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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 ? A. 600 B. 800 C. 1000 D. 1200 E. 1500 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 !
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Last edited by Bunuel on 06 Sep 2012, 10:01, edited 2 times in total.
Edited the question.



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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 ?
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 > \(b=a+5\). Solving for \(x\) > \(x=600\). Answer: A. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps.
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Re: solution required [#permalink]
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B in a minute=x/40 A in a minute=x/60 then, x/40x/60=5
Solving x=600



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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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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 = 1/241/60= 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



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First thing i did was i found out time that is required to B to complete the task alone: 1/241/60=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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\(\frac{1}{A}=\frac{1}{60}\) \(\frac{1}{B}+\frac{1}{A}=\frac{1}{24}\) Get: \(\frac{1}{B}\) \(\frac{1}{B}=\frac{1}{24}\frac{1}{60}=\frac{1}{40}\) Let p be the number of pages produced by A. Let p+5 be the number of pages produced by B. \(24(p + p+5) = 60(p)==> p=10pages\) Answer: 60(p)=600pages
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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 ? A. 600 B. 800 C. 1000 D. 1200 E. 1500 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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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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Re: solution required [#permalink]
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AccipiterQ wrote: Bunuel wrote: 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 ?
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 > \(b=a+5\). Solving for \(x\) > \(x=600\). Answer: A. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps. 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 > \(\frac{1}{b}=\frac{1}{a}+5\). Solving for \(x\) > \(x=600\). Answer: A.
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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 ? A. 600 B. 800 C. 1000 D. 1200 E. 1500 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
Difference 3x  2x = 5 => x = 5 Thus, 120x = 600..(A)
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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 ?
A. 600 B. 800 C. 1000 D. 1200 E. 1500
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 ! Another approach after getting rate for B as rate of B = 1/40 let A prints x pages a minute , then B will print x+5pages a minute A work for 60 minutes and B work for 40 minutes alone and so ,they are able to print same no. of pages thus, 40(x+5) = 60x 40x+200 =60x 20x=200 x=10 thus total no of pages = 60x or 40(x+5) =600 pages Ans A



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Re: solution required [#permalink]
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Bunuel wrote: 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 ?
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 > \(b=a+5\). Solving for \(x\) > \(x=600\). Answer: A. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps. Hi Bunnel, 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



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Re: solution required [#permalink]
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06 Sep 2012, 10:39
kotela wrote: Bunuel wrote: 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 ?
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 > \(b=a+5\). Solving for \(x\) > \(x=600\). Answer: A. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps. Hi Bunnel, 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. Hope it's clear.
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Re: Working together, printer A and printer B would finish the [#permalink]
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25 Sep 2013, 04:09
mbaiseasy wrote: Let p be the number of pages produced by A. Let p+5 be the number of pages produced by B.
\(24(p + p+5) = 60(p)==> p=10pages\)
Answer: 60(p)=600pages How do you come up with these??



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Bunuel wrote: 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 ?
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 > \(b=a+5\). Solving for \(x\) > \(x=600\). Answer: A. Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps. 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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22 Apr 2014, 21:56
A does in one minute =x pages Therefore, in 60 minutes=60x pages
B does in one minute= x+5 In 24 minutes both do 60x pages 24x+24(x+5)=60x X=10 total work=60*10=600 pages



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