Machines Y and Z work at their respective constant rates. If it takes

Question

Machines Y and Z work at their respective constant rates. If it takes machines Y and Z, working together, 12 hours to fill a production order of a certain size, how many more hours does it take machine Y, working alone, to fill the order than it takes machine Z, working alone?

(1) Machine Z, working alone, fills a production order of this size in twice the time than machine Y, working alone, does

(2) Machines Y and Z, working together, fill a production order of this size in two-thirds the time that machine Y, working alone, does

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Kunal1801 · Answer

Given:
Y + Z = 12hours
To find : Y-Z (working alone)
AD/BCE

St.1: If Y takes x hours, accordingly Z takes 2x hours. No particular information about Y or Z given so not alone sufficient

What remains to us is BCE

St.2: Y + Z takes 2/3rd time of Y alone.
A bit confusing but I cracked it this way 
12 x 1.5 we get 18. Checking it backwards as per the statement 2. Y takes 18 so 2/3 of 18 is 12. So we are correct. 
We get Y as 18 but have no idea about Z therefore even B is not sufficient.

Taking both together we get Z as 36

Therefore C is the Answer

Posted from my mobile device

Purnank · Answer

IMO D,

Given - Machines Y and Z work at their respective constant rates (assuming those as y and z per hour respecitively). If it takes machines Y and Z, working together, 12 hours to fill a production order of a certain size (assuming w). 
Therefore, w = 12(y+z) ----- (Eq1)
To calculate, how many more hours does it take machine Y, working alone, to fill the order than it takes machine Z, working alone lets check the given statements,
1st -  Machine Z, working alone, fills a production order of this size in twice the time than machine Y, working alone, does (assuming y worked for t hours)
w = z*2t = y*t
Therefore, y=2z
substituting this in equation 1 we will get,
12(y+z) = z*2t, 
as y=2z,
12*3z = z*2t
t=18 hours,
t' time for machine Z to complete the same order = 2*t = 36Hours, 
So, machine Y doesnt take more time than Z. Hence Sufficient.

2nd - Machines Y and Z, working together, fill a production order of this size in two-thirds the time that machine Y, working alone, does (again assuming same y worked for t hours)
from this the equation we can get,
(2t/3) * (y+z) = y*t
on further solving,
2(y+z) = 3y
y=2z.
This is same we got from statement 1. Hence Sufficient.

But, machine Y doesnt take more time than Z. and question ask us to calculate "how many more hours does it take machine Y, working alone, to fill the order than it takes machine Z, working alone?"
Is there anything i am missing? Experts please advice?

Nidzo · Answer

"...it takes machines Y and Z, working together, 12 hours to fill a production order..." =  \frac{1}{y}+\frac{1}{z}=\frac{1}{12}

(1) Machine Z, working alone, fills a production order of this size in twice the time than machine Y, working alone, does

From this one knows that  z = 2y . Plugging this back into the equation above:

\frac{1}{y}+\frac{1}{2y}=\frac{1}{12}

\frac{3y}{2y^2}=\frac{1}{12}

\frac{3}{2y}=\frac{1}{12}

2y = 36  as we know that  2y = z  then  z = 36  and  y = 18

36 - 18 = 18  hours more.

SUFFICIENT

(2) Machines Y and Z, working together, fill a production order of this size in two-thirds the time that machine Y, working alone, does

Together Y and Z can complete an order in 12 hours, therefore 
 \frac{2}{3}y = 12

y = 18 , plugging this back into  \frac{1}{y}+\frac{1}{z}=\frac{1}{12}  and solving for x.

\frac{1}{18}+\frac{1}{z}=\frac{1}{12}

\frac{1}{z}=\frac{1}{12}-\frac{1}{18}

\frac{1}{z}=\frac{3-2}{36}

z=36

36 - 18 = 18  hours more.

SUFFICIENT

ANSWER D

Shwarma · Answer

Answer D

1. Rz=a/2t, Ry= A/t
(a/2t+a/t)=a/12
You get t an 2t
So you can calculate --> 2t-t -  sufficient

2. 2/3 ty = 12
so you get ty
then you can calculate tz
and you can get tz-tx -  sufficient

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Machines Y and Z work at their respective constant rates. If it takes

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Machines Y and Z work at their respective constant rates. If it takes

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