If Machine X, working alone at its constant rate, performs half of a

Statement 2 is clearly INSUFFICIENT.

Let x = X's rate alone and y = Y's rate alone, implying that the combined rate when X and Y work together = x+y

Statement 1:
Case 1: task = 48 units
Working together at their respective constant rates, machines X and Y perform the task in 6 hours.
Combined rate for X and Y to complete the 48-unit task in 6 hours = \(\frac{work}{time} = \frac{48}{6} = 8\) units per hour, implying that \(x+y = 8\)

If Machine X, working alone at its constant rate, performs half of a certain task and then Machine Y, working alone at its constant rate, performs the rest of the task, the whole task will take a total of 16 hours
Time for X to produce half the 48-unit task at rate of x units per hour \(= \frac{24}{x}\)
Time for Y to produce half the 48-unit task at a rate of y units per hour \(= \frac{24}{y}\)
Since the total time is 16 hours, we get:
\(\frac{24}{x} + \frac{24}{y} = 16\)

Since \(\frac{24}{x} + \frac{24}{y} = 16\) and \(x+y=8\), look for two factors of 24 that sum to 8:
2 and 6
4 and 4
Only the first combination satisfies \(\frac{24}{x} + \frac{24}{y} = 16\):
\(\frac{24}{2 }+ \frac{24}{6 }= 12 + 4 = 16\)

If x=2 and y=6. then the time for X to complete the 48-unit job \(= \frac{work}{rate} = \frac{48}{2} = 24\) hours
If x=6 and y=2. then the time for X to complete the 48-unit job \(= \frac{work}{rate} = \frac{48}{6} = 8\) hours
Since the time for X can be different values, INSUFFICIENT.

Statements combined:
In Case 1, attributing the faster rate to X implies that x=6, with the result that X's time = 8 hours

Case 2: task = 96 units
Working together at their respective constant rates, machines X and Y perform the task in 6 hours.
Combined rate for X and Y to complete the 96-unit task in 6 hours = \(\frac{work}{time} = \frac{96}{6} = 16\) units per hour, implying that \(x+y = 16\)

If Machine X, working alone at its constant rate, performs half of a certain task and then Machine Y, working alone at its constant rate, performs the rest of the task, the whole task will take a total of 16 hours
Time for X to produce half the 96-unit task at rate of x units per hour \(= \frac{48}{x}\)
Time for Y to produce half the 96-unit task ar a rate of y units per hour \(= \frac{48}{y}\)
Since the total time is 16 hours, we get:
\(\frac{48}{x} + \frac{48}{y} = 16\)

Since \(\frac{48}{x} + \frac{48}{y} = 16\) and \(x+y=16\), look for two factors of 48 that sum to 16:
4 and 12
8 and 8
Only the first combination satisfies \(\frac{48}{x} + \frac{48}{y} = 16\):
\(\frac{48}{4}+ \frac{48}{12}= 12 + 4 = 16\)

Attributing the faster rate to X implies that x=12, with the result that X's time to complete the 96-unit job \(= \frac{work}{rate} = \frac{96}{12 }= 8\) hours

In each case, X's time is the same: 8 hours
Thus, the two statements combined are SUFFICIENT.

­Assume there is a total 48u of work that need to be completed.
Machine X did 24u and Machine Y did 24u in a total of 16hrs

Statement1:
They took 6 hours to complete the 48u of work. So in 1 hr, they are doing 8u.
If Machine X did 'a' unit of work then Y did '8-a' in 1hr
So we can form the equation now:

24/a +24/8-a =16
Solving this we will get a=2,6

We don't know if Machine X is doing 2 or 6 unit of work in 1 hr


Statement 2:
Clearly Insufficient

Statement 1+2:
machine X rate > Y's rate
So Machine X is doing 6u of work and Y is doing 2u 

Answer: C
 

Hi GMATGuruNY, how are you just plugging in 48 and 96 units for the values of the total amount of work done by the machines? We don't know enough in this problem to infer a number for the number of units of work completed by the machines. If I were to solve this problem, I wouldn't feel confident plugging in values, especially given that this is a DS problem. Can you explain what is driving that intuition here to just pick two cases for 48 and 96 units and conclude that that is sufficient?

Also, can someone please present a solution for this problem with variables instead of choosing numbers? That would feel much more thorough. I am struggling to understand this one. Devvrata-your solution looks good, but u is still taken as an unknown (you haven't defined u as a specific value; u=1 or 2 or 3, etc.), so I don't understand how you arrived at sufficiency with the statements combined. From my interpretation of your solution, u could be anything. If you could elaborate on your approach, that would be helpful. Thanks in advance!­

 

­If Machine X, working alone at its constant rate, performs half of a certain task and then Machine Y, working alone at its constant rate, performs the rest of the task, the whole task will take a total of 16 hours. How many hours will it take machine X, working alone at its constant rate, to perform the task?

