Machines X and Y can work at their respective constant rates

Question

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!

CAMANISHPARMAR · Accepted Answer

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   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.

EgmatQuantExpert · Answer

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  
  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.

fskilnik · Answer

?\,\,:\,\,{T_X}\,,\,\,{T_Y}\,\,{m{relationship}}\,\,\,\,\,\,\left( {? = {T_X}\mathop 	o \limits^{\Delta \% } {T_Y} = {{{T_Y} - {T_X}} \over {{T_X}}} = {{{T_Y}} \over {{T_X}}} - 1} ight)

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 ight)\,\,{{{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} ight.\,\,\,\,\, \Rightarrow \,\,\,{W_X} = k

{{{W_Y}} \over {{W_X}}} = {2 \over 1}\,\,\,\,\, \Rightarrow \,\,\,\,{{{T_Y}} \over {{T_X}}} = {1 \over 2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{m{SUFF}}.

\left( 2 ight)\,\,{T_Y} = 3{m{h}}\,\,\,\left\{ \matrix{
 \,{m{Take}}\,\,{{m{T}}_{m{X}}} = 3{m{h}}\,\,\,\, \Rightarrow \Delta \% = 0 \hfill \cr 
 \,{m{Take}}\,\,{{m{T}}_{m{X}}} = 4{m{h}}\,\,\,\, \Rightarrow \Delta \% 
e 0 \hfill \cr} ight.

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

Regards, 
Fabio.

nuboard · Answer

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?

nuboard · Answer

Can somebody explain why this is?

LeoN88 · Answer

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

MSIMBA · Answer

Thanks for the simplified explanation

VeritasPrepErika · Answer

So the rule here is that Rate of A + Rate of B = the combined Rate of A and B.

Here, if x is the time taken by Machine X to complete a job, its rate is 1/x. If y is the time taken by Machine Y to complete a job, its rate is 1/y. So the combined Rate of Machines X and Y is 1/x + 1/y.

Let's take this math step by step:

\frac{1}{x}+\frac{1}{y}

Get a common denominator by multiplying first fraction by y/y and second fraction by x/x:

\frac{y}{xy}+\frac{x}{xy}

Add two fractions together:

\frac{y+x}{xy}

BelalHossain046 · Answer

Statement 1: 
Let’s assume:
Total work=6 unit

Y takes 3 days to complete the work ( ie. 2 units per day)

Thus, x&y combined takes 2 days to complete the work (ie 3 units per days)

Let’s calculate UNITS PERDAY

X produces (___ )units per day
Y produces 2 units per day
========================
(X+Y) produce 3 units per day

Thus, X definitely produces 1 unit per day.
In other words, per day production capacity of X (1 unit ) is 50% of production capacity of Y (2 units).

That means, for any work x needs twice as much time as Y needs.

So, statement 1 is SUFFICIENT.

Posted from my mobile device

rushimehta · Answer

As work is constant, for both Machine X and Machine Y,

We can say

RxTx=RyTy i.e. Rx/Ry = Ty/Tx

St 1

(Rx+Ry)(2/3Ty)=W
Since, Work again is same... we can equate this equation with the equation of Machine Y working alone. 
(Rx+Ry)(2/3Ty)=RyTy
Cancelling Ty on both sides. 
(Rx+Ry)(2/3)=Ry
Thus, 
Ry/(Rx+Ry) = 2/3 i.e. Ry=2, Rx =1

As, 
Rx/Ry = Ty/Tx
Rx/Ry = 1/2, thus, Ty/Tx = 1/2

We can now map that the time taken by machine Y is 50% percentage less than that of machine X

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Machines X and Y can work at their respective constant rates

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