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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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06 Oct 2018, 09:29
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20% (01:06) correct 80% (02:06) wrong based on 40 sessions
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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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06 Oct 2018, 09:29
Official Explanation From f(x, y) = 10x/(x+2y)+20y/(2x+y), there are 2 variables (x, y), so it is difficult to find the value of f(x, y). If so, you have to use 0<x<y, ① if x=0, f(0, y) = (10(0))/(0+2y)+20y/(2(0)+y) = 20y/y = 20. ② if x=y, f(x, x) = 10x/(x+2x)+20x/(2x+x) = 10x/3x+20x/3x = 30x/3x = 10. However, from 0<x<y, x is the range between 0 and y, excluding 0 and y, so f(x, y) also becomes the range between 10 and 20, excluding 10 and 20. In other words, the only option that satisfies 10 < f(x, y) < 20 is C, 19. The answer is C.
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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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06 Oct 2018, 09:47
The answer could be A. please find solution attached. Best, G
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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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06 Oct 2018, 10:17
0<x<y f(x,y) =10x/(x+2y) + 20 y /(2x+y)
f(x,y) = 10x(2x+y) + 20y*(x+2y) /(x+2y)*(2x+y)
f(x,y)= 20x2+ 30xy+40y2 /(2x^2 + 5xy +2Y^2) =limiting case will give 10 if x= y
and hence for y>x, only A could be the answer
A



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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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06 Oct 2018, 11:08
f(x,y) = 10x/(x+2y)+20y/(2x+y)= 10(a+b) where a = x/(x+2y) & b=2y/(2x+y) if x= y ; a=1/3 & b= 2/3 hence 10 a = 3.3 & 10 b= 6.6 as x<y.... a<3.3 & b > 6.6.... ofcourse value of a & b are interdependent on x & y 10a+10b is closest to value 10......... will go with 10 Ans B
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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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07 Oct 2018, 10:54
gmatbusters wrote: Weekly Quant Quiz #3 Question No 7 If f(x,y) = \(\frac{10x}{(x+2y)}+\frac{20y}{(2x+y)}\), where 0<x<y, which of the following could be the value of f(x,y)? A. 9 B. 10 C. 19 D. 20 E. 23 Hi chetan2u, BunuelCan you please advise as to what is wrong with below method  f(x,y) = 10  \(\frac{20xy}{(2x^2 + 5xy + 4y^2)}\)  which means the value of f(x,y) is always less than 10. Only option A is correct.
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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f
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07 Oct 2018, 18:06
rahul16singh28 wrote: gmatbusters wrote: Weekly Quant Quiz #3 Question No 7 If f(x,y) = \(\frac{10x}{(x+2y)}+\frac{20y}{(2x+y)}\), where 0<x<y, which of the following could be the value of f(x,y)? A. 9 B. 10 C. 19 D. 20 E. 23 Hi chetan2u, BunuelCan you please advise as to what is wrong with below method  f(x,y) = 10  \(\frac{20xy}{(2x^2 + 5xy + 4y^2)}\)  which means the value of f(x,y) is always less than 10. Only option A is correct. You have missed out on one term in your method .. So \(\frac{20x^2+30xy+40y^2}{2x^2+5xy+2y^2}\)=\(\frac{20x^2+50xy20xy+20y^2+20y^2}{2x^2+5xy+2y^2}\) =\(\frac{10(2x^2+5xy+2y^2)+(2y^22xy)}{2x^2+5xy+2y^2}\)=\(10\frac{2y(yx)}{2x^2+5xy+2y^2}\) 2y(yx) is always positive as y>X so answer becomes  10+(something positive) Thus 9 or 10 can never be the answer. Hope it clarifies where you went wrong..
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Re: If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f &nbs
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07 Oct 2018, 18:06






