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If root(xy) = xy what is the value of x + y?
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11 Jul 2012, 14:05
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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.
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Re: If root(xy) = xy what is the value of x + y?
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11 Jul 2012, 14:22
If \(\sqrt{xy} = xy\) what is the value of x + y? \(\sqrt{xy} = xy\) > \(xy=x^2y^2\) > \(x^2y^2xy=0\) > \(xy(xy1)=0\) > 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.
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Re: If root(xy) = xy what is the value of x + y?
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11 Jul 2012, 22:57
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.
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Re: If root(xy) = xy what is the value of x + y?
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12 Jul 2012, 03:00
Here is my solution
Take square of both sides: (root(xy) )^2 = (xy)^2 xy = (xy)^2 With Option (1): x = 1/2 1/2y = 1/4 * y^2 y = 4 * (1/2) y = 2 Hence, X+Y = 1/2  2 = 5/2.



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Re: If root(xy) = xy what is the value of x + y?
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12 Jul 2012, 03:06



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If √(xy) = xy, what is the value of x + y?
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15 Oct 2014, 16:01



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If √(xy) = xy, what is the value of x + y?
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15 Oct 2014, 20:57
Bunuel wrote: Tough and Tricky questions: Algebra. If √(xy) = xy, what is the value of x + y? (1) x = 1/2 (2) y is not equal to 0
Statement 1: \(X=1/2\).... substitute in main equation
Scenario A: \(Y = 2/1\) is a reciprocal \(\sqrt{xy}=xy\) LHS =\(\sqrt{1/2*2/1}= \sqrt{1}= 1\) RHS =\(1/2*2/1= 1*1 = 1\) Therefore, LHS = RHS
Scenario B: \(Y = 0\) LHS =\(\sqrt{1/2*0}= \sqrt{0}= 0\) RHS =\(1/2*0= 0\) Therefore again, LHS = RHS
Hence, we have two possible solutions, therefore Statement 1 is insufficient
Statement 2 : \(Y\) is not equal to zero But, X could be equal to zero or be the reciprocal of Y, therefore, statement 2 would fall under the same scenarios as statement 1
Hence, we have two possible solutions, therefore Statement 2 is insufficient
Both Statement 1 & 2 together confirms that \(X & Y\) both are not equal to zero, therefore, they have to be reciprocals and since statement gives us the value of \(X=1/2\), we can also the value of \(Y=2/1\) (Reciprocal of X) and thereafter we can find the value of \(X + Y = 1/2 + 2/1 = 5/2\)
Hence, both together are sufficient, therefore C is the correct Answer!



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Re: If √(xy) = xy, what is the value of x + y?
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15 Oct 2014, 21:26
Bunuel wrote: Tough and Tricky questions: Algebra. If √(xy) = xy, what is the value of x + y? (1) x = 1/2 (2) y is not equal to 0 I did something similar to DMMK, except for one thing: I know \(\sqrt{n}\) = \(n\) only when \(n\) = 0 or \(n\) = 1. For all other values of \(n\), the two would not be equal. Knowing this made it much simpler to plug in the statements. I just had to determine if from the statement(s) provided, can I rule out xy = 0 or xy=1. Stament 1: Knowing that x = 1/2, y could equal 0 or 2, and still make the premise true. Either value of y would make x + y a different value. Therefore, it is insufficient. Statement 2: Knowing that y is not equal to 0, x could equal zero or the inverse of y, and still make the premise true. Either value of x would make x + y a different value. Therefore, it is insufficient. Now to evaluate both statements together. The reason we rejected statement 1 by itself was because y could equal one of two possible values. Statement 2 eliminates one of those options. Therefore y must equal 2. And therefore both statements together are sufficient to answer "what is the value of x+y."



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Re: If √(xy) = xy, what is the value of x + y?
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15 Oct 2014, 22:12
Bunuel wrote: Tough and Tricky questions: Algebra. If √(xy) = xy, what is the value of x + y? (1) x = 1/2 (2) y is not equal to 0 let xy=k, thus k^(1/2) =k; squaring both sides we have k=k^2 k(k1)=0 k=0 or 1 i.e. xy=0 or xy=1 st.1 for x=1/2 both y=0 and y=2 satisfy the possible value of xy i.e. 0 or 1 hence not sufficient st.2 y is not equal to zero. clearly not sufficient. as nothing is said about x, therefore x can take any value it can be zero, fraction, integer etc. combining st.1 and st.2 we know that y cannot be equal to zero. thus y=2 and x=1/2 and x+y= 2(1/2)=5/2 hence sufficient.



