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

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

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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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New post 22 Dec 2017, 22:25
adkikani wrote:
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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New post 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

adding total rates:
\(\frac{4}{x}\) + \(\frac{1}{x}\) = 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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adkikani wrote:
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

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


Hi adkikani

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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New post 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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New post 23 Dec 2017, 03:24
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adkikani wrote:
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 :cool:

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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