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Working alone at a constant rate, Machine A can produce 15 widgets in

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Bunuel
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Working alone at a constant rate, Machine A can produce 15 widgets in [#permalink] New post   18 Jan 2019, 01:10
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Working alone at a constant rate, Machine A can produce 15 widgets in 2 minutes. Working alone at a constant rate, Machine B can produce 20 widgets in 3 minutes. If the two machines start working at the same time, after how many minutes will Machine A have produced 40 more widgets than Machine B?

A 12
B 24
C 36
D 48
E 60
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philipssonicare
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Re: Working alone at a constant rate, Machine A can produce 15 widgets in [#permalink] New post   18 Jan 2019, 20:26
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Working alone at a constant rate, Machine A can produce 15 widgets in 2 minutes.
Machine A rate: \(\frac{15 (widgets)}{2 (minutes)}\)

Working alone at a constant rate, Machine B can produce 20 widgets in 3 minutes.
Machine B rate:\(\frac{20 (widgets)}{3 (minutes)}\)

If the two machines start working at the same time, after how many minutes will Machine A have produced 40 more widgets than Machine B?
rate·time=work completed
We have the respective rates from above.
Machine A and B both work for the same time (we'll denote this time with x).
We know the work completed is a difference of 40

Machine A Rate·x - Machine B Rate·x=40 widgets
\(\frac{15 (widgets)}{2 (minutes)}·x - \frac{20 (widgets)}{3 (minutes)}·x=40 (widgets)\)
eliminate units for simplicity. Just remember x will be units of time
\(\frac{15x}{2}-\frac{20x}{3}=40\)
\(\frac{45x}{6}-\frac{40}{6}=40\)
\(\frac{5x}{6}=40\)
5x=240
x=48 minutes

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Re: Working alone at a constant rate, Machine A can produce 15 widgets in [#permalink] New post   18 Jan 2019, 08:09
Bunuel wrote:
Working alone at a constant rate, Machine A can produce 15 widgets in 2 minutes. Working alone at a constant rate, Machine B can produce 20 widgets in 3 minutes. If the two machines start working at the same time, after how many minutes will Machine A have produced 40 more widgets than Machine B?

A 12
B 24
C 36
D 48
E 60


rate for A = 15/2
rate for B = 20/3

solve using answer options we see
that at 48 mins ; A would have made 40 more widgets than B
15*48/2 = 360
20*48/3 = 320

IMO D
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Re: Working alone at a constant rate, Machine A can produce 15 widgets in [#permalink] New post   23 Mar 2023, 06:16
No. of widgets by machine A in 3 minutes = 22.5
i.e. 15 widgets in 2 mins
7.5 widgets in 1 mins
22.5 widgets in 3 mins

we need to find the time when Machine A will have 40 widgets more than Machine B.
so, every 3 mins Machine A will have 2.5 widgets more than Machine B.
thus for 40 widgets:
40/2.5=16 --- 16 times

So the total time will be 16 time 3 mins = 48 minutes.
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Re: Working alone at a constant rate, Machine A can produce 15 widgets in [#permalink] New post   23 Mar 2023, 11:40
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Bunuel wrote:
Working alone at a constant rate, Machine A can produce 15 widgets in 2 minutes. Working alone at a constant rate, Machine B can produce 20 widgets in 3 minutes. If the two machines start working at the same time, after how many minutes will Machine A have produced 40 more widgets than Machine B?

A 12
B 24
C 36
D 48
E 60


We can use the formula:

(difference in work)/(difference in rate) = time

40/(15/2 - 20/3) = time

40/(15/2 - 20/3) = time

40/(45/6 - 40/6) = time

40/(5/6) = time

240/5 = 48 = time

Answer: D
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Re: Working alone at a constant rate, Machine A can produce 15 widgets in [#permalink]
23 Mar 2023, 11:40
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