9. Inequalities

• Theory
  Inequalities Made Easy!
  Inequalities: Tips and hints
  Arithmetic with Inequalities
  Wavy Line Method Application - Complex Algebraic Inequalities
  Solving Quadratic Inequalities: Graphic Approach
  Inequalities - Quadratic Inequalities
  Graphic approach to problems with inequalities
  Inequalities trick
  INEQUATIONS(Inequalities) Part 1
  INEQUATIONS(Inequalities) Part 2
  
  • Questions
    Inequality and absolute value questions from my collection
    DS Questions
    PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Are x and y both positive?


(1) \(2x-2y=1\). Well this one is clearly insufficient. You can do it with number plugging OR consider the following: both x and y are positive means that point (x,y) is in the I quadrant. \(2x-2y=1\) --> \(y=x-\frac{1}{2}\). We know it's an equation of a line and basically the question asks whether this line (all (x,y) points of this line) is only in I quadrant. This line for sure passes I quadrant (for example, x = 1.5 and y = 1) but it cannot entirely be only in I quadrant, so there must be some (x, y) points whose coordinates are not both positive. Not sufficient.


(2) \(\frac{x}{y}>1\) --> 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:

From (1): \(2x-2y=1\), so \(x=y+\frac{1}{2}\)

From (2): \(\frac{x}{y}>1\);

Substitute x into (2): \(\frac{y + \frac{1}{2}}{y}>1\);

\(1 + \frac{1}{2y}>1\);

\(\frac{1}{2y}>0\);

\(y > 0\). \(y\) is positive. Thus, \(x=y+\frac{1}{2}=positive+positive=positive\). So, \(x\) is positive too. Sufficient.


Answer: C.


You can also check GRAPHIC APPROACH below.
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Are x and y both positive?


GRAPHIC APPROACH.

Notice that the question is basically asks whether the point (x, y) is in the first quadrant.


(1) \(2x - 2y = 1\). Draw line \(y=x-\frac{1}{2}\):



Not sufficient.


(2) \(\frac{x}{y} > 1\). Draw line \(\frac{x}{y}=1\). The solutions is the green region:



Not sufficient.


(1)+(2) Intersection is the portion of the blue line which lies in the first quadrant. Sufficient.

Answer: C.


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I got it correct but took approx 2.5 minutes.

stmnt 1 : insufficient by plugging numbers
stmnt 2 :

x/y >1 => not suff as bot x and y can be -ve or both +ve.

combined :

from stmnt 1 we have

x=y+1/2 => x/y = 1+1/2y => suppose x/y is 2 as x/y>1 => 2=1+1/2y = y=1/2 henc x=1

both positive.

Ans C

Am I right in my approach?

I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.

Thanks
-H

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.

Hope it helps.
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Here is my confusion.
Here is how I approached the question
1. 2x-2y=1
so x-y=.5

now x=1, y=.5
or x=1/4, y=-1/4
so can't tell

2. x/y>1
x>y so again can't tell.

Now if we combine both
still the options x=1, x=.5 is true
and so is the option x=1/4, y=-1/4 true

So can't tell hence E. I know this is not the correct answer but what am I missing?

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).
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I found this one easiest to solve by drawing a graph. Clearly 1) and 2) alone are not sufficient as discussed, so what remains to be seen is if 2) adds enough information to 1) to determine if both x and y are positive.

Drawing a quick graph of the line y=x-1/2 we find that the x-intercept of the line is (0.5,0) and the y-intercept is (0,-0.5). From this graph we can clearly see that we don't need to worry about anything in the 4th quadrant (+x/-y is not >1) or the 3rd quadrant (|x|<|y|, therefore x/y is not >1). All that is left is the 1st quadrant, in which x and y are both positive.

Sufficient.

What I have done here is this

1) 2x - 2y = 1
hence x - y = \frac{1}{2} {Dividing both side by 2}
In sufficient

2) [fraction]x > y[/fraction] Alone in sufficient

When (1) + (2) We can say that if X is greater than y than x-y will yield a positive result.

Please correct me if I am wrong

First of all: the question is "are x and Y both positive?" not whether "x-y will yield a positive result".

Next, 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 \(x<y\). 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).

See the complete solution of this problem in my previous post.

Hope it helps.
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1 is not suff, x = 0, y = -1/2

2 is not suff,x and y can be both -ve

Combining both :

x - y = 1/2

and (x - y)/y > 0

so 1/2/y > 0 => y is +ve and because x - y is +ve, x is +ve as well.

So answer is C.
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C is the answer.
Question: Is x > 0 AND y > 0?

