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11 Jul 2012, 14:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:35) correct
34%
(01:38)
wrong
based on 831
sessions
History
Date
Time
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Not Attempted Yet
If \(\sqrt{xy} = xy\) what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out, and when we should use cancelling technique.
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out, and when we should use cancelling technique.
11 Jul 2012, 14:22
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Bookmarks
If \(\sqrt{xy} = xy\) what is the value of x + y?
\(\sqrt{xy} = xy\) --> \(xy=x^2y^2\) --> \(x^2y^2-xy=0\) --> \(xy(xy-1)=0\) --> either \(xy=0\) or \(xy=1\).
(1) x = -1/2 --> either \(-\frac{1}{2}*y=0\) --> \(y=0\) and \(x+y=-\frac{1}{2}\) OR \(-\frac{1}{2}*y=1\) --> \(y=-2\) and \(x+y=-\frac{5}{2}\). Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) \(y\neq{0}\), then from (1) \(y=-2\) and \(x+y=-\frac{5}{2}\). Sufficient.
Answer: C.
As for your solution: you cannot divide by \(xy\) since \(xy\) could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by \(xy\) you assume, with no ground for it, that \(xy\) does not equal to zero thus exclude a possible solution.
Hope it's clear.
\(\sqrt{xy} = xy\) --> \(xy=x^2y^2\) --> \(x^2y^2-xy=0\) --> \(xy(xy-1)=0\) --> either \(xy=0\) or \(xy=1\).
(1) x = -1/2 --> either \(-\frac{1}{2}*y=0\) --> \(y=0\) and \(x+y=-\frac{1}{2}\) OR \(-\frac{1}{2}*y=1\) --> \(y=-2\) and \(x+y=-\frac{5}{2}\). Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) \(y\neq{0}\), then from (1) \(y=-2\) and \(x+y=-\frac{5}{2}\). Sufficient.
Answer: C.
As for your solution: you cannot divide by \(xy\) since \(xy\) could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by \(xy\) you assume, with no ground for it, that \(xy\) does not equal to zero thus exclude a possible solution.
Hope it's clear.
11 Jul 2012, 22:57
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Bookmarks
Responding to a pm:
"If \(\sqrt{XY} = XY\) what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out and when we should use cancelling technique."
For the time being, forget this question. Look at another one.
Which values of x satisfy this equation: \(x^2 = x\)
Let's say we cancel out x from both sides. What do we get? x = 1.
So we get that x can take the value 1.
But is your answer complete? I look at the equation and I say, 'x can also take the value 0.' Am I wrong? No.
x = 0 also satisfies your equation. So why didn't you get it using algebra? It is because you canceled x.
Let me treat this equation differently now.
\(x^2 = x\)
\(x^2 - x = 0\)
\(x * (x - 1) = 0\)
x = 0 OR (x - 1) = 0 i.e. x = 1
Now I get both the possible values that x can take. I do not lose a solution.
When you cancel off a variable from both sides of the equation, you lose a solution so you should not do that. You can cancel off constants of course.
Rule of thumb: Do not cancel off variables. Take them common. In some cases, it may not matter even if you do cancel off but either ways, your answer will not be incorrect of you don't cancel. On the other hand, sometimes, your solution could be incomplete if you do cancel and that's a problem.
"If \(\sqrt{XY} = XY\) what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out and when we should use cancelling technique."
For the time being, forget this question. Look at another one.
Which values of x satisfy this equation: \(x^2 = x\)
Let's say we cancel out x from both sides. What do we get? x = 1.
So we get that x can take the value 1.
But is your answer complete? I look at the equation and I say, 'x can also take the value 0.' Am I wrong? No.
x = 0 also satisfies your equation. So why didn't you get it using algebra? It is because you canceled x.
Let me treat this equation differently now.
\(x^2 = x\)
\(x^2 - x = 0\)
\(x * (x - 1) = 0\)
x = 0 OR (x - 1) = 0 i.e. x = 1
Now I get both the possible values that x can take. I do not lose a solution.
When you cancel off a variable from both sides of the equation, you lose a solution so you should not do that. You can cancel off constants of course.
Rule of thumb: Do not cancel off variables. Take them common. In some cases, it may not matter even if you do cancel off but either ways, your answer will not be incorrect of you don't cancel. On the other hand, sometimes, your solution could be incomplete if you do cancel and that's a problem.
