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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-a ... 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-a ... ml#p653697
4. Are x and y both positive?
(1) \(2x-2y=1\)
(2) \(\frac{x}{y}>1\)
Solution: https://gmatclub.com/forum/inequality-a ... 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-a ... 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-a ... 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-a ... 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-a ... ml#p653789
9. Is n<0?
(1) \(-n=|-n|\)
(2) \(n^2=16\)
Solution: https://gmatclub.com/forum/inequality-a ... 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-a ... ml#p653796
11. Is \(|x+y|>|x-y|\)?
(1) \(|x| > |y|\)
(2) \(|x-y| < |x|\)
Solution: https://gmatclub.com/forum/inequality-a ... ml#p653853
12. Is r=s?
(1) \(-s \leq r \leq s\)
(2) \(|r| \geq s\)
Solution: https://gmatclub.com/forum/inequality-a ... ml#p653870
13. Is \(|x-1| < 1\)?
(1) \((x-1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)
Solution: https://gmatclub.com/forum/inequality-a ... 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-a ... 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-a ... ml#p653697
4. Are x and y both positive?
(1) \(2x-2y=1\)
(2) \(\frac{x}{y}>1\)
Solution: https://gmatclub.com/forum/inequality-a ... 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-a ... 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-a ... 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-a ... 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-a ... ml#p653789
9. Is n<0?
(1) \(-n=|-n|\)
(2) \(n^2=16\)
Solution: https://gmatclub.com/forum/inequality-a ... 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-a ... ml#p653796
11. Is \(|x+y|>|x-y|\)?
(1) \(|x| > |y|\)
(2) \(|x-y| < |x|\)
Solution: https://gmatclub.com/forum/inequality-a ... ml#p653853
12. Is r=s?
(1) \(-s \leq r \leq s\)
(2) \(|r| \geq s\)
Solution: https://gmatclub.com/forum/inequality-a ... ml#p653870
13. Is \(|x-1| < 1\)?
(1) \((x-1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)
Solution: https://gmatclub.com/forum/inequality-a ... 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.
Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.
DS questions in GMAT Focus encompass various types of word problems, such as:
While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."
Check GMAT Syllabus for Focus Edition
You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.
So, you can ignore these question.
DS questions in GMAT Focus encompass various types of word problems, such as:
- Word Problems
- Work Problems
- Distance Problems
- Mixture Problems
- Percent and Interest Problems
- Overlapping Sets Problems
- Statistics Problems
- Combination and Probability Problems
While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."
Check GMAT Syllabus for Focus Edition
You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.
So, you can ignore these question.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
_________________
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ahh..yes...fresh meat
Bunuel
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
Bunuel
I'm getting B for this one
1. (x-1)^2 <= 1
x can be 0 which would make the question no
or x can be 1/2 which would make the answer yes
so 1 is insufficient
2. x^2 - 1 > 0
means x^2>1
so x<-1 or x>1
both of which make the question no
so sufficient
Bunuel
I'm getting c
1. s can be 3 and r can be 3 which makes question yes
s can be 3 and r can be 2 which makes question no
insufficient
2. r can be 3 and s can be 3 makes question yes
r can be 3 s can be 2 makes question no
insufficient
combining:
|r|>=s means
r>=s or r<=-s
and -s<=r<=s means
-s<=r and r<=s
now we have -s<=r and -s>=r so -s = r or s = r
r>=s and r<=s so s = r
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n
answer A
because in number 2 n can be negative or a fraction
(1) n^2 > 16
(2) 1/|n| > n
answer A
because in number 2 n can be negative or a fraction
Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.
Schools: Sloan '14 (S)
Posts: 101
Quality questions as always... Thanks Bunuel! +1
Posts: 1,012
lagomez
(1) (x-1)^2 <= 1
x is 0 to 2.
If x = 2, yes.
If x < 2, No.
(2) x^2 - 1 > 0
x cannot be -1 to 1 i.e. x<-1 or x>1. NSF.
From 1 and 2: x is >1 but <=2. NSF..
E.
gmat620
These questions are from various sources. Couple of questions might be from MGMAT CAT or Gmat Prep, but not more than that.
I'll provide OA in a day or two, after discussions. Tell me if you want the answers for the specific questions earlier than that and I'll mail you.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
Schools: Sloan '14 (S)
Posts: 101
Bunuel
Not sure about this one...
