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655-705 (Hard)|   Algebra|      
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nusmavrik
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 02 Aug 2010, 10:49
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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:

55% (hard)

Question Stats:

52% (01:09) correct 48% (01:05) wrong based on 529 sessions

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Is x = y?

(1) x^2 = xy
(2) x^2 = y^2
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C
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Bunuel
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 04 Aug 2010, 02:30
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nusmavrik
Is x = y?

a) x² = xy

b) x² = y²
Show more
Is \(x=y\)?

(1) \(x^2=xy\) --> \(x^2-xy=0\) --> \(x(x-y)=0\) --> either \(x=0\) (and \(y=any \ value\)) OR \(x=y\). Not sufficient.

(2) \(x^2=y^2\) --> either \(x=y\) or \(x=-y\). Not sufficient.

(1)+(2) \(x^2=xy=y^2\) --> \(x=y\). Sufficient.

Answer: C.

TallJTinChina
Okay I fell in the trap I guess.

I am wondering why you took the first step \(x^2-xy=0\)

Why do we need to set equal to zero?

\(X^2=X*X\)

So I solved for y

\((X*X)/X=Y\)
\(X=Y\)
Show more
When you cancel \(x\), what you are actually doing is dividing both parts of the equation by variable \(x\), thus assuming with no ground for it that this variable does not equal to zero.

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.

nusmavrik
@dwivedys : Here is the trap. Compare previous question with following question. It also has common root (AB=1) but OA is E

Is the product of A and B equal to 1?
1) A*B*A = A
2) B*A*B = B

OA : E
A = 0 or B = 0 or AB = 1. A = 0 and AB = 1 are mutually exclusive. 1) and 2) combined cannot assert AB=1 unless A and B are non zero. However in the first question I had posted x = y is definite solution.

Any one else wants to explain when to take a common solution and when not. May be a second perspective can flush the common solution dilemma ??
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Is the product of A and B equal to 1?

Question: \(ab=1\)?

(1) \(a^2b=a\) --> \(a^2b-a=0\) --> \(a(ab-1)=0\) --> either \(a=0\) (and \(b=any \ value\), including zero) so in this case \(ab=0\neq{1}\) OR \(ab=1\). Two different answers, not sufficient.

(2) \(ab^2=b\) --> \(ab^2-b=0\) --> \(b(ab-1)=0\) --> either \(b=0\) (and \(a=any \ value\), including zero) so in this case \(ab=0\neq{1}\) OR \(ab=1\). Two different answers, not sufficient.

(1)+(2) either \(a=b=0\), so in this case \(ab=0\neq{1}\) and the answer to the question is NO, OR \(ab=1\) and the answer to the question is YES. Two different answers, not sufficient.

Answer: E.

When we consider two statements together we should take the values which satisfy both statements. For this question \(ab=1\) satisfies both statement, but \(a=b=0\) also satisfies both statements. So what you call "common solution" for this question is: \(ab=1\) OR \(ab=0\neq{1}\) and for the first question "common solution" is only \(x=y\).

Hope it's clear.
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 05 Jan 2021, 01:25
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Is x = y?

(1) x^2 = xy
(x^2) - (xy) = 0
x(x-y) = 0
x=0 or x=y
Therefore Statement 1 alone is insufficient.

(2) x^2 = y^2
(x^2) - (y^2) = 0
(x+y) (x-y) = 0
x=y or x= -y
Therefore Insufficient.

(1) and (2) together is sufficient.
Because x=y

Therefore, answer is C.
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 02 Aug 2010, 11:08
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nusmavrik
Is x = y?

a) x² = xy

b) x² = y²
Show more

Is it C?

1) \(x^2 = xy\) => \(x^2-xy=0\) and \(x(x-y)=0\) hence x=0 or x=y so insuff
2) \(x^2=y^2\) => \(x^2-y^2=0\) => \((x-y)(x+y)=0\) => x=y or x=-y hence insuff
1+2: x=y hence suff so C
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 02 Aug 2010, 20:05
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Okay I fell in the trap I guess.

I am wondering why you took the first step \(x^2-xy=0\)

Why do we need to set equal to zero?

\(X^2=X*X\)

So I solved for y

\((X*X)/X=Y\)
\(X=Y\)
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 03 Aug 2010, 02:25
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@dwivedys : Here is the trap. Compare previous question with following question. It also has common root (AB=1) but OA is E

Is the product of A and B equal to 1?
1) A*B*A = A
2) B*A*B = B

OA : E
A = 0 or B = 0 or AB = 1. A = 0 and AB = 1 are mutually exclusive. 1) and 2) combined cannot assert AB=1 unless A and B are non zero. However in the first question I had posted x = y is definite solution.

Any one else wants to explain when to take a common solution and when not. May be a second perspective can flush the common solution dilemma ??
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nusmavrik
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 04 Aug 2010, 07:14
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Bunuel ! This is awesome. Kudos for the explanation.

Bunuel


When we consider two statements together we should take the values which satisfy both statements. For this question \(ab=1\) satisfies both statement, but \(a=b=0\) also satisfies both statements. So what you call "common solution" for this question is: \(ab=1\) OR \(ab=0\neq{1}\) and for the first question "common solution" is only \(x=y\).

Hope it's clear.
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 05 Jan 2021, 11:52
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Bunuel
Is x = y?

(1) x² = xy
(2) x² = y²
Show more

Target question: Is x = y?

Statement 1: x² = xy
NOTE: Although it would be nice to divide both sides of the equation by x (to get x = y), we can't do that because it COULD be the case that x = 0, which means we're dividing by 0 (a big No-No)

Instead, we can subtract xy from both sides to get: x² - xy = 0
Factor: x(x - y)
So, EITHER x = 0 OR x - y = 0

So, it COULD be the case that x = 0 and y = 1, in which case the answer to the target question is NO, x does not equal y
Or, it COULD be the case that x = 1 and y = 1, in which case the answer to the target question is YES, x equals y
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x² = y²
Subtract y² from both sides to get: x² - y² = (x + y)(x - y) = 0
So, EITHER x + y = 0 OR x - y = 0

So, it COULD be the case that x = -1 and y = 1, in which case the answer to the target question is NO, x does not equal y
Or, it COULD be the case that x = 1 and y = 1, in which case the answer to the target question is YES, x equals y
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that EITHER x = 0 OR x - y = 0
Statement 2 tells us that EITHER x + y = 0 OR x - y = 0
Since both statements must be true, it MUST be the case that x - y = 0, which means the answer to the target question is YES, x equals y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 06 Jun 2022, 03:27
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My 2 cents on this:

is x = y? or is x-y=0?

S1: x^2 = xy can be written as x(x-y) = 0

Which means Either x = 0 OR x-y = 0. 2 options so sufficient.

S2: x^2 = y^2 which can be written as (x+y)(x-y) = 0

Which means either x+y = 0 or x-y = 0. 2 options so insufficient.

S1 and S2 together:

Simply plug value for x^2 as y^2 in first equation

--> xy = y^2; cancel out the y terms and we are left with x = y definitive answer.

C
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Is x = y? (1) x^2 = xy (2) x^2 = y^2 14 Mar 2025, 03:15
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