Lesson: Inequalities - Part I

Comment on Inequalities - Part I

Excellent approach

Excellent approach

A ton of practice ques :) Are

A ton of practice ques :) Are inequalities a big portion of the quant section? It's a common question type.

It's is one of the test-makers' favorite topics. Be sure to master all of the concepts and strategies.

I'm a little lost here:

I'm a little lost here:

Given: (y - 3x)/(y - 2x) > 1
Rewrite numerator as: (y - 2x - x)/(y - 2x) > 1
Apply fraction property to get: (y - 2x)/(y - 2x) - x/(y - 2x) > 1 You're referring to my

You're referring to my solution here: https://gmatclub.com/forum/is-y-3x-y-2x-233053.html

I use a often-tested concept that says (a-b)/c = a/c - b/c

For example, (7 - 5)/2 = 7/2 - 5/2

In the original question, we have the fraction (y - 3x)/(y - 2x)

Notice that we can rewrite y -3x as y - 2x - x (both of these expressions are equivalent)

So, we can take the fraction (y - 3x)/(y - 2x) and rewrite it as follows:

(y - 3x)/(y - 2x) = (y - 2x - y)/(y - 2x)
= [(y - 2x) - (y)]/(y - 2x)...I just added brackets
= (y - 2x)/(y - 2x) - y/(y - 2x)...I applied fraction property
= 1 - y/(y - 2x)...I simplified first fraction

Does that help?

Cheers,
Brent

“Target question: Is x² > 1/x

“Target question: Is x² > 1/x ?   We can safely divide both sides of the inequality by x² to get: 1 > 1/x”

My question is how did you get that, meaning 1 > 1/x You're referring to my

You're referring to my solution here: http://www.beatthegmat.com/is-x-2-1-x-t296568.html

Statement 1 tells us that x² > x

When it comes to inequalities, we must be careful to consider whether certain values are NEGATIVE or POSITIVE before we divide both sides of an inequality by that value.

For example, if we divide both sides by a NEGATIVE value, then we must REVERSE the direction of the inequality symbol. Conversely, if we divide both sides by a POSITIVE value, then the direction of the inequality symbol REMAINS as it is.

Since we know that x² is POSITIVE, we can take the inequality x² > x and divide both sides by x² to get: 1 > 1/x

ASIDE: On the left side of the inequality, we have x²/x² = 1.
On the right side of the inequality, we have x/x² = 1/x

Does that help?

Cheers,
Brent

Thanks Brent!

Thanks Brent!

For some reason I got stuck on re-phrasing the main question somehow/for some reason and couldn't work it out correctly. All is good.

Need some help on this one:

Need some help on this one:

https://gmatclub.com/forum/if-x-y-z-0-is-x-135550.html That's a great question.

That's a great question.
Here's my step-by-step solution: https://gmatclub.com/forum/if-x-y-z-0-is-x-135550.html#p1937531

Cheers,
Brent

If abcd ≠ 0, is abcd < 0?

If abcd ≠ 0, is abcd < 0?

1. a/b>c/d
2. b/a>d/c
Sir this sum , I am feeling very difficult to understand.Hence Kindly help me. This is a SUPER TRICKY

This is a SUPER TRICKY question!
My full solution is here: https://gmatclub.com/forum/if-abcd-is-not-equal-to-zero-is-abcd-215779.h...

Cheers,
Brent

If a > 0 and b > 0, is a/b >

If a > 0 and b > 0, is a/b > b/a ?

(1) a = b - 2
(2) a/(4b) =1/5

will u please solve this question correct answer is D but my answer is B. What if we take terms in fractions in statement 1
HERE NOTHING IS WRITTEN THAT THE NUMBER SHOULD BE INTEGER It doesn't matter whether a and b are integers or fractions. The answer will remain D.

Once we recognize that we can rephrase the target question to be "Is b² < a²?" (as I do in my solution), we can then use the fact that a and b are both POSITIVE to rephrase the target question even more.

We'll use a nice property that says, if a and b are positive, and b² < a², then it must be true that b < a. Notice that this rule does not require a and b to be integers.

