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  • Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

     December 14, 2018

     December 14, 2018

     10:00 PM PST

     11:00 PM PST

    Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
  • Free GMAT Strategy Webinar

     December 15, 2018

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    Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

M27-14

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Re: M27-14  [#permalink]

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New post 24 Jun 2018, 09:13
Bunuel wrote:
p2bhokie wrote:

Bunuel Can you please explain statement 'B'? The part I missed was that why didn't you consider the negative of the LHS, and just the +ve?


Absolute value on the left hand side (|x + 10|) cannot be negative: \(|some \ expression|\geq{0}\). Thus 2x + 8, on the right hand side must also be non-negative, which means that \(x \ge -4\). Now, if \(x \ge -4\), then x + 10 will be positive, and we know that if x>0, then |x| = x, so |x + 10| = x + 10.

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Thanks Bunuel. Could you quickly confirm my understanding:
1) Test positive case for |x+10|, i.e. x+10 = 2x+8, i.e. x = 2
2) Test negative case for |x+10|, i.e. -x-10 = 2x+8, i.e. x = -6 => not a solution because of constraint x >= -4
Thus, x = 2 as only solution, right?
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New post 24 Jun 2018, 09:29
gmatprep001 wrote:
Bunuel wrote:
p2bhokie wrote:

Bunuel Can you please explain statement 'B'? The part I missed was that why didn't you consider the negative of the LHS, and just the +ve?


Absolute value on the left hand side (|x + 10|) cannot be negative: \(|some \ expression|\geq{0}\). Thus 2x + 8, on the right hand side must also be non-negative, which means that \(x \ge -4\). Now, if \(x \ge -4\), then x + 10 will be positive, and we know that if x>0, then |x| = x, so |x + 10| = x + 10.

Similar questions to practice:
http://gmatclub.com/forum/is-x-157102.html
http://gmatclub.com/forum/what-is-x-126874.html
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http://gmatclub.com/forum/what-is-the-v ... 61134.html
http://gmatclub.com/forum/what-is-the-v ... 69753.html
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http://gmatclub.com/forum/what-is-the-v ... 27976.html
http://gmatclub.com/forum/is-x-0-1-x-3- ... 27978.html
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Hope this helps.


Thanks Bunuel. Could you quickly confirm my understanding:
1) Test positive case for |x+10|, i.e. x+10 = 2x+8, i.e. x = 2
2) Test negative case for |x+10|, i.e. -x-10 = 2x+8, i.e. x = -6 => not a solution because of constraint x >= -4
Thus, x = 2 as only solution, right?


If you figured that x must be more than or equal to -4, then you could directly get that |x + 10| = x + 10 because if x >= -4, then x + 10 is positive, thus |x + 10| = x + 10 (recall that |a| = a if a >= 0).

If you want to open modulus in conventional way then you should consider two cases: x <= -10 and x >= 10.
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Re: M27-14  [#permalink]

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New post 26 Sep 2018, 06:19
Bunuel wrote:

(x - 3)(2x - 3) < 0 to be true the multiples, x - 3 and 2x - 3 must have different signs.

x - 3 > 0 and 2x - 3 < 0 --> x > 3 and x < 3/2. Both cannot be simultaneously true. So, this case is out.

x - 3 < 0 and 2x - 3 > 0 --> x < 3 and x > 3/2. So, 3/2 < x < 3.


Thanks for the explanation.
This is exactly where I had got myself wrong. (my solution was x<3/2 and X<3)
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Re: M27-14  [#permalink]

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New post 26 Sep 2018, 19:05
Bunuel wrote:
Official Solution:


(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x-\frac{3}{2})(x-3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient.

(2) \(|x+10|=2x+8\). The left hand side (LHS) is an absolute value, which is always non-negative, hence RHS must also be non-negative: \(2x+8 \ge 0\) giving us \(x \ge -4\). Now, for this range \(x+10\) is positive, hence \(|x+10|=x+10\). So, \(|x+10|=2x+8\) can be written as \(x+10=2x+8\), solving for \(x\) gives \(x=2\). Sufficient.


