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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

(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

Re: Inequality and absolute value questions from my collection [#permalink]
25 Aug 2011, 20:53

1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0

Ans:

From the initial equation we get: y*(x^2-6x+9)=0 i.e y=0 or (x-3)^2=0 means either y=0 or (x-3)^2=0 or both A does not give any idea about B states that x^3<0 which means that x is not equal to 3.hence in that case y needs to be zero. so statement 2 itself is sufficient to answer the question.

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