Machine X vs Machine Y

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Machine X vs Machine Y

by kanha81 » Mon Mar 02, 2009 10:49 am
q: Mach. X & Y prod. identical bottles at different constant rates. Mach. X, operating alone for 4 hrs, filled part of the produciton lot; then Mach. Y, operating alone for 3 hrs, filled the rest of lot. how many hrs would it have taken Mach. X operating alone to fill entire production lot?

i) Mach. X produces 30 bottles per min.
ii) Mach X produced twice as many bottles in 4 hrs as Mach. Y produced in 3 hrs.

Is there a simpler way of finding this out than doing the following?

let r(x): rate of Mach. X => r(x)*4=W(x)
r(y): rate of Mach. Y => r(y)*3=W(y)
r(x) <> r(y); t(x)=?

i) r(x)=30 bottles per min
=> W(x)=30*4*60= 7200 bottles
=> Insuff. b/c no information given about entire production lot or relationship between r(x) & r(y)

B,C,E

ii) r(x)=2*r(y)
Total # bottles=W(x) + W(y)=4*r(x) + 3*r(y)=5.5*r(x)

t(x)=5.5*r(x) / r(x) = 5.5 hrs.
=> Suff.

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by cramya » Mon Mar 02, 2009 2:22 pm
Stmt I

No info on what part of the work X does to get its rate.

INSUFF

Stmt II

Machine X produced twice as many bottles in 4 hrs as Mach. Y produced in 3 hrs.

Easiest way is to see that machine X produces the same amount of work in 2 hours that machine Y does in 3.

So machine X needs to work 4+2 = 6 hours

SUFF


No need to solve but just for FYI:

If u were to split 1 work where one part is twice the other it would be 1/3 and 2/3rds

So for Machine x

RATE * TIME = WORK

RATE * 4 = 2/3

RATE = 1/6

1/6 th of work in 1 hour

1 work in 6 hours

Choose B
Last edited by cramya on Mon Mar 02, 2009 2:25 pm, edited 1 time in total.

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by cramya » Mon Mar 02, 2009 2:24 pm
Kanha,

For DS once u set up the equation and know its solvable or data is sufficient u can stop right there and move on. I am sure u wanted to know how to solve if it were a PS.

Regards,
CR

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Re: Machine X vs Machine Y

by sureshbala » Mon Mar 02, 2009 2:35 pm
kanha81 wrote:q: Mach. X & Y prod. identical bottles at different constant rates. Mach. X, operating alone for 4 hrs, filled part of the produciton lot; then Mach. Y, operating alone for 3 hrs, filled the rest of lot. how many hrs would it have taken Mach. X operating alone to fill entire production lot?

i) Mach. X produces 30 bottles per min.
ii) Mach X produced twice as many bottles in 4 hrs as Mach. Y produced in 3 hrs.

Is there a simpler way of finding this out than doing the following?

let r(x): rate of Mach. X => r(x)*4=W(x)
r(y): rate of Mach. Y => r(y)*3=W(y)
r(x) <> r(y); t(x)=?

i) r(x)=30 bottles per min
=> W(x)=30*4*60= 7200 bottles
=> Insuff. b/c no information given about entire production lot or relationship between r(x) & r(y)

B,C,E

ii) r(x)=2*r(y)
Total # bottles=W(x) + W(y)=4*r(x) + 3*r(y)=5.5*r(x)

t(x)=5.5*r(x) / r(x) = 5.5 hrs.
=> Suff.



Folk, when it comes to time and work problems try to replace two persons involved with only of the person and it would be easy to conclude the answer fast. Even when you are dealing with problem solving this would save time.

Statement I is not sufficient as we do not know what part of work is done by machine X (the work is completed by X and Y here)

Satement II. Given that X is twice as fast as Y

In the question it is given that work is completed by X working for 4 hrs and Y working for 3 hrs.

Now X working for 4 hrs is same as Y working for 8 hrs.

So the entire work is done by Y in 8+3 = 11 hrs

Hence the entire work can be done by X in 11/2 = 5hrs 30 min

So B is the answer