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## Comment on

Inequalities - Part II## Just want to make sure I'm

Also on a separate note, it seems that the lessons in the prior chapters kind of build on each other and vice versa and whatnot... (i.e. this video will help me with # property questions and arithmetic questions). Was there a certain reason behind why you ordered your quant video chapters the way they are? Thanks!

## We're told that y < x AND x +

We're told that y < x AND x + 5 < z

So, once we recognize that it is always the case x < x + 5, then we can combine the two inequalities.

I spent A LOT of time thinking about the order in which the topics appear. My goal was to make sure that, if learning concept C requires knowledge about concept A, then concept A must appear before concept C.

For example, we need to know how to collect like terms (i.e., 7x - 2x = 5x) before we can solve basic equations (like 7x - 9 = 2x + 11). So, the video on collecting like terms appears before basic equation solving.

And so on...

## thank you - I love your

## Thanks! I appreciate that.

Thanks! I appreciate that.

## Hi Brent,

For the final example in the video, can we write the relationship between y and w as follows,

y < x < w-5

Deepak

## You're referring to the

You're referring to the question that begins at 4:55 in the video.

Yes, your inequality is good. You just need to keep going, because we're asked to compare y and w (not y and w-5)

Given: y < x

x + 5 < w

You took the second inequality and subtracted 5 from both sides to get: x < w - 5

Then you combined this with the first inequality to get: y < x < w-5

Since we're trying to compare y and w, we need to recognize that w-5 < w for all values of w

So, we can add this last part to your inequality to get:

y < x < w-5 < w, which means we can be certain that y < w

## Hi Brent,

I was looking for some lessons in inequalities with powers and roots. Do you have a video lesson that covers concepts involving these. For eg. x^2 < 81 means -9 < x < 9. What happens if you take the square root on both sides of the original inequality? I'm finding this concept difficult to grasp. Any help is appreciated. Thanks.

## You're referring to quadratic

You're referring to quadratic inequalities, and they're typically pretty tricky questions. Here's the video that covers that concept:

https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

Cheers,

Brent

## Great. Thanks!

## Hi Brent,

I am not able to find answer for this question on beatthegmat. http://www.beatthegmat.com/x-and-y-t276919.html

I know earlier there was your answer on this topic but its not visible now.Can you please give me link of this question's answer.

If x and y are positive integers, is x/y < (x+5)/(y+5) ?

Statement #1: y = 5

Statement #2: x > y

## Hi Neha,

Hi Neha,

I have posted my solution here: https://gmatclub.com/forum/if-x-and-y-are-positive-integers-is-x-y-x-5-y...

Cheers,

Brent

## sir please provide

If x and y are nonzero integers, is x^y < y^x ?

(1) x = y²

(2) y > 2

## Tricky!!

Tricky!!

Here's my full solution: https://gmatclub.com/forum/if-x-and-y-are-nonzero-integers-is-x-y-y-x-16...

Cheers,

Brent

## Is 1/p > r/(r^2 + 2) ?

(1) p = r

(2) r > 0

statement 1

suppose p is +ve and equal to 2

.5>.3

suppose p is -ve and equal to 2

-.5>-.3(incorrect)

statement 2 r>0 insuff

combining statement 1 and 2 -ve case gets eliminated

so c

is this approach correct or is there better way?

## I'd be careful with that

I'd be careful with that approach. You've only shown one case where r (aka p) is positive. That isn't really enough to say that r being positive will result in only one answer to the target question.

Consider this example:

Is x + y > 2?

(1) x = y

(2) x > 0

Statement 1:

Suppose x is positive and equal to 2. So, y = 2, which means x + y > 2

Suppose x is negative and equal to -2. So, y = -2, which means x + y < 2

NOT SUFFICIENT

Statement 2: NOT SUFFICIENT

Statements 1 and 2 COMBINED

Can we say that, since x > 0, we can eliminate the case where x is negative and conclude that the correct answer is C?

No.

It could still be the case that x = y = 0.3, which means x + y < 2

So, the correct answer is actually E

Does that help?

