Inequality and absolute value questions from my collection : GMAT Data Sufficiency (DS) - Page 4
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# Inequality and absolute value questions from my collection

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Inequality and absolute value questions from my collection [#permalink]

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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) $$x^3< 0$$

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) $$\frac{x}{y}>1$$

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 AND inequality-and-absolute-value-questions-from-my-collection-86939-160.html#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: 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) $$\frac{1}{|n|} > n$$

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) $$-s \leq r \leq s$$
(2) $$|r| \geq s$$

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

13. Is $$|x-1| < 1$$?
(1) $$(x-1)^2 \leq 1$$
(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.
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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:13
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

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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:27
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

Combined both and the equation will give x^2 + y^2 = 5a
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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:43
Bunuel wrote:
6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0

Statement 1:

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

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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:48
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
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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:53
h2polo wrote:
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

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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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:54
ichha148 wrote:
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 wrote:
3. Is x^2 + y^2 > 4a?

(1) (x + y)^2 = 9a

(2) (x – y)^2 = a

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)

Last edited by Marco83 on 17 Nov 2009, 09:57, edited 1 time in total.
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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 09:57
Marco83 wrote:
h2polo wrote:
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

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

Nice catch... its getting late.
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17 Nov 2009, 11:24

1) A
2) D
3) E
4) C
5) C
6) C
7) D
8) A
9) C
10) A
11) B
12) C
13) E

Will post my individual solutions in a bit.
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Last edited by sriharimurthy on 18 Nov 2009, 00:03, edited 2 times in total.
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17 Nov 2009, 12:41
Quote:
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.

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17 Nov 2009, 13:01
Quote:
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.

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17 Nov 2009, 13:10
Quote:
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.

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17 Nov 2009, 13:48
Quote:
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.

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17 Nov 2009, 13:54
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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 (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.

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Last edited by sriharimurthy on 18 Nov 2009, 00:07, edited 1 time in total.
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17 Nov 2009, 14:04
Quote:
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.

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17 Nov 2009, 14:29
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Quote:
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.

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17 Nov 2009, 14:36
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Quote:
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.

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17 Nov 2009, 14:48
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Quote:
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.

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Re: Inequality and absolute value questions from my collection [#permalink]

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17 Nov 2009, 17:24
sriharimurthy wrote:
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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Re: Inequality and absolute value questions from my collection [#permalink]

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18 Nov 2009, 00:01
Marco83 wrote:
sriharimurthy wrote:
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

Yes, you are right. I overlooked that.
Thanks.
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Re: Inequality and absolute value questions from my collection [#permalink]

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18 Nov 2009, 09:12
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Bunuel wrote:
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.

just to chime in here your thanks for all this..it's really useful
Re: Inequality and absolute value questions from my collection   [#permalink] 18 Nov 2009, 09:12

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