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If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
 27 Feb 2012, 08:42
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If x, y, and k are positive numbers such that (x/(x+y))(10) + (y/(x+y))(20) = k and if x < y, which of the following could be the value of k? A. 10 B. 12 C. 15 D. 18 E. 30
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Re: If x, y, and k are positive [#permalink]
27 Feb 2012, 08:50
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BANON wrote: If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30 10*\frac{x}{x+y}+20*\frac{y}{x+y}=k10*\frac{x+2y}{x+y}=k10*(\frac{x+y}{x+y}+\frac{y}{x+y})=kFinally we get: 10*(1+\frac{y}{x+y})=kWe know that x<yHence \frac{y}{x+y} is more than 0.5 and less than 10.5<\frac{y}{x+y}<1So, 15<10*(1+\frac{y}{x+y})<20Only answer between 15 and 20 is 18. Answer: D (18)There can be another approach: We have: \frac{10x+20y}{x+y}=k, if you look at this equation you'll notice that it's a weighted average. There are x red boxes and y blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes? k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition. Answer: D (18)
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
27 Feb 2012, 08:56
equation can be simplified to
10y = (x+y)(K-10)
Hence Ans = D as only if k=18, we have a positive ratio that is less than 1 for x/y, {where x<y)
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
14 Apr 2012, 01:19
Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks! 15<10*(1+(y/x+y)<20
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
14 Apr 2012, 01:48
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mymbadreamz wrote: Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks!
15<10*(1+(y/x+y)<20 main funda is y/(x+y) is equal to 1/2 if x=y but when x<y then x+y<2y i.e. y/x+y >1/2 (x,y>0) also the maximum value is x tends to 0 i.e. as x>0 say x be 0.0000...(infinite times )1 then the expression x+y tends to y i.e. y/ (x+y) <1 hence 0.5 < y/(x+y) < 1 => 1+0.5 < 1+y/(x+y) < 1+1 => 10*1.5 < 10*(1+(y/x+y) <2 *10 => 15<10*(1+(y/x+y)<20 hope this helps..!!
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
14 Apr 2012, 02:09
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
23 Apr 2012, 10:24
(x/(x+y))(10) + (y/(x+y))(20)=k this equation is equal to
{(x/(x+y))(10) + (y/(x+y))(10)} + (y/(x+y))(10)=k
{ 10 }+ (y/(x+y))(10)=k
10+(y/(x+y))(10)=k
if x < y but still y< x+y so (y/(x+y))(10) < 10 ==> 10+(y/(x+y))(10) <20, k<20
if x=y again y< x+y but it is certain that (y/(x+y))(10) = 5 ==> 10 +(y/(x+y))(10) =15 k=15
x and y are not equal, therefore k must be bigger than 15 , less than 20
15< k < 20
k= 18 valid answer
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
23 Apr 2012, 17:00
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I like the weighted box answer approach!
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Re: If x, y, and k are positive [#permalink]
14 May 2012, 10:35
Bunuel wrote: BANON wrote: If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = k and if x < y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30 10*\frac{x}{x+y}+20*\frac{y}{x+y}=k10*\frac{x+2y}{x+y}=k10*(\frac{x+y}{x+y}+\frac{y}{x+y})=kFinally we get: 10*(1+\frac{y}{x+y})=kWe know that x<yHence \frac{y}{x+y} is more than 0.5 and less than 10.5<\frac{y}{x+y}<1So, 15<10*(1+\frac{y}{x+y})<20Only answer between 15 and 20 is 18. Answer: D (18)There can be another approach: We have: \frac{10x+20y}{x+y}=k, if you look at this equation you'll notice that it's a weighted average. There are x red boxes and y blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes? k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition. Answer: D (18)Very well solved by Bunuel, this is question no. 148 in OG 12 P.S, which has been solved in a very complicated manner in O.G.
