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Around the World in 80 Questions (Day 4): Does xy = x^2 ?

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Bunuel
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Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:00
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A
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Question Stats:

54% (01:30) correct 46% (01:18) wrong based on 292 sessions
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Does \(xy = x^2\) ?


(1) \(xy = y^2\)

(2) \(x^2 = y^2\)


 


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C
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Muskaan09
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:08
The question can be simplified to is c=y?
A. xy=y^2
-> x= y. Sufficient.
B. Not sufficient because
|x|=|y|.
The signs could be opposite.

Therefore answer would be A.

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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:09
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Does xy = x^2?
Does xy - x^2 = 0?
Does x(y-x) = 0?

1) xy = y^2
=> y(x-y) = 0
=> y = 0 or (x-y) = 0 ---> NOT SUFFICIENT

2) x^2 = y^2
=> (x+y)(x-y) = 0
=> x = y or x=-y ----> NOT SUFFICIENT

Combine: xy = x^2 = y^2 ---> SUFFICIENT [c]
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:11
Both the options alone are sufficient to answer the questions as on substituting the conditions into the equation we get xy = y^2 both times hence answer should be D

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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:15
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Kudos
target is
xy=x^2
x(y-x)=0
x=0 ; x=y
#1
\(xy = y^2\)
y(x-y)=0
y=0 ,x=y
insufficient
#2
\(x^2 = y^2\)
x & y can be of same sign or opposite sign or can be 0
insufficient
from 1 &2
we can say that x=y=0
OPTION C is correct

Bunuel wrote:
Does \(xy = x^2\) ?


(1) \(xy = y^2\)

(2) \(x^2 = y^2\)


 


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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:17
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My Answer: C, Statement (1) and (2) are together sufficient.

Question: xy=x^2 , xy-x^2=0 , x(y-x)=0 so, either x=0 or x=y for xy=x^2.

Statement (1): xy=y^2, xy-y^2 = 0 , y(x-y)=0 so, either y=0 or x=y. Not sufficient.

Statement (2): x^2=y^2, x^2-y^2=0, (x+y)(x-y)=0 so, either x=-y or x=y. Not sufficient.

Statement (1) and (2) together: x=y is the common condition in both the statements. So if both are true then x=y. If x=y, xy=x^2. Hence, both statements together are sufficient.
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:19
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is xy = x^2?

1. xy = y^2 gives us 2 solutions
a. y = 0
b. x = y
insuff

2. x^2 = y^2 gives us 2 solutions
a. x = y
b. x = -y
insuff

both 1 & 2 gives us just 1 solution
x = y
suff

ans C
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:25
Given: xy=x^2

=> y = x

xy = y^2 => y=x
x^2 = y^2 => x=y
Therefore, D – EACH ALONE is sufficient.
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:28
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Bunuel wrote:
Does \(xy = x^2\) ?


(1) \(xy = y^2\)

(2) \(x^2 = y^2\)


 


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for the Around the World in 80 Questions

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\(xy = x^2\)

\(xy - x^2 = 0\)

\(x(y-x) = 0\)

\(x = 0\) or \(x = y\)

(1) \(xy = y^2\)

\(xy - y^2 = 0\)

\(y(x - y) = 0\)

Two possibilities, x = y or y = 0.

If x = y, Does \(xy = x^2\) ? is yes

If y = 0 and x = 4, Does \(xy = x^2\) ? is no

Not sufficient.

(2) \(x^2 = y^2\)

\(x^2 - y^2 = 0\)

\((x+y)(x - y) = 0\)

Two possibilities, x = y or x = -y.

If x = y, Does \(xy = x^2\) ? is yes

If y = 3 and x = -3, Does \(xy = x^2\) ? is no

Not sufficient.

Combined

The common term in both (1) and (2) is x = y.

The statements combined tell us \(xy = x^2\). The answer is Yes.