Here's my full solution: https://gmatclub.com/forum/is-1-p-r-r-86165-20.html#p2148792

Cheers,

Brent

## Hi Brent,

Need your help: https://gmatclub.com/forum/for-integers-x-b-and-c-x-2-bx-c-0-is-x-233688.html

## Question link: https:/

Question link: https://gmatclub.com/forum/for-integers-x-b-and-c-x-2-bx-c-0-is-x-233688...

Here's my full solution: https://gmatclub.com/forum/for-integers-x-b-and-c-x-2-bx-c-0-is-x-233688...

Cheers,

Brent

## and also in this one: https:/

## Tricky!

Tricky!

My solution: https://gmatclub.com/forum/if-x-is-an-integer-and-4-x-100-what-is-x-1063...

Cheers,

Brent

## Hi Brent,

Need your help with this one. The question is from OG 19

If p, s, and t are positive, is |ps – pt| > p(s – t) ?

1. p < s

2. s < t

## My solution: https://gmatclub

My solution: https://gmatclub.com/forum/if-p-s-and-t-are-positive-integer-is-ps-pt-p-...

Cheers,

Brent

## I am not able to comprehend

[y] denotes the greatest integer less than or equal to y. Is d < 1 ?

1. d = y − [y]

2. [d] = 0

## Almost everyone struggles

Almost everyone struggles with these kinds of questions (aka Strange Operators questions).

First it's important to understand what this odd notation means. For example:

[5.1] = 5

[3] = 3

[8.9] = 8

[-1.4] = -2

[-13.6] = -14

Once you understand the implications of the notation, you'll have a better chance of answering the question.

That said, this is a very tricky question.

My solution: https://gmatclub.com/forum/denotes-the-greatest-integer-less-than-or-equ...

Cheers,

Brent

## Hi Brent,

Need your help:

If t and x are integers, what is the value of x ?

1. x²/t² = 4/9

2. x > 0 and t > 0

## Here's my solution: https:/

Here's my solution: https://gmatclub.com/forum/if-t-and-x-are-integers-what-is-the-value-of-...

## Hi Brent,

Came across this question on a practice test. Appreciate it if you could shed some light as to where I went wrong.

What is the value of x?

(1) x + y = 2

(2) xy = z^2 + 1

I came across this question on a practice test.

Of course, 1 and 2 alone are not enough but when combined I got a bit lost.

The way the test worked it out was to substitute the values in both equations and ended up with -z^2 = (y-1)^2

Supposedly, this is adequate to tell us that y is equal to 1 and as such we can derive a value for x.

I get that -z^2 can't be negative and as such (y-1) must be positive or 0. But what can we say with certainty that y=1.

Couldn't y = to any number above 1? It would still be positive wouldn't it?

## Keep in mind that -z² is the

Keep in mind that -z² is the same as -(z²)

Since z² ≥ 0, it must be the case that -(z²) ≤ 0

Similarly, (y-1)² ≥ 0

So, the only time that -(z²) = (y-1)² is when both sides are equal to zero.

Here's my full solution: https://gmatclub.com/forum/what-is-the-value-of-x-1-x-y-2-2-xy-z2-187825...

Cheers,

Brent

## Is mn < 10?

(1) m < 5 and n < 2

(2) 1 < m < 3 and n^2 < 25

Here's another one.

Is there a way for me to combine the two inequalities to get a definitive answer to whether mn > 10? The problem with me is when I try random values (positive, negative, fractions), I tend to get lost in it all and it does take some time.

Can I combine the inequalities in both sets? If so how?

## Question link: https:/

Question link: https://gmatclub.com/forum/is-mn-206826.html

There's no way to combine everything into one nice inequality.

Instead, we must examine some extreme cases to solve this question.

Here's my full solution: https://gmatclub.com/forum/is-mn-206826.html#p2229812

Cheers,

Brent

## Here's another one.

If xy < zy < 0, is y positive?

x < z

x is negative

Could you explain how each statement alone is sufficient?

Here's my approach:

From the question, we know that:

xy < 0 and zy < 0

So if y is positive, then x AND y must share the same negative sign.

(1) x < z

Both could be negative z = -3, x = -4

Both could be positive z = 4, x = 3

Or z being negative, y being positive

So z(+)y(-) < 0 or, z(-)y(+) < 0

Insufficient!

(2) x is negative

If x is negative, then for xy < 0, y MUST be negative

Sufficient!

