Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Inequality and absolute value questions from my collection [#permalink]
16 Nov 2009, 10:33
89
This post received KUDOS
Expert's post
296
This post was BOOKMARKED
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]
01 Dec 2009, 11:57
2
This post received KUDOS
Expert's post
kaptain wrote:
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!
Thanks
This approach is not correct not because we are not told that y is an integer, but because you can not add inequalities like you did.
3(x^2 -4) = y - 2 OR 3(4-x^2) = y - 2, in fact these equation are derived from one and from them only one is right. It's not that we have 3(x^2 -4) = y - 2 AND 3(4-x^2) = y - 2 and we are asked to solve fro unknowns. If it were then your solution would be right.
Re: Inequality and absolute value questions from my collection [#permalink]
02 Dec 2009, 01:05
2
This post received KUDOS
Expert's post
kaptain wrote:
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
Sure when I said that you can not add this way I meant: in this case. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
09 Aug 2012, 16:21
2
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
jayaddula wrote:
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=0. B. x+2=-y-2 --> 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 vise-versa 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.
Hi Bunuel,
I am getting E and just cannot understand D. Please see my solution below - I used number picking.
A. xy<0, x=+ and y=- For this condition choosing different values of x and y (x=2,y=-6: x=3, y=-7)satisfies the given condition in modulus. Hence x=y can be different value or x=- and y=+ - This condition doesn't satisfy the modulus condiotion
B- x>2 and y<2 - As per the above stmt 1 - condition 1, there can be various values for x and y, hence x+y is different.
Hence E. I know I am going wrong some where, please help.
thanks jay
In your example, both pairs give the same value for x+y: 2-6=-4 and 3-7=-4.
We can solve this question in another way:
7. |x+2|=|y+2| what is the value of x+y?
Square both sides: \(x^2+4x+4=y^2+4y+4\) --> \(x^2-y^2+4x-4y=0\) --> \((x+y)(x-y)+4(x-y)=0\) --> \((x-y)(x+y+4)=0\) --> either \(x=y\) or \(x+y=-4\).
(1) xy<0 --> the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=-4\). Sufficient.
(2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=-4\). Sufficient.
Re: Inequality and absolute value questions from my collection [#permalink]
28 Aug 2015, 12:29
2
This post received KUDOS
SauravPathak27 wrote:
Bunuel wrote:
12. Is r=s? (1) -s<=r<=s (2) |r|>=s
This one is tough.
(1) -s<=r<=s, we can conclude two things from this statement: A. s is either positive or zero, as -s<=s; B. r is in the range (-s,s) inclusive, meaning that r can be -s as well as s. But we don't know whether r=s or not. Not sufficient.
(2) |r|>=s, clearly insufficient.
(1)+(2) -s<=r<=s, s is not negative, |r|>=s --> r>=s or r<=-s. This doesn't imply that r=s, from this r can be -s as well. Consider: s=5, r=5 --> -5<=5<=5 |5|>=5 s=5, r=-5 --> -5<=-5<=5 |-5|>=5 Both statements are true with these values. Hence insufficient.
Answer: E.
Can we eliminate B on the basis that it is just a reworded form of Statement A.
My understanding is |x| <1 means -1<x<1. Similarly, -s>=r>=s, which is not possible.
your understanding that |x| <1 means -1<x<1 is correct.
(1) is telling us that \(r\) falls between \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)...... INSUFFICIENT (2) is telling us that \(r\) falls outside \(-s\) and \(s\) INCLUSIVE of \(-s\) and \(s\)........ INSUFFICIENT (1) & (2) together tells us that \(r\) must be equal to either \(-s\) or \(s\) but cannot determine which one........ INSUFFICIENT
Hopefully my freehand below makes it a little clearer.
