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Machine X and machine Y are the only two machines used

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gulsam
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Machine X and machine Y are the only two machines used [#permalink] New post   Updated on: 28 Dec 2019, 07:20
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Machine X and machine Y are the only two machines used to fill bottles of Zap Cola for a certain shipment. X fills each bottle in the shipment with 584 milliliters of cola. Y fills each bottle in a shipment with 594 milliliters of cola. Bottles of Zap Cola in the shipment contain an average of 590 milliliters of cola.


Quantity A
Quantity B
The number of bottles in the shipment filled by machine Y
The number of bottles in the shipment filled by machine X






A.Quantity A is greater
B.Quantity B is greater
C.The two quantities are equal.
D.The relationship cannot be determined from the information given.
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A

Originally posted by gulsam on 28 Dec 2019, 07:07.
Last edited by gulsam on 28 Dec 2019, 07:20, edited 5 times in total.
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gulsam
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   28 Dec 2019, 07:10
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This is a weighted average problem. While it is not possible to know precisely how many bottles each machine fills, it is possible to determine the ratio between them. Consider first what would happen if X fills only one bottle and Y fills only one bottle as well. So the average for the shipment would simply be an average of 584 to 594. On the other hand, if X filled all of the bottles, then the average of the shipment would be 584; if Y filled all of the bottles, the average would be 594.

This simple thought experiment shows that as a machine makes more of a contribution to the shipment, the overall average is pulled closer to that machine. Since the average is closer to the average of Y than it is to the average of X, machine Y must have contributed more to the shipment than X, and thus must have filled more bottles.

In addition, we can get even more definite with this. Since there is a difference of 10 between the average of X and Y, and since the overall average is 6 greater than the average of X (or 4 less than the average of Y), Y must fill 6/10ths of the bottles, and X must fill 4/10ths.

The correct answer is B.
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gulsam
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   29 Dec 2019, 05:16
Carcass wrote:
Please sir follow the rules to format properly a QCQ.

They are at the top of the quant section.

Regards


Thanks, I see
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   02 Jan 2020, 07:38
can someone provide the correct answer to this question ?
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   28 Feb 2020, 17:10
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Hello @Carcass

Can we solve using weighted averages?

x/y = (594-590)/(590-584)...

x/y = 2/3

Hence y is bigger that x in proportion and then the answer will be A.

Ans A
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   29 Feb 2020, 00:07
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FCOCGALVAN wrote:
Hello @Carcass

Can we solve using weighted averages?

x/y = (594-590)/(590-584)...

x/y = 2/3

Hence y is bigger that x in proportion and then the answer will be A.

Ans A


of course :)

here a video by Ron

Efficient Choice of Variables in Word Problems, and Weighted Averages

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Re: Machine X and machine Y are the only two machines used [#permalink] New post   02 Mar 2020, 16:19
@Carcass

How would it be solved usidn weighted averages?

Im doin smthing wrogn cuz im getting A as answer.

Kind regards!
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   03 Mar 2020, 05:54
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584 ---------- X ------------- 590 ------ Y ------ 594

X is 6 and Y is 4

In the average X has a more specific weight

B is the answer
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   03 Mar 2020, 10:43
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584x + 594x = 590
or, x = 0.5
so that X machine = 584x = 584*.5= 292
Y Machine = 594*.5= 297
B is the answer
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   22 Apr 2020, 08:37
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Carcass wrote:
584 ---------- X ------------- 590 ------ Y ------ 594

X is 6 and Y is 4

In the average X has a more specific weight

B is the answer


B - wrong answer!

if y = 4 means that the weight of Y should be higher.
Suppose - we have 10X and 10Y - the mean is (10*584 + 10*594)/20 = 589
The mean = 590 requires that Y should be higher than X. In our case, if X=10, Y=15 that makes mean 590. If Y < X - the mean will always be less than 589.

Answer A is correct.
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   22 Apr 2020, 09:00
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gulsam wrote:
Machine X and machine Y are the only two machines used to fill bottles of Zap Cola for a certain shipment. X fills each bottle in the shipment with 584 milliliters of cola. Y fills each bottle in a shipment with 594 milliliters of cola. Bottles of Zap Cola in the shipment contain an average of 590 milliliters of cola.


