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Re: Running at their respective constant rates, Machine X takes
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22 Apr 2017, 06:25
Engr2012 wrote: jasonfodor wrote: can someone pelase explain how they got from (1/t) + (1/((t2)) = 5/12 TO 5t^234t+24=0 \(\frac{1}{t} + \frac{1}{t2} = \frac{5}{12}\) \(\frac{2t2}{t*(t2)} =\frac{5}{12}\) Cross multiplying and rearranging terms, you get , \(5t^234t+24= 0\) Hope it helps. Note, if you do it this way below, you are going to get the same answer. 1/x + 1/x+2 = 5/12 2x+2/x^2 + 2x = 5/12 12(2x + 2) = 5(x^2 + 2x) 24x + 24 = 5x^2 + 10x 5x^2  14x  24 = 0 (5x + 6)(x4) x + 2 4 + 2 6 Which is answer C



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Re: Running at their respective constant rates, Machine X takes
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05 Jun 2017, 17:23
Bunuel wrote: Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12
1. Do all the calculations for 2w widgets 2. It takes x 4 days more to produce 2w widgets' 3. Together they take 4.8 days to produce 2w widgets 4. 1/y + 1/x=1/4.8 or 1/y + 1/(y+4) =5/24, y=8 . Therefore y+4=12. X takes 12 days to produce 2w widgets
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Re: Running at their respective constant rates, Machine X takes
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08 Jun 2017, 17:06
Bunuel wrote: Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12
We can let x = the number of days it takes Machine X to produce w widgets and y = the number of days it takes Machine Y to produce w widgets. Thus, x = y + 2 or y = x  2. Furthermore, the rate of Machine X is w/x and the rate of Machine Y is 1/y = w/(x  2). We are given that they can produce (5/4)w widgets in 3 days. Thus, we have: 3(w/x) + 3[w/(x  2)] = (5/4)w 3w/x + 3w/(x  2) = (5/4)w Multiplying both sides by 4x(x  2), we have: 12w(x  2) + 12wx = 5wx(x  2) 12(x  2) + 12x = 5x(x  2) 12x  24 + 12x = 5x^2  10x 5x^2  34x + 24 = 0 (5x  4)(x  6) = 0 x = ⅘ or x = 6 However, x can’t be ⅘; if it were, y would be negative. Thus, x must be 6. Since it takes Machine X 6 days to produce w widgets, it will take 12 days to produce 2w widgets. Answer: E
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Running at their respective constant rates, Machine X takes
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11 Oct 2017, 13:37
Set the number of widgets to a preferred number. I choose 20.
(1/y + 2) + (1/y) + 1/3 = 5/4(20) = 25 2/6+y + 3/6+y + 6/6+y = 25 11/25 = 6 + y 11/25  6/1 = y 5/25 = 1/5 = y = machine y takes 5 days to make 25, therefore it takes 4 days to make 20 machine x takes two days more than y to produce 20 widgets; therefore, it takes machine x 2 + 4 = 6 days for 20 widgets
6 days for 20 widgets, 12 days for 40 widgets or 2w therefore, answer (E) 12 days



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Re: Running at their respective constant rates, Machine X takes
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09 May 2018, 14:56
PareshGmat wrote: muhtasimhassan wrote: Hi,
I am new here and I have solved this math correctly. However, here's my problem:
If I consider the number of days required to produce 'W' widgets as 'X' and 'X2' for X and Y respectively, I end up with the equation 5t^2  34t + 24 which yields me 2 solutions, one of which is correct (x=6)
However, when I take 'X' and 'Y' as 'X+2' and 'X', which are the same as above, because either ways, X is taking 2 days longer, I end up with the equation 5x^2  14x  24 which has no solutions (I hope I'm not making some stupid mistake here)
Can someone please clarify this and correct me where I'm wrong. Refer my post above; I did in the same way..... As far as solving the equation, refer below..... \(5x^2  14x  24 = 0\) \(5x^2  20x + 6x  24 = 0\) \(5x(x4) + 6(x4) = 0\) (x4)(5x+6) = 0 x = 4 (Ignore the ve equation) x+2 = 6 pushpitkchere cant fugure out what numbers are multiplied to get 24 and add up to 14. i wanna do factorization as i did in this post, https://gmatclub.com/forum/howmanyrea ... l#p1999687 but simply these figures dont come to my mind and i dont want to use lengthy way through formula b^2 4ac is it some kind of new way of factorization ? \(5x^2  14x  24 = 0\) \(5x^2  20x + 6x  24 = 0\) how to understand this ? \(5x(x4) + 6(x4) = 0\) and this (x4)(5x+6) = 0 how we get x = 4 ? see ya tomorrow thanks!



