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Machines X and Y produced identical bottles at different

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Director
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Machines X and Y produced identical bottles at different [#permalink] New post 24 Nov 2005, 21:27
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Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?

(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours.

If this were a ps question, how would you actually solve it?
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 [#permalink] New post 24 Nov 2005, 21:45
Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?

(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours.

From statement (2),

Work done by X in 4 hours = 2W
Work done by Y in 3 hours = W

3W is the size of the entire production.

It took machine X 4 hours to do 2W. It will take 6 hours for X to do 3W.


Therefore, I pick (B).
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 [#permalink] New post 24 Nov 2005, 23:11
B

Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken machine X operating alone to fill the entire production lot?

(1) Machine X produced 30 bottles per minute.
No information about the amount of work done.. so Insuff

(2) Machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours.

let say McY produces X part in 3 hours, then McX produced 2X in 4 hours
so total production lot = X+2X=3X

time taken by B = 4*3X/2X=6 ....hence Suff
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 [#permalink] New post 24 Nov 2005, 23:43
B

Total work done W = 4x+3y where machine X produce at x bottles per hour and Y produce at y bottles per hour.

1. We can substitute x in below equation but no solution.
Machine X can fill entire lot in W/x hours i.e (4x+3y)/x i.e (4*30*60+3y)/30*60. No solution.

2. It means 4x=2*3y i.e 2x = 3y
Machine X can fill entire lot in W/x hours i.e (4x+2x)/x = 6 hours
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 [#permalink] New post 25 Nov 2005, 00:16
We do not know how big the production lot is.

(1) does not provide any useful information. All we can get, is Machine X produces 7200 bottles in 4 hours.

(2) Assuming Machine Y produces n bottles in 3 hours, then Machine X produces 2n bottles in 4 hours. Thus the lot size is therefore 3n. Machine X will therefore take 3n/2n * 4 = 6 hours to finish the production lot.

Ans: B
  [#permalink] 25 Nov 2005, 00:16
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