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Working alone, printers X, Y, and Z can do a certain [#permalink]
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27 Jan 2007, 16:12
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Working alone, printers X, Y, and Z can do a certain printing job, consisting of a large number of pages, in 12, 15, and 18 hours, respectively. What is the ratio of the time it takes printer X to do the job, working alone at its rate, to the time it takes printers Y and Z to do the job, working together at their individual rates ? (A) 4/11 (B) 1/2 (C) 15/22 (D) 22/15 (E) 11/4
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D
X takes 12 hrs
Y and Z together = (1/15) + (1/18) = 11/90 = 90/11 hrs
ratio = X /( YandZ) = 12 * (11/90) = 22/15



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Indeed D is the correct answer, got 22/15 by the exact same approach
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Rate(X) = 1/12 job/hour or 12 hour/job
Rate(Y) = 1/15 job/hour
Rate(Z) = 1/18 job/hour
Rate(Y + Z) = 1/15 + 1/18 job/hour = (6 + 5)/90 = 11/90 job/hour
This mean that Machine Y and Z can finish 11/90 job in one hour
So how long will will take for Machine X to finish 11/90 job? Rate(X) = 12 hour/job
Time(x) to do 11/90 job = 11/90 job x 12 hour/job = 11 x 12 /90 = 44/30 = 22/15 hours
(D) is the answer.



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Re: Working alone, printers X, Y, and Z can do a certain [#permalink]
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23 Jan 2013, 13:12
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Hi, if somebody could help me what I am doing wrong here, it would be great: 1) I am calculating individual rates for all 3 printer and bring them onto the same denominator. X = 1/12 = 30/360 Y = 1/15 = 24/360 Z = 1/18 = 20/360 2) Comparing the nominators of X with the sum of Y and Z, since they are now comparable. 30/(24+20) = 30/44 = 15/22 The ratio is X to (Y + Z) so it should be fine. This would be answer (C) and not (D). Why should I flip the nominator and denominator here? Thanks for your help in advance



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Re: Working alone, printers X, Y, and Z can do a certain [#permalink]
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23 Jan 2013, 20:02
leventg wrote: Hi, if somebody could help me what I am doing wrong here, it would be great: 1) I am calculating individual rates for all 3 printer and bring them onto the same denominator. X = 1/12 = 30/360 Y = 1/15 = 24/360 Z = 1/18 = 20/360 2) Comparing the nominators of X with the sum of Y and Z, since they are now comparable. 30/(24+20) = 30/44 = 15/22 The ratio is X to (Y + Z) so it should be fine. This would be answer (C) and not (D). Why should I flip the nominator and denominator here? Thanks for your help in advance RATES of X, Y and Z are 30/360, 24/360 and 20/360 Ratio of RATE of X:RATE of Y+Z = 30:44 = 15:22 The question asks for the ratio of TIME TAKEN = 1/15 : 1/22 = 22:15 (Time taken is the inverse of rate)
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Re: Working alone, printers X, Y, and Z can do a certain [#permalink]
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24 Jan 2013, 04:23
Thanks for your fast reply Karishma, As this was still difficult for me to understand, I have created an easy example for better understanding. Let’s assume all printers take 12 hours. So printer Y and Z are doing the same job as printer X twice as fast. X = 1/12 (job/hours) Y = 1/12 (job/hours) Z = 1/12 (job/hours) Y+Z = 2/12 = 1/6 (job/hours) X : (Y+Z) = 1 : 2 => This ratio refers to the output. Regarding Time Taken, X makes in 12 hours 1 job and Y+Z are doing in 6 hours 1 job. So what you are saying is that we are comparing the hours and not the jobs right?And therefore the ratio of X : Y is 12 : 6, which is 2 : 1. Summarizing both steps:X : (Y+Z) = ( 1/12) : ( 2/12) = 1 : 2 => This ratio refers to the output. X : (Y+Z) = (1/ 12) : (1/ 6) = 12: 6 = 2 : 1 => This ratio refers to the time Referring to my example again:X = 12 hours Y+Z = 6 hours Ratio is not 12 : 6 or 2 : 1 because time taken is inverse to rate? Instead the Ratio is (1/12) : (1/6) = (6/12) = 1 : 2 Actually this TIMEISINVERSEAPPROACH is quite difficult to understand. I can apply it but still it is difficult to understand. May be it is just easier to divide 2 fractions. (Divide Y+Z by X).
Last edited by leventg on 24 Jan 2013, 06:04, edited 1 time in total.



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Re: Working alone, printers X, Y, and Z can do a certain [#permalink]
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24 Jan 2013, 04:40
leventg wrote: Thanks for your fast reply Karishma, As this was still difficult for me to understand, I have created an easy example for better understanding. Let’s assume all printers take 12 hours. So printer Y and Z are doing the same job as printer X twice as fast. X = 1/12 (job/hours) Y = 1/12 (job/hours) Z = 1/12 (job/hours) Y+Z = 2/12 = 1/6 (job/hours) X : (Y+Z) = 1 : 2 => This ratio refers to the output. Regarding Time Taken, X makes in 12 hours 1 job and Y+Z are doing in 6 hours 1 job. So what you are saying is that we are comparing the hours and not the jobs right?And therefore the ratio of X : Y is 12 : 6, which is 2 : 1. Summarizing both steps: X : (Y+Z) = (1/12) : (2/12) = 1 : 2 => This ratio refers to the output. X : (Y+Z) = (1/12) : (1/6) = 2 : 1 => This ratio refers to the time.Referring to my example again:X = 12 hours Y+Z = 6 hours Ratio is not 12 : 6 or 2 : 1 because time taken is inverse to rate? Instead the Ratio is (1/12) : (1/6) = (6/12) = 1 : 2 Actually this TIMEISINVERSEAPPROACH is quite difficult to understand. I can apply it but still it is difficult to understand. May be it is just easier to divide 2 fractions. (Divide Y+Z by X). For an intuitive understanding of ratios approach, check out these posts: http://www.veritasprep.com/blog/2011/03 ... ofratios/http://www.veritasprep.com/blog/2011/03 ... osintsd/http://www.veritasprep.com/blog/2011/03 ... problems/
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Re: Working alone, printers X, Y, and Z can do a certain [#permalink]
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05 May 2016, 18:17
Attached is a visual that should help.
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Screen Shot 20160505 at 6.16.09 PM.png [ 100.35 KiB  Viewed 3992 times ]
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Re: Working alone, printers X, Y, and Z can do a certain [#permalink]
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08 Jun 2017, 14:03
prude_sb wrote: Working alone, printers X, Y, and Z can do a certain printing job, consisting of a large number of pages, in 12, 15, and 18 hours, respectively. What is the ratio of the time it takes printer X to do the job, working alone at its rate, to the time it takes printers Y and Z to do the job, working together at their individual rates ?
(A) 4/11 (B) 1/2 (C) 15/22 (D) 22/15 (E) 11/4 ratio of X/(Y+Z) rates=1/4/(1/5+1/6)=15/22 inverting, ratio of X/(Y+Z) times=22/15 D




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