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# Machines X and Y produced identical bottles at different constant rate

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Machines X and Y produced identical bottles at different constant rate [#permalink]

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08 Nov 2011, 20:21
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Question Stats:

61% (01:10) correct 39% (01:16) wrong based on 250 sessions

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Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?

(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours.

[Reveal] Spoiler:
My work:

Bottles = (rateX)4 + (rateY)3 *Did not forget to convert minutes to hours

1) RateX = 30 b/min = 1800 b/hr
Xbottles = 1800(4) = 7200 bottles

Insuff

2) Bottles Y = 1/2 Bottles X

I went with putting them together, getting the total number of bottles produced and backing out the rate for the answer. WRONG.

I have the answer, I just don't understand it.

OPEN DISCUSSION OF THIS QUESTION IS HERE: machines-x-and-v-produced-identical-bottles-at-different-104208.html
[Reveal] Spoiler: OA

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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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08 Nov 2011, 21:39
1
KUDOS
I recommend you to approach the question as mentioned below -

As question is asking to find time taken by Machine X operating alone to fill the entire production lot and we've know that production lot (a) partily filled by X in 4 hrs (b) Rest filled by Y in 3 hrs.
As we dont have any information about Part(b), we need to find or establish a relationship between X and Y so that we can calculate the whole period in terms of X. Fortunately, The relation is clearly given in (2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours.

Hence, B is the answer. No calculation required.

Hope it helps!
Cheers!
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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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09 Nov 2011, 13:00
Thanks. Your explanation is so much clearer.

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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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01 Dec 2011, 00:38
(1) x filled 30 x 4 = 120 bottles, but we do not know the what portion is 120. insufficient.

(2) y = 3b
x = 6b
total 9b
so 2 sufficient.
Ans. B
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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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12 Nov 2014, 14:30
X Y total
Time 4 3 7
rate x y
bottles 4x 3y 4x+3y

from statement 1 - x= 30b/min = 1800b/hr, nothing is known of y here, so insuff
from statement 2 - 4x = 3y * 2, x = 3y/2.
total bottles = 4*3y/2 + 3y = 9y bottles to be manufactured by x alone in how much time is the question.
let the time be T.
so, 3y/2*T = 9y, T = 6 hours.
sufficient.

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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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12 Nov 2014, 18:52
Expert's post
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galaxyblue wrote:
Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?

(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours.

[Reveal] Spoiler:
My work:

Bottles = (rateX)4 + (rateY)3 *Did not forget to convert minutes to hours

1) RateX = 30 b/min = 1800 b/hr
Xbottles = 1800(4) = 7200 bottles

Insuff

2) Bottles Y = 1/2 Bottles X

I went with putting them together, getting the total number of bottles produced and backing out the rate for the answer. WRONG.

I have the answer, I just don't understand it.

There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$. For example when we are told that a man can do a certain job in 3 hours we can write: $$3*rate=1$$ --> $$rate=\frac{1}{3}$$ job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then $$5*(2*rate)=1$$ --> so rate of 1 printer is $$rate=\frac{1}{10}$$ job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then $$3*(2*rate)=12$$ --> so rate of 1 printer is $$rate=2$$ pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is $$rate_a=\frac{job}{time}=\frac{1}{2}$$ job/hour and B's rate is $$rate_b=\frac{job}{time}=\frac{1}{3}$$ job/hour. Combined rate of A and B working simultaneously would be $$rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$ job/hour, which means that they will complete $$\frac{5}{6}$$ job in one hour working together.

3. For multiple entities: $$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}$$, where $$T$$ is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}$$, where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$t_1$$ and $$t_2$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}$$ hours.

BACK TO THE ORIGINAL QUESTION:
Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?

You can solve this question as Karishma proposed in her post above or algebraically:

Let the rate of X be $$x$$ bottle/hour and the rate of Y $$y$$ bottle/hour.
Given: $$4x+3y=job$$. Question: $$t_x=\frac{job}{rate}=\frac{job}{x}=?$$

(1) Machine X produced 30 bottles per minute --> $$x=30*60=1800$$ bottle/hour, insufficient as we don't know how many bottles is in 1 lot (job).

(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours --> $$4x=2*3y$$, so $$3y=2x$$ --> $$4x+3y=4x+2x=6x=job$$ --> $$t_x=\frac{job}{rate}=\frac{job}{x}=\frac{6x}{x}=6$$ hours. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: machines-x-and-v-produced-identical-bottles-at-different-104208.html
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Re: Machines X and Y produced identical bottles at different constant rate   [#permalink] 12 Nov 2014, 18:52
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