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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   95% (hard)

Question Stats: 29% (02:04) correct 71% (02:22) wrong based on 183 sessions

HideShow timer Statistics What is the value of integer $$x$$?

(1) $$2x^2+9 \lt 9x$$

(2) $$|x+10|=2x+8$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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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.

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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.

Hi Bunuel,

Could you please explain the statement A of the solution. How is the root (x-3/2)?

The roots that I am getting are (x-3) and (2x-3).

Please could you help me out with this.

Thanks,
Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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msbcapitalist 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.

Hi Bunuel,

Could you please explain the statement A of the solution. How is the root (x-3/2)?

The roots that I am getting are (x-3) and (2x-3).

Please could you help me out with this.

Thanks,

Yes, $$2x^2 - 9x +9<0$$ can be factored as $$(x - 3)(2x - 3)<0$$ but notice that we can factor 2 out of (2x - 3) and reduce by it to get $$(x-3)(x-\frac{3}{2}) \lt 0$$.

Theory on Inequalities
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
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Bunuel
the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3.
Can you please elaborate on this statement ? How did we reach that conclusion?

Thank you
Math Expert V
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ShravanHemchand wrote:
Bunuel
the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3.
Can you please elaborate on this statement ? How did we reach that conclusion?

Thank you

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Bunuel wrote:
ShravanHemchand wrote:
Bunuel
the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3.
Can you please elaborate on this statement ? How did we reach that conclusion?

Thank you

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?
Math Expert V
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p2bhokie wrote:
Bunuel wrote:
ShravanHemchand wrote:
Bunuel
the "\lt" sign indicates that the solution lies between the roots: 1.5 \lt x \lt 3.
Can you please elaborate on this statement ? How did we reach that conclusion?

Thank you

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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Can you please tell, why are we taking Absolute value as positive?
I have seen in majority of the questions that when we solve Absolute value equation, it gives two solutions. one negative and other positive
For Example: the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html

can you please explain the difference? where we have take both values and where we have to take only positive value
Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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awal_786@hotmail.com wrote:
Can you please tell, why are we taking Absolute value as positive?
I have seen in majority of the questions that when we solve Absolute value equation, it gives two solutions. one negative and other positive
For Example: the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html

can you please explain the difference? where we have take both values and where we have to take only positive value

The absolute value of a number is the value of a number without regard to its sign.

For example, |3| = 3; |-12| = 12; |-1.3|=1.3...

Another way to understand absolute value is as the distance from zero. For example, |x| is the distance between x and 0 on a number line.

From that comes the most important property of an absolute value: since the distance cannot be negative, an absolute value expression is ALWAYS more than or equal to zero.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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Bunuel
Option B can be solved as X

1. x+10=-(2x+8) .....here x= -2/3
2. x+10=2x+8....here x=2
as x is intiger...x=2..sufficient

Is this approach correct?
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(1) 2x2+9<9x2x2+9<9x. Factor quadratics: (x−32)(x−3)<0(x−32)(x−3)<0. The roots are 3232 and 3, the "<<" sign indicates that the solution lies between the roots: 1.5<x<31.5<x<3. Since the only integer in this range is 2, then x=2x=2. Sufficient.

bunuel could u plz explain how are u getting the roots in st1??? how to approach quadratic equation when there is an integer before x^2?
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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.

if $$(x-\frac{3}{2})(x-3) \lt 0$$
then wont x-3/2 < 0
therefore x<3/2

and x-3<0
therefore x<3

then if x is less than 3 and 3/2 at the same time, how does one arrive at x<2
Am I doing this correct?
Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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vdhaval 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.

if $$(x-\frac{3}{2})(x-3) \lt 0$$
then wont x-3/2 < 0
therefore x<3/2

and x-3<0
therefore x<3

then if x is less than 3 and 3/2 at the same time, how does one arrive at x<2
Am I doing this correct?

The correct range is $$1.5 \lt x \lt 3$$, so no you are not doing it right. x < 3 AND x < 1.5 does not make sense. What does it even mean? Can x be 2 since x < 3 or not since x < 1.5. Anyway, please check the links below to study how to derive the ranges for inequalities:

Solving Quadratic Inequalities - Graphic Approach
Inequality tips
Wavy Line Method Application - Complex Algebraic Inequalities

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope this helps.
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Bunuel

His question still stands (the part that's also confusing me too). If (x-3/2)(x-3) < 0, then algebraically (x<3) and (x<3/2). To answer your question, one could think x<3/2 is more limiting than x<3.

Per the answer derived from your solution (x-3/2)(x-3) < 0 when solving for x, the question we ask is

why won't the sign change from x<3/2 to x>3/2? There has to be a negative division in order to change the sign in my knowledge.

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Hi Bunuel,

Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.

Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...

Yes I understand that (x-3) (2x-3)<0 ... I then did the following:

x-3<0 ... so x<3
2x-3<0 ... so 2x<3...x<1.5

How can x be less than 3 and less than 1.5 and thus be an integer 2?

I clearly went wrong somewhere but I don't know where.

Thanks,

Tosin
Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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ttaiwo wrote:
Hi Bunuel,

Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.

Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...

Yes I understand that (x-3) (2x-3)<0 ... I then did the following:

x-3<0 ... so x<3
2x-3<0 ... so 2x<3...x<1.5

How can x be less than 3 and less than 1.5 and thus be an integer 2?

I clearly went wrong somewhere but I don't know where.

Thanks,

Tosin

(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.
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Bunuel wrote:
ttaiwo wrote:
Hi Bunuel,

Thanks for the factoring link...I was having problems factoring quadratics where a was greater than 1.

Anyhow...with my newfound knowledge I attempted this particular question and still came unstuck at one final stage...

Yes I understand that (x-3) (2x-3)<0 ... I then did the following:

x-3<0 ... so x<3
2x-3<0 ... so 2x<3...x<1.5

How can x be less than 3 and less than 1.5 and thus be an integer 2?

I clearly went wrong somewhere but I don't know where.

Thanks,

Tosin

(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 bro. Was cracking my brain all day yesterday on that one.
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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.

Hi Bunuel ,

I understood statement A , however , I'm not able to understand statement 2 , in this , X can hold values from -4<x<2 , then X can be 1 also ??
Please guide ,where I'm going wrong.

I selected E , while taking the gmat club Quant Test , as I got the same answer as yours , in statement 1, however I missed out X is an integer , in that case statement 1 holds true , in statement two , I got values as X=2 and X=-4 , so X hold values from -4 to 2 ?

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Math Expert V
Joined: 02 Sep 2009
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loserunderachiever 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.

Hi Bunuel ,

I understood statement A , however , I'm not able to understand statement 2 , in this , X can hold values from -4<x<2 , then X can be 1 also ??
Please guide ,where I'm going wrong.

I selected E , while taking the gmat club Quant Test , as I got the same answer as yours , in statement 1, however I missed out X is an integer , in that case statement 1 holds true , in statement two , I got values as X=2 and X=-4 , so X hold values from -4 to 2 ?

Does plugging x = 1 makes |x + 10| = 2x + 8 true? |x + 10| = 2x + 8 has only one solution x = 2, I tried to explain (2) in details here: https://gmatclub.com/forum/m27-184490.html#p1419035
_________________ Re: M27-14   [#permalink] 30 May 2018, 22:37

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