If you're certain that statement 1 is not sufficient, you must find some values of a and b that satisfy the given conditions, YET yield different/conflicting answers to the target question. If you try to do so, you will find it is impossible to find such values.

Does that help?

Cheers,
Brent

Hi Brent, I did not

Hi Brent, I did not understand your approach for statement 2. How is y - b > 0, when y could be positive or negative?
If y is negative and b is always negative, then y - b is < 0 because the equation becomes (-y)-(-b) < 0

----------------------------
If b < x < 0 and w < x < y, then which of the following MUST be true?

I. (w + b)/y < 0

II. (y – b)/b < 0

III. (b + w) – (x + y) < 0

(A) II only
(B) III only
(C) I and II only
(D) I and III only
(E) II and III only

Here's a different approach:

I. (w + b)/y < 0
We know that w and b are NEGATIVE, so (w + b) = NEGATIVE
However, we don't know whether y is NEGATIVE or POSITIVE
As such, (w + b)/y can be either POSITIVE or NEGATIVE
So, statement 1 need not be true.

II. (y – b)/b < 0
From the given information, we know that b < y. So, if we subtract b from both sides, we get y - b > 0
In other words, y-b is POSITIVE
Since we also know that b is NEGATIVE, we can see that (y – b)/b = POSITIVE/NEGATIVE = NEGATIVE
So, statement II must be true

III. (b + w) – (x + y) < 0
First rewrite the inequality as: b + w - x - y < 0
The rewrite as: (b - x) + (w - y) < 0
So, statement III can be rewritten as: (b - x) + (w - y) < 0

From the given information, we know that b < x.
If we subtract x from both sides, we get: b - x < 0
In other words, b-x is NEGATIVE

Also, from the given information, we know that w < y.
If we subtract y from both sides, we get: w - y < 0
In other words, w-y is NEGATIVE

This means that (b - x) + (w - y) = NEGATIVE + NEGATIVE = NEGATIVE
So, it must be true that (b - x) + (w - y) < 0

-------------------------------- Given: b < x < 0 and w < x < y

Let's examine parts of each inequality to see that: b < x and x < y
We can now COMBINE b < x and x < y to get: b < x < y
This means that it must be true that b < y

If we take b < y and subtract b from both sides, we get 0 < y - b
This is the same as saying y - b > 0

QUESTION: If y is negative and b is always negative, then y-b is < 0 because the equation becomes (-y)-(-b) < 0

ANSWER: It's possible for y and b to be negative AND for y-b to be positive.
For example, if y = -1 and b = -3, then y - b = (-1) - (-3) = 2 (which is positive)

Does that help?

Cheers,
Brent

Hi Brent, for this question

Hi Brent, for this question (https://gmatclub.com/forum/is-x-2-1-x-239236.html)

I rephrased the question to Is x^3 > 1? I noticed that in your answer explanation you hadn't rephrased it. I was wondering whether my logic in rephrasing it was incorrect? It's always a good idea to check whether rephrasing the target question will make your solution easier.

However, there are two issues with your solution.

First, you broke a rule regarding inequalities when you rephrased the target question.
ORIGINAL target question: Is x² > 1/x?
You took x² > 1/x and multiplied both sides by x to get: x³ > 1
The problem with this is that we don't know whether x is positive or negative.
If x is positive, then our inequality becomes x³ > 1 when we multiply both sides by x.
If x is negative, then our inequality becomes x³ < 1 when we multiply both sides by x

KEY CONCEPT: We must REVERSE the inequality symbol when we multiply or divide both sides of an inequality by a NEGATIVE value.
More on this here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

Second, even IF it were the case that we could successfully rephrase the target question as "Is x³ > 1?", this rephrasing doesn't really help us analyze the two statements:
(1) x² > x
(2) 1 > 1/x

If the statements had an x³ term, then your suggesting rephrasing (Is x³ > 1?) might be of us when we examine those statements. However, there is no x³ expression in either statement.

Does that help?

