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# Machine M, working alone at its constant rate, produces x widgets ever

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

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24 Sep 2013, 08:29
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61% (01:23) correct 39% (01:27) wrong based on 902 sessions

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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 :

[Reveal] Spoiler:
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

[Reveal] Spoiler: OA

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

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24 Sep 2013, 08:46
1
KUDOS
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

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

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25 Sep 2013, 02:40
37
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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.

Hope it's clear.
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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

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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.

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]

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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.

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]

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03 Mar 2014, 21:26
2
KUDOS
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]

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29 Mar 2015, 18:50
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Hi All,

The way that you choose to organize your notes can often impact how much work you have to do to answer a given GMAT question (as well as the difficulty of the work). As such, part of your practice should really involve proper note-taking and working on the most 'efficient' ways for you to take notes, 'set up' your work, etc.

Here, we're told about the rates of 2 machines:
Machine M can produce X widgets in 4 minutes
Machine N can produce Y widgets in 5 minutes

We're then told that each machine works for 20 minutes. We're asked if Machine M produces more widgets than Machine N during that time.

During those 20 minutes, Machine M will produce 5X widgets and Machine N will produce 4Y widgets. In real simple terms, the question asks "is 5X > 4Y?" This is a YES/NO question. A mix of TESTing VALUES and Algebra will help to answer this question.

Fact 1: X > 0.8Y

We can multiply both sides of this inequality by 10....

10X > 8Y

Then divide both sides by 2...

5X > 4Y

Notice how the question asked "is 5X > 4Y?".... Fact 1 tells us that 5X IS greater than 4Y.
Fact 1 is SUFFICIENT

Fact 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.
Fact 2 is INSUFFICIENT

[Reveal] Spoiler:
A

GMAT assassins aren't born, they're made,
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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3505 [3], given: 173 Manager Joined: 08 Oct 2013 Posts: 52 Kudos [?]: 17 [1], given: 16 Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink] ### Show Tags 05 May 2015, 10:39 1 This post received KUDOS 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 Kudos [?]: 17 [1], given: 16 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10111 Kudos [?]: 3505 [1], given: 173 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink] ### Show Tags 06 May 2015, 11:31 1 This post received KUDOS Expert's post 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, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Machine M, working alone at its constant rate, produces x widgets ever [#permalink]

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06 May 2015, 12:08
EMPOWERgmatRichC wrote:
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,
Rich

GMAT assassins aren't born, they're made, - very powerful.

Training to be one

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

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06 May 2015, 20:45
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

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]

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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.

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]

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17 Aug 2015, 01:46
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.

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]

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19 Sep 2015, 12:44
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.

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?

That's circular reasoning. The question we are trying to answer is if $$\frac{x}{4}>\frac{y}{5}$$ and so it cannot be combined with a true statement. I would like to add that, if $$x=2$$ and $$y=3$$, then machine M produces 10 widgets and machine N produces 12 widgets, and therefore, machine M can produce less widgets per 20 minutes than machine N does.

Kr,
Mejia

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

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07 Jan 2016, 09:36
Hi Bunuel,

So basically we are comparing x/4 to 0.8y/4(keeping base same as 4). In second choice, y=x+1. so hence we are now trying to check if x>0.8(x+1) or 0.8x+0.8 I assumed that that 0.8y has to be an integer. Is this assumption correct. I took this assumption and arrived at that x>=0.8x+0.8 for any value where 0.8x+0.8 is an integer. Hence 2 is insufficient. Please tell me if this approach is Ok

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

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12 Jan 2017, 15:49
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Re: Machine M, working alone at its constant rate, produces x widgets ever   [#permalink] 12 Jan 2017, 15:49
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