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# Work-Rate Problem

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Intern
Joined: 06 Aug 2007
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Work-Rate Problem [#permalink]  29 Aug 2010, 19:43
00:00

Difficulty:

25% (medium)

Question Stats:

89% (01:55) correct 11% (01:00) wrong based on 37 sessions
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: Work-Rate Problem [#permalink]  29 Aug 2010, 19:48
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: Work-Rate Problem [#permalink]  29 Aug 2010, 19:58
1
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
Joined: 06 Aug 2007
Posts: 32
Location: Montreal
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Re: Work-Rate Problem [#permalink]  29 Aug 2010, 20:01
Ahhh what a silly mistake! Thank you very much gurpreetsingh
Intern
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Re: Work-Rate Problem [#permalink]  30 Aug 2010, 03:18
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
Manager
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Re: Work-Rate Problem [#permalink]  31 Aug 2010, 04:21
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
Manager
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Re: Work-Rate Problem [#permalink]  31 Aug 2010, 07:26
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
Manager
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Re: Work-Rate Problem [#permalink]  31 Aug 2010, 23:38
(6*2/(6+2))*12 = 18
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Re: Work-Rate Problem [#permalink]  29 Aug 2011, 10:28
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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Manager
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Re: Work-Rate Problem [#permalink]  30 Oct 2011, 10:23
\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
Re: Work-Rate Problem   [#permalink] 30 Oct 2011, 10:23
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# Work-Rate Problem

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