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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: inequalityandabsolutevaluequestionsfrommycollection8693920.html#p6536902. If y is an integer and \(y = x + x\), is \(y = 0\)? (1) \(x < 0\) (2) \(y < 1\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693920.html#p6536953. Is \(x^2 + y^2 > 4a\)?(1) \((x + y)^2 = 9a\) (2) \((x – y)^2 = a\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p6536974. Are x and y both positive?(1) \(2x2y=1\) (2) \(\frac{x}{y}>1\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p6537095. What is the value of y?(1) \(3x^2 4 = y  2\) (2) \(3  y = 11\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p6537316. If x and y are integer, is y > 0? (1) \(x +1 > 0\) (2) \(xy > 0\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p6537407. \(x+2=y+2\) what is the value of x+y?(1) \(xy<0\) (2) \(x>2\), \(y<2\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p653783 AND inequalityandabsolutevaluequestionsfrommycollection86939160.html#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: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p6537899. Is n<0?(1) \(n=n\) (2) \(n^2=16\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p65379210. If n is not equal to 0, is n < 4 ?(1) \(n^2 > 16\) (2) \(\frac{1}{n} > n\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p65379611. Is \(x+y>xy\)?(1) \(x > y\) (2) \(xy < x\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p65385312. Is r=s?(1) \(s \leq r \leq s\) (2) \(r \geq s\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p65387013. Is \(x1 < 1\)?(1) \((x1)^2 \leq 1\) (2) \(x^2  1 > 0\) Solution: inequalityandabsolutevaluequestionsfrommycollection8693940.html#p653886Official answers (OA's) and detailed solutions are in my posts on pages 2 and 3.
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Re: Inequality and absolute value questions from my collection [#permalink]
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18 Aug 2013, 10:58
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0 can we do like this....
stmnt 1 when y=0 then xy=0, well when y not =0 then we have various values of xy as y vary >>>> so insuffiecent
stmnt 2 says x<0 or negative so 6xy=y(x^2 + 9) let take 2 cases x<0 , y<0 and x<0 and y>0 ( )*(  )= ( )* (+) this can`t be true when x<0 , y<0 the equation does`nt hold
() *(+)=(+)*(+) x<0 and y>0 even now the equation does`nt hold true
so the only possible case is x<0 and y=0 only then the equation will remain valid and therefore xy=0 and hence sufficient....... B it is
pleaze point out if i`m wrong in my approach.... thanks



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19 Aug 2013, 01:48



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Re: Inequality and absolute value questions from my collection [#permalink]
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23 Aug 2013, 10:27
11. Is x+y>xy? (1) x > y (2) xy < x
x+y>xy <=> x^2 + 2xy + y^2 > x^2  2xy + y^2 <=> xy >0 1. Insufficient. 2. xy < x <=> x^2  2xy + y^2 < x^2 <=> y(y2x) <0 <=> 0<y<2x (if 2x>0) or 2x<y<0 (if 2x<0) <=> xy >0, Sufficient Pick B



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Re: Inequality and absolute value questions from my collection [#permalink]
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23 Aug 2013, 12:59
Hi Bunuel, sorry to disturb you.
In problem number 3:
3. Is x^2 + y^2 > 4a? (1) (x + y)^2 = 9a (2) (x – y)^2 = a
You've given answer E. But I remember, correct me if I am wrong, that x^2 + y^2 is always >= to 2xy. It this is true, the answer should be A, because the first statement would be sufficient. x^2 + y^2 would be >= 4,5a
Thanks in advance.



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Re: Inequality and absolute value questions from my collection [#permalink]
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25 Aug 2013, 07:20



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Re: Inequality and absolute value questions from my collection [#permalink]
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08 Sep 2013, 04:33
I agree with what you said. But I dont with the interpretation.
In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?



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Re: Inequality and absolute value questions from my collection [#permalink]
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08 Sep 2013, 05:25
ygdrasil24 wrote: I agree with what you said. But I dont with the interpretation.
In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did? Which question are you talking about?
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Re: Inequality and absolute value questions from my collection [#permalink]
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08 Sep 2013, 05:35
Bunuel wrote: ygdrasil24 wrote: I agree with what you said. But I dont with the interpretation.
In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did? Which question are you talking about? Q11 : inequalityandabsolutevaluequestionsfrommycollection8693940.html#p653853



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Re: Inequality and absolute value questions from my collection [#permalink]
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08 Sep 2013, 05:38
ygdrasil24 wrote: Bunuel wrote: ygdrasil24 wrote: I agree with what you said. But I dont with the interpretation.
In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did? Which question are you talking about? Q11 : inequalityandabsolutevaluequestionsfrommycollection8693940.html#p653853In that case I don't understand what you mean. Please elaborate.
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Re: Inequality and absolute value questions from my collection [#permalink]
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08 Sep 2013, 05:44
In B, you showed how x and y should have same sign, for if not, inequality in B doesnt hold. Similarily we see for inequality in A to be true, x,y should be of same sign. I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. What wrong I did?[/quote] Which question are you talking about?[/quote] Q11 : inequalityandabsolutevaluequestionsfrommycollection8693940.html#p653853[/quote] In that case I don't understand what you mean. Please elaborate.[/quote] Question asks Is x+y>xy? From analysis we know this holds good when x,y have same sign ? from B we concluded that and found x,y to have same sign.right ? For me answer should be D, because from statement I, we are sure x and y should have same sign. (x,y) = (3,2) or (2,1) . So we can answer the question . I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. So we are sure of the sign of x,y the same way we were in B. So this should be able to answer the main equality of the question. What wrong I did?[/quote]