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Re: If root(xy) = xy what is the value of x + y?
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04 Jul 2015, 08:43
Why can't x and y both be 1? Thanks



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Re: If root(xy) = xy what is the value of x + y?
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05 Jul 2015, 08:19



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Re: If √(xy) = xy, what is the value of x + y?
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16 Nov 2015, 03:02
Bunuel wrote: Tough and Tricky questions: Algebra. If √(xy) = xy, what is the value of x + y? (1) x = 1/2 (2) y is not equal to 0 Given: √(xy) = xy (xy)^2  xy = 0 xy = 0 or 1  (i) Required: x + y = ? Statement 1: x = 1/2 From the given equation (i), we can have two different values of y. Hence two different values of x + y INSUFFICIENTStatement 2: y is not equal to 0 No information about x INSUFFICIENTCombining Statement 1 and Statement 2: x = 1/2 and y is not equal to 0 From (i), xy cannot be 0 since both x and y are not = 0 Hence xy = 1 y = 2 x + y = \(\frac{5}{2}\) SUFFICIENTOption C



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If √(xy) = xy, what is the value of x + y?
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14 Apr 2017, 07:13
Bunuel wrote: Tough and Tricky questions: Algebra. If √(xy) = xy, what is the value of x + y? (1) x = 1/2 (2) y is not equal to 0 The trick here is from \(\sqrt{xy}=xy\) anyone could easily deduce that \(xy=1\) or \(xy=0\) and just consider one of these cases. Here is my solution: \(\sqrt{xy}=xy \implies \sqrt{xy}xy=0 \implies \sqrt{xy}(\sqrt{xy}1)=0\) Hence we have \(xy=0\) or \(xy=1\). (1) If \(x=\frac{1}{2}\), we need to consider two cases: Case 1: If \(xy=0\implies y=0 \implies x+y = \frac{1}{2}\) Case 2: If \(xy=1 \implies y=2 \implies x+y = \frac{5}{2}\) It's clear that (1) is insufficient. (2) If \(y \neq 0\), we still need to consider two cases: Case 1: If \(xy=0\implies x=0 \implies x+y = y\). If \(y=1 \implies x+y = 1\) If \(y=2 \implies x+y = 2\) Hence, (2) is insufficient. Now combine (1) and (2): Since \(x \neq 0\) and \(y \neq 0\) hence \(xy \neq 0 \implies xy=1\) Since \(x= \frac{1}{2} \implies y = 2 \implies x+y = \frac{5}{2}\) Hence (1) & (2) are sufficient. The answer is C.
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Re: If √(xy) = xy, what is the value of x + y?
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10 Sep 2017, 13:41
Bunuel wrote: Tough and Tricky questions: Algebra. If √(xy) = xy, what is the value of x + y? (1) x = 1/2 (2) y is not equal to 0 rewrite xy= (xy)^2 St 1 Reciprocal property. 1/2(2) =1 [1/2(2)]^2 =1 insuff St 2 Eliminating the possibility of 0 from Y still leaves the possibility of x being 0 or some other numbers which leads to several possibilities. insuff St 1 and St 2 Eliminates possibility of y=2 C



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Re: If √(xy) = xy, what is the value of x + y?
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27 Dec 2017, 00:35
Bunuel wrote: If \(\sqrt{xy} = xy\) what is the value of x + y?
\(\sqrt{xy} = xy\) > \(xy=x^2y^2\) > \(x^2y^2xy=0\) > \(xy(xy1)=0\) > 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. Is it safe to say : x^2 = x only when x = 0 or 1
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Re: If √(xy) = xy, what is the value of x + y?
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27 Dec 2017, 00:39
QZ wrote: Bunuel wrote: If \(\sqrt{xy} = xy\) what is the value of x + y?
\(\sqrt{xy} = xy\) > \(xy=x^2y^2\) > \(x^2y^2xy=0\) > \(xy(xy1)=0\) > 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. Is it safe to say : x^2 = x only when x = 0 or 1 Yes. x^2 = x; x^2 x = 0; x(x  1) = 0; x = 0 or x = 1.
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Re: If √(xy) = xy, what is the value of x + y?
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27 Feb 2018, 10:18
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2



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Re: If √(xy) = xy, what is the value of x + y?
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Re: If √(xy) = xy, what is the value of x + y?
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28 Feb 2018, 01:19
prakash111687 wrote: 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. there are various cases that are possible as following 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. IMO C \(\sqrt{xy} = xy\) 1.x=0 ,y = anything 2.Y=0 , x=anything 3. both x and y 1 or 1 or x=1 and y=1 or vise versa 4.x=1/z and y=z *any integer except 0 lets come to statement 1. x=1/2 y can be 2 or 0 insufficient. 2nd statement y not equal to 0 but don't know about x combining both case with 0 is eliminated only one case left so C is the answer




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