Lesson: Inequalities - Part II

Comment on Inequalities - Part II

Compare y and w
y < x
x+5 < w

----> adding -5 on both sides
x+5-5 < w-5
x < w-5


Now y < x
x < w-5

y < x < w-5

so y < w-5

I did like this . is it correct?... Kindly help
Thanks in advance....
greenlight-admin's picture

You're referring to the task that starts at 4:55.

Your approach is perfect.

If you want to go a little further in comparing y and w, you can recognize that w-5 < w

So, continuing where you left off (y < w-5), we have: y < w-5 < w, which means y < w

Hello I have a question regarding ETS question 166 #4:

Explanation stated ""The length of rectangle A is 10 percent greater than the length of rectangle C , or 1.1ℓ."

I don't know how it came out 1.1 liter. Please let me know
greenlight-admin's picture

Hi jenibae,

Be careful,"ℓ" need not represent liters. It can also be a variable.

In the official solution, they let L = the length of rectangle C (aside: the solution uses lowercase "l", but I'll use capital "L" to avoid confusion)

If the length of rectangle A is 10 percent greater than the length of rectangle C, then the 1.1L = length of rectangle A.

Does that help?

Cheers,
Brent

Thank you for your explanation, however I was more curious how "1.1" equals to " 10% greater than lentgth of rectangle".
greenlight-admin's picture

L = the length of rectangle C

So, 10% of L = (10/100)L = 0.1L

So, if the length of rectangle A is 10% GREATER than the length of rectangle C, then the length of rectangle A = L + 0.1L = 1.1L

More here: https://www.greenlighttestprep.com/module/gre-arithmetic/video/1083 (starting at 7:00 in the video)

Cheers,
Brent

Thank you so much !

Hi Brent,

If the reciprocal of the negative integer x is greater than the sum of y and z, which of the following must be true?
A) x > y+z
B) y and z are positive
C) 1 > x(y+z)
D) 1 < xy + xz
E) 1/x > z-y

If the reciprocal of the negative integer x is greater than y + z then I understand that to be 1/x > y+z.....where 1/x is positive. With that I would choose answer C because one would not change the inequality direction since x is positive. However the official answer given is D; it is stated in the answer explanation the sign changes as x is negative. This seems contradictory to the 1st sentence. Any insight?
Many thanks

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-the-reciprocal-of-the-negative-integ...

You made a small mistake when you wrote "I understand that to be 1/x > y+z.....where 1/x is POSITIVE"

We're told that x is a NEGATIVE integer.
This means 1/x is NEGATIVE
For example, if x = -2, then 1/x = 1/(-2) = -0.5

Here's my full solution: https://gre.myprepclub.com/forum/if-the-reciprocal-of-the-negative-integ...

Cheers,
Brent

Hi Brent,

In this question, https://gre.myprepclub.com/forum/a-b-8949.html, your solution actually says a<b, but then the answer is A. Would you like to clarify?

I solved the eq. this way: I added the two equations to get a + b/10 > b + a/10, I then added the terms and multiplied 10 with a and b on either side.

I got: 10a - a > 10b - b
9a > 9b
a > b

Are these steps correct?

Thanks,
Ketan
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/a-b-8949.html

Good catch! It turns out I accidentally reversed the order of the inequality symbols in both given inequalities.
My edited solution is here: https://gre.myprepclub.com/forum/a-b-8949.html#p37663

Cheers and thanks,
Brent

So I have got the steps right?

Thanks,
Ketan
greenlight-admin's picture

Oops.
Yes, your steps are perfect. Nice work!

Cheers,
Brent

https://gre.myprepclub.com/forum/multiple-values-14584.html
For this problem I added c on both sides and then added the inequalities. I got a+b>0 after adding the inequalities . Is this still a valid way to do it?
greenlight-admin's picture

I think that sounds right.
Just to be sure, can you tell me the steps you took?

https://gre.myprepclub.com/forum/a-b-8949.html#p37663

For this problem when we multiply by 10 for b+a/10<0 we get 10b + a <0 since we want the inequalities to point the same diection, I understand we can multiply by -1 and reverse the sign of the inequality. but could we also rewrite it as 0> 10b+a?
greenlight-admin's picture

Solution link: https://gre.myprepclub.com/forum/a-b-8949.html#p37663

Let's see what happens.
If we do that, we end up with:
10a + b > 0
0 > 10b + a

At this point we might combine them to get: 10b + a < 0 < 10a + b
This means: 10b + a < 10a + b
Subtract b from both sides: 9b + a < 10a
Subtract a from both sides: 9b < 9a
Divide both sides by 9 to get: b < a

Cheers,
Brent

If √a > b > c², which of the following could be true?

I. a > b > c
II. c > b > a
III. a > c > b

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

I wanted to clarify the details on could be problems, my understanding is if it is a COULD BE TRUE problem, then if one solution satisfies the requirements we can accept that condition, if it fails then disregard, right?
greenlight-admin's picture

Question link: https://gmatclub.com/forum/if-a-1-2-b-c-2-which-of-the-following-could-b...

Sorry, I'm not sure I follow your question.

Notice that a = 100, b = 5 and c = 0 satisfy the given condition (√a > b > c²).
These values of a, b and c satisfy statement I (a > b > c), so, we've shown that statement I COULD be true.
These same values of a, b and c DO NOT satisfy statements II and III, but this doesn't necessarily mean that statements II and III cannot be true; it just means that those two statements aren't true for those particular values of a, b and c.

Does that answer your question?

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