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Running at their respective constant rates, Machine X takes
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10 May 2018, 01:16
dave13 wrote: PareshGmat wrote: muhtasimhassan wrote: Hi,
I am new here and I have solved this math correctly. However, here's my problem:
If I consider the number of days required to produce 'W' widgets as 'X' and 'X2' for X and Y respectively, I end up with the equation 5t^2  34t + 24 which yields me 2 solutions, one of which is correct (x=6)
However, when I take 'X' and 'Y' as 'X+2' and 'X', which are the same as above, because either ways, X is taking 2 days longer, I end up with the equation 5x^2  14x  24 which has no solutions (I hope I'm not making some stupid mistake here)
Can someone please clarify this and correct me where I'm wrong. Refer my post above; I did in the same way..... As far as solving the equation, refer below..... \(5x^2  14x  24 = 0\) \(5x^2  20x + 6x  24 = 0\) \(5x(x4) + 6(x4) = 0\) (x4)(5x+6) = 0 x = 4 (Ignore the ve equation) x+2 = 6 pushpitkchere cant fugure out what numbers are multiplied to get 24 and add up to 14. i wanna do factorization as i did in this post, https://gmatclub.com/forum/howmanyrea ... l#p1999687 but simply these figures dont come to my mind and i dont want to use lengthy way through formula b^2 4ac is it some kind of new way of factorization ? \(5x^2  14x  24 = 0\) \(5x^2  20x + 6x  24 = 0\) how to understand this ? \(5x(x4) + 6(x4) = 0\) and this (x4)(5x+6) = 0 how we get x = 4 ? see ya tomorrow thanks! Hi dave13Unfortunately, there will be times when you will have to use the lengthy method Here is a simple workaround which can be used for this problem. We need two numbers whose sum is 14 and product is 24*5 = 120 Primefactorizing 120, we get \(2^3 * 3 * 5\) and there are 4*4 or 16 factors possible. Listing down the factors, we will have 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120. From these factors we need to find a combination which gives a sum of 14, which is possible when x=20 and 6 However, for bigger numbers where is it not possible to list down the factors, it will be difficult to solve a problem using this method. I would recommend that you solve some problems using this formula just in case you can't use the abovegiven method. Hope this helps you!
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Running at their respective constant rates, Machine X takes
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10 May 2018, 09:13
hey pushpitkc will you join on my bumpy ride to understand a few questions regarding my soluton in the post above In my solution above I get the roots 4 and 1.2 are these roots correct ? whereas you get x=20 and 6 through factorization..why ? Also when you say "We need two numbers whose sum is 14 and product is 24*5 = 120" \(5y^2 14y24=0\) and considering that \(5y^2\) is A \(14y\) is B and \(24\) is C so i can conclude you mulptiply C by A > 24 *5 to find the numbers that add up 14 > can you please remind the formula. And if equation were like this \(y^2 14y24=0\) (without 5) so i simply would multiply 24 by 1 ? right ? many thanks and have great weekend p.s looks like I accidently deleted the solution I posted yesterday...



