Inequality and absolute value questions from my collection : DS Retired Questions

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Bunuel

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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.

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:

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

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Updated on: 17 Nov 2009, 10:57

Originally posted by Marco83 on 17 Nov 2009, 10:54.
Last edited by Marco83 on 17 Nov 2009, 10:57, edited 1 time in total.

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ichha148

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

Taking the modulus does not mean rounding up to the nearest integer; it means removing the negative sign if present. |0.8|<0.86

ichha148

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

Nowhere it is said that x and y are non-zero. If x and y are zero, 5a=0, therefore a=0, and the stem is false (x^2+y^2=0)

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17 Nov 2009, 13:41

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

\(6*x*y = x^2*y + 9*y\)
\(6*x = x^2 + 9 = 0\)
\(x^2 - 6*x + 9 = 0\)
\((x-3)^2 = 0\)
\(x = 3\)

St. (1) : y - x = 3
y = 6
Sufficient.

St. (2) : x^3 < 0
Invalid statement. Does not give us value of y.
Insufficient.

Answer : A

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17 Nov 2009, 13:47

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

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

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

St. (1) : (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
Insufficient.

St. (2) : (x - y)^2 = a
x^2 + y^2 - 2xy = a
Insufficient.

St. (1) and (2) together : x^2 + y^2 = 5a
When either x or y is not 0, question stem holds true.
When x and y are both 0, question stem is false.

Hence insufficient.

Answer : E

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

Question Stem : x > 0 ; y > 0 ?

St. (1) : 2x -2y = 1
x = y + 0.5
Equation can be satisfied for both positive and negative values of x and y.
Hence Insufficient.

St. (2) : x/y > 1
Equation can be satisfied when both x and y are either positive or negative.
Hence Insufficient.

St. (1) and (2) together : (y + 0.5)/y > 1
1 + 0.5/y > 1
0.5/y > 1
For this to be true, y must be positive.
If y is positive then x will also be positive.
Hence Sufficient.

Answer : C

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

Question stem : What is the exact value of y?

St. (1) : 3*|x^2 -4| = y - 2
y = 3*|x^2 -4| + 2
From this we can infer that y will be a positive value. That is, y > 0. However, we want to know the exact value of y.
Therefore, Insufficient.

St. (2) : |3 - y| = 11
(a) When (3 - y) > 0 ; 3 - y = 11 ; y = -8.
(b) When (3 - y) < 0 ; - (3 - y) = 11 ; y = 14.
Thus we can see that there are two possible values for y.
Hence Insufficient.

St. (1) and (2) together : y > 0 ; y = 14 or -8.
Obviously since y has to be greater than 0, it cannot be -8. Therefore value of y = 14.
Hence Sufficient.

Answer : C

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

St. (1) : x + 1 > 0
Tells us nothing about y.
Insufficient.

St. (2) : xy > 0
Both x and y can either be positive or negative. Neither x nor y can be 0.
Insufficient.

St. (1) and (2) together :
Since x is an integer and cannot hold the value 0, it has to be greater than 1 in order to satisfy St. (1).
Since we know that x will be positive, y will also have to be a positive integer in order to satisfy St. (2).
Hence Sufficient.

Answer : C

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

Question stem :
Note: Since the equations is symmetrical, there will only be two distinct cases. However, for the sake of explanation, I have illustrated all 4.
(a) When both x and y are greater than - 2 ; x + 2 = y + 2 ; x = y
(b) When both x and y are less than - 2 ; - x - 2 = - y - 2 ; x = y
(c) When x is less than -2 and y is greater than -2 ; - x - 2 = y + 2 ; x + y = - 4
(d) When x is greater than -2 and y is less than -2 ; x + 2 = - y - 2 ; x + y = - 4

St. (1) : xy < 0
This implies that one is negative and the other is positive. Therefore, in order for xy to be less than 0, x cannot be equal to y. Thus in order to satisfy the question stem, it can only be cases (c) and (d).
Thus Sufficient.

St. (2) : x > 2 ; y < 2
Again, this implies that x and y cannot be equal. Thus, in order to satisfy the question stem it can only be cases (c) and (d).
Thus Sufficient.

Answer : D

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

Question stem : Neither a nor b can hold the value 0 ; |a|/|b|=a/b
For condition to be true, both a and b must hold the same sign.

St. (1) : |a*b| = a*b
This condition will be satisfied only when both a and b are either both positive or both negative.
Hence Sufficient.

St. (2) : |a|/|b| = |a/b|
This condition can be satisfied when a and b are same sign as well as opposite sign.
Hence Insufficient.

Answer : A

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Updated on: 18 Nov 2009, 01:07

Originally posted by sriharimurthy on 17 Nov 2009, 14:54.
Last edited by sriharimurthy on 18 Nov 2009, 01:07, edited 1 time in total.

