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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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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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Updated on: 17 Nov 2009, 10:57
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 nonzero. 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
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17 Nov 2009, 13:41
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 \(6*x*y = x^2*y + 9*y\) \(6*x = x^2 + 9 = 0\) \(x^2  6*x + 9 = 0\) \((x3)^2 = 0\) \(x = 3\) 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
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. Answer : D
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17 Nov 2009, 13:54
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. Answer : E
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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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17 Nov 2009, 14:16
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. Answer : C
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17 Nov 2009, 14:43
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. Answer : D
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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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Updated on: 18 Nov 2009, 01:07
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. Answer : C
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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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Re: Inequality and absolute value questions from my collection
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17 Nov 2009, 15:04
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. 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, 08:47
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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Re: Inequality and absolute value questions from my collection
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18 Nov 2009, 08:55
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^22xy+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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Re: Inequality and absolute value questions from my collection
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18 Nov 2009, 09:17
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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