A machine puts c caps on bottles in m minutes. How many hours will it

The key for me is to realise that 1 bottle can only have 1 cap.
(ie bottles can be multiplied with minutes/cap and cancel out leaving only time)

Without that it seems super difficult but actually is a very straightforward calculation.

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.

We can plug in simple numbers

2 caps per 1 min,, then rate= 2 caper/min

Put b= 120 or 60 bottles.. I used 120 bottles

120= 2 * t.....>t=60 min =1 hr

Scanning the answer choice.. Eliminate A & D as you need to divide minute by 60.
Eliminate E as CM should be together.

Substitute number in choice B & C

In bm/60c= (120*1)/ (2 *60) = 1

Answer: B

A machine puts c caps on bottles in M minutes. A bottle can only have 1 cap, so this means that in M minutes, the machine will cap C bottles

\(\frac{M minutes}{C bottles}\)

How many hours will it take to put caps on B bottles? This is a comparison ratio:

\(\frac{M minute}{C bottles}\) * \(\frac{1 hour}{60 minutes}\) = \(\frac{X hours}{B bottles}\)

This becomes:

\(\frac{(M minute)(B bottles)(1 hour)}{(C bottles)(60 minutes)}\) = \(x hours\)

Simplify to:

\(\frac{mb}{60c}\) = \(x hours\)

This question's only difficulty is seeing that they want you to do a comparison between c bottles and b bottles. They intentionally made the letters confusing for those that could only see b should stand for bottles.

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

Rate Time = Work
C/M M C

C/M x = B

So, (RT = W); X*C/M = B; BM/C this is in Hours, therefore BM/C/60= BM/C60

my approach:

we have : c caps --> m minutes
therefore b caps --> \(\frac{(b*m)}{c}\) minutes

And : 60 minutes = 1 hour
so: \(\frac{(b*m)}{c}\) minutes = \(\frac{(b*m)}{(c*60)}\) hours

Answer: B.

rate of caps on bottles in per m minute ; c/m
and in 1 hour caps being put on bottles at given rate ; c*60/m
w = r*t
b = c*60/m * t
t = bm/60c
option B