First I reduced the given equation (divided out the y) and solved for x:
6*x*y = x^2*y + 9*y
6*x = x^2 + 9
0 = x^2 - 6*x + 9
0 = (x-3)^2
x = 3
Statement 1:
y-x=3
y-3=3
y=6
xy=3*6=18
SUFFICIENT
Statement 2:
x^3<0
We have no idea what the value of y is from this statement. The only thing that made me look twice was the face that if x^3 is true, then x should be a negative value... did I calculate the value of x incorrectly above?
INSUFFICIENT
ANSWER: A.
Schools: Sloan '14 (S)
Posts: 101
Bunuel
Another way of looking at the problem is to ask, is x<0? Because if it is, then we know that y is zero. The only case in which y will not be zero is if x is positive.
Statement 1:
x<0... answers my question above.
SUFFICIENT
Statement 2:
y<1
Because y is an integer, it must be one of the following values: 0, -1, -2, -3...
BUT |x| + x can never be a negative value. The lowest value that it can be is 0.
Hence, y can never be negative and the only possible value it can be then is 0.
SUFFICIENT
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
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
Marco83
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.
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Schools: Sloan '14 (S)
Posts: 101
4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1
Statement 1:
2(1)-2(1/2)=1 , x,y are both positve
2(1/2)-2(-1/2)=1 x is positive, y is negative
INSUFFICIENT
Statement 2:
Either (x,y) are both positive or both negative
INSUFFICENT
Statement 1 and 2:
With both requirements x must be greater than y and satisfy this equation: 2x-2y=1
2(1)-2(1/2)=1 , x,y are both positve and x>y
2(1/2)-2(-1/2)=1 x is positive, y is negative and x>y
Answer: E
(1) 2x-2y=1
(2) x/y>1
Statement 1:
2(1)-2(1/2)=1 , x,y are both positve
2(1/2)-2(-1/2)=1 x is positive, y is negative
INSUFFICIENT
Statement 2:
Either (x,y) are both positive or both negative
INSUFFICENT
Statement 1 and 2:
With both requirements x must be greater than y and satisfy this equation: 2x-2y=1
2(1)-2(1/2)=1 , x,y are both positve and x>y
2(1/2)-2(-1/2)=1 x is positive, y is negative and x>y
Answer: E
12. Is r=s?
(1) -s<=r<=s
(2) |r|>=s
E – for this - both can be true or false when 0< r < 1
For example , take r as 0.8
S = 0.86 i.e. -0.86 < = 0.8 < = 0.86
|0.8|>= 0.86 i.e. 1 >= 0.86
Combining , any values can be taken , on values > =1 , both r and s
will be same
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a
C is the answer
Combined both and the equation will give x^2 + y^2 = 5a
_________________
(1) -s<=r<=s
(2) |r|>=s
E – for this - both can be true or false when 0< r < 1
For example , take r as 0.8
S = 0.86 i.e. -0.86 < = 0.8 < = 0.86
|0.8|>= 0.86 i.e. 1 >= 0.86
Combining , any values can be taken , on values > =1 , both r and s
will be same
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a
C is the answer
Combined both and the equation will give x^2 + y^2 = 5a
_________________
Always tag your question
Schools: Sloan '14 (S)
Posts: 101
Bunuel
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.
Schools: Sloan '14 (S)
Posts: 101
Bunuel
Statement 1:
Nothing about y... INSUFFICIENT
Statement 2:
two equations, two unknowns... INSUFFICIENT
Statements 1 and 2:
From x +1 > 0 and the fact that x must be an integer, we know that x must be [0,1,2,3...]
Because we know that xy > 0, we know that x cannot be 0... therefore y must be a positive integer!
SUFFICIENT
ANSWER: C.
4)
I) 2x-2y=1 so y=x-1/2 NS
II)x/y>0 so x and y have the same sign and the modulus of x has to be larger than the modulus of y NS
Together, to satisfy both clues needs to be larger than 1/2 and x becomes larger than 0; the stem is true, therefore C
I) 2x-2y=1 so y=x-1/2 NS
II)x/y>0 so x and y have the same sign and the modulus of x has to be larger than the modulus of y NS
Together, to satisfy both clues needs to be larger than 1/2 and x becomes larger than 0; the stem is true, therefore C
h2polo
Your last choice of numbers: x=1/2, y=-1/2 does not satisfy clue I, because 2*(1/2)-2*(-1/2)=2, not 1
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