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Inequality and absolute value questions from my collection [#permalink]
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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: http://gmatclub.com/forum/inequalityan ... ml#p6536902. If y is an integer and \(y = x + x\), is \(y = 0\)? (1) \(x < 0\) (2) \(y < 1\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6536953. Is \(x^2 + y^2 > 4a\)?(1) \((x + y)^2 = 9a\) (2) \((x – y)^2 = a\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6536974. Are x and y both positive?(1) \(2x2y=1\) (2) \(\frac{x}{y}>1\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537095. What is the value of y?(1) \(3x^2 4 = y  2\) (2) \(3  y = 11\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537316. If x and y are integer, is y > 0? (1) \(x +1 > 0\) (2) \(xy > 0\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537407. \(x+2=y+2\) what is the value of x+y?(1) \(xy<0\) (2) \(x>2\), \(y<2\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p653783 AND http://gmatclub.com/forum/inequalityan ... l#p11117478. \(a*b \neq 0\). Is \(\frac{a}{b}=\frac{a}{b}\)?(1) \(a*b=a*b\) (2) \(\frac{a}{b}=\frac{a}{b}\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p6537899. Is n<0?(1) \(n=n\) (2) \(n^2=16\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65379210. If n is not equal to 0, is n < 4 ?(1) \(n^2 > 16\) (2) \(\frac{1}{n} > n\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65379611. Is \(x+y>xy\)?(1) \(x > y\) (2) \(xy < x\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65385312. Is r=s?(1) \(s \leq r \leq s\) (2) \(r \geq s\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p65387013. Is \(x1 < 1\)?(1) \((x1)^2 \leq 1\) (2) \(x^2  1 > 0\) Solution: http://gmatclub.com/forum/inequalityan ... ml#p653886Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.PLEASE READ THE WHOLE DISCUSSION BEFORE POSTING A QUESTION.
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23 Oct 2014, 01:21
Bunuel wrote: 10. If n is not equal to 0, is n < 4 ? (1) n^2 > 16 (2) 1/n > n
Question basically asks is 4<n<4 true.
(1) n^2>16 > n>4 or n<4, the answer to the question is NO. Sufficient.
(2) 1/n > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.
Answer: A. hi m a little confused here. How is A sufficient isn't n^2>16 > n^216>0 > (N^24^2) >0 > n+4>0 or n4>0 > n> 4 or n>4.....how did u get n<4.... kindly correct me if m wrong



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23 Oct 2014, 01:23



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11 Nov 2014, 07:04
Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=0 > 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.
(2) x^2  1 > 0 > x<1 or x>1. Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.
Answer: E. I was hoping if we could compare the above question (in terms of asked ranges and the range that we calculate) with the below one:Bunuel wrote: earnit wrote: Thoughtosphere wrote: If x is not equal to 0, is x less than 1?
(1) x/x< x
(2) x > x
Will really appreciate if answer is supported by explanation.
1  x / x < x This would mean that 1 < x < 0 & 0 < x <infinite Thus 1 is insufficient
2  x > x  This would mean that negative infinite values < x < 0 This is also not sufficient.
Combining 1 & 2  the overlapping region is 1 < x < 0
We can conclude that x < 1
So C Solving the Question is not much of a problem as deciding whether our solved range: \(1 < x < 0\) is SUFFICIENT to answer the asked range: \(1 < x < 1\) I need to understand that: Is our answer sufficient to conclude that YES x lies between (1,1) even though our solution was that x lies between (1,0) ? The question asks whether 1 < x< 1. We got that 1 < x < 0. So, the answer is YES: x is definitely from the range (1, 1). WHAT IS THE UNDERLYING REASON FOR DISCARDING A GIVEN RANGE IN ONE QUESTION AND A GIVEN RANGE IN OTHER QUESTION? Seems to be a Classic case of DS.



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13 Nov 2014, 06:21
Marco83 wrote: 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 stmt1> 2x2y = 1 xy=1/2 (1)(1/2)=1/2 {both x and y positive} (1/2)(1)=1/2 {both x and y negative} Not Sufficient stmt2> x/y >1 if this is true then both x and y has to be of same sign. Not Sufficient combining 1 n 2: condition x/y > 1 will eliminate the (1/2)(1)=1/2 {both x and y negative values} as (1/2)/(1) is not greater than 1 thus we left with (1)(1/2)=1/2 {both x and y positive} hence [C]
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09 Jan 2015, 14:34
Hi Bunuel,
Can you please explain the Highlighted part in your solution below.
Many Thanks
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=0. 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 nonnegative only. Hence if xy is negative 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.



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11 Jan 2015, 11:35
bhatiavai wrote: Hi Bunuel,
Can you please explain the Highlighted part in your solution below.
Many Thanks
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=0. 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 nonnegative only. Hence if xy is negative 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. Easier approach is given here: inequalityandabsolutevaluequestionsfrommycollection86939160.html#p1111747Hope it helps.
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09 Feb 2015, 00:21
4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
S1 2(xy)=1, xy=1/2 X could be +ve and y ve. Or the obvious case: x and y could both be +ve. AD S2 x and y have same signs. Both could be ve and still have a fraction bigger than 1. B
S1+S2 X and Y cant both be ve, because, though it would comply with the 2nd statements restrictions, it wouldn't with the 1st. i.e, inserted in S1's equation, the RHS would be a ve number not 1/2. So only left option is that both be +ve. E
C it is then
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09 Feb 2015, 08:14
Bunuel wrote: 9. Is n<0? (1) n=n (2) n^2=16
Answer: C. Great question. I looked at S1 and as usual thought of special numbers, but then proceeding with 0=0 did not make sense to me because I thought "0 is neither +ve nor ve so why negate it and then find absolute value". Is there some logic i'm missing or is it just a 'possible calculation'? Thank you



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09 Feb 2015, 10:15
Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=0 > 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.
(2) x^2  1 > 0 > x<1 or x>1. Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.
Answer: E. About S1: can i say x1<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me.



