I have a question.
Statement 1) x + 2|y| = 0, which can be written as
x + 2y = 0
x - 2y = 0
For both these equations to satisfy, shouldn't x and y be both zero? Doesn't this make stmt 1 sufficient?
Same with Statement 2.
2)y + 2|x| = 0
absolute value
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straight answer =C
stmt 1 and 2 are not SUFF alone.
Stmt1) variables can have either zero or non zero values,
stmt 2) same as above
but when we combined them , x,y can only have 0 value in order to satisfy the equation
stmt 1 and 2 are not SUFF alone.
Stmt1) variables can have either zero or non zero values,
stmt 2) same as above
but when we combined them , x,y can only have 0 value in order to satisfy the equation
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Here's a fun proof to show that if x + 2|y| = 0 and y + 2|x| = 0 then |x| + |y| = 0.
x + 2|y| = 0 and y + 2|x| = 0
=> x + y + 2|x| + 2|y| = 0 //add the two equations to each other
=> -1/2(x + y) = |x| + |y| //rearrange the equation
-1/2(x + y) <= |-1/2(x + y)| //since absolute value of a number is never less than the number
=> -1/2(x + y) <= 1/2(|x + y|) //rearrange inequality
1/2(|x + y|) <= 1/2(|x| + |y|) //since |x + y| is always less than or equal to |x| + |y| (see below for proof of this)
1/2(|x| + |y|) <= |x| + |y| //since 1/2 of a non-negative number is always less than the whole number
By combining all these inequalities, we get...
-1/2(x + y) <= 1/2(|x + y|) <= 1/2(|x| + |y|) <= |x| + |y|
1/2(|x| + |y|) = |x| + |y| //since -1/2(x + y) = |x| + |y| and 1/2(|x| + |y|) is between these two values
=> 0 = 1/2(|x| + |y|) //rearrange the equation
Therefore: 0 = |x| + |y| //rearrange the equation
Well, to me, it's a fun proof, but you should probably use one of the previous solutions on the actual gmat so you don't waste time.
For all x and y, is |x + y| <= |x| + |y|?
There exists x and y such that |x + y| > |x| + |y| //Assume the opposite and look for a contradiction
|x + y|^2 > (|x| + |y|)^2 //square two positive numbers and the inequality will hold
(x + y)^2 > x^2 + 2|x||y| + y^2 //since a square of a number is always positive
x^2 + 2xy + y^2 > x^2 + 2|x||y| + y^2 //algebra
xy > |x||y| //subtract x^2 + y^2 from both sides
xy > |xy|
But this is a contradiction since a number (xy) is never greater than its absolute value (|xy|).
Therefore, our original statement is true: for all x and y, |x + y| <= |x| + |y|
x + 2|y| = 0 and y + 2|x| = 0
=> x + y + 2|x| + 2|y| = 0 //add the two equations to each other
=> -1/2(x + y) = |x| + |y| //rearrange the equation
-1/2(x + y) <= |-1/2(x + y)| //since absolute value of a number is never less than the number
=> -1/2(x + y) <= 1/2(|x + y|) //rearrange inequality
1/2(|x + y|) <= 1/2(|x| + |y|) //since |x + y| is always less than or equal to |x| + |y| (see below for proof of this)
1/2(|x| + |y|) <= |x| + |y| //since 1/2 of a non-negative number is always less than the whole number
By combining all these inequalities, we get...
-1/2(x + y) <= 1/2(|x + y|) <= 1/2(|x| + |y|) <= |x| + |y|
1/2(|x| + |y|) = |x| + |y| //since -1/2(x + y) = |x| + |y| and 1/2(|x| + |y|) is between these two values
=> 0 = 1/2(|x| + |y|) //rearrange the equation
Therefore: 0 = |x| + |y| //rearrange the equation
Well, to me, it's a fun proof, but you should probably use one of the previous solutions on the actual gmat so you don't waste time.
For all x and y, is |x + y| <= |x| + |y|?
