Machine M, working alone at its constant rate, produces x widgets ever : Data Sufficiency (DS)

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

24 Sep 2013, 08:46

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The easiest way to look at the question (imo) is to say that;

Total produced= 20*(X/4)+20*(X/5) => T=5X+4Y

1) If x>0.8y

then T=5*(4y/5)+4Y => T=4y+4y since x is bigger then 0.8y T=(a number higher then 4)y +4y

Sufficient

2) y= x+1

then T=5X+4X+4 => Total M=5X Total N= 4x+4, if x=1 then no if x=1000 then yes

Insufficient

B

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

03 Mar 2014, 01:40

Bunuel wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

The rate of M = x/4 widgets per minute.
The rate of N = y/5 widgets per minute.

The question basically asks whether x/4 > y/5 (if per minute M produces more widgets than N, then M obviously produces more widgets than N, in 20, 30, ... or in any time period. So, we can compare the rates per 1 minute).

(1) x > 0.8y --> x > 4/5*y --> x/4 > y/5. Sufficient.

(2) y = x + 1. The question becomes: is x/4 > (x+1)/5, or is x>4. We don't know that. Not sufficient.

Answer: A.

Hope it's clear.

Bunuel,

I am not able to understand statement B. Can you please elaborate...

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

03 Mar 2014, 01:43

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Mountain14 wrote:

Bunuel wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

The rate of M = x/4 widgets per minute.
The rate of N = y/5 widgets per minute.

The question basically asks whether x/4 > y/5 (if per minute M produces more widgets than N, then M obviously produces more widgets than N, in 20, 30, ... or in any time period. So, we can compare the rates per 1 minute).

(1) x > 0.8y --> x > 4/5*y --> x/4 > y/5. Sufficient.

(2) y = x + 1. The question becomes: is x/4 > (x+1)/5, or is x>4. We don't know that. Not sufficient.

Answer: A.

Hope it's clear.

Bunuel,

I am not able to understand statement B. Can you please elaborate...

The question asks: is x/4 > y/5 ?

(2) says: y = x + 1. Substitute y = x + 1 into the question: is x/4 > (x+1)/5? --> is x > 4? Since we cannot answer this question, then this statement is not sufficient.

Hope it's clear.

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

03 Mar 2014, 21:26

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M = x/4
N = y/5

5x/20>4y/20 -->
5x>4y?

I.
x>.8y
x>4y/5
5x>4y
Suff

II.
y=x+1
5x>4x+4 ?

x=1/20
1/4 > 1/5 + 4 --> no
x=5
25>24 --> yes

Insuff

A

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

05 May 2015, 10:39

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IMHO, this is a fairly easy question but the language will deceive you and you will mark the wrong answer unless you muscle the algebra. I solved by intuition and got it wrong.

So basically the question is, "Is the work done by Machine M is greater than the work done by N" (in 20 min)

Machine M: T= 4 min, W=x hence rate = x/4
Machine N: T= 5 min, W=y hence Rate = y/5

M will do more work if its rate is more than N so is x/4 > y/5?
x > 4/5(y) or is x > 0.8(y) ? - this is the Question

Statement 1 is a direct answer

Statement 2 becomes:
x/4 > y/5?
x/4 > (x+1)/5?
x/4 > (x+1)/4+1? - this is a general property of ratios. the inequality is true if x<4. Since we don't know the value of x, state2 is insufficient.

Answer A

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

06 May 2015, 11:31

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Hi AjChakravarthy,

You bring up a number of important points in your post that are worth emphasizing:

1) Many DS questions ARE actually pretty easy, so you shouldn't take any chances when it comes to solving them - do the necessary work and get those points!

2) A Test Taker's "instinct" when dealing with DS question can often be incorrect, so you have to do enough work to PROVE that your instinct is correct.

3) Taking the time to "rewrite" the question can often lead to shortcuts later on in the work.

GMAT assassins aren't born, they're made,
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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

06 May 2015, 20:45

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violetsplash wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

1) x > 0.8y
2) y = x + 1

My question :

1st statement tells that x/y < 4/5 => this means x's rate is more than y hence its sufficient to answer that machine M produce more widgets than machine N in that time

2nd statement says that y = x + 1
so if x = 2 widgets
then y = 3 widgets
this means that in 20 minutes x will produce 10 widgets and in same time y will produce 12 widgets so we get a definite answer hence its sufficient

Thats why I chose D but the OA is A

Can someone please explain ?

Instead of picking numbers on this question, one might try to sit back and work out the logic of statement 2. It is quite simple once you get down to it.

