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√(x1) > x3
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26 Oct 2019, 03:10
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
√(x1) > x3
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 > (x5)(x2) 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<(x5)(x2) 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=(x5)(x2) 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



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29 Oct 2019, 20:39
vanam52923 wrote: Here you go vanam52923: https://gmatclub.com/forum/howmanypeo ... l#p2393262
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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29 Oct 2019, 20:55
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/letsbethe ... l#p2393283
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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30 Oct 2019, 03:22
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
√(x1) > x3
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 > (x5)(x2) 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<(x5)(x2) 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=(x5)(x2) 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.
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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30 Oct 2019, 03:23
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.
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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02 Nov 2019, 03:06
VeritasKarishma 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
√(x1) > x3 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 > (x5)(x2) 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<(x5)(x2) 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=(x5)(x2) 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, Thank you for your reply! A follow up question here: 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: √(x1) > x3 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



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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04 Nov 2019, 20:29
Lionila wrote: VeritasKarishma 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
√(x1) > x3 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 > (x5)(x2) 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<(x5)(x2) 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=(x5)(x2) 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, Thank you for your reply! A follow up question here: 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: √(x1) > x3 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.
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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08 Nov 2019, 04:09
Dear Karishma,
Thank you once again for your answer!
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



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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14 Nov 2019, 19:23
Lionila wrote: Dear Karishma,
Thank you once again for your answer!
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.
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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01 Dec 2019, 05:48
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



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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02 Dec 2019, 20:57
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 copypaste the exact question. There are typos here.
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18 Dec 2019, 07:59



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22 Dec 2019, 02:31
Aviral1995 wrote: Here you go Aviral1995: https://gmatclub.com/forum/giventwopo ... l#p2428550
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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22 Dec 2019, 02:37
VeritasKarishma wrote: Hi, Karishma I’ll have an exam in three days. What do you advice to recap the most important part of Quant? Thanks in advance. Posted from my mobile device



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22 Dec 2019, 02:57
Aviral1995 wrote: https://gmatclub.com/forum/johnwantstoconstructacircularswimmingpoolinhisplotwhichis311206.html VeritasKarishma please help me with this question! Aviral1995, here you go: https://gmatclub.com/forum/johnwantst ... l#p2428605
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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22 Dec 2019, 03:07
lacktutor wrote: VeritasKarishma wrote: Hi, Karishma I’ll have an exam in three days. What do you advice to recap the most important part of Quant? Thanks in advance. Posted from my mobile deviceHey 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.
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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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30 Dec 2019, 12:28
VeritasKarishmais 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/inthefigur ... 21194.htmlPlease explain



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30 Dec 2019, 12:29
VeritasKarishmais 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/inthefigur ... 21194.htmlPlease explain



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Re: Veritas Prep PS Forum Expert  Karishma  Ask Me Anything about Math
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30 Dec 2019, 20:38
Aviral1995 wrote: VeritasKarishmais 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/inthefigur ... 21194.htmlPlease explain Absolutely. You don't need to calculate anything in this question. Check: https://gmatclub.com/forum/inthefigur ... l#p2432203
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