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

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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?
[Reveal] Spoiler: OA

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
4
KUDOS
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
In your algorithm, by mistake you converted minutes into hours by multiplying minutes with 60, you should have divided minutes by 60.

Please give a kudo if you like my explanation.
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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
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Expert's post
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.

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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

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.

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

$$= b * \frac{m}{60c}$$

$$= \frac{bm}{60c}$$

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

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13 Oct 2015, 01:15
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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.
A machine puts c caps on bottles in m minutes. How many hour   [#permalink] 04 Nov 2015, 14:48
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