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If (x-7)^2=-|y-5|, xy=?

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If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 00:13
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[GMAT math practice question]

If \((x-7)^2=-|y-5|, xy=?\)

\(A. 5\)
\(B. 7\)
\(C. 12\)
\(D. 35\)
E. cannot be determined

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Re: If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 07 Feb 2019, 19:28
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MathRevolution wrote:
[GMAT math practice question]

If \((x-7)^2=-|y-5|, xy=?\)

\(A. 5\)
\(B. 7\)
\(C. 12\)
\(D. 35\)
E. cannot be determined



A square and a modulus can never be NEGATIVE


So least value of a square or modulus is 0..

\((x-7)^2=-|y-5|\) ..
Here -|y-5| will never be positive as |y-5| is greater than or equal to 0..
A square on left side tells us that (x-7)^2=0, or x=7..
And -|y-5|=0 means y =5
Therefore xy=7*5=35

D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 00:30
If y>5, then -|y-5| = -(y-5) = 5-y
\(x^2+49-14x = 5-y\)
\(x^2+44-14x+y = 0\)........(1)
If y<5, then -|y-5| = -(-y+5) = y-5
\(x^2+49-14x = y-5\)
\(x^2+54-14x = y\)..........(2)
(1) in (2) gives
\(2x^2+98-28x = 0\)
\(x^2+49-14x = 0\)
\((x-7)^2 = 0\)
x = 7
If x = 7, then -|y-5| = 0; y = 5
xy = 35

D is the answer.
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Re: If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 01:29
MathRevolution wrote:
[GMAT math practice question]

If \((x-7)^2=-|y-5|, xy=?\)

\(A. 5\)
\(B. 7\)
\(C. 12\)
\(D. 35\)
E. cannot be determined



|y-5|= sqrt ( y-5)^2

and
(x-7)^2 = sqrt ( (x-5)^2
squaring both sides
(x-7) = (x-5)^2
solve
y^2-10y+32-x=0-- ( 1)

seeing answer option
x*y=35
x=7 & y = 5
satisfies the eqn ( 1)
so IMO C
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If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 03:57
I assume

x=7 because (x-7)=0, therefore x= 7 when you rearrange

then
7= -|y-5|
7= -(y-5)
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 (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 04:14
MathRevolution wrote:
[GMAT math practice question]

If \((x-7)^2=-|y-5|, xy=?\)

\(A. 5\)
\(B. 7\)
\(C. 12\)
\(D. 35\)
E. cannot be determined


We can start with plugging in the values for x and y

Only D when plugged in the \((x-7)^2=-|y-5|\), 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 (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 04:19
Albs wrote:
I assume

x=7 because (x-7)=0, therefore x= 7 when you rearrange

then
7= -|y-5|
7= -(y-5)
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 Albs

the 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= -|y-5|

7 ! = - 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 (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 05:43
MathRevolution wrote:
[GMAT math practice question]

If \((x-7)^2=-|y-5|, xy=?\)

\(A. 5\)
\(B. 7\)
\(C. 12\)
\(D. 35\)
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 (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 06:40
I have a query.

|y-5| will always be +ve an (x-7)^2 = -|y-5| can't be possible as LHS is a square number?
So how can we find a deterministic answer in such case?


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Re: If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 06:41
rish2708 wrote:
I have a query.

|y-5| will always be +ve an (x-7)^2 = -|y-5| 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 (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 05 Feb 2019, 09:28
MathRevolution wrote:
[GMAT math practice question]

If \((x-7)^2=-|y-5|, xy=?\)

\(A. 5\)
\(B. 7\)
\(C. 12\)
\(D. 35\)
E. cannot be determined


Since Square of any number is >=0; \(A^2>=0\)
\((x-7)^2=A,-|y-5|=B\)
since A cant be negative ,So B=0
y=5 and x=7
X*Y=35
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Re: If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 07 Feb 2019, 17:29
=>
\((x-7)^2=-|y-5|\)
\(=> (x-7)^2+|y-5| = 0\)
\(=> x = 7\) and \(y = 5\)
This yields \(xy = 35\).

Therefore, the answer is D.
Answer: D
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Re: If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 07 Feb 2019, 18:28
MathRevolution

Sorry 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

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Re: If (x-7)^2=-|y-5|, xy=?  [#permalink]

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New post 09 Feb 2019, 15:32
KanishkM wrote:
Albs wrote:
I assume

x=7 because (x-7)=0, therefore x= 7 when you rearrange

then
7= -|y-5|
7= -(y-5)
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 Albs

the 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= -|y-5|

7 ! = - 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 (x-7)^2 = -|y-5| or (7-7)^2 = -|y-5|.
Thus -|y-5| = 0 and we have y = 5.

x*y = 7*5 = 35.
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Re: If (x-7)^2=-|y-5|, xy=?   [#permalink] 09 Feb 2019, 15:32
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