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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]

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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, 13:14, edited 1 time in total.
Edited the question and added the OA

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Re: work/rate [#permalink]

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New post 13 Oct 2009, 12:33
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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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).

Answer: D.
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New post 13 Oct 2009, 17:39
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]

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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.

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%.

Answer: D
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Re: Work-Rate problem [#permalink]

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New post 07 Jan 2010, 02:24
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]

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Answer is D...

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]

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New post 07 Jan 2010, 12:30
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]

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New post 12 Jan 2010, 03:10
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]

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New post 12 Jan 2010, 19:07
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?
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....

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New post 13 Jan 2010, 03:17
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%

Hence the answer is D

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New post 12 Feb 2010, 22:35
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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New post 13 Feb 2010, 00:17
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?
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

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Working alone at its own constant rate, a machine seals k cartoons in 8 hours, and working at its own constant rate a second machine seals k cartoons in 4 hours. if 2 machines each working at its own constant rate and for the same period of time together sealed a certain no of cartoons. what percentage of the cartoons were sealed by the machine working at the faster rate?

A. 25%
B. 33 1/3%
C. 50%
D. 66 2/3%
E. 75%

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Re: working alone at its own constant rate [#permalink]

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anilnandyala wrote:
working alone at its own constant rate, a machine seals k cartoons in 8 hours, and working at its own constant rate a second machine seals k cartoons in 4 hours. if 2 machines each working at its own constant rate and for the same period of time together sealed a certain no of cartoons. what percentage of the cartoons were sealed by the machine working at the faster rate?
a 25%
b 33 1/3%
c 50%
d 66 2/3%
e 75%

i am missing something in this, please explain thanks in advance


As second one works twice as fast as the first, working together second will do 2/3 of work and the first 1/3 of work (their ratio 1/2).

Answer: D.
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Re: working alone at its own constant rate [#permalink]

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anilnandyala wrote:
working alone at its own constant rate, a machine seals k cartoons in 8 hours, and working at its own constant rate a second machine seals k cartoons in 4 hours. if 2 machines each working at its own constant rate and for the same period of time together sealed a certain no of cartoons. what percentage of the cartoons were sealed by the machine working at the faster rate?
a 25%
b 33 1/3%
c 50%
d 66 2/3%
e 75%

i am missing something in this, please explain thanks in advance


or think of it this way:
First machine seals k cartons in 8 hours.
Second machine seals k cartons in 4 hours i.e. 2k cartons in 8 hours.

If they both are working for the same period of time i.e. 8 hours, they together seal 3k cartons. Faster machine (second one) seals 2k out of these 3k so it seals 2/3 = 66.66% of the total cartons.
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Re: working alone at its own constant rate [#permalink]

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I solved it this way:

Step 1:

In one hour, the amount of work completed by Machine 1 is 1/8 and the amount of work completed by Machine 2 is ¼ i.e. Machine 2 works twice as fast as Machine 1.

(You can get the answer in Step 1 only. Anyhow, you will understand how to get the answer in the Step 1 by using a shortcut in Step 3)


Step 2:

In two hours, the amount of work completed by Machine 1 is 2/8 and the amount of work completed by Machine 2 is 2/4.

⇨ Total work completed in 2 hours by 2 machines = (2/8)+(2/4) => (1/4) + (1/2) => 0.25+0.50 = 0.75.
⇨ So the total amount of work completed by both the machines is 75 %. Out of the 75%, Machine 1 completes 50% of the work and Machine 2 completes 25% of the work.


Step 3:

Only 25% of the work is left. So the remaining work has to be divided between Machine 1 and Machine 2. As we know, Machine 2 can do the work twice as fast as Machine 1.

For example, if Machine 1 can do 10% of the work, Machine 2 can do 20% of the work in the same time. So, together they will complete 30% of the work. So from this example, we can understand that any work can be divided into 3 parts, and if Machine 1 completes 1 part of the work, and Machine 2 completes 2 parts of the work in the same time, irrespective of the quantum of work.

Now, the pending work is 25%. Let’s divide the work into 3 equal parts i.e. 8.33% per part. So, Machine 1 will complete 8.33% of the work whereas Machine 2 will complete 16.66% of the work, which is remaining.

So the total work completed by Machine 1 = 25%+8.33% = 33.33%
The total work completed by Machine 2 = 50% + 16.66% = 66.66%

So, the answer is (D)

We can apply Step 3 in Step 1 to get the solution faster, avoiding all the unnecessary calculations of Step 2 and Step 3.

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New post 23 May 2011, 21:34
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if each one has worked for 8 hrs then number of cartoons = k + 2k = 3k.
Mac B produces 2k of these.

hence % = 2k/3k = 66.66%
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I solved the problem directly
Rate one machine = k/8
Rate of the other machine = k/4
Total k/8+ k/4 = 3k/8
Faster seal to total seal = k/4 x 8/k = 66.33
Ans. D
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Re: working alone at its own constant rate [#permalink]

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New post 06 Jan 2012, 00:17
As the second machine works twice as fast as does the first one, the second machine will do twice as much work as does the first one. Together, if they complete a job, the second machine has worked twice as much as has the first machine.
Thus, work done by second machine relative to the total work done is 2/3 => 66.67%
Answer: D
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Re: working alone at its own constant rate   [#permalink] 06 Jan 2012, 00:17

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