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09 Feb 2011, 15:24
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
58% (01:25) correct
42%
(01:38)
wrong
based on 296
sessions
History
Date
Time
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Not Attempted Yet
09 Feb 2011, 15:45
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Is x = y?
(1) \(x^2 + y^2 = 2xy\):
Re-arrange:
Recognize that we have the square of the difference of two numbers (x and y):
Sufficient.
(2) \(4^{(x^2+y^2)}=1\).
For the above equation to hold true, the power of 4 must be zero:
The sum of two non-negative values(x^2 and y^2) is zero if both \(x\) and \(y\) are zero:
Sufficient.
Answer: D.
(1) \(x^2 + y^2 = 2xy\):
Re-arrange:
- \(x^2-2xy+y^2=0\);
Recognize that we have the square of the difference of two numbers (x and y):
- \((x-y)^2=0\);
\(x-y=0\);
\(x=y\).
Sufficient.
(2) \(4^{(x^2+y^2)}=1\).
For the above equation to hold true, the power of 4 must be zero:
- \(x^2+y^2=0\);
The sum of two non-negative values(x^2 and y^2) is zero if both \(x\) and \(y\) are zero:
- \(x=y=0\).
Sufficient.
Answer: D.
General Discussion
15 Jan 2016, 19:40
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Is x = y?
(1) x^2 + y^2 = 2xy.
(2) 4^(x^2+y^2)=1
In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 variables. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2),
for 1), (x-y)^2=0에서 x-y=0, x=y, which is yes and sufficient.
For 2), 4^(x^2+y^2)=1=4^0, x^2+y^2=0, x=y=0, which is yes and sufficient. Therefore the answer is D.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
Is x = y?
(1) x^2 + y^2 = 2xy.
(2) 4^(x^2+y^2)=1
In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 variables. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2),
for 1), (x-y)^2=0에서 x-y=0, x=y, which is yes and sufficient.
For 2), 4^(x^2+y^2)=1=4^0, x^2+y^2=0, x=y=0, which is yes and sufficient. Therefore the answer is D.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
