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Running at their respective constant rates, Machine X takes

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Running at their respective constant rates, Machine X takes [#permalink] New post 18 Mar 2014, 00:39
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

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Question: 173
Category: Algebra; Applied problems
Page: 85
Difficulty: 600


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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 18 Mar 2014, 00:39
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SOLUTION

Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

For work problems one of the most important thin to know is rate*time=job \ done.

Let the time needed for machine X to produce w widgets be t days, so the rate of X would be rate=\frac{job \ done}{time}=\frac{w}{t};

As "machine X takes 2 days longer to produce w widgets than machines Y" then time needed for machine Y to produce w widgets would be t-2 days, so the rate of Y would be rate=\frac{job \ done}{time}=\frac{w}{t-2};

Combined rate of machines X and Y in 1 day would be \frac{w}{t}+\frac{w}{t-2} (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: 3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4} --> \frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}.

\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12} --> reduce by w --> \frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}.

At this point we can either solve quadratic equation: 5t^2-34t+24=0 --> (t-6)(5t-4)=0 --> t=6 or t=\frac{4}{5} (which is not a valid solution as in this case t-2=-\frac{6}{5}, the time needed for machine Y to ptoduce w widgets will be negatrive value and it's not possible). So t=6 days is needed for machine X to produce w widgets, hence time needed for machine X to produce 2w widgets will be 2t=12 days.

OR try to substitute the values from the answer choices. Remember as we are asked to find the time needed for machine X alone to produce 2w widgets then the answer should be 2t among answer choices: E work - 2t=12 --> t=6 --> \frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}.

Answer: E.
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Running at their respective constant rates, Machine X takes [#permalink] New post 20 Mar 2014, 00:52
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Time taken by Machine Y = t days

Time taken by Machine X = (t+2) days

Rate of Machine X = \frac{w}{t+2}
Rate of Machine Y = \frac{w}{t}

\frac{5w}{4} widgets produced in 3 days; so \frac{5w}{12} produced in 1 day

Equation setup will be

\frac{w}{t+2} + \frac{w}{t} = \frac{5w}{12}

Solving,

5t^2 - 14t - 24 = 0

t = 4

Time taken by Machine X = t+2 = 6 for w

So for 2w; it will be 12

Answer = E

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Last edited by PareshGmat on 14 Aug 2014, 02:01, edited 1 time in total.
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 23 Mar 2014, 07:13
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SOLUTION

Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

For work problems one of the most important thin to know is rate*time=job \ done.

Let the time needed for machine X to produce w widgets be t days, so the rate of X would be rate=\frac{job \ done}{time}=\frac{w}{t};

As "machine X takes 2 days longer to produce w widgets than machines Y" then time needed for machine Y to produce w widgets would be t-2 days, so the rate of Y would be rate=\frac{job \ done}{time}=\frac{w}{t-2};

Combined rate of machines X and Y in 1 day would be \frac{w}{t}+\frac{w}{t-2} (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: 3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4} --> \frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}.

\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12} --> reduce by w --> \frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}.

At this point we can either solve quadratic equation: 5t^2-34t+24=0 --> (t-6)(5t-4)=0 --> t=6 or t=\frac{4}{5} (which is not a valid solution as in this case t-2=-\frac{6}{5}, the time needed for machine Y to ptoduce w widgets will be negatrive value and it's not possible). So t=6 days is needed for machine X to produce w widgets, hence time needed for machine X to produce 2w widgets will be 2t=12 days.

OR try to substitute the values from the answer choices. Remember as we are asked to find the time needed for machine X alone to produce 2w widgets then the answer should be 2t among answer choices: E work - 2t=12 --> t=6 --> \frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}.

Answer: E.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 16 May 2014, 06:22
QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 16 May 2014, 08:32
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gciftci wrote:
QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?


Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

Hope it helps.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 27 May 2014, 14:16
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Alternate Solution without algebra using POE.

If it takes 3 days to produce 5/4 widgets, it will take 3*(4/5) = 12/5 days to produce 1 widget.

If two machines were working at the same rate it would take, 2*12/5 i.e. approx 5 days days for x to produce 1 widget. Since machine X is slower it will take more than 5 days to produce 1 widget and subsequently more than 10 days to produce 2 widgets. The only answer more than 10 is 12 days. Answer E
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Re: Running at their respective constant rates, Machine X takes [#permalink] New post 10 Aug 2014, 01:25
Back tracking method:
We have to find the number of days it takes Machine X alone to produce 2w widgets.
Answer options are as below:

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12

We are given that X and Y produce (5/4)w in 3 days and also given that X take 2 days more than Y.
So to find corresponding value of X and Y use the answer options.
First half the value of X to find number of days it takes Machine X to complete *W* widgets and find the value of Y by reducing 2 from all answer options.

--> X to complete W widgets
(A) X=2 and Y= 2-2 =0
(B) X=3 and Y= 3-2 =1
(C) X=4 and Y= 4-2 =2
(D) X=5 and Y= 5-2 =3
(E) X=6 and Y= 6-2 =4

Now we have all the values of X and Y. Find out the combination X and Y, whose values will produce (5/4)w for 3 days of work.

3(1/6+1/4)=>5/4-->Option (E) gives this value so correct answer is (E)-->12.
Re: Running at their respective constant rates, Machine X takes   [#permalink] 10 Aug 2014, 01:25
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