Question 1155818: a candy company knows that it will take one candy wrapping machine 40 minutes to fill an order. after the machine has been working for 15 minutes the machine operators start another candy wrapping machine on the same order. with both machines running they finish the order 15 minutes later. how long would it take the second machine working alone to fill the order?
Found 3 solutions by josgarithmetic, jim_thompson5910, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Lets say we had 6000 candy bars. We can have any number of bars. I'm picking an arbitrarily large value that I have found to work. This value won't produce any fractional or decimal amounts when we divide later on (I found 6000 through trial and error more or less).
If it takes 40 minutes for machine A to wrap all 6000 bars, then machine A's rate is 6000/40 = 150 bars per minute. In other words, after one minute has passed by, machine A is able to wrap 150 bars.
After machine A has worked for 15 minutes, it is able to wrap 15*150 = 2250 bars. After another 15 minutes, machine A wraps another 2250 bars. After 30 minutes, machine A has wrapped 2250+2250 = 4500 bars.
There are 6000-4500=1500 bars left over for machine B to work on. It does this in 15 minutes. Machine B's rate is 1500/15 = 100 bars per minute.
Now let's give the entire job (of wrapping 6000 bars) to machine B only. We don't know how long it will take for machine B to get the job done all by itself, so we'll make x equal to that time value.
x = time it takes for machine B to get the job done all by itself
Assuming machine B's rate is held the same, it wraps 100 bars per minute, so we can say,
(rate)*(time) = total number of bars needed to be wrapped
(100 bars per min)*(x minutes) = 6000 bars
100x = 6000
100x/100 = 6000/100 ... divide both sides by 100
x = 60 minutes = 1 hour
Answer: 1 hour
This is assuming you want machine B to wrap all 6000 bars without machine A's help.
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Another approach:
We start off with x total bars to wrap. Machine A can wrap them all in 40 minutes, so its rate is x/40 bars per minute. After 30 minutes, it has wrapped 30(x/40) = 30x/40 = 3x/4 bars in total. Leaving x-3x/4 = 4x/4-3x/4 = x/4 bars left over for machine B to handle.
Machine B wraps x/4 bars in 15 minutes, so its rate is
(x/4 bars) divided by (15 minutes) = (x/4)/15 = (x/4)*(1/15) = x/60
the units are bars per minute
Let y be the time it takes machine B to do the job of wrapping all the bars by itself.
(rate)*(time) = number of bars
(x/60)*y = x
(1/60)*y = 1 ... divide both sides by x, they cancel
y = 60 minutes = 1 hour
We get the same answer as before. The only real difference is that we're using an unknown total number of bars (x) rather than some numeric amount. The first approach is probably a more common way to do it, in terms of doing it in the real world and if you haven't had much experience with algebra. I'm including the second approach to show how algebra ties in.
Edit: The tutor @josgarithmetic has an interesting approach, though unfortunately solving the provided equation for x leads to a negative value. This is one way to see that the equation is not valid. You should get a positive x solution. A negative time value does not make sense.
Edit2: Now I realize the typo. The 1/20 should be 1/40. Other than that, everything is perfect. The equation  solves to x = 60.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The (corrected) solution by tutor @josgarithmetic is a typical good formal algebraic approach.
If a formal algebraic solution is not required, you can solve the problem informally, using basically the same calculations, with just a few quick mental calculations.
If I were to solve this problem by the quickest possible method, I would do it something like this.
The first machine works for a total of 30 minutes; since it takes that machine 40 minutes to do the job alone, in those 30 minutes it does 3/4 of the job.
That means the second machine, which works for 15 minutes, does 1/4 of the job.
And that means it would take the second machine 60 minutes to do the whole job alone.
Note if you finish the formal algebraic solution suggested by tutor @josgarithmetic, those are exactly the same calculations you do.
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