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

Question Stem : Is n negative?

St. (1) : -n = |-n|
Let -n = A ; therefore the statement becomes : A = |A|.
This can only be valid when A is positive (or equal to 0). This in turn means that n must be negative (or equal to 0).
-n=|-n| also for n=0, hence not sufficient.

St. (2) : n^2 = 16
n = ±4.
Thus Insufficient.

Answer : C

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

Question Stem : n # 0 ; -4 > n > 4 (excluding 0)

St. (1) : n^2 > 16
(n - 4)*(n +4) = 0 Therefore boundary conditions are 4 and - 4. Thus we can write it as : n < -4 and n > 4.
Sufficient.

St. (2) : 1/|n| > n
This condition will be valid for all n < 1 excluding 0.
Thus it will be impossible to tell whether |n| < 4.
Hence Insufficient.

Answer : A

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

Question Stem : Is the distance between X and -Y greater than the distance between X and Y?
Note : Using number line approach.

St. (1) : |x| > |y|
OR, The distance between X and the origin is greater than the distance between Y and the origin.

Now we can have two cases :

(a) When X is positive : In this case, X > Y for the above condition to be true.

_________|_________|_________|_________|_________|__________
_________________-Y______Origin______Y........ Region of X.........

Thus we can see that the distance between X and - Y will always be greater than the distance between X and Y. Hence question stem is true.

(b) When X is negative : In this case, X < -Y for the statement to be true.

_________|_________|_________|_________|_________|__________
.... Region of X.....-Y_____Origin_______Y___________

Thus we can see that the distance between X and -Y will always be less than the distance between X and Y. Hence, question stem becomes false.

Due to conflicting statements, St (1) becomes Insufficient.

St. (2) : |x-y| < |x|
OR, the distance between X and Y is less than the distance between X and the origin.

Now again let us consider the different cases :

(a) When X is positive : For the statement to be true and for X to be positive, X must be greater than Y/2. For any value of X less than Y/2 the statement will become false. The statement will be true for any value greater than Y/2.

Thus we see that only one case comes into the picture. Now let us see how it relates to the question stem.

_________|_________|____;____|_________|________
________-Y____Origin__y/2...Y...... Region of X....

Thus we can see that the distance between X and - Y will be greater than the distance between X and Y for all values of X > Y/2. Thus question stem is true.
St. (2) is sufficient.

Answer : B

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

St. (1) : -s <= r < = s
Clearly Insufficient.

St. (2) : |r| >= s
When r > 0 ; r >= s.
When r < 0 ; -r >= s ; r <= -s
Therefore, this statement can be rewritten as : -s >= r >= s
Insufficient.

St. (1) and (2) : -s <= r < = s ; -s >= r >= s
For both statements to be simultaneously valid, r must be equal to s.
Hence Sufficient.

Answer : C

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

Question Stem : Is |x-1| < 1 ?
When x > 1 ; x - 1 < 1 ; x < 2.
When x < 1 ; -x + 1 < 1 ; x > 0.
Thus it can be written as : 0 < x < 2.

St. (1) : (x-1)^2 <= 1
x^2 + 1 - 2x <= 1
x^2 - 2x <= 0
x(x - 2) <= 0 ; Thus boundary values are 0 and 2.
Therefore statement can be written as : 0 <= x <= 2.
Since the values are inclusive of 0 and 2, it cannot give us the answer.
Insufficient.

St. (2) : x^2 - 1 > 0
(x + 1)*(x - 1) > 0
Statement can be written as x > 1 and x < -1.
Thus it is possible for x to hold values which make the question stem true as well as false.
Insufficient.

St. (1) and (2) : 0 <= x <= 2 ; x > 1 and x < -1
Thus combined, the statements become : 1 < x <= 2.
Since it is inclusive of 2, it will give us conflicting solutions for the question stem.
Hence Insufficient.

Answer : E

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Quote:

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

Question Stem : Is n negative?

St. (1) : -n = |-n|
Let -n = A ; therefore the statement becomes : A = |A|.
This can only be valid when A is positive. This in turn means that n must be negative.
Thus Sufficient.

-n=|-n| also for n=0, hence not sufficient.

Everything else is as you suggested, therefore C

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Quote:

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

-n=|-n| also for n=0, hence not sufficient.

Everything else is as you suggested, therefore C

Yes, you are right. I overlooked that.
Thanks.

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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.

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2. If y is an integer and y = |x| + x, is y = 0?

Notice that from \(y=|x|+x\) it follows that y cannot be negative:
If \(x>0\), then \(y=x+x=2x=2*positive=positive\);
If \(x\leq{0}\) (when x is negative or zero) then \(y=-x+x=0\).

(1) \(x<0\) --> \(y=|x|+x=-x+x=0\). Sufficient.

(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.

Answer: D.

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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.

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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.

Answer: C.