What am I missing out here?

## I'm having a hard time trying

I'm having a hard time trying to figure out what you mean by this:

(1) x < z

Both could be negative z = -3, x = -4

Both could be positive z = 4, x = 3

Or z being negative, y being positive

So z(+)y(-) < 0 or, z(-)y(+) < 0

Insufficient!

Can you elaborate?

For statement 2, you wrote:

(2) x is negative

If x is negative, then for xy < 0, y MUST be negative

If x is negative, AND xy < 0, then y is POSITIVE.

Here's my full solution: https://gmatclub.com/forum/if-xy-zy-0-is-y-positive-1-x-z-2-x-is-negativ...

Cheers,

Brent

## Hi could you help with this

Is x > 1?

1) (x + 1)/(|x| - 1) > 0

|x| when opened up could be x or -x

Hence, (x+1)/(x-1) OR (x+1)/(-x-1) = -(x+1)/(x+1) = -1 .....since -1 cannot be >0, hence |x| cannot be -x and hence, it is ONLY positive x

now (X+1)/(x-1) > 0 is given....but how does this lead to an answer that x>1? what am I missing here. How do I resolve the solution to the absolute value above? thanks

## We know that (x + 1)/(|x| - 1

We know that (x + 1)/(|x| - 1) > 0

So, there are two possible cases:

case i: (x + 1) and (|x| - 1) are both positive

case ii: (x + 1) and (|x| - 1) are both negative

If we know that x is positive (as you have suggested), then we can eliminate case ii

This means case i is true. That is, (x + 1) and (|x| - 1) are both positive

Since (|x| - 1) must be positive, we know that |x| - 1 > 0

Add 1 to both sides to get: |x| > 1

This means EITHER x > 1 OR x < -1

Since we know x is POSITIVE, we can eliminate the possibility that x < -1, which means it must be the case that x > 1

Does that help?

Cheers,

Brent

## Hi Brent,

I just want to understand one thing, while taking reciprocal do we have to flip signs ? is there any rule for that . Could you please show me an example to illustrate

## If both fractions have the

If both fractions have the SAME SIGN, then we can take: a/b < x/y, and rewrite it as: b/a > y/x

For example, since 2/5 < 6/7, it is also true that 5/2 > 7/6

Likewise, since -11/2 < -5/3, it is also true that -2/11 > -3/5

If both fractions have DIFFERENT SIGNS, then the positive fraction will always be greater than the negative fraction.

Finally, if we don't know anything about the signs of the fractions, then we can't make any conclusions about their reciprocals.

Cheers,

Brent

## Hi Brent,

I just wanted to confirm if my understanding is correct

Take x^2/y < 3y

In this case we cannot we cannot take y of the LHS to RHS since we don't know the sign of y.

If Y is negative then the inequity sign changes. so in that case can i come to the conclusion that multiplying one side of the equation with a negative number can change the sign?

## In order to x^2/y < 3y get

In order to take x^2/y < 3y and get rid of the y on the left side, we must multiply both sides by y.

However, since we don't know whether y is positive or negative, we don't know the direction of the inequality symbol in the resulting inequality.

Your second question is: Can I come to the conclusion that multiplying one side of the equation with a negative number can change the sign?

First of all, multiplying just one side by a negative number totally changes the nature of the inequality. So, that should be avoided.

That said, to answer your question, multiplying ONE side by a negative number may or may not cause the inequality symbol to reverse its direction.

For example, take: 3 < 10

If we multiply the left side by -1, then we get -3, and it is true that -3 < 10 (direction doesn't change)

Likewise, take: -4 < 7

If we multiply the left side by -1, then we get 4, and it is true that 4 < 7 (direction doesn't change)

Conversely, take: -8 < 7

If we multiply the left side by -1, then we get 8, and it is true that 8 > 7 (here, we reversed the signs)

Does that help?

