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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 06 Oct 2018, 10:29
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C
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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
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C
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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 06 Oct 2018, 10:29
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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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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 06 Oct 2018, 10:47
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The answer could be A.

please find solution attached.

Best,
G
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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 06 Oct 2018, 11:17
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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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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 06 Oct 2018, 12:08
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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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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 07 Oct 2018, 11:54
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gmatbusters

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
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Hi chetan2u, Bunuel

Can 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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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 07 Oct 2018, 19:06
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rahul16singh28
gmatbusters

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, Bunuel

Can 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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You have missed out on one term in your method ..
So \(\frac{20x^2+30xy+40y^2}{2x^2+5xy+2y^2}\)=\(\frac{20x^2+50xy-20xy+20y^2+20y^2}{2x^2+5xy+2y^2}\)
=\(\frac{10(2x^2+5xy+2y^2)+(2y^2-2xy)}{2x^2+5xy+2y^2}\)=\(10-\frac{2y(y-x)}{2x^2+5xy+2y^2}\)
2y(y-x) 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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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 14 Aug 2021, 10:59
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GMATBusters

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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Nice solution. But wouldn't you have to guarantee tat the function is monotonic in x in the interval [0,y] ? If it is not, your solution was a gamble.
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If f(x,y) = [m][fraction]10x/(x+2y)[/fraction]+[fraction]20y/(2x+y)[/f 03 Oct 2024, 05:35
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