve the above problem;i found at the rate at which the machine produces a bottle in the said minutes hence c/m.time eqauls work done over rate;in that case,b/c/60m=60bm/c...following my algorithm,i got option A for an answer yet the problem holds B for an answer.what could possibly go wrong? A machine puts c caps in m minutes per bottle. How many hours will it take to put caps on b bottles? I think this wording is more clear Cheers J Hi there, IMHO the wording is already clear since 1 cap can be per bottle only it cannot be the case that 3 caps are on 1 bottle. 'C' Caps > 'm' minutes 1 Cap/bottle > m/c minutes b bottle > mb/c minutes mb/60 c
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A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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25 Jun 2014, 21:24
Either we can fix it with the help of a small formula which is : (M1 * D1 * H1)/ W1 = (M2 * D2 * H2)/ W2 where M , D and H stands for Men, Days and Hours invested for each of the 2 Works respectively.
So here it goes: M1= 1 (as we are not given the value) D1 = m/60 (in hrs) H1 = 1 (as we are not given the value) and W1 = c
and
M2 = 1 (as we are not given the value) D2 = x (in hrs) (to be found out) H2 = 1 (as we are not given the value) and W2 = b (as given)
So, (1 *m/60 *1)/c = (1* x* 1)/b => (m/60)/c = x/b Therefore, => x= bm/60c.
Answer B.
P.S. Please let me know if anything went wrong as this is my first post as a solution.



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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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25 Jun 2014, 21:59
Time to put a cap on 1 bottle is m/c minutes Time to put caps on b bottles is bm/c minutes = bm/(60*c) hours
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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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25 Jun 2014, 22:23
Time required for c caps = m minutes OR Time required for c caps = \(\frac{m}{60}\) Hrs So time required for 1 cap \(= \frac{m}{60} * \frac{1}{c}\) \(= \frac{m}{60c}\) So time required for b caps \(= b * \frac{m}{60c}\) \(= \frac{bm}{60c}\) Answer = B
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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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04 Nov 2015, 14:48
If machine caps c bottles in m minutes, rate per minute is c/m. Rate per hour will be 60c/m. To cap b bottles, machine will need b/(60c/m)=bm/60c hours.



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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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Re: A machine puts c caps on bottles in m minutes. How many hour [#permalink]
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12 Jun 2017, 19:46
chiccufrazer1 wrote: A machine puts c caps on bottles in m minutes. How many hours will it take to put caps on b bottles?
A. 60bm/c B. bm/60c C. bc/60m D. 60b/cm E. b/60cm bottling rate=c/m minutes to cap b bottles=b/(c/m)=bm/c hours to cap b bottles=(bm/c)/60=bm/60c B




Re: A machine puts c caps on bottles in m minutes. How many hour
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