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Machine X can produce 3 widgets every 2 minutes and Machine Y produces

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Machine X can produce 3 widgets every 2 minutes and Machine Y produces  [#permalink]

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Machine X can produce 3 widgets every 2 minutes and Machine Y produces 5 widgets every 3 minutes. If both the machines work together for 25 minutes what part of the production will be finished by Machine X?

(A) 2/19
(B) 3/19
(C) 4/19
(D) 5/19
(E) 9/19

Originally posted by dhruv09arora on 16 Sep 2018, 22:43.
Last edited by Bunuel on 16 Sep 2018, 23:21, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Machine X can produce 3 widgets every 2 minutes and Machine Y produces  [#permalink]

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New post 17 Sep 2018, 06:44
dhruv09arora wrote:
Machine X can produce 3 widgets every 2 minutes and Machine Y produces 5 widgets every 3 minutes. If both the machines work together for 25 minutes what part of the production will be finished by Machine X?

(A) 2/19
(B) 3/19
(C) 4/19
(D) 5/19
(E) 9/19


Machine X ..
3 widgets in 2 minutes, so 3/2 widget in 1 minute....
Therefore 25*3/2=75/2 in 25 minutes

Machine Y
5 widgets in 3 minutes, so 5/3 widget in 1 minute....
Therefore 25*5/3=125/3 in 25 minutes

Total widgets = (75/2+125/3)=(225+250)/6=475/6
Machine X's work = (75/2)/(475/6)=(75*6)/2*475 = (25*3*6)/(25*19*2)=9/19

E
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Re: Machine X can produce 3 widgets every 2 minutes and Machine Y produces  [#permalink]

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New post 17 Sep 2018, 07:18
Machine X can produce 3 widgets every 2 minutes and Machine Y produces 5 widgets every 3 minutes. If both the machines work together for 25 minutes what part of the production will be finished by Machine X?

(A) 2/19
(B) 3/19
(C) 4/19
(D) 5/19
(E) 9/19


Work per minute of x : 3/2 and y : 5/3

Combined rate shall be {(3/2)+(5/3)}: 19/6
in 25 mins work would be (19/6)*25

X work in 25 mins (25 * x rate); 25* (3/2)

Part of work done by X in 25mins for combined work done :

[ (6/19)*(1/25) * (3/2)* 25)
=> 9/19
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Re: Machine X can produce 3 widgets every 2 minutes and Machine Y produces  [#permalink]

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New post 17 Sep 2018, 09:48
1
dhruv09arora wrote:
Machine X can produce 3 widgets every 2 minutes and Machine Y produces 5 widgets every 3 minutes. If both the machines work together for 25 minutes what part of the production will be finished by Machine X?

(A) 2/19
(B) 3/19
(C) 4/19
(D) 5/19
(E) 9/19


Efficiency of Machine X = 3/2 widgets per minute
Efficiency of Machine Y = 5/3 widgets per minute

So, In 6 Minutes X Produces 3/2*6 = 9 widgets
And In 6 Minutes X Produces 5/3*6 = 10 widgets

Thus, in 6 minutes Machine X and Y produces 19 widgets and X's contribution is 9/19 widgets, Answer must be (E)
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Machine X can produce 3 widgets every 2 minutes and Machine Y produces  [#permalink]

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New post 17 Sep 2018, 10:07
dhruv09arora wrote:
Machine X can produce 3 widgets every 2 minutes and Machine Y produces 5 widgets every 3 minutes. If both the machines work together for 25 minutes what part of the production will be finished by Machine X?

(A) 2/19
(B) 3/19
(C) 4/19
(D) 5/19
(E) 9/19

One approach: find (Part X : Part Y)
Then (Part X : Whole X+Y)

I. Change the time, find (part: part) ratio, then (part: whole)
Use ANY easier time period to find PART : PART ratio

Relative portions that X and Y finish will not change as long as time worked is the same
\(X\)'s part of total production is: \(\frac{X}{X+Y}\)

(1) Change the time period.
• X's output + Y's output = total output

• Total output for each is:

\(X\): \(r_{x}*T=W_{x}\)

\(Y\): \(r_{y}*T=W_{y}\)

Same time worked? Then number of widgets each makes depends ONLY on their rates.

(2): compare rates with a time that is easier than 25 minutes
• Use LCM of 6 minutes to get an identical time period worked, which will yield
relative portions of total work each produces

\(X\)'s rate in 6 minutes: \(\frac{3w}{2min}=\frac{9w}{6min}\)

\(Y\)'s rate in 6 minutes: \(\frac{5w}{3min}=\frac{10w}{6min}\)

• The part of production finished by X?

\(\frac{X}{(X+Y)}=\frac{9}{9+10}=\frac{9}{19}\)

Answer E

II. Harder: TOTAL output in 25 minutes
In 25 minutes, X finishes \(\frac{X_{w}}{Total_{w}}\)

• \(X\), total output:

\(W_{x}=r_{x}*25\)

\(W_{x}=(\frac{3w}{2min}*25min)=\frac{75}{2}w\)

• \(Y\), total output

\(W_{y}=r_{y}*25\)

\(W_{y}=(\frac{5w}{3min}*25min)=\frac{125}{3}w\)


• Overall total output (X + Y):

\((X+Y)=(\frac{75}{2}w+\frac{125}{3}w)=\)

\((\frac{225}{6w}+\frac{250w}{6}w)=\frac{475}{6}w\)

• X finishes what part of total production?

\(\frac{X}{Total}=\frac{(\frac{75}{2}w)}{(\frac{475}{6}w)}=\)

\((\frac{75}{2}w*\frac{6}{475}w)=\) \(\frac{225w}{475w}\)

Divide by 25:

X finishes \(\frac{9}{19}\) of production

Answer E
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Re: Machine X can produce 3 widgets every 2 minutes and Machine Y produces  [#permalink]

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New post 18 Sep 2018, 17:52
dhruv09arora wrote:
Machine X can produce 3 widgets every 2 minutes and Machine Y produces 5 widgets every 3 minutes. If both the machines work together for 25 minutes what part of the production will be finished by Machine X?

(A) 2/19
(B) 3/19
(C) 4/19
(D) 5/19
(E) 9/19


Machine X’s rate is 3/2, and Machine Y’s rate is 5/3. Their combined rate is (3/2 + 5/3). We can create the expression:

(3/2)/(3/2 + 5/3)

Multiplying the expression by 6/6 we have:

9/(9 + 10) = 9/19

No matter the amount of time they work together, Machine X will always produce 9/19 of the combined output.

Answer: E
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Re: Machine X can produce 3 widgets every 2 minutes and Machine Y produces &nbs [#permalink] 18 Sep 2018, 17:52
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