Cheers,
Brent

Hi Brent,

Hi Brent,

Yes that helps! Thank you :)

Hi Brent, Q link : https:/

Hi Brent, Q link : https://gmatclub.com/forum/are-x-and-y-both-positive-1-2x-2y-1-2-x-y-63377.html

How can the answer be C for this? My answer is E

Are x and y both positive?
(1) 2x - 2y = 1
(2) x/y > 1

the statements each are not sufficient so lets try directly to combine both,
2x = 1 + 2y and x > y
x = y + 0.5 and x>y

if we test some values to satisfy both conditions,
y=2, x = 2.5.....x and y are +ve
y=-2, x = -1.5.....x and y are -ve

Since we get both results 1 and 2 are not sufficient.
Answer E? What's wrong in my approach? The problem here is that we cannot take statement 2 (x/y > 1) and rewrite it as x > y

KEY CONCEPTS:
If we multiply both sides of an inequality by a POSITIVE value, then the direction of the inequality symbol remains THE SAME.
If we multiply both sides of an inequality by a NEGATIVE value, then the direction of the inequality symbol is REVERSED.
For more on this, watch: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

You have taken x/y > 1 and multiplied both sides by y to get: x > y
HOWEVER, since we don't know whether y is POSITIVE or NEGATIVE, we can't be certain about the direction of the inequality symbol.

For example, if x = 3 and y = 2, then the inequality x/y > 1 becomes 3/2 > 1, which is true.
In this case, your inequality x > y is true (since 3 > 2)

Conversely, if x = -3 and y = -2, then the inequality x/y > 1 becomes -3/-2 > 1, which is ALSO true.
In this case, your inequality x > y is NOT true (since -3 < -2)

Here's my full solution: https://gmatclub.com/forum/are-x-and-y-both-positive-1-2x-2y-1-2-x-y-633...

Cheers,
Brent

You proved me wrong again! I

You proved me wrong again! I missed the trick with inequalities due to negative division. Thanks Brent!

Hello! Could you please explain why we cannot perform ca-cb>0 here? and get c(a-b)>0 and thus prove for sure that a>b?

https://gmatclub.com/forum/is-a-b-103316.html thanks beforehand! KEY CONCEPTS:
If we divide both sides of an inequality by a NEGATIVE number, we must REVERSE the direction of the inequality symbol.
If we divide both sides of an inequality by a POSITIVE number, the direction of the inequality symbol REMAINS the same.

You are correct to say that, if ca - cb > 0 then it is also true that c(a - b) > 0
HOWEVER, we cannot then divide both sides of the inequality by c to get (a - b) > 0
We can't do this because we don't know whether c is positive or negative.

Consider these two cases that satisfy the inequality c(a - b) > 0
case i: c = 1, a = 3 and b = 2. In this case, a > b
case ii: c = -1, a = 2 and b = 3. In this case, a < b

Does that help?

Cheers,
Brent

Yes that makes sense now!

Yes that makes sense now!
One last question, would you say always be cautious if we were to do divide or multiply a variable, in this case c, in inequalities because we are not entirely sure if that variable is +/- .. Exactly! Whenever you find

Exactly! Whenever you find yourself multiplying or dividing an INEQUALITY by a variable, you should always pause and ask "Do I know whether this variable is positive or negative?"

http://www.veritasprep.com

http://www.veritasprep.com/blog/2015/10/gmat-tip-of-the-week-trick-or-treat/

hi brent, is the answer to this question c? Yes, the correct answer is C

Cheers,
Brent

If 1 > 1 - ab > 0, which of

If 1 > 1 - ab > 0, which of the following must be true?

As your solution notes, this is simplified to:

0 < ab < 1
I understand that a/b must share the same sign and so a/b MUST be positive. But is the inverse true? That is, is b/a 100% TRUE? I'm entirely sure what you

I'm entirely sure what you mean by "inverse"

If 0 < ab, then we know that a/b > 0
If 0 < ab, then we know that b/a > 0
If 0 < a/b, then we know that ab > 0
If 0 < b/a, then we know that ab > 0
If 0 < b/a, then we know that a/b > 0
If 0 < a/b, then we know that b/a > 0

Does that help?

