Quote:
If \(x\), \(y\), and \(k\) are positive numbers such that \(\frac{x}{(x + y)}(10) + \frac{y}{(x + y)}(20) = k\) and if \(x < y\), which of the following could be the value of \(k\)?
A. 10
B. 12
C. 15
D. 18
E. 30
Here the equation is:
\(\frac{x}{(x + y)}(10) + \frac{y}{(x + y)}(20) = k\)
-->\(\frac{10x+20y}{x+y}=k\)
-->\(k=\frac{10x+10y+10y}{x+y}\)
-->\(k=\frac{10x+10y}{x+y}+\frac{10y}{x+y}\)
-->\(k=\frac{10(x+y)}{x+y}+\frac{10y}{x+y}\)
-->\(k=10+\frac{10y}{x+y}\)
^^(we'll put the value of the variable in this above equation)
Condition:
\(x>0\)
\(y>0\)
\(k>0\)
and
\(y>x\)
Let's make some creativity of the condition:
\(y>x\):
\(y=1; x=0\) (\(x \) can't be \(0\) because the condition says that \(x>0\); but what if we let \(x=0\))
Then \(k\) equals to 20. But, in the condition says that x>0, so, k must be less than 20.
Example:
\(k=10+\frac{10×1}{0+1}\)
-->\(k=10+10\)
-->\(k=20\) (the upper one if \(x=0\))
So, if \(y=1; x=0.000001\):
\(k=10+\frac{10×1}{0.000001+1}\)
-->\(k=10+9.99999000\)
-->\(k=19.9999\)
So, \(k\) is definitely less than 20.
So, E is out.Let, \(x=y\) (but in condition \(y>x\))
\(k=10+\frac{10×1}{1+1}\)
-->\(k=15 \)
So, C is out.Let, \(x>y\) (but in condition \(y>x\))
\(k=10+\frac{10×1}{4+1}\)
-->\(k=12\)
So, C is out.If you put the condition as \(x>y\) (this is the reverse condition of actual condition-in actual condition it says \(y>x)\), the result of \(k\) will be less than 15.
So, A and B out.Note: Be serious about the condition. GMAC doesn't put the ''condition'' randomly.The correct choice DThanks__