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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  [#permalink]

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New post Updated on: 13 Aug 2018, 06:00
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Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #4

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.

Option choices:
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.





To read the article: Solve Time and Work Problems Efficiently using Efficiency Method!

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Originally posted by EgmatQuantExpert on 23 May 2018, 04:16.
Last edited by EgmatQuantExpert on 13 Aug 2018, 06:00, edited 1 time in total.
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Re: Machines X and Y can work at their respective constant rates  [#permalink]

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New post 26 May 2018, 13:57
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Question Stem:-
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?
Rephrase:-
Machines X and Y can work at their respective constant rates of 1/x and 1/y respectively (where x and y indicate time taken to complete one job).

\(\frac{y}{x}-1\) = ? i.e can we get the value of \(\frac{y}{x}\)

Kindly note:-
When both work together means they work at a rate of \(\frac{1}{x}\)+\(\frac{1}{y}\) = \(\frac{x+y}{xy}\)
Time taken when both machines work together = \(\frac{xy}{x+y}\)


Lets evaluate both the statements:-
Statement (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.

Therefore from (1) \(\frac{xy}{x+y}\) = \(\frac{2y}{3}\)
which implies \(\frac{x}{x+y}\) = \(\frac{2}{3}\)
or\(\frac{y}{x} = \frac{1}{2}\)
Since we have got the value of \(\frac{y}{x}\) Statement 1 is sufficient; Not lets check statement 2 [Cross B,C & E] Evaluate between A & D:-

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

There is no mention about the rate of machine X hence statement (2) is insufficient. Hence the correct answer is option A.
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Re: Machines X and Y can work at their respective constant rates  [#permalink]

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New post 27 May 2018, 05:32

Solution



Given:
    • Machine X and Y work at their respective constant rates to manufacture a certain production unit

To find:
    • When both machines are working alone, the time taken by machine Y is what percentage more/less than that of machine X

Approach and Working:
If we assume the time taken by machine X and Y individually to complete the work is \(t_1\) and \(t_2\) respectively, then
    • Time \(t_2\) is more than time \(t_1\) by (as a percentage) = \(\frac{(t_2 – t_1)}{t_1} * 100\)
      o This can be written as = (\(\frac{t_2}{t_1}\) – 1) * 100
[Note that if \(t_2\) is less than \(t_1\), then the above expression will be equal to (1 – \(\frac{t_2}{t_1}\)) * 100
    • Hence, if we can find the value of the ratio \(\frac{t_2}{t_1}\), we can find out the required percentage

Analysing Statement 1
    • As per the information given in statement 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.
      o The time taken by both of them together to complete the job = \(\frac{t_1t_2}{(t_1 + t_2)}\)
    • Given that, \(\frac{t_1t_2}{(t_1 + t_2)}\) = \(\frac{2}{3} t_2\)
      o Simplifying, we can write \(\frac{t_2}{t_1}\) = \(\frac{2}{1}\)
Hence, statement 1 is sufficient enough to answer the question

Analysing Statement 2
    • As per the information given in statement 2, Machine Y, working alone, fills a production order of twice the size in 6 hrs
      o From this statement, we can say \(t_2\) = 3 hours
      o But we cannot conclude any relationship between \(t_1\) and \(t_2\)
Hence, statement 2 is not sufficient to answer the question

Hence, the correct answer is option A.

Answer: A

Important Observation


    • When we are calculating by what percentage one element is more/less than the other element, we only need to find out the ratio between those elements.

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Re: Machines X and Y can work at their respective constant rates  [#permalink]

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New post 30 Oct 2018, 03:15
EgmatQuantExpert 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.

\(?\,\,:\,\,{T_X}\,,\,\,{T_Y}\,\,{\rm{relationship}}\,\,\,\,\,\,\left( {? = {T_X}\mathop \to \limits^{\Delta \% } {T_Y} = {{{T_Y} - {T_X}} \over {{T_X}}} = {{{T_Y}} \over {{T_X}}} - 1} \right)\)

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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Re: Machines X and Y can work at their respective constant rates  [#permalink]

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New post 26 May 2019, 17:05
Kindly note:-
When both work together means they work at a rate of 1/x+1/y = (x+y)/xy

Can somebody explain why this is?
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Re: Machines X and Y can work at their respective constant rates  [#permalink]

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New post 26 May 2019, 17:06
nsato wrote:
Kindly note:-
When both work together means they work at a rate of 1/x+1/y = (x+y)/xy


Can somebody explain why this is?
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Machines X and Y can work at their respective constant rates  [#permalink]

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New post 26 May 2019, 17:41
nsato wrote:
nsato wrote:
Kindly note:-
When both work together means they work at a rate of 1/x+1/y = (x+y)/xy


Can somebody explain why this is?


let d1 be time taken by x and y together
d2 by x alone and
d3 by y independently.

(x+y)d1 = xd2 = yd3......................(A)

We have to find d3/d2 somehow...

Statement 2.

y*6 = 2(x+y)*d1, clearly insufficient.

Statement 1.
(x+y) (2/3) *t = y*t, solving this we get x/y as 1/2.

Putting x/y into equation (A)
we can deduce the relationship between-
1. d1 and d2
2. d1 and d3, so sufficient.

nsato if you understand equation (A) then rest of the sum will be easy for you. Check: https://gmatclub.com/forum/gmat-math-book-87417.html
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Machines X and Y can work at their respective constant rates   [#permalink] 26 May 2019, 17:41
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