IMO C
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:30
Is xy=x^2
x^2 -xy =0
x(x-y)=0
either x=0 or x=y

(1) xy=y^2
y^2 -xy =0
y(y-x)=0
either y=0 or x=y
Not sufficient

(2) x^2=y^2
x^2-y^2 =0
(x-y)(x+y)=0
either x=y or x=0 and y=0
Sufficient

Therefore B.
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:32
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Asked: Does \(xy = x^2\) ?

xy - x^2 = 0
x(y-x)=0
Either x=0
or y=x

(1) \(xy = y^2\)
xy - y^2 =0
y(x-y)=0
Either y=0; xy may or may not be = x^2
or x=y; xy = x^2
NOT SUFFICIENT

(2) \(x^2 = y^2\)
x^2 - y^2 = 0
(x-y)(x+y) =0
Either x=y; xy=x^2
or x=-y; xy may or may not be = x^2
NOT SUFFICIENT

(1) + (2)
(1) \(xy = y^2\)
(2) \(x^2 = y^2\)
xy = x^2 = y^2
SUFFICENT

IMO C
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:32
The Answer is A.
The question basically asks if the value of x=y
In statement 1 if xy= y(square) which essentially means that x=y. Hence sufficient
In statement 2 you cannot know the absolute value when two terms are squared. It could be positive or negative. Hence insufficient
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:33
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Does xy=x^2?
xy-x^2 =0
x(y-x)=0
so it has 2 answers x=0 or y=x


(1) xy=y^2
xy-y^2=0
y(x-y)=0
y=0 or x =y
so if y =0 hence xy is not equal to x^2, but if x = y then xy=x^2
A and D are out

(2) x2=y2
x= +y/-y
suppose x=y hence x^2=x^2
but if x = -y hence y=-x then -x^2 is not equal to x^2
Hence B is out
from 1 and 2 we found that x=y so the answer is c
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:38
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Does xy=x^2

(1) xy=y^2
(2) x^2=y^2


case i:
xy = y^2
y= 0 or x = y

case ii:
x^2 = y^2
thus x = y, x = -y
thus insufficient

on combining x = y
thus x*y = x^2
hence c
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 08:50
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is x(y-x)=?

s1) y(x-y)=0; y=0 or x=y
hence x(y-x)=-x^2 or 0 (NS)

s2) x=y or -y
=> x^2=yx or-yx (NS)

S1+S2) dividing
y/x=1;
y=x; xy=x^2 (Suff)

C)
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 09:12
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answer is c
xy-x2=0
so x=0 or x=y (tio prove
1. xy -y2 =0
y=0 or x=y .not suff
2. x2-y2=0
x=y and x=-y .not suff
both suff
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 09:14
1
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Does xy=x^2 ?

(1) xy=y^2

(2) x^2=y^2
_____________________
Let's start with the question .
xy-y^2= 0 ?
y*(x-y) = 0 ?

(1) y*(x-y)=0
so , it could be as follows:
y=0 or x=y

If y=0, we dont know the answer to xy=x^2.
because , we dont know x.
So, there are 2 alternative answers, (1) is not sufficient.

(2) (x-y)(x+y)=0
x=y and x=-y
in this case, we can infer that one of the x and Y or both are zero. Still not sufficient.

When we combine , we can get x=y which is sufficient to answer the question .

C is our winner.
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 09:19
Statement 1: xy=y^2
Solving it we get x=y, which satisfies the equation xy=x^2

Statement 2: x^2=y^2

Solving it we get x=y, which also satisfies the equation xy=x^2

Therefore, both statement alone is sufficient (D)

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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 09:23
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Bunuel wrote:
Does \(xy = x^2\) ?


(1) \(xy = y^2\)

(2) \(x^2 = y^2\)


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

Win over $20,000 in prizes: Courses, Tests & more

 



Question asks: xy=x^2
that mean we need to prove either x=0 or x=y.
1. xy = y^2
So x=y or y=0.
We cannot be sure that x=y. Hence this is insufficient.

2. x^2 = y^2
x=y, or x=-y
Hence insufficient.

Combining both, we have x=y common. Hence sufficient.
Answer is C!
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink] New post   20 Jul 2023, 09:28
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Bunuel wrote:
Does \(xy = x^2\) ?


(1) \(xy = y^2\)

(2) \(x^2 = y^2\)


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

Win over $20,000 in prizes: Courses, Tests & more

 



S1:- xy=y^2
y(x-y)=0
y=0 or x=y
if x=y then xy=x^2
but if y=0, then xy=0 and we don't know about x^2.
So, not sufficient.

S2:- x^2= y^2
x=y or x=-y
if x=y then xy=x^2
but if x=-y, then xy=-x^2
So, not sufficient.

Combining both statements, we get x=y, and therefore xy=x^2

So, correct choice is C.
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Re: Around the World in 80 Questions (Day 4): Does xy = x^2 ? [#permalink]
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