Answer: D


Hi Bunuel
Thanks for your explanations
I have a real problem with factorizing quadratics containing "a>1 or a<1" coefficient (ax2+bx+c).
To me it's a real headache & nightmare, making me sometimes to use formula to find the roots which is absolutely time consuming and under the pressure of the test, mostly doesn't work at all.
I would be grateful if you share me anything on "how to factorize a quadratic equation with a non-1 coefficient for x2"
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Re: M27-14  [#permalink]

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New post 26 Sep 2018, 19:48
1
RamSep wrote:
Bunuel wrote:
Official Solution:


(1) \(2x^2+9 \lt 9x\). Factor quadratics: \((x-\frac{3}{2})(x-3) \lt 0\). The roots are \(\frac{3}{2}\) and 3, the "\(\lt\)" sign indicates that the solution lies between the roots: \(1.5 \lt x \lt 3\). Since the only integer in this range is 2, then \(x=2\). Sufficient.

(2) \(|x+10|=2x+8\). The left hand side (LHS) is an absolute value, which is always non-negative, hence RHS must also be non-negative: \(2x+8 \ge 0\) giving us \(x \ge -4\). Now, for this range \(x+10\) is positive, hence \(|x+10|=x+10\). So, \(|x+10|=2x+8\) can be written as \(x+10=2x+8\), solving for \(x\) gives \(x=2\). Sufficient.


Answer: D


Hi Bunuel
Thanks for your explanations
I have a real problem with factorizing quadratics containing "a>1 or a<1" coefficient (ax2+bx+c).
To me it's a real headache & nightmare, making me sometimes to use formula to find the roots which is absolutely time consuming and under the pressure of the test, mostly doesn't work at all.
I would be grateful if you share me anything on "how to factorize a quadratic equation with a non-1 coefficient for x2"


Factoring Quadratics
Solving Quadratic Equations


7. Algebra



For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: M27-14  [#permalink]

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New post 27 Sep 2018, 22:33
Bunuel

For Statement B : |x+10|=2x+8 ,we can consider x+10= 2x+8 and x+10 = - [2x+8] and this gives 2 values 2 and -6 and so we can say B is not sufficient. I also understand your and other justification how B can be correct as well but why I am posting question is because I have seen other problems and solution on our club where when given mod of X , we consider both values of X while doing calculations. So want to understand when to consider both values + and - and when to consider only + value ?
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New post 27 Sep 2018, 22:48
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NAvinash wrote:
Bunuel

For Statement B : |x+10|=2x+8 ,we can consider x+10= 2x+8 and x+10 = - [2x+8] and this gives 2 values 2 and -6 and so we can say B is not sufficient. I also understand your and other justification how B can be correct as well but why I am posting question is because I have seen other problems and solution on our club where when given mod of X , we consider both values of X while doing calculations. So want to understand when to consider both values + and - and when to consider only + value ?


When you solve the way you did, you should plug back the roots to confirm that they do in fact satisfy the equation. If you plug x = -6 you'd get that LHS = 4 and RHS = -4. 4 ≠ -4. So, x = -6 does not satisfy |x + 10| = 2x + 8.


Or if you adopt more conventional way you'd have:

If x < -10, then x + 10 < 0, so |x + 10| = -(x + 10) and in this case we'd have -(x + 10) = 2x + 8 --> x = -6. You should discard this solution because x = -6 is not in the range we are considering (x < -10).

If x >= -10, then x + 10 >= 0, so |x + 10| = x + 10 and in this case we'd have x + 10 = 2x + 8 --> x = 2. This solution is valid because x = 2 is in the range we are considering (x >= -10).

So, |x + 10| = 2x + 8 has only one solution: x = 2.

Hope it's clear.
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New post 02 Oct 2018, 19:28
Thanks for the last post/reply Bunuel. This helps. I will prefer first part of answer as it is easy to apply in exam. I will solve LHS RHS using the roots obtained as one of the easiest and quick trial way. Thanks.
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New post 09 Oct 2018, 04:17
Hello Bunuel
I have doubt in second statement need your help. I have solved it by taking two condition
1) when X>=-10
in this case equation became
x+10=2x+8
X=2
case 2) when X<-10
-X-10=2X+8
X=-3
hence Two solution not sufficient, please help me to get this right.
Thank you.
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New post 09 Oct 2018, 04:25
1
Praveenksinha wrote:
Hello Bunuel
I have doubt in second statement need your help. I have solved it by taking two condition
1) when X>=-10
in this case equation became
x+10=2x+8
X=2
case 2) when X<-10
-X-10=2X+8
X=-3
hence Two solution not sufficient, please help me to get this right.
Thank you.


Plug back x = -3 does it satisfy the equation?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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New post 09 Oct 2018, 04:29
Bunuel got it, thanks man.
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Re: M27-14 &nbs [#permalink] 09 Oct 2018, 04:29

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