If it helps.... Throw me a Kudos
Attachments
IMG_20150829_22738.jpg [ 43.72 KiB | Viewed 336 times ]
_________________
If you found my post useful, please consider throwing me a Kudos... Every bit helps
Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 02:15
1
This post received KUDOS
Expert's post
gmat620 wrote:
Bunuel, thanks for the questions. Please provide the OA's too. It would be great if you can provide them soon. I am having my GMAT this week, so kinda tensed and impatient. Also, I am yet to give my MGMAT CAT's, so tell me whether should I solve the questions on the forum because if the questions are from the MGMAT CAT's or Gmat Prep then it may overestimate my result. I would appreciate your response. Thanks once again.
These questions are from various sources. Couple of questions might be from MGMAT CAT or Gmat Prep, but not more than that.
I'll provide OA in a day or two, after discussions. Tell me if you want the answers for the specific questions earlier than that and I'll mail you. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:07
1
This post received KUDOS
Expert's post
Marco83 wrote:
4) I don’t get the two clues; they seem to be mutually exclusive
Yes there was a typo in 4. Edited. Great job Marco83. Even though not every answer is correct, you definitely know how to deal with this kind of problems. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 09:34
1
This post received KUDOS
Bunuel wrote:
5. What is the value of y? (1) 3|x^2 -4| = y - 2 (2) |3 - y| = 11
Statement 1:
Two equations, two unknowns... INSUFFICIENT
Statement 2:
|3 - y| = 11 (3-y)=11 or (3-y)=-11 y=-8, 14
INSUFFICIENT
Statements 1 and 2:
y must be 14 because 3|x^2 -4| can never be a negative value (no matter what you plug in for x, you will get a positve value because of the absolute value signs).
SUFFICIENT
ANSWER: C.
Last edited by h2polo on 17 Nov 2009, 09:54, edited 1 time in total.
Re: Inequality and absolute value questions from my collection [#permalink]
17 Nov 2009, 13:43
1
This post received KUDOS
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.
Re: Inequality and absolute value questions from my collection [#permalink]
18 Nov 2009, 16:19
1
This post received KUDOS
2
This post was BOOKMARKED
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 \(6*x*y = x^2*y + 9*y\):
\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). 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) \(y-x=3\). If y is not 0, x must be 3 and y-x 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.
Re: Inequality and absolute value questions from my collection [#permalink]
22 Dec 2009, 13:09
1
This post received KUDOS
Bunuel wrote:
4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1
(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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 time-consuming and comfortable for you personally.
One of the approaches: \(2x-2y=1\) --> \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) --> \(\frac{x-y}{y}>0\) --> substitute x --> \(\frac{1}{y}>0\) --> \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.
Answer: C.
1 and 2 are insuff for the above mentioned (by Bunuel) reasons, now taking 1 and 2 together, here's another approach: 2x-2y=1 2(x-y)=1 (x-y)=1/2
y(x/y -1)=1/2 now we know that from option 2 x/y>1
so, y(x/y -1)=1/2====> y[(a value >1) - 1]=1/2 so, [y*(+ve value)]=1/2 hence y= +ve
now since y=+ve and (x/y)>1, we have x=+ve therefore, option c. _________________
Re: Inequality and absolute value questions from my collection [#permalink]
21 May 2010, 02:06
1
This post received KUDOS
Expert's post
ManishS wrote:
I am a little confused on this one . Can the answer be E??
From A: 2x-2y=1 => x-y= 0.5 INSF
From B x/y > 1 => x > y INSF
From A & B x-y =0.5 and x > y
If x = -0.5 and y = -1 then x > y and x - y = (-0.5) - (-1) = -0.5 + 1 = 0.5 Hence both x and y can be negative
If x= 1 and y = 0.5 then x > y and x- y = 1 -0.5 = 0.5 Hence both x and y can be positive
Ans = E ??
Problem with your solution is that the red part is not correct.
\(\frac{x}{y}>1\) does not mean that \(x>y\). If both x and y are positive, then \(x>y\), BUT if both are negative, then \(x<y\).
From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative).