You want to know whether x/4 > y/5. (x and y are good positive integers so no complications.)
So if x is greater than y, certainly x/4 will be greater than y/5 because x is divided by a smaller number.
If x is smaller than y, then it depends on how much smaller. If x is only slightly smaller than y, then it is possible that x/4 > y/5.

2nd statement says that y = x + 1
So x is 1 smaller than y. Now the problem is that you don't know the values of x and y so you don't know whether this 1 is huge in comparison to x and y or little. Therefore, you cannot say whether x/4 will be smaller or y/5.

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

16 Aug 2015, 22:03

Bunuel wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

The rate of M = x/4 widgets per minute.
The rate of N = y/5 widgets per minute.

The question basically asks whether x/4 > y/5 (if per minute M produces more widgets than N, then M obviously produces more widgets than N, in 20, 30, ... or in any time period. So, we can compare the rates per 1 minute).

(1) x > 0.8y --> x > 4/5*y --> x/4 > y/5. Sufficient.

(2) y = x + 1. The question becomes: is x/4 > (x+1)/5, or is x>4. We don't know that. Not sufficient.

Answer: A.

Hope it's clear.

Hi Banuel: Is the following reasoning correct/possible to identify statement (2) as sufficient? Reasoning as follows:

Because we know x > 4, it must be at least 5 assuming you can't have a fraction of a widget (i.e. you can't produce 4.5 widgets every 4 minutes). Given that y = x + 1 which yields 6 in this case, then 5/4 > 6/5 is in fact true. Furthermore, this inequality stays true as the value of x increases. Thus, the statement could be sufficient?

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

17 Aug 2015, 01:46

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tigrr49 wrote:

Bunuel wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

The rate of M = x/4 widgets per minute.
The rate of N = y/5 widgets per minute.

The question basically asks whether x/4 > y/5 (if per minute M produces more widgets than N, then M obviously produces more widgets than N, in 20, 30, ... or in any time period. So, we can compare the rates per 1 minute).

(1) x > 0.8y --> x > 4/5*y --> x/4 > y/5. Sufficient.

(2) y = x + 1. The question becomes: is x/4 > (x+1)/5, or is x>4. We don't know that. Not sufficient.

Answer: A.

Hope it's clear.

Hi Banuel: Is the following reasoning correct/possible to identify statement (2) as sufficient? Reasoning as follows:

Because we know x > 4, it must be at least 5 assuming you can't have a fraction of a widget (i.e. you can't produce 4.5 widgets every 4 minutes). Given that y = x + 1 which yields 6 in this case, then 5/4 > 6/5 is in fact true. Furthermore, this inequality stays true as the value of x increases. Thus, the statement could be sufficient?

No, that's not correct. We can say, for example, that machine M produces 0.5 widgets every 4 minutes, this would mean that to produce 1 widget it needs 8 minutes.

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

17 Jun 2018, 03:21

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violetsplash wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

(1) x > 0.8y
(2) y = x + 1

Both machines work for the same amount of time.
To produce more widgets than N, M must work faster than N.

Question rephrased:
Is M faster than N?

Statement 1: x > 0.8y
Let y=10, implying that x>8.
Since y=10, N produces 10 widgets every 5 minutes, implying that N's rate = 10/5 = 2 widgets per minute.
Since x>8, M produces MORE than 8 widgets every 4 minutes, implying that M's rate = (more than 8)/4 = MORE than 2 widgets per minute.
Thus, M is faster than N.
SUFFICIENT.

Statement 2: y = x+1
Case 1: x=1, y=2
Since y=2, N produces 2 widgets every 5 minutes, implying that N's rate = 2/5 widget per minute.
Since x=1, M produces 1 widget every 4 minutes, implying that M's rate = 1/4 widget per minute.
In this case, N is faster than M.

Case 2: x=100, y=101
Since y=101, N produces 101 widgets every 5 minutes, implying that N's rate = 101/5 ≈ 20 widgets per minute.
Since x=100, M produces 100 widgets every 4 minutes, implying that M's rate = 100/4 = 25 widgets per minute.
In this case, M is faster than N.

Since N is faster than M in Case 1 but slower than M in Case 2, INSUFFICIENT.

