Machines K, M, and N, each working alone at its constant rate, produce

Straight to point: Let's use some sense here instead of calculations
Given N produces 1 piece in 2 minutes--> 30 in 1 hour; K,M 1 in x,y mins

1)x < 1.5 mins
----> 1 piece in < 1.5 mins
----> 40 pieces in < 60min (1.5*40=60 min = 1hr.)
---->40+30=70 pieces only K and N

2)y<1.2
if 1.5mins>>40 pieces
then 1.2 mins ---> produces even more pieces(>40) than K---->>> M and N produces >70 pieces in 1 hour


[color=#00a651]Therefore Answer is D each alone is suff.
HOPE this HELPS

THANKS

Rate of m/c K for producing 1 widget= \(\frac{1}{x}\)
Rate of m/c M for producing 1 widget= \(\frac{1}{y}\)
Rate of m/c N for producing 1 widget= \(\frac{1}{2}\)

We can approach this one in two ways:
A) Either check for whether the 3 machines working together can produce 50 widgets in less than 1 hour(as question states)
B) Or check for what number of widgets are produced in 1 hour by the 3 machines working together.

Getting answers of any of the two conditions above will fulfil our requirement.
A) Solving this is tricky as we need values of both x and y - on an initial thought. However, let's try
St. 1: x < 1.5, Checking for the slowest speed of K.
Time taken by K to complete 1 widget(job done alone) = 1.5min
Time taken by K to complete 50 widgets = 50*1.5 = 75min
Time taken by N to complete 1 widget(job done alone) = 2min
Time taken by N to complete 50 widgets = 50*2 = 100min

Thus, in 1 minute, K completes = \(\frac{1}{75}\) of the work
in 1 minute, N completes = \(\frac{1}{100}\) of the work

Therefore, in 1 minute, together they complete = \((\frac{1}{75} + \frac{1}{100})\) of the work(lets say 1/T, where T is the time taken together by K and N)
\(\implies \frac{1}{60} + \frac{1}{100} = \frac{1}{T}\)
T = \(\frac{300}{7}\) ~43 minutes(actually would be less)
With only 2 machines - K and N - 50 widgets are completed in less than 1 hour.

SUFFICIENT.

St. 2: y < 1.2, Checking for the slowest speed of M.(Though this is not required to be done as similar approach as that applied in St. 1 is applicable)
Time taken by M to complete 1 widget(job done alone) = 1.2min
Time taken by M to complete 50 widgets = 50*1.5 = 60min
Time taken by N to complete 1 widget(job done alone) = 2min
Time taken by N to complete 50 widgets = 50*2 = 100min

Thus, in 1 minute, M completes = \(\frac{1}{60}\) of the work
in 1 minute, N completes = \(\frac{1}{100}\) of the work

Therefore, in 1 minute, together they complete = \((\frac{1}{60} + \frac{1}{100})\) of the work(lets say 1/T, where T is the time taken together by M and N)
\(\implies \frac{1}{60} + \frac{1}{100} = \frac{1}{T}\)
T = \(\frac{300}{8}\) ~37.xx minutes(again would be less)
With only 2 machines - M and N - 50 widgets are completed in less than 1 hour.

SUFFICIENT.

Personally, I like second option(B) better.

B) \(60* (\frac{1}{x} + \frac{1}{y} + \frac{1}{2}) < 50\)
St. 1: x < 1.5
Now, any value of x that is less than 1.5 means the machine K works faster - the lesser the value of x the faster the machine K works. Let's check for x = 1.5
\(\frac{60}{1.5} + \frac{60}{y} + \frac{60}{2} < 60 \)
\(40 + \frac{60}{y} + 30\) < 50

No matter what LHS is always > 70.

SUFFICIENT.

St. 2: y < 1.2
Applying same logic and checking for y = 1.2
\(\frac{60}{x} + \frac{60}{1.2} + \frac{60}{2} < 60 \)
\(\frac{60}{x} + 50 + 30\) < 50

Again LHS is always > 80.

SUFFICIENT.

Answer D.

Sachin19 Genoa2000 HTH.

