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

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

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02 Jul 2011, 23:53
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/machines-x-an ... 04208.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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03 Jul 2011, 04:34
siddhans wrote:
Please explain in detail? If anyone has studied from Manhattan . Can you please solve using their method?

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.

U can plug in the values using option B .
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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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03 Jul 2011, 04:39
so we have here Total Work Done = 4X+3Y ie 4 hours of X rate + 3 hours of Y rate
let the net rate be r. so 7 hours of r = 4x+3y
ie 7r=4x+3y

now statement 1 : gives me rate of x...it is not sufficient since it doesnt say anything about y
statement 2 : 4 hours of x = 3 hours of y
ie 4x=3y
substituting in original equation we get => 7r =8x or x=(7/8)r

now 7 hours of r -----> full work
p hours of x -----> full work

therefore 7r = px but x=(7/8)r
so, 7r = 7/8 * r * p
or p = 8 hours...so the statement is sufficient.

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

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03 Jul 2011, 10:15
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KUDOS
fivedaysleft wrote:
statement 2 : 4 hours of x = 2( 3 hours of y)

Shouldn't it be twice?

Let the rate of m/c X be x bottles/hour
Let the rate of m/c Y be y bottles/hour

So in 4 hours m/c X will produce = 4x bottles
and in 3 hours m/c Y will produce = 3y bottles.
Total bottles = 4x+3y ----(1)

Stmt1: Machine X produced 30 bottles per minute. i.e x=30bottles/min but rate of y is not given. Insufficient.

Stmt2: Machine X produced twice as many bottles in 4 hours as machine Y produced in 3 hours.

4x = 2 * 3y
y=2/3 x
Substituting in (1), 4x+ 3 * 2/3 x = 6x total number of bottles.

Now rate of m/c X is x bottles in 1 hour
so 6x bottles in 6 hours.

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

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03 Jul 2011, 10:21
yeah...my bad...but since it DS...it didnt do THAT much harm...the concept is fine though right?

any other mistake, kindly point out!
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How charged with punishments the scroll,
I am the master of my fate :
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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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05 Dec 2017, 02:51
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: http://gmatclub.com/forum/machines-x-an ... 04208.html
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Re: Machines X and Y produced identical bottles at different constant rate   [#permalink] 05 Dec 2017, 02:51
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