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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Target question: Are x and y both positive?

Statement 1: 2x - 2y = 1
There are several pairs of numbers that satisfy this condition. Here are two:
Case a: x = 1 and y = 0.5, in which case x and y are both positive
Case b: x = -0.5 and y = -1, in which case x and y are not both positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x/y > 1
This tells us that x/y is positive. This means that either x and y are both positive or x and y are both negative. Here are two possible cases:
Case a: x = 4 and y = 2, in which case x and y are both positive
Case b: x = -4 and y = -2, in which case x and y are not both positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2
Statement 1 tells us that 2x - 2y = 1.
Divide both sides by 2 to get: x - y = 1/2
Solve for x to get x = y + 1/2

Now take the statement 2 inequality (x/y > 1) and replace x with y + 1/2 to get:
(y + 1/2)/y > 1
Rewrite as: y/y + (1/2)/y > 1
Simplify: 1 + 1/(2y) > 1
Subtract 1 from both sides: 1/(2y) > 0
If 1/(2y) is positive, then y must be positive.

Statement 2 tells us that either x and y are both positive or x and y are both negative.
Now that we know that y is positive, it must be the case that x and y are both positive
Since we can now answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Solution:

We need to determine whether x and y are both positive.

Statement One Alone:

2x – 2y = 1

Simplifying statement one we have:

2(x – y) = 1

x – y = ½

The information in statement one is not sufficient to determine whether x and y are both positive. For instance if x = 1 and y = ½, x and y are both positive; however if x = -1/2 and y = -1, x and y are not both positive. We can eliminate answer choices A and D.

Statement Two Alone:

x/y > 1

Using the information in statement two, we see that x and y can both be positive or both be negative. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using the information in statements one and two we know that x – y = ½ and that x/y > 1. Isolating x in the equation we have: x = ½ + y. We can now substitute ½ + y for x in the inequality x/y > 1 and we have:

(1/2 + y)/y > 1

(1/2)/y + y/y > 1

1/2y + 1 > 1

1/2y > 0

Thus, y must be greater than zero. Also, since x/y is greater than one, x also must be greater than zero.

Answer: C
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Video solution from Quant Reasoning starts at 17:28
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

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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