The problem with this task is that if you follow your habit and go with algebraic approach, i.e 1/x+1/y+1/2 >50/60 , and merge LHS into one long fraction, the whole expression becomes a mess and you may trick yourself into picking C. I followed this approach and realized about 6 minutes in after plugging numbers that this is a dead-end, and tried to think logically to arrive to answer, which took me 8 minutes in total. Question to GMAT tutors: How do we decide under time constraint when the expression should be simplified without even looking into two statements and when we have to think thoroughly first and apply logic?

Can we take decimal values?

If the question is silent about integer/non-integer related property of any measure then YES you can always take those measures in Decimals.

However in some cases, e.g. number of men, number of machines, number of pens etc. it's understood thvalues must be Integers. So just use logical ability to treat any measure as per question definition.

lakshya14

Machine N produces one widget every 2 minutes, so N produces 30 widgets in an hour. Statement 1 alone tells us K is faster than N (it takes less time for K to produce a widget than it takes N), so machine K will definitely make more than 30 widgets per hour, and K and N together will make more than 60 per hour. So Statement 1 is sufficient even if machine M is useless.

Similarly, Statement 2 tells us machine M is faster than N, so M alone makes more than 30 widgets per hour, and M and N together make more than 60 per hour, and Statement 2 alone is also sufficient. So the answer is D.

Given: Machines K, M, and N, each working alone at its constant rate, produce 1 widget in x, y, and 2 minutes, respectively.
Asked: If Machines K, M, and N work simultaneously at their respective constant rates, does it take them less than 1 hour to produce a total of 50 widgets?

Machines K, M, and N work simultaneously at their respective constant rates would produce number of widgets in 1 minute = (1/x + 1/y + 1/2)
Time taken to produce 50 widgets = 50/{60(1/x + 1/y + 1/2)} hours

(1) x < 1.5
Machines K, M, and N work simultaneously at their respective constant rates would produce number of widgets in 1 minute = (1/x + 1/y + 1/2) >= 1/1.5 + 1/2 = 2/3 + 1/2 = 7/6 widgets
Time taken to produce 50 widgets <= 50*6/7 = 300/7 = 42 6/7 minutes < 1 hour
SUFFICIENT

(2) y < 1.2
Machines K, M, and N work simultaneously at their respective constant rates would produce number of widgets in 1 minute = (1/x + 1/y + 1/2) >= 1/1.2 + 1/2 = 5/6 + 1/2 = 8/6 = 4/3 widgets
Time taken to produce 50 widgets <= 50*3/4 = 150/4 = 37.5 minutes < 1 hour
SUFFICIENT

IMO D

This question shows us something important: always skim the statements in Data Sufficiency questions!

Statement 2 gives the "tightest constraint" of the statements since the accepted range for y (in Statement 2) lies within the accepted range for x (in Statement 2).
If you can find Statement 2 to be sufficient in this question, Statement 1 will automatically be sufficient. Saving time!

We can make this inference because their value are equally unknown with both rates being \(\frac{1}{x}\) and \(\frac{1}{y}\).

Another way of thinking about this question: N produces 1 widget each 2 minutes, so in 60 minutes (1 hour) N will produce 30 widgets. Our goal is to produce 50 widgets in one hour, so apart from N production, we need 20 more widgets.

20 more widgets in 60 minutes is equivalent to 1 widget each 3 minutes. Any adicional machine that pairs with N and produces at least 1 widget each 3 minutes give us a definitive awnser to the question.

Cleary, we can see that statments 1 and 2 give us a definitive yes, both are faster than 3 minutes per widget.

Does that make sense? Hope so!

Here already N takes 2 min to make a widget. So in 1 hour, it will make 30 widgets. so, others have to make 50-30=20 widgets. hence, rate of M and K are given in conditions to be more than N. Hence, both will individually make more than 30 widgets in one hour. hence, total will always be more than 50 widgets.

GMATinsight
I don't get it... what if x is .00001 minutes and y is .00001 minutes, --> so then machine K and M are each only producing .0006 widgets in an hour 30+2*(.0006) is less than 50

Hi woohoo921
Thanks for your query.


Your question indicates one of two possibilities: either you are confused about how to correctly translate the question or about how to correctly apply the rate-time concept.

I will try to give you clarity on both simultaneously. At the end, I will show you how even your example (x = 0.00001) works well in GMATinsight’s approach.


