Machine M and Machine N working alone at their constant rate

Question

Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate

Bunuel · Accepted Answer

Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate.

If the rate of M is 2,000 nails per hour and the rate of N is 4,000 nails per hour, then to produce 6,000 nails, M needs 3 hours and to produce 8,000 nails N, needs 2 hours. In this case M worked longer than N.

If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.

Not sufficient.

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate. In the time M needs to produce 6,000 nails, N can produce 12,000 nails, thus it can produce 8,000 nails in less time than M can produce 6,000 nails. Sufficient.

Answer: B.

PiyushK · Answer

Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

Let N is number of nails machine N produced.
Let M is number of nails machine M produced.
And x and y are respective time each machine took to produce 6000 and 8000 nails.

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate

Given N= M + 2000
Mx = 6000 ------(1)
Ny = 8000 ==> (M+2000)y=8000 ---------(2)
Solve 1 and 2 to find ratio of x/y.

(\frac{6000}{x}+2000)y=8000  
We can not reduce this equation in  x/y  form. Thus, (1) insufficient.

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate.
N=2M

Mx = 6000 --------(1)
Ny = 8000 ==> 2My=8000 -------(2)

2y\frac{6000}{x}=8000 
Solve for  x/y ; we can calculate x:y. Thus, statement (2) is sufficient.

shishir16 · Answer

Hi Bunuel,

Why are we rounding off the number of hours here?
Am i missing something in the question, why arent we considering minutes, for which im getting the first statement as being sufficient.
Thanks

Bunuel · Answer

Cannot follow what you mean...

Where do we round the number of hours? Also, in my solution for (1) there are an example given which gives two different answers to the question, which means that this statement is not sufficient.

earnit · Answer

The highlighted part: If the question is to produce 6000 nails then definitely M needs 1 hour but N with a rate of 8000 nails per hour would need less than an hour right?

Please clarify if possible.
Thank you.

Bunuel · Answer

How does N, at the rate of 8,000 nails per hour, need more than an hour to produce 8000 nails? It will need exactly 1 hour.

Again, the question asks: did machine M (which produced 6000 nails) work longer than machine N (which produced 8000 nails)?

If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.

earnit · Answer

Got it. 
Its not the common number that the machines are expected to produce. 
That part slipped out when exploring the options.

Key take away: if there is a common number required to produce then even (1) would be sufficient. But not in the above case.

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

Machine M and Machine N working alone at their constant rates, non stop, produced 6000 and 8000 nails respectively. Did machine M work longer than machine N?

(1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate

(2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate
 
In the original condition, r1*t1=6000, r2*t2=8000 and throw some transformation we have t1=6000/r1, t2=8000/r2 and 6000/r1>8000/r2, 3r2>4r1?, therefore the question is all about comparing r1 and r2. 
We have 4 variables (r1,r2,t1,t2), 2 equations (r1*t1=6000, r2*t2=8000) therefore we need 2 more equations and therefore C is likely the answer. Using both 1) & 2) together, r2=r1+2000, r2=2r1 gives us r1=2000, r2=4000 and thus C is the answer. But such trivial conditions are rarely the answer, so we might try it separately again.

Using 1), 2)separately, (from Common mistake type 4(A), in case of 2) substituting r2=2r1 to 3r2>4r1? gives us 3*2r1>4r1?, 6>4. Therefore the answer is yes, and the condition is sufficient. Therefore the answer is B. (when looking at conditions 1), 2) if one is given by value and the other is given by ratio, the one with ratio is usually the answer. That's why we just calculated using con 1) here. )

DeD · Answer

Here is my approach

Is 6000/(rate of m = rm) > 8000 / (rate of n = rn) ? <--> Is 6/rm > 8/rn? <--> Is 6 rn > 8 rm ?

St. 1: rn = rm + 2000 --> Is 6 rm + 12000 > 8 rm ? Here we get a condition, i.e. rm < 6000 that is not verified a priori --> NOT SUFF

St. 2: rn = 2 rm --> Is 12 rm > 8 rm ? Yes because rm >0 --> SUFF

B

sayan640 · Answer

Statement 1 :-
Say , Machine M produces x nails / hour and machine N produces (x +2000 ) nails / hour.
Machine M worked for = 6000/x hours.
Machine N worked for = 8000 / (x + 2000) hours.
The denominator is not same . We cant tell whether 6000/x > 8000/(x+2000)
It depends on the value of x.

Statement 2:-
Say , Machine M produces x nails / hour and machine N produces 2*x nails / hour.
So, Machine M worked for = 6000/x hours.
Machine N worked for = 8000 /2*x hours = 4000/x hours

Since the denominator is fixed , we can say YES , machine M work longer than machine N.

Take a fixed denominator for example , x = 2
6000 / 2 = 3000
4000/2 = 2000 
3000 > 2000

Please give me kudo s if you liked my explanation.

dine5207 · Answer

Very well curated question from GMAT !!

Option 2 is easily sufficient , and coming option 1 is trap as we don't consider below scenario or hardly skip this part.
 
If the rate of M is 6,000 nails per hour and the rate of N is 8,000 nails per hour, then to produce 6,000 nails, M needs 1 hours and to produce 8,000 nails N, needs 1 hour. In this case M did not work longer than N.

Dumsy_1711 · Answer

Machine M :  
Work = 6000 nails
Let time = m
Rate =  \frac{6000}{m}

Machine N : 
Work = 8000 nails
Let time = n
Rate =  \frac{8000}{n}

Question: Is m>n?

1:

(\frac{8000}{n}) * 1 - (\frac{6000}{m}) * 1 = 2000
4m-3n = mn

Not sufficient

2:

\frac{8000}{n} = 2* (\frac{6000}{m})
m=\frac{3n}{2}
m>n

Sufficient.

KarishmaB · Answer

Responding to a pm: Simply note that work done by machines M : N = 6000:8000 = 3:4 Now if the ratio of rates of work of M : N were 3:4, then they would have worked for the same time (because work done directly varies with rate when time is held constant) If this ratio is less than 3/4, then machine M has worked longer. If this ratio is more than 3/4, machine M has worked for less time. (1) Machine N produces 2000 more nails than machine M in one hour when each machine work at its constant rate We don't know their relative rate of work. N's rate is 2000 nails/hr more than M's rate. But we don't know the actual rate of either. If M's rate is 1000 nails/hr, N's rate is 3000 nails/hr and the ratio of their rates is far less than 3/4. So M has worked for far longer time. If M's rate instead is 50,000 nails/hr, N's rate is 52000 nails/hr, then the ratio is more than 3/4 and hence M has worked for less time. Not sufficient. (2) Machine N produces twice as much as machine M in one hour when each machine work at its constant rate This directly tells us that the ratio of their rates is 1/2 which is less than 3/4. Hence M has worked longer. We can answer with a definite yes. Sufficient alone Answer (B) Here is a video discussion on work-rate using Ratios: https://youtu.be/88NFTttkJmA

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Machine M and Machine N working alone at their constant rate

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