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work machine A & B

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by canuckclint » Sun Oct 26, 2008 4:59 pm
Machine A, working alone at a constant rate, can complete a certain production lot in x hours. Machine B, working alone at a constant rate, can complete 1/5 of the same production lot in y hours. Machines A and B, working together, can complete 1/2 of the same production lot in z hours. What is the value of y in terms of x and z?


5x – 10z

2xz


2xz

5x - 10z


5xz

x + z


xz

x + z


xz

x + 2z
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Last edited by canuckclint on Sun Oct 26, 2008 6:28 pm, edited 1 time in total.
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by 2009wish » Sun Oct 26, 2008 5:45 pm
A takes x hrs to complete one work

B takes 5y hrs to complete one work

together they take 2z hrs

hence 1/x + 1/5y = 1/ 2z (as rate is inversely proportional to time taken)

solving we get :

y= 2xy / 5 (x - 2z)
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by canuckclint » Sun Oct 26, 2008 6:26 pm
2009wish wrote:A takes x hrs to complete one work

B takes 5y hrs to complete one work

together they take 2z hrs

hence 1/x + 1/5y = 1/ 2z (as rate is inversely proportional to time taken)

solving we get :

y= 2xy / 5 (x - 2z)
Sorry but y cannot be in the answer.
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by vishubn » Sun Oct 26, 2008 6:51 pm
hence 1/x + 1/5y = 1/ 2z (as rate is inversely proportional to time taken)

solving we get :

y= 2xy / 5 (x - 2z)

I think this is a genuine typo.... ans is B !

numerator is not 2xy but instead it is 2xz


Vishu
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by cramya » Sun Oct 26, 2008 8:18 pm
1/x+1/5y = 1/2z

1/5y = 1/2z - 1/x

1/5y = (x-2z) / 2xz

5y = 2xz/x-2z
y = 2xz/5x-10z

B)
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by Stockmoose16 » Sun Oct 26, 2008 8:49 pm
cramya wrote:1/x+1/5y = 1/2z

1/5y = 1/2z - 1/x

1/5y = (x-2z) / 2xz

5y = 2xz/x-2z
y = 2xz/5x-10z

B)
The answer should be "B"... make a chart

-------------------RATE ------ TIME ------- DISTANCE
Machine A:-------1/x ------ Z----------- Z/X

Machine B: ------(1/5)/y -----Z ----------- Z/5Y

-----------------------------------------------1/2

Z/X + Z/5Y =1/2

Simplifies to:

5ZY/5YX + ZX/5YX =1/2

Simplifies to:

(5ZY + ZX)/5YX =1/2

Simplifies to:

10ZY +2ZX=5YX

Simplifies to:

5YX -10YZ=2ZX

Simplifies to:

Y(5X-10Z)=2ZX

Simplifies to:

5x-10Z=2ZX/Y

Multiply the right side of the equation by the reciprical (Y/2ZX)

1/Y=(5X-10Z)/2ZX

Reciprocate the equations:

Y= 2ZX/ (5X-10Z)

The answer is B.

Out of curiosity, what level did MGMAT list this question at?
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by logitech » Sun Oct 26, 2008 11:17 pm
Since it appeared as the 3rd question, it should be around 600 :lol:
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by RHINO » Fri Mar 12, 2010 3:57 pm
Can someone please explain how you go from 1/2z-1/x to x-2z/2xz? I don't understand those steps.

Thanks...
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by missrochelle » Wed Aug 11, 2010 9:15 am
cramya wrote:1/x+1/5y = 1/2z

1/5y = 1/2z - 1/x

1/5y = (x-2z) / 2xz

5y = 2xz/x-2z
y = 2xz/5x-10z

B)
Can some explain the "reciprocate the equation" rule. I don't remember reading that in any books.... that you can just flip equations in order to get rid of the denominator. does this hold true in all equations?
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