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Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 10:33
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Expert's post
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Guys I didn't forget your request, just was collecting good questions to post.
So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a few, which are quite tricky. Please provide your explanations along with the answers.
1. If \(6*x*y = x^2*y + 9*y\), what is the value of xy? (1) \(y – x = 3\) (2) \(x^3< 0\)
Re: Inequality and absolute value questions from my collection [#permalink]
19 Dec 2010, 02:12
Bunuel wrote:
4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1
(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.
(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.
(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.
One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.
Answer: C.
Hi
I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient.
Consider putting the values as 1) x=1 y=1/2 ,now x>y and x-y=1/2.In this case both x and y are positive Now2) x=-1 and y=-1/2,now again x>y and x-y=-1/2.In this case both are negative.
Hence insufficient to deduce whther x and y are both positive.
Re: Inequality and absolute value questions from my collection [#permalink]
19 Dec 2010, 02:27
Expert's post
Eshika wrote:
Bunuel wrote:
4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1
(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.
(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.
(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.
One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.
Answer: C.
Hi
I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient.
Consider putting the values as 1) x=1 y=1/2 ,now x>y and x-y=1/2.In this case both x and y are positive Now2) x=-1 and y=-1/2,now again x>y and x-y=-1/2.In this case both are negative.
Hence insufficient to deduce whther x and y are both positive.
Please share your inputs on the same.
OA for this question is C, not E.
The red part in your reasoning is not correct: if x=-1 and y=-1/2 then y>x.
Here is the logic for C:
From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).
When we consider two statement together:
From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.
OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.
(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.
(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.
Bunuel...here why cant we use squaring on both sides of the statements. _________________
(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.
(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.
Bunuel...here why cant we use squaring on both sides of the statements.
I done't understand what you mean. Can you please SHOW what you mean. _________________
(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.
(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.
Bunuel...here why cant we use squaring on both sides of the statements.
I done't understand what you mean. Can you please SHOW what you mean.
well i was just wanted to know why cant we do this
1. x + y = 3sqrt a 2. x - y = sqrt a
then solve the 2 equations to get x and y. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
07 Feb 2011, 18:06
lagomez wrote:
Bunuel wrote:
5. What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11
(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.
(2) |3 - y| = 11:
y<3 --> 3-y=11 --> y=-8 y>=3 --> -3+y=11 --> y=14
Two values for y. Not sufficient.
(1)+(2) y>=2, hence y=14. Sufficient.
Answer: C.
So here in statement 1 we are not at all concerned with |X^2 -4| _________________
Re: Inequality and absolute value questions from my collection [#permalink]
08 Feb 2011, 03:33
Expert's post
ajit257 wrote:
lagomez wrote:
Bunuel wrote:
5. What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11
(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.
(2) |3 - y| = 11:
y<3 --> 3-y=11 --> y=-8 y>=3 --> -3+y=11 --> y=14
Two values for y. Not sufficient.
(1)+(2) y>=2, hence y=14. Sufficient.
Answer: C.
So here in statement 1 we are not at all concerned with |X^2 -4|
Yes. We just have that left hand side is some absolute value and as absolute value is always non-negative then right hand side must also be non-negative. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
08 Aug 2011, 10:20
Quote:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0 First, devide the whole equation by y, and 6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3 (1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient (2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient B
Thanks ImJun for the above solution.
I just could not understand how option B was sufficient. I had left out on an important bit, (x-3)^2 will have both positive and negative roots. _________________
(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.
(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.
Answer: E.
duh! missed out on the 0 and thought C was the ans.
Is this what you call the ZIP trap.. if so, then i sure did get zipped.. dang _________________
Re: Inequality and absolute value questions from my collection [#permalink]
08 Aug 2011, 18:53
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0 First, devide the whole equation by y, and 6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3 (1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient (2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient B
There is only one answer from the equation that you formed above from the question stem. x^2 - 6xy +9 = 0
Re: Inequality and absolute value questions from my collection [#permalink]
08 Aug 2011, 20:54
manishgeorge wrote:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0 First, devide the whole equation by y, and 6*x=x^2+9 => x^2-6*x+9=0 => (x-3)^2=0, x=3, -3 (1) We know x could be 3 oro -3, let's put these two numbers in and see, if x=3, y =6, xy= 18,... and if x=-3, y =0, xy= 0. We can't know for sure what the value of that is. .... Insufficient (2)Now we know X^3 is less than 0, then x must be less than 0 too. that means it has to be -3, then the answer comes out! .... Sufficient B
There is only one answer from the equation that you formed above from the question stem. x^2 - 6xy +9 = 0
X =3
There for statement 1 is sufficient
Answer is A
read Bunuel's and ImJun's post for this problem.
He writes that we can't divide both sides by 'y' since that would mean that we are assuming that y is not equal to 0.
The question prompt asks us the value of xy
Sol. would be:
If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
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