Statement 1: 2x - 2y = 1 => 2(x - y) = 1 => x - y = 1/2
This just tells us that the difference is positive. But this can be true for cases when both x and y are positive, and when both x and y are negative.
For instance, x = 1.5, y = 1 => x - y = 0.5; also, x = -1, y = -1.5 => x - y = 0.5. Thus, INSUFFICIENT.

Statement 2: x/y > 1
This just tells us that x and y have the same sign. That is, both are positive or both are negative. INSUFFICIENT.

Combining these statements, we can use the same numbers used in Statement 1 to find out that both the cases together do not work for negative numbers.
For instance, x = -1, y = -1.5 => x - y = 0.5. However, x/y < 1. This violates statement 2.

Thus, the combination of the given statements tells us that x and y both have to be positive. => x > 0 AND y > 0. SUFFICIENT.

Statement (1): x-y = 1/2. We can have x=1,y=1/2. Can also have x=0,y=-1/2. Insufficient.
Statement (2): x/y>1. We can have x=3,y=2. Can also have x=-3,y=-2. Insufficient.

Combining both,
(y+1/2)/y > 1
=> 1/2y>0
=> y>0

Also as x/y>1, x must be>0. Sufficient.

C it is.
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Are x and y both positive?

1) 2x-2y=1
2(x-y)=1
x-y=1/2
-->3/4-1/4=1/2....YES
-->-1/4-(-3/4)=1/2...NO
INSUFFICIENT

2) x/y>1
This just means that x and y have the same sign. They're either both positive or both negative.
INSUFFICIENT

1&2)
x=1/2+y

(1/2+y)/y>1
y/2 + 1 > 1
y/2 > 0 which means that Y is greater than 0. And since both x and y have the same sign, both x and y are Positive. YES.

Answer is C.

1. x-y = 1/2
This means that the distance between x and y is 1/2 unit and that x is greater than y.
But x and y could be positive such as x=5 and y=4.5, OR
x and y could be both negative such as x=-4 and y=-4.5

INSUFFICIENT.

2. x/y > 1
This shows that x and y must be positive meaning they are either both (+) or both (-).
ex) x/y = 5/2 OR x/y = -5/-2 = 5/2 still > 1

INSUFFICIENT.

Combine.
Let x = 5 and y=9/2: 5/(9/2) = 10/9 > 1 - This means when x and y are both positive it could be a solution to x/y > 1
Let x = -4 and y=-9/2: -4/(-9/2) = 8/9 < 1 - This means when x and y are negative it could not be a solution to x/y > 1

Thus, SUFFICIENT that x and y are both positive.

Answer: C

Hello Bunuel,
Request you to please provide your comments on the doubt posted here-

Usually, whenever I see combining an inequality and equation, I substitute the value of one of the variable in the inequality and then analyze the effect.
So, going by that approach;

x-y=1/2 ---(1)
x/y>1 --(2)
Substituting the value of x in equation(2)

(y+1/2)/y>1

Lets assume that y is positive-

(y+1/2) > y

1/2>0 --This means that our assumption is true since 1/2 is greater than Zero. Hence, y > 0

Now, Lets assume that y is negative-

Now, here I'm stuck, I know that multiplying by a negative number changes the sign of the inequality.
I'm sure that the sign will be changed but what would be the resulting equation. I mean, do we need to replace y with "-y" in the whole equation. Please clarify. Which of the following would be correct then

a) y+1/2 <y
b) y+1/2 < -y
c) -y+1/2 < -y

Please help.
Thanks

Refer to the highlighted portion : Actually you don't have to take 2 cases at this point: The expression you have is : \(\frac{y+0.5}{y}>1 \to 1+\frac{0.5}{y}>1 \to \frac{1}{y}>0\)--> Hence, y>0.

As for your doubt, if y is negative, we cross-multiply it and get : \(y+0.5<y \to 0>0.5\), which is absurd.

If y is negative, then -y would be positive, and for multiplying a positive quantity, you don't need to flip signs. So , yes expression a is correct.
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St 1) 2x-2y = 1 => 2 (x-y) = 1 => x-y =1/2 => all this tells us is that x > y (could be positive or negative) == hence INSUFF

St 2) x/y > 1 => we don't know if y is (+) or (-) . So we have two cases:

if y positive, then x>y; if y negative, then x<y (again INSUFF)

Combining 1) and 2) we get x>y (from 1) ...which means y is positive (from 2)

Hence, if y is positive, and x >y, then x is also positive. SUFF!!

Hope this was reasoned properly.