Inequality and absolute value questions from my collection

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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: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653690

2. If y is an integer and y = |x| + x, is y = 0?
(1) x < 0
(2) y < 1

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-20.html#p653695

3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653697

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653709

5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653731

6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653740

7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653783

8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653789

9. Is n<0?
(1) -n=|-n|
(2) n^2=16

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653792

10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653796

11. Is |x+y|>|x-y|?
(1) |x| > |y|
(2) |x-y| < |x|

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653853

12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653870

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Solution: inequality-and-absolute-value-questions-from-my-collection-86939-40.html#p653886

Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.

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.

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.

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=-2.
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 positive only. Hence if xy is not positive 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.

3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(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.

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.

6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0

(1) x+1>0 --> x>-1. As x is an integer x can take the following values 0,1,2,... But we know nothing about y. Not sufficient.

(2) xy>0. x and y have the same sign (both positive OR both negative) and neither x nor y is zero. Not sufficient.

(1)+(2) x is positive, as from (1) it's 0,1,2.. and from (2) x is not zero. Hence xy to be positive y also must be positive. Sufficient.

Answer: C.

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.

Answer: D.

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.

Answer: A.

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.)

Answer: B.

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.

13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

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.

Answer: E.

This approach is not correct not because we are not told that y is an integer, but because you can not add inequalities like you did.

3(x^2 -4) = y - 2 OR 3(4-x^2) = y - 2, in fact these equation are derived from one and from them only one is right. It's not that we have 3(x^2 -4) = y - 2 AND 3(4-x^2) = y - 2 and we are asked to solve fro unknowns. If it were then your solution would be right.

Hope it's clear.

Sure when I said that you can not add this way I meant: in this case.

ahh..yes...fresh meat

1. x < 0
you will always get x minus itself so always 0

2. y < 1
y is an integer so y<=0
y can't be negative because x minus itself is always zero

answer d

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

Yes there was a typo in 4. Edited. Great job Marco83. Even though not every answer is correct, you definitely know how to deal with this kind of problems.

Statement 1:

Two equations, two unknowns... INSUFFICIENT

Statement 2:

|3 - y| = 11
(3-y)=11 or (3-y)=-11
y=-8, 14

INSUFFICIENT

Statements 1 and 2:

y must be 14 because 3|x^2 -4| can never be a negative value (no matter what you plug in for x, you will get a positve value because of the absolute value signs).

SUFFICIENT

ANSWER: C.

Question Stem gives us :

(a) If x > 0 ; y = 2x
(b) If x < 0 ; y = 0

St. (1) : x < 0
Sufficient.

St. (2) : y < 1
Since y is an integer and y cannot be less than 0 (question stem part b) therefore y must be 0.
Sufficient.

Answer : D