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Re: Running at their respective constant rates, Machine X takes
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11 May 2018, 10:59
dave13 wrote: hey pushpitkc will you join on my bumpy ride to understand a few questions regarding my soluton in the post above In my solution above I get the roots 4 and 1.2 are these roots correct ? whereas you get x=20 and 6 through factorization..why ? Also when you say "We need two numbers whose sum is 14 and product is 24*5 = 120" \(5y^2 14y24=0\) and considering that \(5y^2\) is A \(14y\) is B and \(24\) is C so i can conclude you mulptiply C by A > 24 *5 to find the numbers that add up 14 > can you please remind the formula. And if equation were like this \(y^2 14y24=0\) (without 5) so i simply would multiply 24 by 1 ? right ? many thanks and have great weekend p.s looks like I accidently deleted the solution I posted yesterday... Hey dave13The roots(in the previous solution) are x = 4 and \(5x = 6\) > \(x = \frac{6}{5} = 1.2\) You have solved the problem correctly by using the formula and arrived at the right solution. Have a great weekend!
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Re: Running at their respective constant rates, Machine X takes
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15 Aug 2018, 09:11
Running at their respective constant rates, Machine X takes 2 dazys longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12 It can be very easily solved using efficiency method. So the question says, X+Y together produces 1.25 widgets in 3 days Thus, efficiency of X and Y to produce 1.25 widget in 3 days= 3/1.25=41.6% (eqn 1) Now we are asked to find the days taken by X to produce 2w widgets. Lets go to the answers. Consider C: 8 days. thus, X takes 8 days to produce 2w widgets==>4 days to produce w widgets thus X's efficiency= 1/4=25% So Y efficiency would be (41.625)=16.66% . (from eqn 1) This implies: X would take 4 days (i.e 100/25=4) and Y would take 6 days (i.e 100/16.66=6) but the question says the opposite: X should take 2 more days. So now consider option E: 12 days. So 12 days for 2w widgets. So 6 days for w widget. So X's efficiency= 100/6=16.66% = 6 days (i.e 100/16.66=6) ==> Y's efficiency= 41.6616.66=25%= 4 days (i.e 100/25=4) So this is correct as X would take 6 days and Y would take 4 days; So 64=2 days. Hence option E is the correct answer.
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Re: Running at their respective constant rates, Machine X takes
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16 Aug 2018, 04:04
Bunuel wrote: Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12
Let # days required for machine Y to produce w widgets be y days Hence rate of machine Y = w/y widgets per day Hence rate of machine X = w/y+2 widgets per day given that both machines together can produce 5w/4 widgets in 3 days Hence rate at which both machines working together = (5w/4)/3 We get w/(y+2) + w/y = (5w/4)/3 solving this for y we get, y = 4 days Hence # of days machine X takes to produce w widgets = 4 + 2 = 6 days Therefore # of days machine X takes to produce 2w widgets = 12 days Answer E. Thanks, GyM
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Re: Running at their respective constant rates, Machine X takes
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24 Mar 2019, 05:24
Dear Bunuel First of all, I have to say that I have a deep respect for your work on this forum and for the brilliant explanations you provide to the rest of the community. Nevertheless, in this question, I have found your answers disappointing. In my opinion it is not rational to expect an exam taker to find the solution to the equation \(5t2−34t+24=0\) in less than 100 seconds. I have tried different paths, but I think that all of them would take even the brightest minds on this forum more than 2 minutes to solve. Path 1: Decomposition \(5t2−34t+24=0\) > Let's find two numbers that multiply to 120 and sum up to 34Mmmm, at first sight, it doesn't seem easy, let's move on. Path 2: Formula Let's turn our equation into \((34+\sqrt{(34^24*5*24)})*\frac{1}{2*5}\) Damn, goog luck obtaining the square root of the number behind \(34^24*5*24\), let's see what else can we do... Path 3: Backsolving So, best case scenario, I have already spent about 1 minute and half and, so far, I got no clue about how to resolve this one. At least I know that \(\frac{1}{t}+\frac{1}{(t−2)}=\frac{5}{12}\) so let's see if the ol' reliable backsolving can help to eliminate a couple of questions. option c) \(\frac{1}{8}+\frac{1}{(8−2)}=\frac{5}{12}\) ? Wait a second, this is not that easy, I have to add up fractions and I just got 30 seconds left. Let's better go for a lucky a shot and move on onto the next question... So, I don't know if I have explained myself clearly enough, but my complain here is that the explanations provided are not realistic with time and skills constraints that an exam taker will be facing at an exam. There must be an easier way to solve this problem, otherwise, it seems like the question is out of scope. P.S. Sorry for my poor english



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Re: Running at their respective constant rates, Machine X takes
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11 May 2019, 18:11
Bunuel wrote: Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12 Given:X has a rate of w/(d+2) Y has a rate of w/d X+Y have a rate of w/d + w/(d+2) * 3 days = 5w/4, or... w/d + w/(d+2) = 5w/12 To solve: 2d+4 = ? (if it takes X d+2 for 1w, it takes twice that for 2w) Strategy: plugin answers C) 8 = 2d+4, 4=2d, 2=d (w/2 + w/4) = 5w/12 ? (6w)/8 = 5w/12 ? 3w/4 ≠ 5w/12 9w/12 ≠ 5w/12 (the rate is TOO BIG, so we need to INCREASE the time to decrease rate) D) 10 = 2d+4, 6=2d, 3=d (w/3 + w/5) = 5w/12 ? 8w/15 ≠ 5w/12 (rate is still TOO big) E) on the actual test I wouldn't even check E, but for the sake of completeness: 12 = 2d+4, 8=2d, 4=d (w/4 + w/6) = 5w/12 ? 10w/24 = 5w/12 ? 5w/12 = 5w/12 (MATCH, this is our answer)




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