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Machines X and Y can work at their respective constant rates

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Machines X and Y can work at their respective constant rates

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Machines X and Y can work at their respective constant rates to manufacture a certain production unit. If both are working alone, then the time taken by machine Y is what percentage more/less than that of machine X?

(1) Machines X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone, does.
(2) Machine Y, working alone, fills a production order of twice the size in 6 hrs.

OA A

Source: e-GMAT

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BTGmoderatorDC wrote:
Machines X and Y can work at their respective constant rates to manufacture a certain production unit. If both are working alone, then the time taken by machine Y is what percentage more/less than that of machine X?

(1) Machines X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone, does.
(2) Machine Y, working alone, fills a production order of twice the size in 6 hrs.
Statement 1:
Time and rate have a RECIPROCAL relationship.
Since X and Y together take 2/3 as long as Y alone, X and Y together work 3/2 as fast as Y alone.
Thus, if Y's rate alone = 2 units per hour, then X and Y's combined rate = (3/2)(2) = 3 units per hour, implying that X's rate alone = 3-2 = 1 unit per hour.
Since Y works twice as fast as X, Y's time will be 1/2 X's time.
In other words:
Y's time will be 50% less than X's time.
SUFFICIENT.

Statement 2:
No information about X's rate or time.
INSUFFICIENT.

The correct answer is A.

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BTGmoderatorDC wrote:
Machines X and Y can work at their respective constant rates to manufacture a certain production unit. If both are working alone, then the time taken by machine Y is what percentage more/less than that of machine X?

(1) Machines X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone, does.
(2) Machine Y, working alone, fills a production order of twice the size in 6 hrs.
Source: e-GMAT
$$?\,\,:\,\,{T_X}\,,\,\,{T_Y}\,\,{\rm{relationship}}\,\,\,\,\,\,\left( {? = {T_X}\mathop \to \limits^{\Delta \% } {T_Y} = {{{T_Y} - {T_X}} \over {{T_X}}} = {{{T_Y}} \over {{T_X}}} - 1} \right)$$
Important: the ratio of time taken (for any given job) is the inverse of the ratio of the work done (for any given time).

$$\left( 1 \right)\,\,{{{T_{X \cup Y}}} \over {{T_Y}}} = {2 \over 3}\,\,\,\,\, \Rightarrow \,\,\,\,\,{{{W_{X \cup Y}}} \over {{W_Y}}} = {3 \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,{W_{X \cup Y}} = 3k \hfill \cr
\,{W_Y} = 2k \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,{W_X} = k$$
$${{{W_Y}} \over {{W_X}}} = {2 \over 1}\,\,\,\,\, \Rightarrow \,\,\,\,{{{T_Y}} \over {{T_X}}} = {1 \over 2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,{T_Y} = 3{\rm{h}}\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,{{\rm{T}}_{\rm{X}}} = 3{\rm{h}}\,\,\,\, \Rightarrow \Delta \% = 0 \hfill \cr
\,{\rm{Take}}\,\,{{\rm{T}}_{\rm{X}}} = 4{\rm{h}}\,\,\,\, \Rightarrow \Delta \% \ne 0 \hfill \cr} \right.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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BTGmoderatorDC wrote:
Machines X and Y can work at their respective constant rates to manufacture a certain production unit. If both are working alone, then the time taken by machine Y is what percentage more/less than that of machine X?

(1) Machines X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone, does.
(2) Machine Y, working alone, fills a production order of twice the size in 6 hrs.

OA A

Source: e-GMAT
Say the rate of Machine X is x hours per unit and that of Machine Y is y hours per unit.

Let's take each statement one by one.

(1) Machines X and Y, working together, complete a production order of the same size in two-thirds the time that machine Y, working alone, does.

Part of work done by Machines X and Y, working together = 1/x + 1/y
Part of work done by Machines Y, working alone = 1/y

=> 2/3(1/x + 1/y) = 1/y
= x/y = 1/2.

Time taken by machine Y is [(y - x)/x]*100% = [(2 - 1)/1]*100% = 100% more than that of machine X. Suffcient.

(2) Machine Y, working alone, fills a production order of twice the size in 6 hrs.

No information about Machine X is given. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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