TRANSLATE AND SOLVE:
Machines K, M, and N, each working alone at its constant rate, produce 1 widget in x, y, and 2 minutes, respectively.

This statement tells us three things:

• Machine K can produce 1 widget in x minutes,
• Machine Y can produce 1 widget in y minutes, and
• Machine Z can produce 1 widget in 2 minutes.

These are NOT the rates at which the machines work. Rate is the work done per unit time- this has to be calculated!

• You straight used x = 0.00001 as the rate of machine K, and this is where you went wrong. Instead, x is just the number of minutes in which machine K makes 1 widget.
• The correct rate of producing widgets for machine K = (1/x) widget per minute. That is in one minute, machine K can produce 1/x widget.

With the correct rate, if you consider x = 0.00001, then machine K can produce 1/0.00001 widget in a minute.

• That is, K can produce 1/0.00001 (= \(10^5\)) widgets in a minute.
• See! Machine K alone can produce such a huge number of widgets and that too in just one MINUTE!
  
  • Forget about an hour (60 mins) for which 3 machines work; the 1-minute output of just one machine is way higher than the 50 widgets they had to make.

All of this happened because x and y are in the denominators of the rate expressions for machines K and M. And thus, the smaller the values of x and y, the larger the number of widgets they can produce in one minute.

Now, at this point, you can also understand the reason behind considering the maximum value for x (that is 1.5) in the given explanation. Because of x being in the denominator, the maximum value of x will give us the minimum possible number of widgets produced by machine X in a minute. And if our minimum produce from all three machines combined comes out to be 50+, then we are sure that the answers from each statement is a sure YES.


TAKEAWAY:
Always make sure that you find the correct rates from whatever information you are given. In this question, you were given number of minutes to produce 1 widget, and you had to find the number of widgets they could produce in 1 minute.


Hope this helps!

Best,
Aditi Gupta
Quant expert, e-GMAT

 
­
I'll answer this, if still relevant.

What is x? The number of minutes it takes a machine to produce 1 widget.

If the number of minutes is small, the machine is fast. The smaller the time to produce 1 widget, the more widgets it can produce in an hour.

If x = 0.0001, that means the machine is amazingly fast. It takes the machine a very small fraction of a minute to produce a widget. It can produce 10,000 widgets in just 1 minute. In an hour, it will produce 60 times that.


If x = 1, the machine will produce 60 widgets in 1 hour (60 minutes).
If x = .1, the machine will produce 600 widgets in 1 hour (60 minutes).
 

­Here is the video solution to this problem: 
https://youtu.be/3zQ20VEdSa8

­Here is a detailed video solution for this interesting question. 
See how this question can be solved in the test environment using "Owning the Dataset" approach.



Let us know if you have any questions.­

Would the below understanding be correct?

	Rate	Time	Work
K	1/x	x mins	1 widget
M	1/y	y mins	1 widget
N	1/2	2 mins	1 widget
K+M+N	1/x + 1/y + 1/2 = (2y+2x+xy)/ 2xy		50 widgets

Therefore Time= 50 / (2y+2x+xy)/2xy = 100xy/(2x+2y+xy). We need to check if this is below 60 mins.

100xy/(2x+2y+xy) < 60
On cross multiplication and further simplification we need to check if xy < 3x+3y.

(1) x<1.5. Let x =1 then is y < 3+3y? which becomes is -3 < 2y? which becomes is -3/2 <y ? Since y is time and can't be negative hence answer is Yes. Suff
(2) Similar logic for statement 2, hence Suff
Answer D

Would the below understanding be correct?

	Rate	Time	Work
K	1/x	x mins	1 widget
M	1/y	y mins	1 widget
N	1/2	2 mins	1 widget
K+M+N	1/x + 1/y + 1/2 = (2y+2x+xy)/ 2xy		50 widgets

Therefore Time= 50 / (2y+2x+xy)/2xy = 100xy/(2x+2y+xy). We need to check if this is below 60 mins.

100xy/(2x+2y+xy) < 60
On cross multiplication and further simplification we need to check if xy < 3x+3y.

(1) x<1.5. Let x =1 then is y < 3+3y? which becomes is -3 < 2y? which becomes is -3/2 <y ? Since y is time and can't be negative hence answer is Yes. Suff
(2) Similar logic for statement 2, hence Suff
Answer D