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# Machines X and Y produce bottles at their respective constant rates.

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Machines X and Y produce bottles at their respective constant rates. [#permalink]

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29 Aug 2010, 20:43
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Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?

(A) 8
(B) 12
(C) 15
(D) 18
(E) 24
[Reveal] Spoiler: OA
Intern
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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29 Aug 2010, 20:48
My answer to this question was 8

X -> k bottles in 6 hours -> 1/6
Y -> k bottles in 2 hours -> 1/2

Working together -> 2/3

if they produce k bottles in 2/3 hrs, they will produce 12 * 2/3 bottles in 8 hours.

Could you please point out where I am mistaken??
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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29 Aug 2010, 20:58
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KUDOS
Safiya wrote:
My answer to this question was 8

X -> k bottles in 6 hours -> 1/6
Y -> k bottles in 2 hours -> 1/2

Working together -> 2/3

if they produce k bottles in 2/3 hrs, they will produce 12 * 2/3 bottles in 8 hours.

Could you please point out where I am mistaken??

X - k bottle in 6 hours => In 1 hour - k/6 bottles
Y - k bottle in 2 hours => In 1 hour - k/2 bottles

Working together in 1 hour - 2k/3

working together in 3 hours - 2k bottles

2k bottles in 3 hours
=> 2k *6 bottles = 12 bottles in 3*6 = 18 hours
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Intern
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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29 Aug 2010, 21:01
Ahhh what a silly mistake! Thank you very much gurpreetsingh
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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30 Aug 2010, 04:18
X in 6 hours: k bottles
Y in 2 hours: k bottles
Y in 6 hours: 3k bottles

X and Y in 6 hours: 4k bottles

12k bottles: 12/4 = 3 (6 hour periods)
hence 18 hours
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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31 Aug 2010, 05:21
try considering the below simple solution.

we have two different time periods resulting in the same quantity of the product (K).

X working for 6 hours, produces K bottles
in those 6 hours Y produces 3K bottles as it produces K in 2 hours.

so in first 6 hours K+3K = 4K
in 12 hours 4K+4k
in 18 hours 4K+4K+4K = 12K
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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31 Aug 2010, 08:26
Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?

formula is work = rate x time

for machine (X) k= r X 6
for machine (Y) k= r X 2
now take a value for k ( for ease which is a multiple of both mac X & y)
take k=12
so rate of machine (x) is 2
so rate of machine (y) is 6

now work is 12k = 12x12 = 144
two machines working siman so rate = 6+2 =8
work = rate X time
144 = 8 X time
so time = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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01 Sep 2010, 00:38
(6*2/(6+2))*12 = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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29 Aug 2011, 11:28
since the question asks number of HOURS, so we should use this method.

x,y working together to produce k bottles = 6*2/6+2 = 3/2 hrs
12k bottles = 3*12/2 = 18
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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30 Oct 2011, 11:23
$$\frac{1}{6}+\frac{1}{2}=\frac{1}{T}$$

$$T=\frac{3}{2}$$ for k bottles

and $$\frac{3}{2}*12=18$$ for 12K bottles
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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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20 Feb 2015, 10:34
Safiya wrote:
Machines X and Y produce bottles at their respective constant rates. Machine X produces k bottles in 6 hours and machine Y produces k bottles in 2 hours. How many hours does it take machines X and Y , working simultaneously , to produce 12k bottles?

(A) 8
(B) 12
(C) 15
(D) 18
(E) 24

x rate = k/6
y rate = k/2

k/6 + k/2 = 12k / T

solving T = 18

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Re: Machines X and Y produce bottles at their respective constant rates. [#permalink]

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20 Feb 2015, 13:30
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Expert's post
Hi All,

This is an example of a Work Formula question with a minor 'twist.' The "math" in these questions can be done in a few different ways, but since we have 2 machines working on a task together, without any major "twists" (e.g. one of them stops working at a certain point), we can use the Work Formula:

Work = (A)(B)/(A+B) where A and B are the respective rates of the two machines to do the same task individually.

We're told:
1) Machine X can produce K bottles in 6 hours.
2) Machine Y can produce K bottles in 2 hours.

(6)(2)/(6+2) = 12/8 = 1.5 hours to produce K bottles when working together.

The minor 'twist' is that the question asks how long it takes to produce 12K bottles (not K bottles), so we have to multiply this result by 12...

(1.5 hours)(12) = 18 hours.

[Reveal] Spoiler:
D

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Re: Machines X and Y produce bottles at their respective constant rates.   [#permalink] 20 Feb 2015, 13:30
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