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Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 10:33
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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) \(x^3< 0\)
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 07:39
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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. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 09:10
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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.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 11:26
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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.
(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.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 14:48
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11. Is |x+y|>|x-y|? (1) |x| > |y| (2) |x-y| < |x|
To answer this question you should visualize it. We have comparison of two absolute values. Ask yourself when |x+y| is more then than |x-y|? If and only when x and y have the same sign absolute value of x+y will always be more than absolute value of x-y. As x+y when they have the same sign will contribute to each other and x-y will not.
5+3=8 and 5-3=2 OR -5-3=-8 and -5-(-3)=-2.
So if we could somehow conclude that x and y have the same sign or not we would be able to answer the question.
(1) |x| > |y|, this tell us nothing about the signs of x and y. Not sufficient.
(2) |x-y| < |x|, says that the distance between x and y is less than distance between x and origin. This can only happen when x and y have the same sign, when they are both positive or both negative, when they are at the same side from the origin. Sufficient. (Note that vise-versa is not right, meaning that x and y can have the same sign but |x| can be less than |x-y|, but if |x|>|x-y| the only possibility is x and y to have the same sign.)
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 07:47
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2. If y is an integer and y = |x| + x, is y = 0? (1) x < 0 (2) y < 1
Note: as \(y=|x|+x\) then \(y\) is never negative. For \(x>{0}\) then \(y=x+x=2x\) and for \(x\leq{0}\) then (when x is negative or zero) then \(y=-x+x=0\).
(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.
(2) \(y<1\), as we concluded y is never negative, and we are given that \(y\) is an integer, hence \(y=0\). Sufficient.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 08:17
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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.
Is |x-1| < 1? Basically the question asks is 0<x<2 true?
(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.
(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 11:34
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8. a*b#0. Is |a|/|b|=a/b? (1) |a*b|=a*b (2) |a|/|b|=|a/b|
|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.
(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.
(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 15:42
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12. Is r=s? (1) -s<=r<=s (2) |r|>=s
This one is tough.
(1) -s<=r<=s, we can conclude two things from this statement: A. s is either positive or zero, as -s<=s; B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s. But we don't know whether r=s or not. Not sufficient.
(2) |r|>=s, clearly insufficient.
(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well. Consider: s=5, r=5 --> -5<=5<=5 |5|>=5 s=5, r=-5 --> -5<=-5<=5 |-5|>=5 Both statements are true with these values. Hence insufficient.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 16:51
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lagomez wrote:
Just curious if my thinking is correct.
on the 2nd part I get y = -8 and y =14 Then I substituted the values into the first equation: 3|x^2-4|=-10 the answer will never give -10/3
do the same for 14 3|x^2-4|=12 x = 0
using my methodology I also got C, but is my thinking correct?
Well you can even not calculate for x. Statement 1 says that y must be greater than or equal to 2. Statement 2 gives 2 values of y -8 OR 14. Combining we get that y=14.
Remember we are asked to determine the value of y not x. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
20 Nov 2009, 13:45
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tihor wrote:
Bunuel, two questions: shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2? secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2? does this not translate into x<=0 or x<=2?
thank you very much for all the questions and solutions.
Thank you very much for this catch. +1. There was a typo. So you are right with the first one: |x-1| < 1 means 0<x<2. Already edited the post.
As for the second one: x(x-2)<=0 means 0<=x<=2, if you plug the values from this range you'll get the values less than or equal to 0. If you plug the values less than 0 or more than 2 you'll get only positive values.
x(x-2) is "smiling" parabola, and the intersections with X-axis are at x=0 and x=2, the range between will be below X-axis.
Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 06:38
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3) I) (x+y)^2=9a x^2+y^2=9a-2xy NS II) (x-y)^2=a x^2+y^2=a+2xy NS Together 2(x^2+y^2)=10a x^2+y^2=5a If either x or y are larger than 0, the stem would be true, but if they’re both zero the stem is false, hence E
4) I don’t get the two clues; they seem to be mutually exclusive
5) I) 3|x^2-4|=y-2 either y=3x^2-10 or y=14-3x^2 NS II) |3-y|=11 either y=-8 or y=14 NS Together -8=3x^2-10 so 3x^2=2 ok 14=3x^2-10 so 3x^2=28 ok, hence E
6) I) x+1>0 so x={0, 1, 2, …} NS II) xy>0 so x and y have the same sign and none of them is zero NS Together, x={1, 2, 3, ..} and y has the same sign, hence C
7) |x+2|=|y+2| either x+2=y+2 or x+2=-y-2 (the other two combinations can be transformed into these by multiplying by -1) Reordering: x-y=0 or x+y=-4 I)xy<0, hence x and y have different signs and none of them is zero. The only possibility is x+y=-4 S II) x>2, y<2 hence x#y. The only possibility is x+y=-4 S, therefore D
8)a*b#0, hence a and b are both non-zero I) |a*b|=a*b a and b have the same sign and the stem is always true S II) |a|/|b|=|a/b| this is true regardless of the values of a and b, and nothing can be said about the stem NS, therefore A
9) I) –n=|-n| n<=0 NS II) n^2=16 n=+/-4 NS Together n=-4 therefore C
10)n#0 I) n^2>16, so |n|>4 S II) 1/|n|>n true for n<-1 NS, therefore A
11) Plugging in numbers I get B, but there’s no rime or reason to my solution
12) I) –s<=r<=s obviously NS. Since s>=-s, s is either positive or zero II)|r|>=s obviously NS Together: I) tells us that s>=0; II) tells us that r>=s or r<=-s. The only case in which I and II are simultaneously satisfied is r=s, therefore C
13) x=(0:2) with 0 and 2 excluded I) (x-1)^2<=1, hence x=[0:2] with 0 and 2 included, hence NS II) x^2-1>0 x<-1 or x>1. For x=1.5 the stem is true, for x=3 it is false, hence NS Together, for x=1.5 the stem is true, for x=2 it is false, hence E
Is |x-1| < 1? Basically the question asks is -2<x<2 true?
(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 2! Else it would be sufficient. So not sufficient.
(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that -2<x<2 is true. Not sufficient.
Answer: E.
Bunuel, two questions: shouldnt |x-1| < 1 be 0<x<2? and not -2<x<2? secondly, how does this happen: x(x-2)<=0 --> 0<=x<=2? does this not translate into x<=0 or x<=2?
thank you very much for all the questions and solutions.
gmatclubot
Re: Inequality and absolute value questions from my collection
[#permalink]
20 Nov 2009, 13:19
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