Assume Machine X and Machine Y take x and y hours, respectively, to complete the task alone. Thus, Machine X needs x/2 hours to perform half of the task, and Machine Y needs y/2 hours to complete the remaining half. Therefore, we are given that x/2 + y/2 = 16, leading to x + y = 32.

(1) Working together at their respective constant rates, machines X and Y perform the task in 6 hours.

This implies that 1/x + 1/y = 1/6. Substituting y = 32 - x, we get 1/x + 1/(32 - x) = 1/6, which leads to the quadratic equation x^2 - 32x + 192 = 0. Solving this results in x = 8 or x = 24. Therefore, either x = 8 and y = 24, or x = 24 and y = 8. Not sufficient.

We could avoid all the algebra above if you had spotted that Machines X and Y are indistinguishable in this scenario — there is not a single piece of information that differentiates Machine X from Machine Y. Consequently, we cannot differentiate between x and y. This means that the equation 1/x + 1/(32 - x) = 1/6 would yield two potential sets of values for x and y: x < y and x > y, because the machines are interchangeable in their roles. Without additional details, we cannot conclusively determine which completion time belongs to which machine — unless, of course, x equals y, which is clearly not the case here.

(2) Machine X has a faster rate than machine Y.

This statement only implies that x < y, which is clearly insufficient on its own.

(1)+(2) Since from (2) x < y, then x = 8 and y = 24. Sufficient.­

Answer: C.

Hope it helps.­

 

Here, the question stem asks for a value: the number of hours it takes X to perform the task alone.
If we test two different options for the task and get a certain number of hours for X in the first case but a different number of hours for X in the second case, then we have insufficient information to determine the number of hours for X.
But if the time for X in both cases is the SAME, the implication is that the size of the task is irrelevant: the number of hours for X will remain constant, whether the task is 48 units, 96 units, or 48000 units.

When implementing this approach, we should aim to test values that make the math easy.
For a work/rate problem, it is generally wise to select values divisible by the given rates and/or times.
Here, we are given two times -- 16 hours in the prompt, 6 hours in Statement 1 -- so I first tested a task of 48 units (the LCM of 16 and 6) and then a task of 96 units (the next greatest multiple of 16 and 6).




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­Thank you Bunuel for the explanation, it makes more sense now. @GMATGuruNY-makes sense on your approach too with getting two different values for X and concluding it's insufficient, I guess I'm just a bit hesitant to plug in numbers for value-based DS questions in general, and am still not understanding how I'd arrive at sufficiency with your approach when combining the statements together. Regardless, I think Bunuel's explanation cleared up my doubts, thanks!

 

­The logic behind the testing-cases approach is as follows:
When the statements are combined, the time for X alone is 8 hours, no matter what value we assign to the task.

In my earlier post, Case 1 (task=48) and Case 2 (task=96) both yielded a time of 8 hours for X alone.

Case 3: task = 240 units

Again, let x = the rate for X and y = the rate for Y, implying that their combined rate = x+y.

Working together at their respective constant rates, machines X and Y perform the task in 6 hours.
Combined rate for X and Y to complete the 240-unit task in 6 hours = \(\frac{work}{time} = \frac{240}{6} = 40\) units per hour, implying that \(x+y = 40\).

If Machine X, working alone at its constant rate, performs half of a certain task and then Machine Y, working alone at its constant rate, performs the rest of the task, the whole task will take a total of 16 hours.
Time for X to produce half the 240-unit task at rate of x units per hour \(= \frac{120}{x}\)
Time for Y to produce half the 240-unit task at a rate of y units per hour \(= \frac{120}{y}\)
Since the total time is 16 hours, we get:
\(\frac{120}{x} + \frac{120}{y} = 16\)

Since \(\frac{120}{x} + \frac{120}{y} = 16\) and \(x+y=40\), look for two factors of 120 that sum to 40:
10 and 30
This combination satisfies both of the equations above.
Since X has the faster rate, x=30 and y=10.
At a rate of 30 units per hour, the time for X alone to complete the entire 240-unit task = \(\frac{work}{rate} = \frac{240}{30} = 8\) hours.

When the statements are combined, the time for X alone is 8 hours, no matter value is assigned to the task.
If the task = 48, X's time alone is 8 hours.
If the task = 96, X's time alone is 8 hours.
If the task = 240, X's time alone is 8 hours.
Since the time is THE SAME in every case, the two statements combined are SUFFICIENT.
Regardless of the size of the task, X's time alone will be 8 hours.


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