There exists x and y such that |x + y| > |x| + |y| //Assume the opposite and look for a contradiction
|x + y|^2 > (|x| + |y|)^2 //square two positive numbers and the inequality will hold
(x + y)^2 > x^2 + 2|x||y| + y^2 //since a square of a number is always positive
x^2 + 2xy + y^2 > x^2 + 2|x||y| + y^2 //algebra
xy > |x||y| //subtract x^2 + y^2 from both sides
xy > |xy|
But this is a contradiction since a number (xy) is never greater than its absolute value (|xy|).
Therefore, our original statement is true: for all x and y, |x + y| <= |x| + |y|
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|x| + |y| = 0 translates into x = y = 0 as addition of two positive values can be zero only when each is zero.
statements are not sufficient individually that's easy to see.
however combining the two statement and getting an indicative answer is something that needs a little skill.
I have to practice more.
statements are not sufficient individually that's easy to see.
however combining the two statement and getting an indicative answer is something that needs a little skill.
I have to practice more.
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Since most people have figured out that required condition is |x|=|y|=0, but struggle with combining the statements, here's how to go about it:vikramaditya234 wrote:|x| + |y| = 0 translates into x = y = 0 as addition of two positive values can be zero only when each is zero.
statements are not sufficient individually that's easy to see.
however combining the two statement and getting an indicative answer is something that needs a little skill.
I have to practice more.
1)x + 2|y| = 0
=> x= -2|y|
=> |x| = |-2|y|| (take absolute value on both sides)
=> |x| = 2|y|
2)y + 2|x| = 0
In the same way as above,
y = -2|x|
=> |y| = 2|x|
=> |y| = 2 (2|y|) = 4|y| (substituting the equation above)
=> |y| (1-4)=0
=> |y| = 0.
Hence |x| = 2|y| = 0 also.
So we get |x|=|y|=0. Sufficient. And C is correct
Cheers!
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|x| + |y| = 0 means that |x| and |y| should individually = 0prashant misra wrote:i am still not able to understand why the answer is C.how have been the two options combined.can anyone explain
Statement 1:
x + 2|y| = 0
x = -2|y|
Putting in the main equation, 2|y| + |y| = 3|y| may be = 0 or may not be =0, Not sufficient.
Similarly for statement 2, Not sufficient.
If we combine Statement 1 and Statement 2:
From Statement 1:
x = -2|y|
From Statement 2:
y = -2|x|
Hence y = -4|y|,
if y <0
y = -4(-y) = 4y => 3y = 0, hence x and y individually = 0
if y > 0
y = -4y => 5y = 0 again x and y individually = 0
Hence in either case, it is sufficient.
Hence both is required and hence C
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|x| and |y| must be NONNEGATIVE.colakumarfanta wrote:Is |x| + |y| = 0?
1)x + 2|y| = 0
2)y + 2|x| = 0
Thus, for the sum of |x| and |y| to be 0, |x| and |y| themselves must each be equal to 0.
Question rephrased: Does x=y=0?
Statement 1: x = -2|y|.
It's possible that x=y=0.
It's possible that y=1 and x=-2.
INSUFFICIENT.
Statement 2: y = -2|x|.
It's possible that x=y=0.
It's possible that x=1 and y=-2.
INSUFFICIENT.
Statements combined:
Substituting y=-2|x| into x = -2|y|, we get:
x = -2| -2|x| |
x = -2 |2x|
x = -4|x|.
Since |x| must be NONNEGATIVE, |x|≥0.
If |x| > 0, then both sides of the equation above can safely be divided by |x|:
x/|x| = -4.
Not possible:
If x>0, then x/|x| = 1.
If x<0, then x/|x| = -1.
Since it is not possible that |x|>0, |x| = 0.
Since y = -2|x|, y = -2*0 = 0.
Thus, x=y=0.
SUFFICIENT.
The correct answer is C.
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The answer is C because when you combine A and B the example values should satisfy both A and B and 0,0 is the only set which satisfy both A and B.
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Is |x| + |y| = 0
1)x + 2|y| = 0 -> |x| = 2|y|
2)y + 2|x| = 0 -> |x| = 0.5|y|
The 2 statements contradict each other, unless y = x = 0
Therefore |X| + |Y| = 0 TRUE
1)x + 2|y| = 0 -> |x| = 2|y|
2)y + 2|x| = 0 -> |x| = 0.5|y|
The 2 statements contradict each other, unless y = x = 0
Therefore |X| + |Y| = 0 TRUE
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