When we consider two statement together:
From (1): \(2x-2y=1\) --> \(x=y+\frac{1}{2}\)
From (2): \(\frac{x}{y}>1\) --> \(\frac{x}{y}-1>0\) --> \(\frac{x-y}{y}>0\) --> substitute \(x\) from (1) --> \(\frac{y+\frac{1}{2}-y}{y}>0\)--> \(\frac{1}{2y}>0\) (we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) --> \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive.
OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too.
Re: Inequality and absolute value questions from my collection [#permalink]
23 Feb 2012, 11:00
1
This post received KUDOS
Expert's post
shankar245 wrote:
Quote:
4. Are x and y both positive? (1) 2x-2y=1 (2) x/y>1
(1) 2x-2y=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. 2x-2y=1 --> y=x-1/2, 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.
In the first statement from 2x-2y=1 --> we can sat x-y=1/2
So it cud be 8.5-8 or 0.25 - (-0.25) HOw can we say both x and y are positive?
similarily statement 2
x/y>1 =>x>y
how can we be sure x and y have the same sign we can have 8>7 or 8>-8
Bunel can you pls xplain..or am i missing sumtin fundamental?
For (1) we are not saying that x and y are positive, we saying that if they were both positive then it would mean that the line y=x-1/2 is only in I quadrant, which is not possible. So from (1) we cannot say that x and y are both positive.
Or consider following numbers: x=2, y=1.5 and x=-2, y=-2.5.
For (2). First of all you can not multiply x/y>1 by y and write x>y, since you don't know the sign of y: if it's positive then x>y but if it's negative then when multiplying by negative number you should flip the sing of the inequality and write x<y. Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.
Next, since x/y>1 then x/y>0, which means that x and y are either both positive or both negative, hence they have the same sign.
Is |x-1| < 1? Basically the question asks is 0<x<2 true?
(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=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.
Can you please explain : --> how did u derive --> "x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive" ???
Here is what i did
Question: |x-1| < 1 critical point x>1 or x<1
when x>1 then (x-1)<1 x<2 when x<1 then -(x-1)<1 -x<0 therefore x>0 to prove 0<x<2 --? [understood this]
(A)(x-1)^2 <= 1 x^2 - 2x + 1 <= 1 x (x-2) <= 0 [can you please explain after this who did you derive 0<=x<=2 ===?????]
my thoughts [what m i doing wrong??] when (x = 0) then x-2<=0 therefoe x<=2 when (x-2 = 0) then x<=0 ????? i am confused here ?????
(B) x^2-1>0 therefore we have x>1 or x<-1 - not sufficient [understood this]
Re: Inequality and absolute value questions from my collection [#permalink]
25 Aug 2012, 06:26
1
This post received KUDOS
Bunuel wrote:
SOLUTIONS:
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 \(6*x*y = x^2*y + 9*y\):
\(y*(x^2-6x+9)=0\) --> \(y*(x-3)^2=0\). 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) \(y-x=3\). If y is not 0, x must be 3 and y-x 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.
OR:
\(y-x=3\) --> \(x=y-3\) --> \(y*(x-3)^2=y*(y-3-3)^2=y(y-6)^2=0\) --> either \(y=0\) or \(y=6\) --> if \(y=0\), then \(x=-3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. 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.
The answer to this one is C right? B alone is not sufficient. _________________
Encourage me by pressing the KUDOS if you find my post to be helpful.
Re: Inequality and absolute value questions from my collection [#permalink]
04 Oct 2012, 03:00
1
This post received KUDOS
Expert's post
carcass wrote:
Bunuel I have very clear the 1 statement but not so much the secon one : 1/|n|< n this is true only for negative value. So we could have i. e. : -1 or -6 so insuff but how we you arrive to this conclusion:
Given: 1/|n| > n. Now, 1/|n| is always positive (because of the absolute value in the denominator), so if n is negative then this inequality will always hold true: 1/|n|=positive>n=negative.
Hope it's clear.
P.S. Complete solution for 1/|n| > n is n<0 or 0<n<1. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...