If f(x,y) = 10x/(2x + 3y) + 20y/(3x + 2y), where 0

Question

If  f(x,y) = \frac{10x}{2x+3y}+\frac{20y}{3x+2y} , where  0<x<y , which of the following could be the value of  f(x,y) ?

A. 8
B. 10
C. 12
D. 14
E. 16

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Oppenheimer1945 · Answer

An out of the way approach:
f(x,y) = 10x/(2x + 3y) + 20y/(3x + 2y)

when x is very large , f(x,y)=10/2=5
when y is very large, f(x,y)=20/2=10

The ans should be somewhere b/w 5 & 10 (exclusive).

A) 8

Regor60 · Answer

As X increases towards Y starting at 0, the first part progresses from 0 to a maximum of 2.

At the same time, the right side progresses from 10 to 4.

So the range of the sum of both is from 10 to 6 as X progresses from 0 to Y, putting an upper bound of 10.

However, this only occurs at the disallowed value of X=0.

As X is increased from 0, the right side value drops more quickly from 10 than the left side value increases towards 2, rendering the sum less than 10 progressively towards 6.

The only value that fits is 8.

Posted from my mobile device

samarpan.g28 · Answer

Given, 0<x<y.
Case 1: If x is very small compared to y, we get f(x,y)<10.
Case 2: If x has a value close to y but slightly lesser than y, then we get a value less than or equal to 6.
That means the value of f(x,y) lies somewhere between 6 to 9.something. Option (A) is correct.

Wadree · Answer

but in scenario 1..........if x is very large y is also larger so how can we not assume y in da calculation?........btw i understood scenario 2 where y is very large...

Bunuel · Answer

Check this reply: https://gmatclub.com/forum/if-f-x-y-10x ... l#p3347443

Wadree · Answer

bruh thanks i get it now.....

If f(x,y) = 10x/(2x + 3y) + 20y/(3x + 2y), where 0 < x < y, which of

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