GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Aug 2018, 04:33

### GMAT Club Daily Prep

#### 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

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

# Inequality and absolute value questions from my collection

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

05 Apr 2018, 12:24
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.

Hi Bunuel, I solved it as below
(1) 2x-2y=1 -> x-y= 1/2. This means x>y by 1/2 but x can be 1/2 and y=0 so this is insufficient
(2) x/y=1 -> x>y which is no different from (1) so insufficient again

But I didn't follow your approach on combining the 2 statements, how did you get x to be substituted by 1/y? The fact that x is positive is already proved by (1) isn't it?

The red parts are 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 $$x<y$$. What you are actually doing when writing $$x>y$$ from $$\frac{x}{y}>1$$ is multiplying both parts of inequality by $$y$$: never multiply (or reduce) an inequality by variable (or by an 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. So from (2) $$\frac{x}{y}>1$$, we can only deduce that $$x$$ and $$y$$ have the same sigh (either both positive or both negative).

Also, from (1) we can only say that x > y (because x = y + 1/2 = y + positive number) but we cannot say that x is positive. For example, consider x = -2 and y = -2.5.

Finally, when we substitute $$x = y + \frac{1}{2}$$ into $$\frac{x}{y}>1$$ we'll get:

$$\frac{y + \frac{1}{2}}{y}>1$$;

$$1 + \frac{1}{2y}>1$$;

$$\frac{1}{2y}>0$$;

$$y > 0$$.

If $$y > 0$$, then $$x > y > 0$$.

Hope it helps.
_________________
Intern
Joined: 09 Dec 2013
Posts: 29
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

17 Apr 2018, 23:32
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

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 cannot say for sure that 0<x<2 is true. Not sufficient.

Still not clear on this one.
Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient?
DS Forum Moderator
Joined: 27 Oct 2017
Posts: 675
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

18 Apr 2018, 01:15
Is |x-1| <1?
Since Absolute value function is always non negative, we can square both sides,

We get is (x-1)^2<1?

Statement 1 is (x-1)^2< = 1

If (x-1)^2<1: answer to question is yes

(x-1)^2= 1: hence answer is No,

So Statement 1 is NOT SUFFICIENT.

Statement2: Question stem (x-1)^2<1?

Can be reduced to is x(x-2)<0

Or is 0<x<2

Now St2: x^2-1>0 gives x >1 or x <-1

Now x can be -2, answer to question stem is No

Or 1.5 answer to question stem is yes.

Hence NOT SUFFICIENT.

Combining both Statement 1 & 2, we get

X can be 2, answer to question stem- NO

Or 1.5 answer to question stem is yes.

Buttercup3 wrote:
Bunuel wrote:
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

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 cannot say for sure that 0<x<2 is true. Not sufficient.

Still not clear on this one.
Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient?

_________________
Manager
Joined: 30 Jul 2014
Posts: 151
GPA: 3.72
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

27 Jun 2018, 07:14
Bunuel wrote:
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.

I forgot the instruction by the time, I started assessing option B. Y is an integer.
_________________

A lot needs to be learned from all of you.

Manager
Joined: 30 Jul 2014
Posts: 151
GPA: 3.72
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

27 Jun 2018, 07:33
Bunuel wrote:
5. What is the value of y?
(1) 3|x^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 --> 3-y=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 have done a silly mistake while solving this problem, in my mind I assumed that the statement B is an inequality rather than equal sign.
_________________

A lot needs to be learned from all of you.

Intern
Joined: 02 Nov 2017
Posts: 22
Location: India
Concentration: General Management, Finance
Schools: ISB
GMAT 1: 690 Q50 V35
GPA: 3.31
WE: General Management (Energy and Utilities)
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

12 Jul 2018, 00:47
sriharimurthy wrote:
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

$$6*x*y = x^2*y + 9*y$$
$$6*x = x^2 + 9 = 0$$
$$x^2 - 6*x + 9 = 0$$
$$(x-3)^2 = 0$$
$$x = 3$$

St. (1) : y - x = 3
y = 6
Sufficient.

St. (2) : x^3 < 0
Invalid statement. Does not give us value of y.
Insufficient.

solving the equation gives either y=0 or x=3
(1) y=x+3
so if y=0, x=-3, xy=0
if x=3 y =6, xy=18
insufficient
(2) x^3<0
means x<0, so y=0
and xy=0
sufficient (B)
Manager
Joined: 16 Jan 2018
Posts: 60
Location: New Zealand
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

15 Jul 2018, 03:00
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.

Hi Bunnuel, I didnt quite understand the part where you said "B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less than -2 and the other is more than -2."

why can't x & y be -2 & -2 here?
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

15 Jul 2018, 06:56
blueviper 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.

Hi Bunnuel, I didnt quite understand the part where you said "B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less than -2 and the other is more than -2."

why can't x & y be -2 & -2 here?

You might find the following solution simpler: https://gmatclub.com/forum/inequality-a ... l#p1111747

Hope it helps.
_________________
Intern
Joined: 09 Sep 2017
Posts: 1
Location: India
WE: Information Technology (Consulting)
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

31 Jul 2018, 14:56
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Can someone clarify whether |a/b|=|a|\|b| exists as a property?
Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re: Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

31 Jul 2018, 20:49
arunjohn43 wrote:
Bunuel wrote:
8. a*b#0. Is |a|/|b|=a/b?
(1) |a*b|=a*b
(2) |a|/|b|=|a/b|

|a|/|b|=a/b is true if and only a and b have the same sign, meaning a/b is positive.

(1) |a*b|=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence |a|/|b|=a/b. Sufficient.

(2) |a|/|b|=|a/b|, from this we can not conclude whether they have the same sign or not. Not sufficient.

Can someone clarify whether |a/b|=|a|\|b| exists as a property?

Yes, $$|\frac{a}{b}|=\frac{|a|}{|b|}$$ and $$|ab|=|a|*|b|$$ are generally true.
_________________
Intern
Joined: 12 Sep 2016
Posts: 4
Inequality and absolute value questions from my collection  [#permalink]

### Show Tags

17 Aug 2018, 23:00
In case anybody struggles with #4.
I found it easier to think about (1+2) as 2 cases:
Stmt 1:
Step 1) 2x-2y =1
Step 2) 2(x-y) = 1
Step 3) x-y = 1/2
At this point its easy to think of cases when either x or y are negative or both are positive, for example x = 0 y = (-1/2) or x = 1 and y = 1/2, Hence, NS - BCE

Stmt 2: x/y > 1
From this statement it should be clear that both x and y are either negative or positive. NS - CE

Now Stmt 1 and 2 combined:
consider 2 cases from stmt 2 split:
case 1 - both positive:
x/y > 1 => (multiply both sides by "y", since "y" > 0 we keep the sign)
x > y, we can still satisfy the condition from stmt 1. Example x = 1 and y = 1/2

case 2 - both negative:
x/y > 1 => (multiply both sides by "y", since "y" < 0 we flip the sign)
x < y, no we can no longer satisfy the condition from stmt 1. x - y = 1/2.
(x - y) has to be positive. We cannot subtract something bigger (y) from something smaller (x) and get positive number. Hence, case 2 does not work. (1+2) combined Sufficient - Answer is C
Inequality and absolute value questions from my collection &nbs [#permalink] 17 Aug 2018, 23:00

Go to page   Previous    1   2   3   4   5   6   7   8   9   10   11   [ 212 posts ]

Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.