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Intern  B
Joined: 29 Oct 2017
Posts: 10

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Dear Karishma,

First of all thank you for offering this forum!

I am struggling with inequalities that require squaring such as a>b where I cannot be sure whether a+b is <> or = 0.

One example

√(x-1) > x-3

I know that a>= 0 but I do not know whether a+b is >< or = 0
I also know that x>=1

I believe that I need to solve this question considering each case as follows:

case a+b>0 --> I square and do not flip the sign!
I get: 0 > (x-5)(x-2)
solution 2<x<5
I test a number from this range and the original inequality holds true so I conclude that this is one correct result.
Is it correct that I need to test the range? I assume this because a+b>0 is only a hypothesis and not a fact.

case a+b<0 --> I square and flip the sign because at least one side is negative
I get: 0<(x-5)(x-2)
solution: x>5 and 1=<x<2
I test a number in each range. The range x>5 does not hold the inequality true. But the second range 1=<x<2 does. How is this possible?
1) a+b can only be > or < than 0 so I assume that only the first or the second range can be valid, or not?
2) Also within the same case a+b<0, why is 1=<x<2 valid and x>5 invalid?

case a+b = 0
I get: 0=(x-5)(x-2)
solution x = 2 or 5
I test both roots but only 2 is valid.

Overall I conclude that 1=<x<5 is the solution.

Thanks for commenting/clarifying on my approach, solution, assumptions and inferences.

Thanks upfront!
Lionila
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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vanam52923 wrote:
hi ,plz clear this doubt,

https://gmatclub.com/forum/how-many-peo ... l#p2379587

thanks a lot Here you go vanam52923:

https://gmatclub.com/forum/how-many-peo ... l#p2393262
_________________
Karishma
Veritas Prep GMAT Instructor

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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Ramya16 wrote:
Hi Karishma,

Please can you help solve this question.
Let S be the set of all positive integers having at most 4 digits and such that each of the digits is 0 or 1. What is the greatest prime factor of the sum of all the numbers in S ?

A. 11
B. 19
C. 37
D. 59
E. 101

PS41661.01

Hey Ramya16,

Check here:
https://gmatclub.com/forum/let-s-be-the ... l#p2393283
_________________
Karishma
Veritas Prep GMAT Instructor

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

### Show Tags

Lionila wrote:
Dear Karishma,

First of all thank you for offering this forum!

I am struggling with inequalities that require squaring such as a>b where I cannot be sure whether a+b is <> or = 0.

One example

√(x-1) > x-3

I know that a>= 0 but I do not know whether a+b is >< or = 0
I also know that x>=1

I believe that I need to solve this question considering each case as follows:

case a+b>0 --> I square and do not flip the sign!
I get: 0 > (x-5)(x-2)
solution 2<x<5
I test a number from this range and the original inequality holds true so I conclude that this is one correct result.
Is it correct that I need to test the range? I assume this because a+b>0 is only a hypothesis and not a fact.

case a+b<0 --> I square and flip the sign because at least one side is negative
I get: 0<(x-5)(x-2)
solution: x>5 and 1=<x<2
I test a number in each range. The range x>5 does not hold the inequality true. But the second range 1=<x<2 does. How is this possible?
1) a+b can only be > or < than 0 so I assume that only the first or the second range can be valid, or not?
2) Also within the same case a+b<0, why is 1=<x<2 valid and x>5 invalid?

case a+b = 0
I get: 0=(x-5)(x-2)
solution x = 2 or 5
I test both roots but only 2 is valid.

Overall I conclude that 1=<x<5 is the solution.

Thanks for commenting/clarifying on my approach, solution, assumptions and inferences.

Thanks upfront!
Lionila

When you have inequalities such as
x > y
in which x or y may be positive or negative, you cannot square both sides. You do not know whether the inequality will flip or not.

e.g.

4 > 3
You can square without flipping the inequality sign.

4 > -3
You can square without flipping the inequality sign.

4 > -6
When you square, the inequality sign flips.

-4 > -5
When you square, the inequality sign flips.

When both sides are positive, squaring does not change the inequality sign.
When both sides are negative, squaring flips the inequality sign.
When one is positive, one is negative, you do not know whether the inequality sign will be flipped or not.

As for this question, send me the link of the actual question and I will let you know how to solve it.
_________________
Karishma
Veritas Prep GMAT Instructor

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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I apologise for the delayed replies guys. I have been travelling for the past couple of weeks. Rest assured, I will get to any queries you post here, sooner or later.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 29 Oct 2017
Posts: 10

### Show Tags

Lionila wrote:
Dear Karishma,

First of all thank you for offering this forum!

I am struggling with inequalities that require squaring such as a>b where I cannot be sure whether a+b is <> or = 0.