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10 Feb 2015, 06:00
deeuk wrote: Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=0 > 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.
(2) x^2  1 > 0 > x<1 or x>1. Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.
Answer: E. About S1: can i say x1<=1 after rooting the given and putting it in modulus? I recall reading about squaring being used to remove modulus, so I thought to reverse that. Please correct me. Yes, that;s correct. We CAN take the square root from both sides of (x1)^2 <= 1 because both sides are nonnegative. For more on inequalities check here: inequalitiestipsandhints175001.htmlHope it helps.
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29 May 2015, 09:51
Bunuel wrote: 9. Is n<0? (1) n=n (2) n^2=16
(1) n=n, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.
(2) n^2=16 > n=4 or n=4. Not sufficient.
(1)+(2) n is negative OR n equals to zero from (1), n is 4 or 4 from (2). > n=4, hence it's negative, sufficient.
Answer: C. Bunuel Cant get thru stmt 1. Opening the mod we get either n=n or n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.?



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30 May 2015, 04:15
Bunuel wrote: sinhap07 wrote: Bunuel wrote: 9. Is n<0? (1) n=n (2) n^2=16
(1) n=n, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.
(2) n^2=16 > n=4 or n=4. Not sufficient.
(1)+(2) n is negative OR n equals to zero from (1), n is 4 or 4 from (2). > n=4, hence it's negative, sufficient.
Answer: C. Bunuel Cant get thru stmt 1. Opening the mod we get either n=n or n=n. First case gives null so am left with n=0 and hence sufficient. Where am I wrong in opening up the mod.? First of all n = n, so we are given that n = n, which implies that n <= 0. How did we arrive at n = n?



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lagomez wrote: Bunuel wrote: 1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
This one was quite tricky and was solved incorrectly by all of you.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. would you think this is a 700+ question? Hi Bunuel, Can you please comment if my approach is correct? Given: 6*x*y = x^2*y + 9*y, which is reduced to y*(x3)^2 = 0.......(1) Statement [1] yx=3 or y = x+3 sub'ing y in (1) (x+3)(x3)(x3) = 0 (x^29)(x3)=0 so, x^2 = 9 or x = +/3 and x=3 therefore, y = 6 or 0 for x = 3 and 3 respectively..........Not sufficient Statement [2]....says x < 0, so (x3)^2 will never be = 0, y has to be it.....therefore xy=0 so sufficient! So my question to you is whether I handled [1] properly? and whether substituting y into the original problem equation and arriving at x = +/ 3 is correct? So far, I've learnt that Sqrt(x) will only have +x as solution as GMAT doesn;t consider ve solution, but x^2 =25 will have +/5 as solution so we should consider both solutions for x^2 = 25 in our analysis.....Did I get this right? Please opine. Thanks!



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01 Jun 2015, 01:59
rohitd80 wrote: lagomez wrote: Bunuel wrote: 1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
This one was quite tricky and was solved incorrectly by all of you.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.
Never multiply (or reduce) inequality by variable (or expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. would you think this is a 700+ question? Hi Bunuel, Can you please comment if my approach is correct? Given: 6*x*y = x^2*y + 9*y, which is reduced to y*(x3)^2 = 0.......(1) Statement [1] yx=3 or y = x+3 sub'ing y in (1) (x+3)(x3)(x3) = 0 (x^29)(x3)=0 so, x^2 = 9 or x = +/3 and x=3 therefore, y = 6 or 0 for x = 3 and 3 respectively..........Not sufficient Statement [2]....says x < 0, so (x3)^2 will never be = 0, y has to be it.....therefore xy=0 so sufficient! So my question to you is whether I handled [1] properly? and whether substituting y into the original problem equation and arriving at x = +/ 3 is correct? So far, I've learnt that Sqrt(x) will only have +x as solution as GMAT doesn;t consider ve solution, but x^2 =25 will have +/5 as solution so we should consider both solutions for x^2 = 25 in our analysis.....Did I get this right? Please opine. Thanks! Yes, everything tis correct. When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is: \(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2; Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and 3.
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01 Jun 2015, 02:27
Bunuel wrote: 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. Bunuel, Thanks for keeping me honest After adding [1] + [2] to get x^2 + y^2 = 5a, I jumped off the cliff with Answer = C If the question has stated explicitly that x,y,a are Non zero....the answer would have been C ? Many thanks!



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01 Jun 2015, 02:43
rohitd80 wrote: Bunuel wrote: 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. Bunuel, Thanks for keeping me honest After adding [1] + [2] to get x^2 + y^2 = 5a, I jumped off the cliff with Answer = C If the question has stated explicitly that x,y,a are Non zero....the answer would have been C ? Many thanks! ___________________ Yes.
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Re: Inequality and absolute value questions from my collection [#permalink]
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01 Jun 2015, 20:13
Bunuel wrote: 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. Hi Bruel, For (1): Because (x + y)^2 <= (1^2+1^2)(x^2+y^2) => x^2+y^2 >= 4.5a > 4a. So A is sufficient? Could you help me find out whether there is any mistake in my solution? Thanks Bruel,




Re: Inequality and absolute value questions from my collection
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01 Jun 2015, 20:13



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