Cheers,
Brent

Should I be worried if I am

Should I be worried if I am finding it relatively ok to tackle the 650 - 800 problems and struggling with 500 - 650 problems? That's certainly a unique

That's certainly a unique predicament!
I'd wait to see how this plays out on an official practice test to see if there's cause for alarm.
Have you taken any practice tests yet?

Cheers,
Brent

I am confused about the direction of inequality when dividing by a negative number.

10<12

So when we divide this by -2, wont we get ......5>6 ....why have you retained the signs of -5 > -6? Wont the minuses on both sides cancel out at the SAME TIME as when we reverse the sign of the inequality from < to >?

Thanks I'm not sure what you mean

I'm not sure what you mean when you write "Wont the minuses on both sides cancel out at the SAME TIME as when we reverse the sign of the inequality from < to >?"

We know that 10 < 12. However, we also know that 5 < 6 (you are suggesting that 5 > 6, which is not true)

You might want to review the lesson on multiplying and dividing positive and negative numbers: https://www.gmatprepnow.com/module/gmat-arithmetic/video/1060

10 ÷ (-2) = -5, and 12 ÷ (-2) = -6

NOTE: Many problems arise if we go with your approach and say that 10 ÷ (-2) = 5
Since we already know that 10 ÷ 2 = 5, it should come as a surprise that 10 ÷ (-2) AND 10 ÷ 2 both equal 5

Cheers,
Brent

Hi Brent, what I meant to ask

Hi Brent, what I meant to ask is - what happens when we divide both sides by '-2'..........10/-2 < 12/-2?

Would that be -5 < -6 or would that be -5>-6? I mean I know that -5 is logically greater than -6....but what I dont understand is that when do you decide to switch the sign of the inequality from -5<-6 to -5>-6 when you divide 10<12 both sides by -2?

thanks This is a property of

This is a property of inequalities.

If we take the inequality 3 < 5 and multiply both sides by 2, we get: 6 < 10.
This resulting inequality, 6 < 10, is valid.

If we take the inequality 15 < 45 and divide both sides by 3, we get: 5 < 9.
This resulting inequality, 5 < 9, is valid.

If we take the inequality 3 < 5 and multiply both sides by -2, we get: -6 < -10.
This resulting inequality,-6 < -10 is NOT valid.

If we take the inequality 15 < 45 and divide both sides by -3, we get: -5 < -9.
This resulting inequality, -5 < -9, is NOT valid.

So, to counter-act the effects of multiplying or dividing both sides by a NEGATIVE value, we must REVERSE the direction of the inequality sign.

Does that help?

Cheers,
Brent

yes, thanks for clarifying.

yes, thanks for clarifying.

Hi Brent, Please check out this question.
https://gmatclub.com/forum/if-x-y-and-x-0-then-which-of-the-following-must-be-true-292333.html

In your solving method you point out that D is the answer but then our fellas from gmatclub mark E as the correct answer. Which one is it? Thank you!

Keet it real!
Alex Let's show why statement III is NOT true.

III. 1/(x+1) < 1/(y+1)

We're told that y < x < 0.
If we let x = -0.5 and let y = -2, statement III becomes: 1/(-0.5 + 1) < 1/(-2 + 1)
Simplify: 1/0.5 < 1/-1
Simplify more: 2 < -1. NOT TRUE

I already showed that statements I and II are true, so the correct answer is D

Cheers,
Brent

For the integers m, n, r, and

For the integers m, n, r, and s, if m + n = 250 and m > n, is (m – r) > (s – n)?

(1) 250 > r + s
(2) m + r + s = 375
For this question, how did you add r and n without knowing if they are positive or negative? When it comes to solving inequalities, we can safely ADD (or SUBTRACT) any values to (from) both sides, regardless of whether the variables are positive or negative.

HOWEVER, when it comes to MULTIPLYING or DIVIDING both sides of an inequality, we need to know whether the variables are positive or negative.

This is covered in the above video (from 2:45 to 3:45)

Cheers,
Brent

inequalities is one of the

inequalities is one of the toughest section for me, especially the DS section. Can you give any other tips to simplify them. I'm having difficulty rephrasing questions and inserting variables. You're not alone. Most

You're not alone. Most students find inequality questions difficult.