One example

√(x-1) > x-3
I know that a>= 0 but I do not know whether a+b is >< or = 0
I also know that x>=1

I believe that I need to solve this question considering each case as follows:

case a+b>0 --> I square and do not flip the sign!
I get: 0 > (x-5)(x-2)
solution 2<x<5
I test a number from this range and the original inequality holds true so I conclude that this is one correct result.
Is it correct that I need to test the range? I assume this because a+b>0 is only a hypothesis and not a fact.

case a+b<0 --> I square and flip the sign because at least one side is negative
I get: 0<(x-5)(x-2)
solution: x>5 and 1=<x<2
I test a number in each range. The range x>5 does not hold the inequality true. But the second range 1=<x<2 does. How is this possible?
1) a+b can only be > or < than 0 so I assume that only the first or the second range can be valid, or not?
2) Also within the same case a+b<0, why is 1=<x<2 valid and x>5 invalid?

case a+b = 0
I get: 0=(x-5)(x-2)
solution x = 2 or 5
I test both roots but only 2 is valid.

Overall I conclude that 1=<x<5 is the solution.

Thanks for commenting/clarifying on my approach, solution, assumptions and inferences.

Thanks upfront!
Lionila

When you have inequalities such as
x > y
in which x or y may be positive or negative, you cannot square both sides. You do not know whether the inequality will flip or not.

e.g.

4 > 3
You can square without flipping the inequality sign.

4 > -3
You can square without flipping the inequality sign.

4 > -6
When you square, the inequality sign flips.

-4 > -5
When you square, the inequality sign flips.

When both sides are positive, squaring does not change the inequality sign.
When both sides are negative, squaring flips the inequality sign.
When one is positive, one is negative, you do not know whether the inequality sign will be flipped or not.

As for this question, send me the link of the actual question and I will let you know how to solve it.

Hi Karishma,

4 > -3
You can square without flipping the inequality sign.

I know this is true from the values but doesn't this violate the basic rule in inequalities "If multiplying with a negative number, you must flip the sign"? The sign does not flip here even though we multiply the inequality with (-3).

As for the question: √(x-1) > x-3
This is not OG and I cannot find the source anymore. I understand if you don't want to go through it then but if you find the time, I would greatly appreciate it.

Many greetings
Lionila
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

### Show Tags

Lionila wrote:
Lionila wrote:
Dear Karishma,

First of all thank you for offering this forum!

I am struggling with inequalities that require squaring such as a>b where I cannot be sure whether a+b is <> or = 0.

One example

√(x-1) > x-3
I know that a>= 0 but I do not know whether a+b is >< or = 0
I also know that x>=1

I believe that I need to solve this question considering each case as follows:

case a+b>0 --> I square and do not flip the sign!
I get: 0 > (x-5)(x-2)
solution 2<x<5
I test a number from this range and the original inequality holds true so I conclude that this is one correct result.
Is it correct that I need to test the range? I assume this because a+b>0 is only a hypothesis and not a fact.

case a+b<0 --> I square and flip the sign because at least one side is negative
I get: 0<(x-5)(x-2)
solution: x>5 and 1=<x<2
I test a number in each range. The range x>5 does not hold the inequality true. But the second range 1=<x<2 does. How is this possible?
1) a+b can only be > or < than 0 so I assume that only the first or the second range can be valid, or not?
2) Also within the same case a+b<0, why is 1=<x<2 valid and x>5 invalid?

case a+b = 0
I get: 0=(x-5)(x-2)
solution x = 2 or 5
I test both roots but only 2 is valid.

Overall I conclude that 1=<x<5 is the solution.

Thanks for commenting/clarifying on my approach, solution, assumptions and inferences.

Thanks upfront!
Lionila

When you have inequalities such as
x > y
in which x or y may be positive or negative, you cannot square both sides. You do not know whether the inequality will flip or not.

e.g.

4 > 3
You can square without flipping the inequality sign.

4 > -3
You can square without flipping the inequality sign.

4 > -6
When you square, the inequality sign flips.

-4 > -5
When you square, the inequality sign flips.

When both sides are positive, squaring does not change the inequality sign.
When both sides are negative, squaring flips the inequality sign.
When one is positive, one is negative, you do not know whether the inequality sign will be flipped or not.

As for this question, send me the link of the actual question and I will let you know how to solve it.

Hi Karishma,

4 > -3
You can square without flipping the inequality sign.

I know this is true from the values but doesn't this violate the basic rule in inequalities "If multiplying with a negative number, you must flip the sign"? The sign does not flip here even though we multiply the inequality with (-3).

As for the question: √(x-1) > x-3
This is not OG and I cannot find the source anymore. I understand if you don't want to go through it then but if you find the time, I would greatly appreciate it.