Cheers,

Brent

## Hi Brent,

question https://gmatclub.com/forum/is-n-136517.html

Perhaps it is a stupid question, but I can not wrap my mind around it. We are asked if n>6. For this to be true some possible values of n must be 6.00001, 6.1, 6.15, 6.2, 7, 8, 9 and so on. So to answer this question with certainty we must be able to prove that n can take one of those values. Statement 1 tells us that n>6.25, so it definitely is telling us that n can be 6.26, 6.27, 6.28 and so on. But what about values between 6 and 6.25, with statement 1 we cannot prove that n can be between 6 and 6.25 How can this statement alone be sufficient then? What am I missing?

Thanks a bunch!

## Question link: https:/

Question link: https://gmatclub.com/forum/is-n-136517.html

It's important to keep reminding ourselves of the target question: Is n > 6?

You're correct to say that Statement 1 tells us that n > 6.25, and that this means n can be 6.26, 6.27, 6.28 etc.

For all of the possible values of n (6.26, 6.27, 6.28 etc.), the answer to the target question is always "YES, n IS greater than 6"

This means statement 1 is sufficient.

Your question: "But what about values between 6 and 6.25?"

My answer: If n > 6.25, then n cannot have a value between 6 and 6.25, so we must focus on the values that n CAN have, and these possible values are defined by statement 1.

Does that help?

Cheers,

Brent

## If I were to change the

## That's a great example!

That's a great example!

## Hi Brent,

I have a question about these two questions https://gmatclub.com/forum/5n-2-12-and-7n-5-44-n-must-be-between-which-numbers-215173.html and https://gmatclub.com/forum/if-3x-7-y-and-x-y-2-which-if-the-following-must-be-true-232927.html

Why in one instance we add two equations and in another question, we simply solve for n. When I saw question https://gmatclub.com/forum/5n-2-12-and-7n-5-44-n-must-be-between-which-numbers-215173.html I immediately started to add them up (I made sure first both equations have signs looking in the same direction, multiplied equation 7n - 5 < 44 by (-1) and got 5-7n>-44). I got an answer n<39/2, which I did not see in the offered answer choices. Am I understanding it correctly that we need to add two equations together only when we have two variables in an equation? And if it is just one variable we just solve for that variable? Did I get it right? Or is there another explanation when we need to use the technique of adding two equations together?

Thank you.

## Question link: https:/

Question link: https://gmatclub.com/forum/5n-2-12-and-7n-5-44-n-must-be-between-which-n...

By adding the two inequalities, you got n < 39/2 (in other words, n < 19.5)

So, it must be true that n < 19.5. HOWEVER, doesn't help us, since all of the answer choices tell us that n < 19.5.

So, we must try a new approach.

Sometimes one technique will be better than another technique. At other times, both techniques are fine.

For example, we can answer the 2nd questions (https://gmatclub.com/forum/if-3x-7-y-and-x-y-2-which-if-the-following-mu...) using BOTH strategies.

Here's my solution that simply combines the inequalities: https://gmatclub.com/forum/if-3x-7-y-and-x-y-2-which-if-the-following-mu...

## Hi Brent,

Can you please explain in this question https://gmatclub.com/forum/if-a-and-b-are-integers-such-that-5-a-1-and-b-2-x-then-which-232925.html Why exactly you are changing equation from 5 ≥ a > 1 to 1 < a ≤ 5, when I look it the original equation I see that a > 1 and b ≥ -2 and it seems like signs both look in the same direction which means we can add them together and get a + b≥ -1.

I guess my question is how do you know that you need to use a<= 5 portion of the equation to add to b and not a>1 portion?

Thank you.

## Question link: https:/

Question link: https://gmatclub.com/forum/if-a-and-b-are-integers-such-that-5-a-1-and-b...

You're correct to conclude that a + b ≥ -1, but that's not what the question is asking. We need to find possible value of a - b.

If we want to ADD the inequalities, we must first change the b (in b ≥ -2) into -b, so that we'll get a-b on one side of the inequality.

So, take b ≥ -2 and multiply both sides by -1 to get: -b ≤ 2

How do we combine this with 5 ≥ a > 1?

It may help to first rewrite this as: 1 < a ≤ 5

At this point, we can see that, in order to get a - b, we must focus on the a ≤ 5 part of the inequality.

Take:

-b ≤ 2

a ≤ 5

Add them to get: a - b ≤ 7

So, a - b CANNOT equal 8. Answer: E

Does that help?

Cheers,

Brent

## Perfect explanation! Thanks!

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