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

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

Last edited by Marco83 on 17 Nov 2009, 09:57, edited 1 time in total.

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

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

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 13:16

Quote:

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.

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 13:54

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.

Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 14:29

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.

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.

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

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

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

Answer: C.

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, 16:43

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.

Answer: C.

Just curious if my thinking is correct.

on the 2nd part I get y = -8 and y =14 Then I substituted the values into the first equation: 3|x^2-4|=-10 the answer will never give -10/3

do the same for 14 3|x^2-4|=12 x = 0

using my methodology I also got C, but is my thinking correct?

Re: Inequality and absolute value questions from my collection [#permalink]
01 Dec 2009, 10:58

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.

Answer: C.

Bunuel, I tried to solve this in another way.

1) 3|x^2 -4| = y - 2 if (x^2 -4) is positive, we can rewrite above as 3(x^2 -4) = y - 2 => 3x^2-y = 10 -> Eqn. 1 if (x^2 -4) is negative, we can rewrite above as 3(4-x^2) = y - 2 => -3x^2-y = -14 -> Eqn. 2 Adding equation 1 and 2, we get -2y = -4 or y = 2. So (A) as the answer is tempting.

I know this is not correct and carries the assumption that y is an integer which is not the case here.

If y indeed were an integer in the question, do you think the above approach had any problems ? I am a little confused because every inequality problem appears to have a different method for solving it!

Re: Inequality and absolute value questions from my collection [#permalink]
01 Dec 2009, 23:05

Bunuel, you are correct. The key is understanding that the two equations are an 'OR' (either one is true depending on whether x^2-4 is positive or negative) and not an 'AND' (both are correct).

You mentioned that inequalities cannot be added 'the way' I did. I believe you are not saying that we cannot add inequalities. I saw an interesting discussion here - > http://www.beatthegmat.com/combining-in ... 21610.html (Sorry for the cross posting, but this may be of use to someone confused like me!)

+1 from me.

cheers

gmatclubot

Re: Inequality and absolute value questions from my collection
[#permalink]
01 Dec 2009, 23:05

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