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If (x7)^2=y5, xy=?
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05 Feb 2019, 01:13
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42% (01:32) correct 58% (01:48) wrong based on 120 sessions
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[GMAT math practice question] If \((x7)^2=y5, xy=?\) \(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined
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Re: If (x7)^2=y5, xy=?
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07 Feb 2019, 20:28
MathRevolution wrote: [GMAT math practice question]
If \((x7)^2=y5, xy=?\) \(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined A square and a modulus can never be NEGATIVE So least value of a square or modulus is 0.. \((x7)^2=y5\) .. Here y5 will never be positive as y5 is greater than or equal to 0.. A square on left side tells us that (x7)^2=0, or x=7.. And y5=0 means y =5 Therefore xy=7*5=35 D
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If (x7)^2=y5, xy=?
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05 Feb 2019, 01:30
If y>5, then y5 = (y5) = 5y \(x^2+4914x = 5y\) \(x^2+4414x+y = 0\)........(1) If y<5, then y5 = (y+5) = y5 \(x^2+4914x = y5\) \(x^2+5414x = y\)..........(2) (1) in (2) gives \(2x^2+9828x = 0\) \(x^2+4914x = 0\) \((x7)^2 = 0\) x = 7 If x = 7, then y5 = 0; y = 5 xy = 35 D is the answer.
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Re: If (x7)^2=y5, xy=?
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05 Feb 2019, 02:29
MathRevolution wrote: [GMAT math practice question]
If \((x7)^2=y5, xy=?\) \(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined y5= sqrt ( y5)^2 and (x7)^2 = sqrt ( (x5)^2 squaring both sides (x7) = (x5)^2 solve y^210y+32x=0 ( 1) seeing answer option x*y=35 x=7 & y = 5 satisfies the eqn ( 1) so IMO C



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If (x7)^2=y5, xy=?
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05 Feb 2019, 04:57
I assume
x=7 because (x7)=0, therefore x= 7 when you rearrange
then 7= y5 7= (y5) 7= y+5 y= 2
Therefore x*y= 14
Is it E based on my assumption above? I don't think so but made sense to me.



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Re: If (x7)^2=y5, xy=?
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05 Feb 2019, 05:14
MathRevolution wrote: [GMAT math practice question]
If \((x7)^2=y5, xy=?\) \(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined We can start with plugging in the values for x and y Only D when plugged in the \((x7)^2=y5\), gives equal values 1 * 5 or 5 * 1 1 * 7 or 7 * 1 3 * 4 or 4 * 3 or 2 * 6 or 6 * 2
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Re: If (x7)^2=y5, xy=?
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05 Feb 2019, 05:19
Albs wrote: I assume
x=7 because (x7)=0, therefore x= 7 when you rearrange
then 7= y5 7= (y5) 7= y+5 y= 2
Therefore x*y= 14
Is it E based on my assumption above? I don't think so but made sense to me. Hi Albsthe highlightedvalue when substituted back in the bold part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case. so you got y = 2 lets put it back in 7= y57 ! =  7 You have to search for different cases for x and y to satisfy (x−7)^2=−y−5 But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good.
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If (x7)^2=y5, xy=?
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05 Feb 2019, 06:43
MathRevolution wrote: [GMAT math practice question]
If \((x7)^2=y5, xy=?\) \(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined
\({\left( {x  7} \right)^2} =  \left {y  5} \right\,\,\,\,\,\left( * \right)\) \(? = xy\) \(\left. \matrix{ {\left( {x  7} \right)^2} \ge 0\,\,\, \hfill \cr  \left {y  5} \right \le 0 \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{\left( {x  7} \right)^2} = 0 =  \left {y  5} \right\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{ \,x  7 = 0 \hfill \cr \,y  5 = 0 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 \cdot 7 = 35\) The correct answer is therefore (D). We follow the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: If (x7)^2=y5, xy=?
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05 Feb 2019, 07:40
I have a query.
y5 will always be +ve an (x7)^2 = y5 can't be possible as LHS is a square number? So how can we find a deterministic answer in such case?
Regards, Rishav



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Re: If (x7)^2=y5, xy=?
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05 Feb 2019, 07:41
rish2708 wrote: I have a query.
y5 will always be +ve an (x7)^2 = y5 can't be possible as LHS is a square number? So how can we find a deterministic answer in such case?
Regards, Rishav Got it.. only 0 it can satisfy!! Thanks



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Re: If (x7)^2=y5, xy=?
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05 Feb 2019, 10:28
MathRevolution wrote: [GMAT math practice question]
If \((x7)^2=y5, xy=?\) \(A. 5\) \(B. 7\) \(C. 12\) \(D. 35\) E. cannot be determined Since Square of any number is >=0; \(A^2>=0\) \((x7)^2=A,y5=B\) since A cant be negative ,So B=0 y=5 and x=7 X*Y=35



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Re: If (x7)^2=y5, xy=?
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07 Feb 2019, 18:29
=> \((x7)^2=y5\) \(=> (x7)^2+y5 = 0\) \(=> x = 7\) and \(y = 5\) This yields \(xy = 35\). Therefore, the answer is D. Answer: D
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Re: If (x7)^2=y5, xy=?
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07 Feb 2019, 19:28
MathRevolutionSorry to say but your answer gives no additional knowledge and thus is not really useful in this case. The point to observe is that a perfect square has been equated a negative absolute value function. We must note that a perfect square can never be negative. It can only achieve a value of zero or positive. Same is the case with absolute value function. This, to ensure that both the functions yield the same answer, we must equate both the function to zero. This, x=7 and y=5 XY = 35 Posted from my mobile device



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Re: If (x7)^2=y5, xy=?
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09 Feb 2019, 16:32
KanishkM wrote: Albs wrote: I assume
x=7 because (x7)=0, therefore x= 7 when you rearrange
then 7= y5 7= (y5) 7= y+5 y= 2
Therefore x*y= 14
Is it E based on my assumption above? I don't think so but made sense to me. Hi Albsthe highlightedvalue when substituted back in the bold part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case. so you got y = 2 lets put it back in 7= y57 ! =  7 You have to search for different cases for x and y to satisfy (x−7)^2=−y−5 But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good. If x = 7, then we have (x7)^2 = y5 or (77)^2 = y5. Thus y5 = 0 and we have y = 5. x*y = 7*5 = 35.
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Re: If (x7)^2=y5, xy=?
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