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# Working together, Machine A and Machine B can produce a total of 200

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Study Buddy Forum Moderator
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Working together, Machine A and Machine B can produce a total of 200 [#permalink]

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22 Dec 2017, 21:34
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Working together, Machine A and Machine B can produce a total of 200 widgets in 4 hours. How many hours would it take Machine A, working alone, to produce 200 widgets?

(1) Working alone, Machine B takes 5 hours to produce 50 widgets.

(2) Machine A can produce 4 widgets in the same amount of time it takes Machine B to produce 1 widget.
[Reveal] Spoiler: OA

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Re: Working together, Machine A and Machine B can produce a total of 200 [#permalink]

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22 Dec 2017, 22:25
Working together, Machine A and Machine B can produce a total of 200 widgets in 4 hours. How many hours would it take Machine A, working alone, to produce 200 widgets?

(1) Working alone, Machine B takes 5 hours to produce 50 widgets.

(2) Machine A can produce 4 widgets in the same amount of time it takes Machine B to produce 1 widget.

A+B = 50 widgets = 1 hour
A = 200 widgets = x hours

(1) B = 50 widgets = 4 hours
B = 50/4 widgets = 1 hour
We know
A+B = 50 widgets = 1 hour
Then A = 150/4 = 1 hour
4 hr = 150
4/3 hr = 50
16/3 hr = 200 widgets
Sufficient

(2) We know
A+B = 50 widgets = 1 hour
A will produce 40 widgets in 1 hr as B will produce 10 widgets in 1 hr (Machine A can produce 4 widgets in the same amount of time it takes Machine B to produce 1 widget)

Hence A = 40 = 1 hr
200 = 5 hr
Sufficient

D

Please correct me if i am wrong
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Re: Working together, Machine A and Machine B can produce a total of 200 [#permalink]

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22 Dec 2017, 22:45
niks18 amanvermagmat

What is wrong in below approach for St 2:

A = R * T
200 = $$\frac{200}{4}$$ * 4 (A + B)
4 = $$\frac{4}{x}$$ * x (Only A)
1 = $$\frac{1}{x}$$ * x (Only B) we are given same time x for each A and B

$$\frac{4}{x}$$ + $$\frac{1}{x}$$ = 50 or x = $$\frac{1}{10}$$
time = Amount of work / rate ie 200 / (1/10) ie 2000
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Re: Working together, Machine A and Machine B can produce a total of 200 [#permalink]

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23 Dec 2017, 01:34
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niks18 amanvermagmat

What is wrong in below approach for St 2:

A = R * T
200 = $$\frac{200}{4}$$ * 4 (A + B)
4 = $$\frac{4}{x}$$ * x (Only A)
1 = $$\frac{1}{x}$$ * x (Only B) we are given same time x for each A and B

$$\frac{4}{x}$$ + $$\frac{1}{x}$$ = 50 or x = $$\frac{1}{10}$$
time = Amount of work / rate ie 200 / (1/10) ie 2000

Sorry but i could not comprehend your approach

This is a very simple question that do not require any calculation. For DS you need to simplify the question stem first and then take a shot at the statements

Let A's rate be $$r_1$$ widget/hr and B's rate be $$r_2$$ widget/hr. So in $$4$$ hours number of widgets produced by them -

$$4r_1+$$$$4r_2=200=>$$ $$r_1+$$$$r_2=50$$-----------(1) we need to find $$r_1$$. so we need either the value of $$r_2$$ or relationship between $$r_1$$ & $$r_2$$

Statement 1: Directly provides the value of $$r_2=\frac{50}{5}=10$$. Hence sufficient

Statement 2: Directly gives the relationship between $$r_1$$ & $$r_2$$. Hence sufficient.

For the sake of calculation, let B produces $$1$$ widget in $$t$$ hours so $$r_2=\frac{1}{t}=>t=\frac{1}{r_2}$$

also $$r_1=\frac{4}{t}=>r_1=4r_2$$ (substituting the value of $$t$$)

Hence we have $$r_1$$$$+r_2=50=>4r_2+r_2=50=>r_2=10$$

so $$r_1=40$$. Hence time taken by A to produce $$200$$ widgets $$= \frac{200}{40}=5$$ hrs

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Re: Working together, Machine A and Machine B can produce a total of 200 [#permalink]

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23 Dec 2017, 02:36
Thanks niks18 for your two cents.

I followed the same approach using variable x instead of t in my earlier solution.
Then I tried to add individual rates of A and B as you did here:

Quote:
For the sake of calculation, let B produces $$1$$ widget in $$t$$ hours so $$r_2=\frac{1}{t}=>t=\frac{1}{r_2}$$

also $$r_1=\frac{4}{t}=>r_1=4r_2$$ (substituting the value of $$t$$)

Hence we have $$r_1$$$$+r_2=50=>4r_2+r_2=50=>r_2=10$$

In your approach you substituted r2 for r1 and I simply added the individual fractions (rates) to combine rates as 50.
I added the rates in terms of time and got final time as 1/10 hrs.
Let us say in your example we do not substitute r2 for r1 and since we have a single variable t we still have
1/t + 4/t = 50 or t = 1/10 hrs. Note that our final requirement is to find t (as per Q stem)

In spite of intermediate steps coinciding why are final answers different?
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Re: Working together, Machine A and Machine B can produce a total of 200 [#permalink]

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23 Dec 2017, 03:24
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Thanks niks18 for your two cents.

I followed the same approach using variable x instead of t in my earlier solution.
Then I tried to add individual rates of A and B as you did here:

Quote:
For the sake of calculation, let B produces $$1$$ widget in $$t$$ hours so $$r_2=\frac{1}{t}=>t=\frac{1}{r_2}$$

also $$r_1=\frac{4}{t}=>r_1=4r_2$$ (substituting the value of $$t$$)

Hence we have $$r_1$$$$+r_2=50=>4r_2+r_2=50=>r_2=10$$

In your approach you substituted r2 for r1 and I simply added the individual fractions (rates) to combine rates as 50.
I added the rates in terms of time and got final time as 1/10 hrs.
Let us say in your example we do not substitute r2 for r1 and since we have a single variable t we still have
1/t + 4/t = 50 or t = 1/10 hrs. Note that our final requirement is to find t (as per Q stem)

In spite of intermediate steps coinciding why are final answers different?

aahhh, now I got your approach

ya so your method is perfectly fine and time taken by A for producing 4 widgets will be $$t=\frac{1}{10}=0.1$$ hrs

so time taken by A to produce $$200$$ widgets $$= \frac{200}{4}*0.1=5$$ hrs

what you missed in your approach is that $$\frac{1}{10}$$ is the time taken by A to produce $$4$$ widgets, hence time to produce only $$1$$ widget will be $$\frac{1}{10*4}$$

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Re: Working together, Machine A and Machine B can produce a total of 200   [#permalink] 23 Dec 2017, 03:24
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