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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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New post 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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New post 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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New post 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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New post 03 Jul 2011, 10:15
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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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New post 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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Re: Machines X and Y produced identical bottles at different constant rate [#permalink]

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New post 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.

Answer: B.

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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