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A

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

23 Sep 2018, 09:03

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violetsplash wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

(1) x > 0.8y
(2) y = x + 1

\(M\,\,\, - \,\,\,4\,\,{\text{min}}\,\,\, - \,\,\,x\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,M\,\,\, - \,\,\,20\,{\text{min}}\,\,\, - \,\,\,5x\,\,{\text{widgets}}\)
\(N\,\,\, - \,\,\,5\,\,{\text{min}}\,\,\, - \,\,\,y\,\,{\text{widgets}}\,\,\,\,\, \Rightarrow \,\,\,\,\,N\,\,\, - \,\,\,20\,{\text{min}}\,\,\, - \,\,\,4y\,\,{\text{widgets}}\)

\(5x\mathop > \limits^? \,4y\)

\(\left( 1 \right)\,\,\,x > \frac{4}{5}y\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,5} \,\,\,\,\,5x > 4y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,\,\,y = x + 1\,\,\,\,\left\{ \begin{gathered}\\
\,\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\, \hfill \\\\
\,\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {5,6} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\, \hfill \\ \\
\end{gathered} \right.\)

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

Regards,
Fabio.

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

19 Dec 2020, 16:28

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Video solution from Quant Reasoning (starts at 0:19:48):
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

25 Mar 2021, 08:25

Bunuel wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

The rate of M = x/4 widgets per minute.
The rate of N = y/5 widgets per minute.

The question basically asks whether x/4 > y/5 (if per minute M produces more widgets than N, then M obviously produces more widgets than N, in 20, 30, ... or in any time period. So, we can compare the rates per 1 minute).

(1) x > 0.8y --> x > 4/5*y --> x/4 > y/5. Sufficient.

(2) y = x + 1. The question becomes: is x/4 > (x+1)/5, or is x>4. We don't know that. Not sufficient.

Answer: A.

Hope it's clear.

I am wondering where I went wrong here.

If we test various values for x: 5,6,7,8,100, etc. It seems that we can definitively say that 5x > 4y...? What am I missing.

L

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

25 Mar 2021, 19:46

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Hi CEdward,

You can TEST VALUES to prove that Fact 1 is SUFFICIENT (and the answer to the question is 'ALWAYS YES'). However, the values that you would TEST for Fact 2 include inconsistent results (sometimes the answer to the question is "YES" and sometimes it's "NO"). For example:

2) Y = X + 1

IF....
X = 1
Y = 2
5(1) is NOT > 4(2) and the answer to the question is NO.

IF...
X = 10
Y = 11
5(10) IS > 4(11) and the answer to the question is YES.
Thus, Fact 2 is INSUFFICIENT.

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

08 Jul 2021, 09:01

VeritasKarishma wrote:

violetsplash wrote:

Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes. If machines M and N working simultaneously at their respective constant rates for 20 minutes, does machine M produce more widgets than machine N in that time ?

1) x > 0.8y
2) y = x + 1

My question :

1st statement tells that x/y < 4/5 => this means x's rate is more than y hence its sufficient to answer that machine M produce more widgets than machine N in that time

2nd statement says that y = x + 1
so if x = 2 widgets
then y = 3 widgets
this means that in 20 minutes x will produce 10 widgets and in same time y will produce 12 widgets so we get a definite answer hence its sufficient

Thats why I chose D but the OA is A

Can someone please explain ?

Instead of picking numbers on this question, one might try to sit back and work out the logic of statement 2. It is quite simple once you get down to it.

You want to know whether x/4 > y/5. (x and y are good positive integers so no complications.)
So if x is greater than y, certainly x/4 will be greater than y/5 because x is divided by a smaller number.
If x is smaller than y, then it depends on how much smaller. If x is only slightly smaller than y, then it is possible that x/4 > y/5.

2nd statement says that y = x + 1
So x is 1 smaller than y. Now the problem is that you don't know the values of x and y so you don't know whether this 1 is huge in comparison to x and y or little. Therefore, you cannot say whether x/4 will be smaller or y/5.

How do we decide that for the question we should not waste our time in substituting numbers?

L

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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

08 Jul 2021, 16:41

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Hi swim2109,

Many GMAT questions (in both the Quant and Verbal sections) become significantly easier to solve if you re-organize the information that you've been given. With this DS question, by doing a little bit of set-up work before you get down to the two Facts, you can then use a mix of basic Algebra and TESTing VALUES to quickly prove what the correct answer is. None of that work is particularly time-consuming as long as you choose to reorganize the information in the initial prompt (which is also not particular time-consuming or difficult). To maximize your performance on Test Day, you need to hone a variety of different skills - and look for the 'clues' in each prompt to help you to quickly determine which skills will be fastest and easiest to use for that prompt.

GMAT assassins aren't born, they're made,
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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

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