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Re: Value of k, given x,y and k are positive integers [#permalink]
30 Sep 2012, 14:22
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ausadj18 wrote: Can someone show me the shortcut to solving number 148 from the OG:
If x,y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and x <y, which of following could be value for k?
a) 10
b) 12
c)15
d)18
e) 30 \frac{10x}{x+y}+\frac{20y}{x+y}=\frac{10(x+y)+10y}{x+y}=10+\frac{10y}{x+y}. \frac{10y}{x+y}>\frac{10y}{2y}=5 and \,\,\frac{10y}{x+y}<\frac{10y}{y}=10, therefore 10+5<k<10+10 or 15<k<20. Answer D.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
10 Dec 2012, 04:35
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\frac{10x}{x+y}+ \frac{20y}{x+y}=k
10x + 20y = kx + ky
(10-k)x=(k-20)y
\frac{x}{y}=\frac{k-20}{10-k}
(A) \frac{-10}{0} Out! x and y are non-zeroes.
(B) \frac{x}{y}=\frac{-8}{-2}=\frac{4}{1} Out! x should be less than y.
(C) \frac{-5}{-5} Out! x should not be equal to y.
(D) \frac{-2}{-8}=\frac{1}{4} Bingo!
Answer: D
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
11 Dec 2012, 04:10
I use this approach:
X=1
Y=2
1/3*10 + 2/3*20
10/3+40/3=
50/3--> 16...... Te only outcome greater than 15 is 18 (D). 30 is too much. am I correct? thanks
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
15 Dec 2012, 00:05
fabrizio1983 wrote: I use this approach:
X=1
Y=2
1/3*10 + 2/3*20
10/3+40/3=
50/3--> 16...... Te only outcome greater than 15 is 18 (D). 30 is too much. am I correct? thanks The expression is all about recognizing that it represents "weighted averages" and that k could have a range of values. In this case, the range of values is 15 < k < 20. Perhaps this link would be helpful: Weighted Averages Pattern on the GMAT
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Question to OG 12th Edition PS 148. [#permalink]
29 Jan 2013, 03:47
Hey guys, this is my first post here. Please pardon me if I am doing anything wrong  I am currently preparing for the test and I stumbled upon the task in the title. It is like \frac{x}{x+y} * 10 + \frac{y}{x+y} * 20 = k ; y>x What is a possible value of k? The OG has a very complicated and time consuming explanation for this and I hope that it can be solved more quickly. I also don't know if it is really that difficult.. Can anyone help? Thank you in advance!
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Re: Question to OG 12th Edition PS 148. [#permalink]
29 Jan 2013, 04:02
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Re: Question to OG 12th Edition PS 148. [#permalink]
29 Jan 2013, 04:05
0haxx wrote: Hey guys, this is my first post here. Please pardon me if I am doing anything wrong I am currently preparing for the test and I stumbled upon the task in the title. It is like \frac{x}{x+y} * 10 + \frac{y}{x+y} * 20 = k ; y>x What is a possible value of k? The OG has a very complicated and time consuming explanation for this and I hope that it can be solved more quickly. I also don't know if it is really that difficult.. Can anyone help? Thank you in advance! The above equation can be written as \frac{10x}{x+y} + \frac{20y}{x+y} = k\frac{10x + 20y}{x+y} = k\frac{10x + 10y}{x+y} + \frac{10y}{x+y} = k10 + \frac{10y}{x+y} = kNow since x<y try to come up with a relation between x and y so that we can eliminate the fraction \frac{10y}{x+y} = kIn this case 4x = y or x = y/4 then the fraction will become 10y/(y/4 + y) = 10y/5y/4 = 8. So the final answer would be 10 + 8 = 18. We are trying to eliminate the fraction because we need a integer value for k. Please give a kudo if you like my explanation. Please ask if you did not understood something.
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Re: Question to OG 12th Edition PS 148.
[#permalink]
29 Jan 2013, 04:05
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