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Working together, printer A and printer B would finish the [#permalink]

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Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600 b. 800 c. 1000 d. 1200 e. 1500

answer: A I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\); Printer A alone would finish the task in 60 minutes --> \(60a=x\); Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

Re: Working together, printer A and printer B would finish the [#permalink]

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18 Mar 2014, 06:00

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Rate A= X Rate B= X+5

Work(A)=> X * 60 = 60X

Rate(A+B) * 24 = Work

(2X+5) * 24 = 60X

X=10
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Re: Working together, printer A and printer B would finish the [#permalink]

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15 May 2014, 10:02

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Easiest way to do this: Machine A and B can do the task in 24 minutes thus Rate of A and B = 1/24. Now given A can do the task in 60 minutes therefore Rate of A= 1/60. We know that Rate of A and B = Rate of A + Rate of B therefore Rate of B= Rate of A and B - Rate of A = 1/24-1/60= 1/40. Now we know that Rate of B = 1/40 thus B can do the work in 40 minutes.

Let pages printed per minute by A = x, given that pages printed by B per minute is 5 more than that of A Pages printed by B per minute = x+5 Now Complete task is done by A in 60 minutes therefore total number of pages printed by A = x * 60 Also Complete task is done by B in 40 minutes therefore total number of pages printed by B = (x+5) * 40 therefore x * 60 = (x+5) * 40 therefore x=10 thus the total number of pages in task = x*60 = 10*60 = 600

Re: Working together, printer A and printer B would finish the [#permalink]

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08 Sep 2012, 04:51

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First thing i did was i found out time that is required to B to complete the task alone: 1/24-1/60=3/120=1/40. Then i looked at the information which states that the rate of B is 5+ page than that of A so, lets say x is the number of pages printed by A per minute, so the task consists of 60*x or 40*(x+5) pages. I can make an equation: 60x=40(x+5), 20x=200, x=10, total number of pages is 60*10=600 or 40*15=600

Answer is A.

It is clear but it took me about 3 min to do it, does it because i am doing it slow or i am using longer route?
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Re: Working together, printer A and printer B would finish the [#permalink]

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03 Mar 2013, 23:40

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gauravnagpal wrote:

Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

total time taken by B = 24 * 60 / (60 -24) = 40 min.

A take 60 min. B takes 40 min to complete a task.

Now, divide the values given in option (in Ans) to get the rate per min.

option A: 600 / 10 = 60 & 600/40 = 15...> this satisfies the condition given in question stem that printer B prints 5 pages a minute more than printer A ? . therefore A
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Re: Working together, printer A and printer B would finish the [#permalink]

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21 Sep 2013, 10:31

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Ta = 60min Ra = 1/Ta = 1/60 Rb = 1/Tb

Combined task completion time 24min. =Ra + Rb =1/60 + 1/Tb = 1/24 Tb = 40 min.

Ra = X/Ta Rb = X/Tb

Ra + 5 = Rb X/Ta + 5 = X/Tb X/60 + 5 = X/40 X=600 Ans.
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Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600 b. 800 c. 1000 d. 1200 e. 1500

answer: A I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\); Printer A alone would finish the task in 60 minutes --> \(60a=x\); Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

I'm curious about the 1/a+1/b=1/24 solution as well. I started out trying that way and got stuck at the same point as the other fellow. Is there any way to solve it once you have 1/60+1/40 for their combined rates? Or is that just a dead end?

edit: Also, question #2:

Shouldn't the equation be 24(1/a+1/b)=x; ? Since you have 24 minutes in which the machines are working at their individual rates, doing 1/a and 1/b of the job per minute? I don't get how one can know when to arbitrarily use "a" instead of '1/a" or "b" instead of "1/b"

In this case we'd have: The rate of printer A = 1/a pages per minute, where a is the time to print 1 page. The rate of printer B = 1/b pages per minute, where b is the time to print 1 page.

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(\frac{1}{a}+\frac{1}{b})=x\); Printer A alone would finish the task in 60 minutes --> \(60*\frac{1}{a}=x\); Printer B prints 5 pages a minute more than printer A --> \(\frac{1}{b}=\frac{1}{a}+5\).

Re: Working together, printer A and printer B would finish the [#permalink]

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18 Jul 2016, 04:29

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gauravnagpal wrote:

Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

Okay..this is how I did it.. Let the task(number of pages) be 120x(LCM of all numbers given in the problem)..

A and B take 24 minutes to complete it..thus, pages per min = 5x A's pages per minute = 2x B's pages per minute = 3x

.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600 b. 800 c. 1000 d. 1200 e. 1500

answer: A I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\); Printer A alone would finish the task in 60 minutes --> \(60a=x\); Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600 b. 800 c. 1000 d. 1200 e. 1500

answer: A I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\); Printer A alone would finish the task in 60 minutes --> \(60a=x\); Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

I have always one confusion in rate and work problems can you please clarify this?,

When do we add rates i.e what you did above...... \(24(a+b)=x\);

and when do we divide by rates i.e something . rate =(Job done/ time)

Regards

Srinath

You can denote rate directly by some variable (a in the solution) or express rate as a reciprocal of time. For example, say printer A needs t minutes to print 1 page and printer B needs m minutes to print 1 page, then the rate of printer A would be job/time=1/t pages per minute and the rate of printer B would be 1/m pages per minute (rate is a reciprocal of time, so 1/t=a and 1/m=b). In this case the equation would be 24(1/t+1/m)=x.

.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600 b. 800 c. 1000 d. 1200 e. 1500

answer: A I know this question is relatively symol if make an equation in one vaibale ... I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60 combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here kindly see where am I losing the track !

Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\); Printer A alone would finish the task in 60 minutes --> \(60a=x\); Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

I'm curious about the 1/a+1/b=1/24 solution as well. I started out trying that way and got stuck at the same point as the other fellow. Is there any way to solve it once you have 1/60+1/40 for their combined rates? Or is that just a dead end?

edit: Also, question #2:

Shouldn't the equation be 24(1/a+1/b)=x; ? Since you have 24 minutes in which the machines are working at their individual rates, doing 1/a and 1/b of the job per minute? I don't get how one can know when to arbitrarily use "a" instead of '1/a" or "b" instead of "1/b"

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