Many greetings
Lionila

Lionel, the point is that when you have x > y such that x is positive and y negative, you CANNOT square it because you do not know the relation that will hold after squaring.
In case of 4 > -3, the same relation will hold 16 > 9
But in case of 4 > -6, the opposite relation will hold 16 < 36
So you CANNOT say what happens when you square both sides so you cannot square them. Will x^2 > y^2 hold or x^2 < y^2 hold, we don't know.

You can square when both are positive or both are negative since you know the relation that will hold after squaring. The sign stays the same when both are positive and the sign flips when both are negative (you can think of this as first multiplying both sides by -1 which flips the inequality sign and makes both sides positive. Then you square both sides the usual way)

I don't have a problem with the question. What I would like is the complete question so that I can see what is given what is asked and what the options are. That can tell us how best to solve the question.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 29 Oct 2017
Posts: 10

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Dear Karishma,

Unfortunately, I don't have the answer choices BUT I think I understood where the fault in my logic was. For some reason I thought that a+b can only be negative or positive. But because it is an inequality, there is a range of possibilities for x. I am quite confident now that the solution for that range is 1 =<x<5.

Am I off in my thinking?

Best,
Lionila
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

### Show Tags

1
Lionila wrote:
Dear Karishma,

Unfortunately, I don't have the answer choices BUT I think I understood where the fault in my logic was. For some reason I thought that a+b can only be negative or positive. But because it is an inequality, there is a range of possibilities for x. I am quite confident now that the solution for that range is 1 =<x<5.

Am I off in my thinking?

Best,
Lionila

Yes Lionila, the correct range is 1 <= x < 5.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 02 Nov 2017
Posts: 32
Location: India
GPA: 3.87

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Machine A can do a job in 24 hours at a constant rate and machine A do the job in 8 hours. Machine B can do the same job at a 2/3 constant rate of machine A. If machine B does the rest job, what is the amount of hours done by machine B alone?

A. 12hrs B. 16hrs C. 24hrs D. 28hrs E. 32hrs
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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indu1954 wrote:
Machine A can do a job in 24 hours at a constant rate and machine A do the job in 8 hours. Machine B can do the same job at a 2/3 constant rate of machine A. If machine B does the rest job, what is the amount of hours done by machine B alone?

A. 12hrs B. 16hrs C. 24hrs D. 28hrs E. 32hrs

indu1954:
Please copy-paste the exact question. There are typos here.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  S
Joined: 13 Apr 2019
Posts: 208
Location: India
Schools: Kellogg '22
GPA: 3.85

### Show Tags

Manager  S
Joined: 13 Apr 2019
Posts: 208
Location: India
Schools: Kellogg '22
GPA: 3.85

### Show Tags

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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Aviral1995 wrote:

Here you go Aviral1995:
https://gmatclub.com/forum/given-two-po ... l#p2428550
_________________
Karishma
Veritas Prep GMAT Instructor

Director  D
Joined: 25 Jul 2018
Posts: 712

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Hi, Karishma
I’ll have an exam in three days. What do you advice to recap the most important part of Quant?

Posted from my mobile device
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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Aviral1995 wrote:
https://gmatclub.com/forum/john-wants-to-construct-a-circular-swimming-pool-in-his-plot-which-is-311206.html

Aviral1995, here you go:
https://gmatclub.com/forum/john-wants-t ... l#p2428605
_________________
Karishma
Veritas Prep GMAT Instructor

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

### Show Tags

lacktutor wrote:

Hi, Karishma
I’ll have an exam in three days. What do you advice to recap the most important part of Quant?

Posted from my mobile device

Hey lacktutor,

In the last 3 days, all you should do is review your formulas and your error log, if you have one. Nothing else is required. Don't get entangled in anything new. A calm mind could be the biggest contributor to a great score, much more than 2 days of study.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  S
Joined: 13 Apr 2019
Posts: 208
Location: India
Schools: Kellogg '22
GPA: 3.85

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is there any way to determine in less time that condition A would not work rather than by calculating and then determining

https://gmatclub.com/forum/in-the-figur ... 21194.html

Manager  S
Joined: 13 Apr 2019
Posts: 208
Location: India
Schools: Kellogg '22
GPA: 3.85

### Show Tags

is there any way to determine in less time that condition A would not work rather than by calculating and then determining

https://gmatclub.com/forum/in-the-figur ... 21194.html

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10467
Location: Pune, India

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Aviral1995 wrote:

is there any way to determine in less time that condition A would not work rather than by calculating and then determining

https://gmatclub.com/forum/in-the-figur ... 21194.html

Absolutely. You don't need to calculate anything in this question. Check:
https://gmatclub.com/forum/in-the-figur ... l#p2432203
_________________
Karishma
Veritas Prep GMAT Instructor Re: Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math   [#permalink] 30 Dec 2019, 20:38

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# Veritas Prep PS Forum Expert - Karishma - Ask Me Anything about Math  