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16 Nov 2009, 11:33
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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) \(x^3< 0\)
Solution: https://gmatclub.com/forum/inequality-a ... ml#p653690
2. If y is an integer and \(y = |x| + x\), is \(y = 0\)?
(1) \(x < 0\)
(2) \(y < 1\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653695
3. Is \(x^2 + y^2 > 4a\)?
(1) \((x + y)^2 = 9a\)
(2) \((x – y)^2 = a\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653697
4. Are x and y both positive?
(1) \(2x-2y=1\)
(2) \(\frac{x}{y}>1\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653709
Graphic approach: https://gmatclub.com/forum/inequality-a ... l#p1269802
5. What is the value of y?
(1) \(3|x^2 -4| = y - 2\)
(2) \(|3 - y| = 11\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653731
6. If x and y are integer, is y > 0?
(1) \(x +1 > 0\)
(2) \(xy > 0\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653740
7. \(|x+2|=|y+2|\) what is the value of x+y?
(1) \(xy<0\)
(2) \(x>2\), \(y<2\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653783 AND https://gmatclub.com/forum/inequality-a ... l#p1111747
8. \(a*b \neq 0\). Is \(\frac{|a|}{|b|}=\frac{a}{b}\)?
(1) \(|a*b|=a*b\)
(2) \(\frac{|a|}{|b|}=|\frac{a}{b}|\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653789
9. Is n<0?
(1) \(-n=|-n|\)
(2) \(n^2=16\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653792
10. If n is not equal to 0, is |n| < 4 ?
(1) \(n^2 > 16\)
(2) \(\frac{1}{|n|} > n\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653796
11. Is \(|x+y|>|x-y|\)?
(1) \(|x| > |y|\)
(2) \(|x-y| < |x|\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653853
12. Is r=s?
(1) \(-s \leq r \leq s\)
(2) \(|r| \geq s\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653870
13. Is \(|x-1| < 1\)?
(1) \((x-1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653886
Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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) \(x^3< 0\)
Solution: https://gmatclub.com/forum/inequality-a ... ml#p653690
2. If y is an integer and \(y = |x| + x\), is \(y = 0\)?
(1) \(x < 0\)
(2) \(y < 1\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653695
3. Is \(x^2 + y^2 > 4a\)?
(1) \((x + y)^2 = 9a\)
(2) \((x – y)^2 = a\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653697
4. Are x and y both positive?
(1) \(2x-2y=1\)
(2) \(\frac{x}{y}>1\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653709
Graphic approach: https://gmatclub.com/forum/inequality-a ... l#p1269802
5. What is the value of y?
(1) \(3|x^2 -4| = y - 2\)
(2) \(|3 - y| = 11\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653731
6. If x and y are integer, is y > 0?
(1) \(x +1 > 0\)
(2) \(xy > 0\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653740
7. \(|x+2|=|y+2|\) what is the value of x+y?
(1) \(xy<0\)
(2) \(x>2\), \(y<2\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653783 AND https://gmatclub.com/forum/inequality-a ... l#p1111747
8. \(a*b \neq 0\). Is \(\frac{|a|}{|b|}=\frac{a}{b}\)?
(1) \(|a*b|=a*b\)
(2) \(\frac{|a|}{|b|}=|\frac{a}{b}|\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653789
9. Is n<0?
(1) \(-n=|-n|\)
(2) \(n^2=16\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653792
10. If n is not equal to 0, is |n| < 4 ?
(1) \(n^2 > 16\)
(2) \(\frac{1}{|n|} > n\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653796
11. Is \(|x+y|>|x-y|\)?
(1) \(|x| > |y|\)
(2) \(|x-y| < |x|\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653853
12. Is r=s?
(1) \(-s \leq r \leq s\)
(2) \(|r| \geq s\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653870
13. Is \(|x-1| < 1\)?
(1) \((x-1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)
Solution: https://gmatclub.com/forum/inequality-an ... ml#p653886
Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
18 Nov 2009, 08:39
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SOLUTIONS:
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 let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
This one was quite tricky and was solved incorrectly by all of you.
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.
Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.
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 let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(y-x=3\). If y is not 0, x must be 3 and y-x to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=-3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
This one was quite tricky and was solved incorrectly by all of you.
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.
Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.
18 Nov 2009, 12:26
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7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2
This one is quite interesting.
First note that |x+2|=|y+2| can take only two possible forms:
A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.
When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.
(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.
(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.
Answer: D.
(1) xy<0
(2) x>2 y<2
This one is quite interesting.
First note that |x+2|=|y+2| can take only two possible forms:
A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.
When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.
(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.
(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.
Answer: D.