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08 Sep 2013, 05:50
ygdrasil24 wrote: Question asks Is x+y>xy? From analysis we know this holds good when x,y have same sign ? from B we concluded that and found x,y to have same sign.right ? For me answer should be D, because from statement I, we are sure x and y should have same sign. (x,y) = (3,2) or (2,1) . So we can answer the question . I took (x,y) as ( 3,2) , (1,0), (1,0) , (2,1) . In this set, the inequality in A holds in 1st and last case, clearly , x,y should be same sign. '0' is not a positive number so 1,0 is ruled out. So we are sure of the sign of x,y the same way we were in B. So this should be able to answer the main equality of the question. What wrong I did? So, you question is why (1) is not sufficient??? (1) x > y. This does NOT mean that x and y have the same sign. Consider this: if x=2 and y=1, then x+y=1<3=xy and we have a NO answer.
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Re: Inequality and absolute value questions from my collection [#permalink]
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20 Sep 2013, 06:23
Quote: 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.
I guess the answer is E here. Because what we get from 1 is that y>=2. However, when we try to use that in (2) we are still hanging between y<3 and y>=3. Hence we have y=8 and y=14. Kindly let me know if I am missing something.



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20 Sep 2013, 06:27
gmatter0913 wrote: Quote: 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.
I guess the answer is E here. Because what we get from 1 is that y>=2. However, when we try to use that in (2) we are still hanging between y<3 and y>=3. Hence we have y=8 and y=14. Kindly let me know if I am missing something. y cannot be 8 (y cannot be lees than 2) because in this case the first statement would be violated: absolute value cannot equal to a negative number. If y=8, then 3x^2 4 = y  2 = 8  2 = 10, which cannot be true. Hope it's clear.
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Re: Inequality and absolute value questions from my collection [#permalink]
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20 Sep 2013, 06:59
Yes, it makes sense.
But, what is wrong with my approach?
My approach is as follows:
Looking at option 1
3 *x^2  4 = y  2
LHS is +ve or 0 Therefore, y>=2 Not Sufficient to answer the value of y.
Looking at option 2
3y=11 is the given information
We have two cases here. Case 1: When 3y<0 (y>3) => the given information 3y=11 becomes (3y)=11 => So, y=14 Case 2: When 3y>0 (y<3) => the given information 3y=11 becomes 3y=11 => So, y=8
This option says that if you know that y < 3 then the value of y is 8 y>3 then the value of y is 14
This is still insufficient information because we still don't know whether y>3 or y<3
Combining option 1 and 2
From option we know that y>=2. So, let us see if this helps to boil down between y>3 or y<3.
We are still not sure as Option 1 says 'y' could be >3 or <3 (y>=2)
Please help me understand where I have gone wrong??



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20 Sep 2013, 07:12
gmatter0913 wrote: Yes, it makes sense.
But, what is wrong with my approach?
My approach is as follows:
Looking at option 1
3 *x^2  4 = y  2
LHS is +ve or 0 Therefore, y>=2 Not Sufficient to answer the value of y.
Looking at option 2
3y=11 is the given information
We have two cases here. Case 1: When 3y<0 (y>3) => the given information 3y=11 becomes (3y)=11 => So, y=14 Case 2: When 3y>0 (y<3) => the given information 3y=11 becomes 3y=11 => So, y=8
This option says that if you know that y < 3 then the value of y is 8 y>3 then the value of y is 14
This is still insufficient information because we still don't know whether y>3 or y<3
Combining option 1 and 2
From option we know that y>=2. So, let us see if this helps to boil down between y>3 or y<3.
We are still not sure as Option 1 says 'y' could be >3 or <3 (y>=2)
Please help me understand where I have gone wrong?? From (1) y>=2, From (2) y=8 or y=14. Question: what does y equals to???
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20 Sep 2013, 07:26
Sorry to bother you on this silly problem, but I am still not convinced.
I am rephrasing the problem removing all the Math, so that I can focus on the part I am getting confused at.
What is the value of y? 1. y>=2 2. If y<3 then the value of is 8, if y>3 then the value of y is 14 (think we have two groups here)
The problem is I am not sure whether I should use the information from Option 1 (y>=2) to find out which group (from option 2) it belongs to or should I say y>=2 hence it has to be 14.
What is wrong in using the information to solve which group?



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20 Sep 2013, 07:42
gmatter0913 wrote: Sorry to bother you on this silly problem, but I am still not convinced.
I am rephrasing the problem removing all the Math, so that I can focus on the part I am getting confused at.
What is the value of y? 1. y>=2 2. If y<3 then the value of is 8, if y>3 then the value of y is 14 (think we have two groups here)
The problem is I am not sure whether I should use the information from Option 1 (y>=2) to find out which group (from option 2) it belongs to or should I say y>=2 hence it has to be 14.
What is wrong in using the information to solve which group? Even if we use your approach: is y<3??? NO. So, we have the second group > y=14.
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20 Sep 2013, 07:47
That's not true, right? Because Option 1 says y>=2 (y can be equal to 2 also) If I say y=2 then it belongs to 1st group (If y<3 then the value of y is 8) If I say y>3 then it belongs to 2nd group (If y>3 then the value of y is 14)



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20 Sep 2013, 08:04
That is exactly where my confusion is.. Quote: What is the value of y? 1. y>=2 2. If y<3 then the value of y is 8, if y>3 then the value of y is 14 (think we have two groups here)
When using Option 1 in Option 2, You are looking at comparing with the result of the groups (8 or 14). As y>=2, it has to be 14. I am looking at the conditions of the groups ( y< 3 or y>3). As y>=2 I cannot decide the group. Hence Option 1 and Option 2 together are not sufficient. What is wrong in this approach?




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