Lesson: Inequalities - Part I

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Comment on Inequalities - Part I

Excellent approach

A ton of practice ques :) Are inequalities a big portion of the quant section?
gmat-admin's picture

It's a common question type. So, you should become proficient with inequalities.

Hi Brent,

If abcd ≠ 0, is abcd < 0?

1. a/b > c/d
2. b/a > d/c

the OA is C. I'm not able to understand how? Please help with a simple explanation
gmat-admin's picture

Mitch provides an awesome solution here (2nd post): http://www.beatthegmat.com/gmat-advanced-quant-t291008.html

Let me know if you need me to clarify any steps.


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
gmat-admin's picture

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

I use a well-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?


“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 
gmat-admin's picture

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?


How does x² > 1/x get re-phrased into 1 > 1/x?
1) x² > 1/x
x² x²

2) 1> 1/x (isn’t 1/x divided by x2 equal to 1/x3 ?)

gmat-admin's picture

In my solution (at http://www.beatthegmat.com/is-x-2-1-x-t296568.html), I never rephrased x² > 1/x to get 1 > 1/x

I did, however, rephrase statement 1 (x² > x) to get the equivalent inequality: 1 > 1/x

Does that help?

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:

gmat-admin's picture

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


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.
gmat-admin's picture

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...


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
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-a-0-and-b-0-is-a-b-b-a-224007.html#p1725459

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?


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

Answer: E
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-b-x-0-and-w-x-y-236379.html

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?


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