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

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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?

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% _________________

Bhushan S. If you like my post....Consider it for Kudos

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).

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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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.

Cheers! JT........... If u like my post..... payback in Kudos!!

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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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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? please update.

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....

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? please update.

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

Re: Working alone at its own constant rate, a machine seals k [#permalink]

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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%

Re: Working alone at its own constant rate, a machine seals k [#permalink]

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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

60/90 is the answer
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Re: Working alone at its own constant rate, a machine seals k [#permalink]

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18 Mar 2014, 11:59

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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]

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11 Oct 2016, 02:59

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