I'm not sure what I can add that hasn't already been stated in the two videos on Inequalities (the video above and this one: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...)

If I were you, I'd first review the video above and then start answering linked questions in the reinforcement activities box (starting with the easiest questions) and slowly work your way to upper-level questions.

I hope that helps.

Cheers,
Brent

Hi Brent,

Hi Brent,

question https://gmatclub.com/forum/if-z-is-an-odd-integer-is-300z-208299.html

Do you have a video or maybe a post explaining how to solve inequality where a variable is squared for example 1<x^2<36. Since x could be +/-1 and +/-6. How can we approach it in a way it makes sense logically? I vaguely remember from school about flipping signs, however, now that I've watched your video that explains that we flip the sign when we divide by a negative number, but in this particular example we do not actually divide anything by a negative number but still have to flip signs I. I cannot make a logical sense out of it. An explanation of how to properly approach/solve/understand such inequalities would be very helpful!

Many thanks! I don't have a video that specifically addresses this concept.
Instead, we can use a bit of number sense.

If 1 < x² < 36, we can see that, if x = -1 or 1, then x² = 1, and x = -6 or 6, then x² = 36

So, if 1 < x² < 36, then EITHER -6 < x < 1 OR 1 < x < 6

In general, if a < x² < b (such that a ≥ 0, and b ≥ 0), then EITHER -√b < x < √a OR √a < x < √b

Does that help?

Cheers,
Brent

Hi Brent,

Hi Brent,

question https://gmatclub.com/forum/is-1-x-1-y-271297.html

Can you please check if my logic is sound when combining statements 1&2 together. We are told that x<y and x>0, from x<y we can conclude that x-y<0, if we multiply both sides by (-1) we can conclude that y-x>0, from here we can conclude that y>x and because we are told that x>0, then y is also positive, hence xy>0, y-x>0, hence the whole fraction (y-x)/xy is greater than zero. Is it a reasonable approach?

Thanks. That approach definitely works. Nice work.

My only comment would be that you take a few extra steps to conclude that y is positive.
Once we know that x < y and 0 < x (from statements 1 and 2), we can combine them to get: 0 < x < y, which means y is positive.

That said, your extra steps probably only took a few extra seconds :-)
I just wanted to show another way to reach the same conclusion that you reached.

Cheers,
Brent

Hi Brent,

Hi Brent,

question https://gmatclub.com/forum/in-the-figure-above-is-ab-bc-note-figure-not-necessarily-drawn-224125.html

This is a very weird question. How do you know to treat the graph as a question about the length of the sections AB, CD and so on? When I saw the question, I thought we are shown values A, B, C, and D on a number line and we are asked whether the value of A*B is greater than the value of B*C (and the order of the values must be preserved in order to solve the problem). Is it totally wrong to understand it this way? As such I thought that if A,B,C,D are values and we are given that AB>CD, then the only way this could be true is if 0 is on the far right side and the values A, B, C, D must all be negative, because there is no way A*B can be larger than C*D if all values A, B, C, D are positive. It also can be that 0 is between values of B and C, this way condition that AB>CD is also met. So hence I concluded that in order for statement 1 to be true either all values are negative, or values A, B are negative and values C and D are positive, as such I concluded that A*B will always be greater than B*C, and thought statement 1 is sufficient. What is wrong with this approach? I would really appreciate your comment on this question, as I find it very intriguing.

Many Thanks! Good point!! The question is somewhat ambiguous.
The intent of the question is to compare the lengths of line segments (not to compare the products AB and BC)

To avoid this ambiguity, an official GMAT question would add some words to explain this. For example:
"Is LINE SEGMENT AB longer than LINE SEGMENT BC?"

Cheers,
Brent

https://gmatclub.com/forum/is

https://gmatclub.com/forum/is-x-3-2-1-2-3-x-1-x-3-2-x-x-62992.html

For this question I get that : Rephrase to |x-3| = 3-x?
1) x-3 = 3-x and x = 3
2) -(x-3) = 3-x , so -x+3 = 3-x and x=anything

How do you then get to the step that x <= 3? apart from thinking about the range from (2)