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# Working alone at its own constant rate, a machine seals k

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Working alone at its own constant rate, a machine seals k [#permalink]  13 Oct 2009, 11:13
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Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

A. $$25%$$

B. $$33\frac{1}{3}%$$

C. $$50%$$

D. $$66\frac{2}{3}%$$

E. $$75%$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Feb 2012, 12:14, edited 1 time in total.
Edited the question and added the OA
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Re: work/rate [#permalink]  13 Oct 2009, 11:33
work will be done A(8hrs):B(4hrs) = 1:2
for first 2 hrs = 2 x (1/8+1/4) = 3/4 work is done
remaining work 1/4th will be done in 1:2 ratio
thus A will do 1/12 th work
B will do 1/6 work

Hence fatser m/c B will do = 1/4+1/4+1/6 = 2/3 = 66 2/3%
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Re: work/rate [#permalink]  13 Oct 2009, 14:28
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Expert's post
bhushan252 wrote:
work will be done A(8hrs):B(4hrs) = 1:2
for first 2 hrs = 2 x (1/8+1/4) = 3/4 work is done
remaining work 1/4th will be done in 1:2 ratio
thus A will do 1/12 th work
B will do 1/6 work

Hence fatser m/c B will do = 1/4+1/4+1/6 = 2/3 = 66 2/3%

No need for lengthy calculations: since the second one works twice as fast as the first, working together second will do 2/3 of the job (their ratio 1/2).

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Re: work/rate [#permalink]  13 Oct 2009, 16:39
if someone wants to double check and keep things simple, lets assume that they work for 8 hours and k=10, so the guy who takes 8 hours complets 10 and the guy who takes 4 hours complets 20 cartons

so faster guy % = 100* 20/30 = 66 2/3 %
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Re: Work-Rate problem [#permalink]  07 Jan 2010, 00:57
If I am understanding this correctly, you could do the following:

Let's make k = 32 cartons. This means Machine A is doing this at a rate of 32 cartons/8 hours = 4 cartons/hour. Likewise, Machine B is working at a rate of 32 cartons/4 hours = 8 cartons/hour. Machine B is working at a faster rate!

Let's multiply them by the same time to see their output; let's use 10 hours.

Machine A: (4 cartons per hour)*(10 hours) = 40 cartons
Machine B: (8 cartons per hour)*(10 hours) = 80 cartons
Total: 120 cartons

Now it's just a percentage of the total.

Machine A is is doing 40/120 = 1/3 of the work, and Machine B is doing 2/3, or a little over 66%.

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Re: Work-Rate problem [#permalink]  07 Jan 2010, 01:24
Thanks for the update.

I have thought a general way.

Lets X machine seal K cartons in 8 hours : K/8

Lets Y machine seal K cartons in 4 Hours : K/4

Keeping in mind that they both sealed same period of time - Machine Y seals 2K cartons in 8 hours.

so K + 2K = 3K total carton sealed.
So machine Y sealed 2K/3K cartons that's 66%
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Re: Work-Rate problem [#permalink]  07 Jan 2010, 11:09
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1st Machine:
8 hrs --> K
1 hr ---> K/8

2nd Machine:
4hrs --> K
1 hr ---> K/4

Both Machines together:
1 hr --> K/4 + K/8 = 3K/8

Faster Machine (2nd) % $$= \frac{K/4 }{3K/8} * 100= \frac{K*8}{4*3K} * 100 = \frac{2}{3} * 100 = 66\frac{2}{3} %$$
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Re: Work-Rate problem [#permalink]  07 Jan 2010, 11:30
Slow machine: K in 8 hours
Fast machine: K in 4 hours => 2K in 8 hours

Therefore, in 8 hours, with both machines working, there would be 3K.

Fast machine's % of output is 2K/3K = 2/3.
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Re: Work-Rate problem [#permalink]  12 Jan 2010, 02:10
the key here is that it doesn't specify what period of time they were working together. So it must be that for any period of time the percentage will be the same. So you should pick a time period that is convenient for quick calculations.

machine 1 can do the job in 8 hours
machine 2 can do the job in 4 hours

lets choose the time period to be 8 hours since this will conveniently give us easy numbers for k: After 8 hours, machine one does k amount of work and machine two does 2k amount of work.

What percentage of the total work does the faster machine do? (machine 2 is the faster one.)

percentage of work that machine 2 does = (machine 2 work) / (total work) = 2k / (k + 2k) = 2k/3k = 2/3 = 66.6%
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Re: Work-Rate problem [#permalink]  12 Jan 2010, 18:07
vinayaksatapute wrote:
22.Working alone at its own constant rate, a machine seals k cartons in 8 hours,
and working alone at its own constant rate, a second machine seals k cartons in
4 hours. If the two machines, each working at its own constant rate and for the
same period of time, together sealed a certain number of cartons, what percent
of the cartons were sealed by the machine working at the faster rate?

25%
33 1/3%
50%
66 2/3%
75%

Is the answer 50% for this question?

Specify an output for machine A. For simplicity I picked 8 units in 8 hours. Thus, A makes 1 unit per hour.
Since unit B works at twice the speed it puts out 2 units/hour.
Total 3 units are being produced, 2 of them by machine B.
2/3= 66.66....
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Re: Work-Rate problem [#permalink]  13 Jan 2010, 02:17
let the machine 1 do X work in 1 hour
Then the machine 2 would do 2X work in 1 hour

So the machine 2 would do 2X/3X*100=66.66%

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Re: Work-Rate problem [#permalink]  12 Feb 2010, 21:35
2nd mechine is twice as fast as 1st machine, so if 1st seals 1 box, the 2nd can seal 2 in the same time.

Overall 2nd= 2/(2+1)=2/3=66.667%
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Re: Work-Rate problem [#permalink]  12 Feb 2010, 23:17
vinayaksatapute wrote:
22.Working alone at its own constant rate, a machine seals k cartons in 8 hours,
and working alone at its own constant rate, a second machine seals k cartons in
4 hours. If the two machines, each working at its own constant rate and for the
same period of time, together sealed a certain number of cartons, what percent
of the cartons were sealed by the machine working at the faster rate?

25%
33 1/3%
50%
66 2/3%
75%

Is the answer 50% for this question?

If both work for 4 hours then
A sealed k/2 cartoons
B sealed k cartoons so B is 2 times faster then A.
total cartoons sealed is k+k/2 = 3k/2
%age of B = k/(3k/2) * 100 = 66 2/3% hence D
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  09 Feb 2012, 20:31
Ans is D
One can seal k cartons in 8hours and the other can seal k cartons in 4 hours so it can seal 2k cartons in 8 hours
so total both can seal 3k cartons in 8 hours by working independently.
So for the faster one it is 2k/3k =200/3%
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  23 Feb 2012, 07:23
Agree with D

rate of the first machine is k/4
rate of the second machine is k/8

easy to compute that it is 66 2/3 from here
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  18 Mar 2014, 08:22
One machine's rate = 2 Other machine's rate
y=2x

And consider whole lot of 90 then

x+y will produce 90

3x will produce 90

x=30

then 2x=60

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Re: Working alone at its own constant rate, a machine seals k [#permalink]  18 Mar 2014, 11:59
The second machine seals twice the number of cartons in the same time, so it is twice as fast. Therefore it will do twice the work that the first machine does. Divide 100% into two parts such that one is double the other to get (D) as the answer.
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Re: Working alone at its own constant rate, a machine seals k   [#permalink] 18 Mar 2014, 11:59
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