Quantity A
Quantity B
The number of bottles in the shipment filled by machine Y
The number of bottles in the shipment filled by machine X



ASIDE--------------------
Key concept: If we combine EQUAL amounts of two populations, then the average of those two populations COMBINED will equal the AVERAGE of those two populations
For example, let's say solution A is 30% salt, and solution B is 40% salt.
If we combine EQUAL amounts of solution A and solution B, the concentration of the resulting mixture = (30 + 40)/2 = 35
So the resulting mixture is 35% salt.
------------------------

GIVEN:
Machine X bottles contain (on average) 584 ml
Machine Y bottles contain (on average) 594 ml

If we were to combine an EQUAL number of bottles from each machine, then the average volume of the COMBINED population = (584 + 594)/2 = 589 ml

GIVEN: The COMBINED shipment has an average volume of 590 ml
Since 590 is closer to Machine Y's average volume (of 594 ml) than it is to Machine X's average volume (of 584 ml), we know that the COMBINED shipment must contain more bottles from Machine Y.

Answer: A

The above concepts are described in greater detail in the following video:
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   22 Apr 2020, 09:03
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FCOCGALVAN wrote:
@Carcass

How would it be solved usidn weighted averages?

Im doin smthing wrogn cuz im getting A as answer.

Kind regards!


You're right. The correct answer is A. It looks like the original poster added the wrong answer.
I've now corrected the original post.
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   09 Jul 2022, 03:06
Can someone say what is wrong with this calculation.

Total volume in shipment if n is number of bottles= 590n.
Number of bottles filled by X = 590n/584= 1.010273573n
Number of bottles filled by Y = 590n/594= 0.993265993n
Since n must be a positive integer B is greater.
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   09 Jul 2022, 06:41
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Adewale wrote:
Can someone say what is wrong with this calculation.

Total volume in shipment if n is number of bottles= 590n.
Number of bottles filled by X = 590n/584= 1.010273573n
Number of bottles filled by Y = 590n/594= 0.993265993n
Since n must be a positive integer B is greater.


Presumably, n is the TOTAL number of bottles in the shipment (including bottles filled by machine X and bottles filled by machine Y).
So, 590n = the TOTAL volume of cola in the entire shipment.

Now you need to ask yourself, "What does it mean when I write: X = 590n/584= 1.010273573n?"
Hint: Notice that 1.010273573n is greater than n.
So, what does 1.010273573n mean?
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   09 Jul 2022, 09:33
Okay Brent. I understand the question but why did the calculation lead me there?
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   14 Aug 2022, 15:55
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First of all, we can just see that 590 is close to the amount 594 than it is of 584, so without any calculations, the answer is clearly machine Y (A). However, I didn't see this at first, so here is the calculation I did:

X = fraction of bottles filled by machine X
Y = fraction of bottles filled by machine Y = 1-X

584X + 594Y = 590 --> 584X + 594(1-X) = 590
584X - 594X + 594 = 590
-10X = -4
X = 4/10 (meaning 4/10 bottles are filled by machine X), which means Y = 6/10
So more bottles are filled by machine Y --> answer A
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   16 Jul 2024, 08:49
Weighted average works here. :)
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   27 Oct 2024, 11:38
FCOCGALVAN wrote:
Hello @Carcass

Can we solve using weighted averages?

x/y = (594-590)/(590-584)...

x/y = 2/3

Hence y is bigger that x in proportion and then the answer will be A.

Ans A



I diidnot Get it??
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   26 Dec 2024, 18:27
1
I think it is also possible to go:

(584x + 594y)/(x+y) = 590
590x + 590y = 584x + 594y
4y=6x
x=(4/6)y

y is bigger
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Re: Machine X and machine Y are the only two machines used [#permalink] New post   29 Jan 2026, 11:39
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Re: Machine X and machine Y are the only two machines used [#permalink]
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