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Re: Running at their respective constant rates, Machine X takes [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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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