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Re: Inequality and absolute value questions from my collection
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17 Nov 2009, 10:13
4. Are x and y both positive? (1) 2x2y=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: 2x2y=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



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Re: Inequality and absolute value questions from my collection
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17 Nov 2009, 10:27
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 C is the answer 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
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17 Nov 2009, 10:43
Bunuel wrote: 6. If x and y are integer, is y > 0? (1) x +1 > 0 (2) xy > 0
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.



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Re: Inequality and absolute value questions from my collection
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17 Nov 2009, 10:48
4) I) 2x2y=1 so y=x1/2 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
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17 Nov 2009, 10:53
h2polo wrote: 4. Are x and y both positive? (1) 2x2y=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: 2x2y=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 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
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17 Nov 2009, 14:01
Quote: 4. Are x and y both positive? (1) 2x2y=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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17 Nov 2009, 14:10
Quote: 5. What is the value of y? (1) 3x^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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Re: Inequality and absolute value questions from my collection
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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. Answer : A
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Re: Inequality and absolute value questions from my collection
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17 Nov 2009, 15:36
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. Answer : C
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Re: Inequality and absolute value questions from my collection
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17 Nov 2009, 15:48
Quote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0 Question Stem : Is x1 < 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) : (x1)^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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Re: Inequality and absolute value questions from my collection
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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
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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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Re: Inequality and absolute value questions from my collection
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18 Nov 2009, 17:43
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=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: 3x^24=10 the answer will never give 10/3 do the same for 14 3x^24=12 x = 0 using my methodology I also got C, but is my thinking correct?



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18 Nov 2009, 19:45
Awesome stuff Bunuel! Hats off to you dude. +5 from me.



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Re: Inequality and absolute value questions from my collection
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01 Dec 2009, 11:58
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=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) 3x^2 4 = y  2 if (x^2 4) is positive, we can rewrite above as 3(x^2 4) = y  2 => 3x^2y = 10 > Eqn. 1 if (x^2 4) is negative, we can rewrite above as 3(4x^2) = y  2 => 3x^2y = 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! Thanks



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Re: Inequality and absolute value questions from my collection
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13 Dec 2009, 19:52
lagomez wrote: Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
I'm getting B for this one 1. (x1)^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 hi how would mod(1x)<1 would resolve, i mean the interval of x



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Re: Inequality and absolute value questions from my collection
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14 Dec 2009, 01:21
ISBtarget wrote: hi
how would mod(1x)<1 would resolve, i mean the interval of x x1<1: x1 switch sign at x=1, which means that we should check for the two ranges: A. x<1 > x+1<1 > x>0; And B. x>=1 > x1<1 > x<2; Hence x1<1 can be rewritten as 0<x<2.
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Re: Inequality and absolute value questions from my collection
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22 Dec 2009, 13:55
Bunuel wrote: 7. x+2=y+2 what is the value of x+y? (1) xy<0 (2) x>2 y<2
This one is quite interesting.
First note that x+2=y+2 can take only two possible forms:
A. x+2=y+2 > x=y. This will occur if and only x and y are both >= than 2 OR both <= than 2. In that case x=y. Which means that their product will always be positive or zero when x=y=2. B. x+2=y2 > x+y=4. This will occur when either x or y is less then 2 and the other is more than 2.
When we have scenario A, xy will be positive only. Hence if xy is not positive we have scenario B and x+y=4. Also note that viseversa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.
(1) xy<0 > We have scenario B, hence x+y=4. Sufficient.
(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=4. Sufficient.
Answer: D. hey Bunuel!! first i would like to thank you for posting such wonderful questions.. regarding a question that you posted above, i got a small doubt.. x+2=y+2 so lets say x+2=y+2=k (some 'k') now x+2=k =====> x+2=+/ k and x+2= +k, iff x>2 x+2= k, iff x<2 also we have y+2=k =====> y+2=+/ k and y+2= +k, iff y>2 y+2= k, iff y<2 so x+2=y+2 ===> x=y , iff (x>2 and y>2) or (x<2 and y<2)eq1 and x+y=4, iff (x<2 and y>2) or (x>2 and y<2)eq2 now coming to the options, 1) xy<0 i.e., (x=ve and y=+ve) or (x=+ve and y=ve) (x=ve and y=+ve): this also means that x and y can have values, x=1 and y= some +ve value. so eq2 cannot be applied, x+y#4. if x=3 and y=some +ve value, x+y=4. two cases. data insuff. (x=+ve and y=ve): this also means that x and y can have values, x=+ve value and y=1.so eq2 cannot be applied, x+y#4. if x=some +ve value and y=3, x+y=4. two cases. data insuff. 2)x>2,y>2 for this option too we cannot judge the value of x+y, with the limits of x and y being different in the question and the answer stem. so data insuff. so i have a doubt that, why the answer cannot be E?? plz point out if i made any mistake..



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Re: Inequality and absolute value questions from my collection
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22 Dec 2009, 18:50
Hi Bunuel, Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0. Bunuel wrote: 4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
(1) 2x2y=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. 2x2y=1 > y=x1/2, 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 timeconsuming and comfortable for you personally.
One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.
Answer: C.



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Re: Inequality and absolute value questions from my collection
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22 Dec 2009, 19:09
lionslion wrote: Hi Bunuel, Is (1/2y) > 0 or (1/y) >0. While I was solving I am getting (1/2y)>0.
I dropped 2, as (1/2y) > 0 and (1/y) >0 are absolutely the same (you can multiply both sides of inequality by 2 and you'll get 1/y>0). What is important that you can get that y>0 from either of them. Hope it's clear.
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