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# Inequality and absolute value questions from my collection

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Intern
Joined: 08 Nov 2009
Posts: 45
Re: Inequality and absolute value questions from my collection  [#permalink]

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17 Nov 2009, 10:48
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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17 Nov 2009, 10:53
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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Updated on: 17 Nov 2009, 10:57
1
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)

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

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17 Nov 2009, 13:41
2
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, 13:47
2
Quote:
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.

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17 Nov 2009, 13:54
1
2
Quote:
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.

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

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17 Nov 2009, 14:01
2
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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17 Nov 2009, 14:10
2
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, 14:16
1
Quote:
6. If x and y are integer, is y > 0?
(1) x +1 > 0
(2) xy > 0

St. (1) : x + 1 > 0
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.

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17 Nov 2009, 14:43
2
3
Quote:
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.

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

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17 Nov 2009, 14:48
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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Updated on: 18 Nov 2009, 01:07
1
2
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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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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17 Nov 2009, 15:04
1
1
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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17 Nov 2009, 15:36
1
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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Re: Inequality and absolute value questions from my collection  [#permalink]

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17 Nov 2009, 15:48
2
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, 18:24
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, 01:01
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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Posts: 55188
Re: Inequality and absolute value questions from my collection  [#permalink]

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18 Nov 2009, 08:47
31
1
46
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.

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18 Nov 2009, 08:55
35
1
83
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.

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

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18 Nov 2009, 09:17
16
49
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.

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Re: Inequality and absolute value questions from my collection   [#permalink] 18 Nov 2009, 09:17

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