Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

13 Oct 2009, 12:13

7

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

74% (02:22) correct
26% (01:45) wrong based on 306 sessions

HideShow timer Statistics

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 % _________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

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!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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% _________________

If you like my post, a kudos is always appreciated

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]

Show Tags

09 Feb 2012, 21:31

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]

Show Tags

18 Mar 2014, 09:22

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 _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

Show Tags

18 Mar 2014, 12:59

1

This post received KUDOS

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. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

They say you get better at doing something by doing it. then doing it again ... and again ... and again, and you keep doing it until one day you look...