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22 Sep 2004, 06:47
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here is another one, with explanation please
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22 Sep 2004, 07:04
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Sorry, here is the explanation.
From the equation we have
10X + 20Y = K(X+Y)
=>10(X+2Y)=K(X+Y)
=>[(X+Y)+Y]/(X+Y) = K/10
=>1 + Y/(X+Y) = K/10
=>Y/(X+Y) = (K-10)/10
Now, lets use the answer choices and plug-in values for K. Take A for ex(10), we would then have Y=0. What we need to determine is that X<Y.
put B (12), we have 10Y=2(X+Y) =>5Y=X+Y =>4Y=X. Clearly X>Y.
put D(18), we have 10Y=8(X+Y) =>10Y=8X+8Y =>2Y=8X =>Y=4X.Clearly X<Y
put C(15), we have 10Y=5(X+Y) =>Y=X.
put E(30), we have 10Y=20(X+Y) => Y=2X+2Y =>-Y=X.
So, Ans is D. Sorry made a subtle typo error last time when I posted it as B. Is should read D.
From the equation we have
10X + 20Y = K(X+Y)
=>10(X+2Y)=K(X+Y)
=>[(X+Y)+Y]/(X+Y) = K/10
=>1 + Y/(X+Y) = K/10
=>Y/(X+Y) = (K-10)/10
Now, lets use the answer choices and plug-in values for K. Take A for ex(10), we would then have Y=0. What we need to determine is that X<Y.
put B (12), we have 10Y=2(X+Y) =>5Y=X+Y =>4Y=X. Clearly X>Y.
put D(18), we have 10Y=8(X+Y) =>10Y=8X+8Y =>2Y=8X =>Y=4X.Clearly X<Y
put C(15), we have 10Y=5(X+Y) =>Y=X.
put E(30), we have 10Y=20(X+Y) => Y=2X+2Y =>-Y=X.
So, Ans is D. Sorry made a subtle typo error last time when I posted it as B. Is should read D.
22 Sep 2004, 09:08
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Agree with D. If you solve the equation, you get (k-20)/(10-K) = x/y. Since x<y, you need to show that (k-20)/(10-K) is a positive fraction less than 1.
Only D satisfies this condition.
Only D satisfies this condition.
