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Re: Machine X and machine Y are the only two machines used [#permalink]
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]
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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]
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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]
@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]
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Expert Reply
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]
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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]
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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]
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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]
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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]
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]
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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]
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]
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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]
Weighted average works here. :)
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