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If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)
[Reveal] Spoiler: OA

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Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A
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Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A


Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks

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Re: M08-16 [#permalink]

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Expert's post
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A


Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks


When \(x \le 2\) then x-3 (expression in modulus) is less than 0, thus \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).
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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

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Bunuel wrote:
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A


Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks


When \(x \le 2\) then x-3 (expression in modulus) is less than 0, thus \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).


Oh Ok. Logically makes sense. Thanks

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Re: M08-16 [#permalink]

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New post 08 Aug 2015, 23:47
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buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A


Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks


Bunuel,

I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered |x-3| ?
We should then only consider the positive option (x-3)
I chose (x-3) for the first term and ended up as option B

Am i missing anything?
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sagar2911 wrote:
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A


Hi Bunuel

I understand the properties but unable to comprehend the highlighted part.

It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with
\(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\)

Pls help understand this

Thanks


Bunuel,

I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered |x-3| ?
We should then only consider the positive option (x-3)
I chose (x-3) for the first term and ended up as option B

Am i missing anything?


Because the question says "each expression under the square root is greater than or equal to 0", we need to understand that if x-3 is 0, then 2-x is a negative number, so the root of 2 - x is root of -1 what breaks the rule stated in the question.

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Re: M08-16 [#permalink]

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New post 15 Aug 2015, 00:33
roneileonel wrote:
sagar2911 wrote:

Bunuel,

I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered |x-3| ?
We should then only consider the positive option (x-3)
I chose (x-3) for the first term and ended up as option B

Am i missing anything?


Because the question says "each expression under the square root is greater than or equal to 0", we need to understand that if x-3 is 0, then 2-x is a negative number, so the root of 2 - x is root of -1 what breaks the rule stated in the question.


Thank you dude! This helps a lot!
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Re: M08-16 [#permalink]

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New post 11 Feb 2016, 09:41
schak2rhyme wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Next: as x≤2 then |x−3| becomes |x−3|=−(x−3)=−x+3.

In the excerpt from the solution above why is |x−3| not >=0 instead of <=.Please explain.



The question says the expression cant be negative .. not roots .. x=-3 because the whole expression root of 2-x should be positive ..even if x is negative the first expression remains positive .. am i correct ?

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New post 16 Jul 2016, 21:32
I think this is a high-quality question and I agree with explanation. Brilliant Question !
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New post 01 Sep 2016, 21:09
Please let me know if I did this correctly alternatively.

\(\sqrt{x^2 - 6x + 9}\) , solving (x-3)*(x-3) , x = 3,

plug in 3 for all x, becomes answer A

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New post 02 Sep 2016, 03:58
xnthic wrote:
Please let me know if I did this correctly alternatively.

\(\sqrt{x^2 - 6x + 9}\) , solving (x-3)*(x-3) , x = 3,

plug in 3 for all x, becomes answer A


No, that's not correct. Why are you equating \(\sqrt{x^2 - 6x + 9}\) to 0? Why should x be 3? As shown in the solution x must be less than or equal to 2 for \(\sqrt{2-x}\) to be defined.
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New post 26 Sep 2016, 12:42
This is a really great question. Thanks Bunuel :-D

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I think this is a high-quality question and I agree with explanation.

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New post 02 Jul 2017, 05:44
Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A

Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x-3 to be positive. the quantity under the root is (x-3)^2, which is always positie irrespective of what value x takes.

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prateeksaxena55 wrote:
Bunuel wrote:
Official Solution:

If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\)?

A. \(\sqrt{2 - x}\)
B. \(2x - 6 + \sqrt{2 - x}\)
C. \(\sqrt{2 - x} + x - 3\)
D. \(2x - 6 + \sqrt{x - 2}\)
E. \(x + \sqrt{x - 2}\)


Absolute value properties:

When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\);

Another important property to remember: \(\sqrt{x^2}=|x|\).

Back to the original question:

\(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3\).

Now, as the expressions under the square roots are more than or equal to zero than \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

\(|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}\).


Answer: A

Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x-3 to be positive. the quantity under the root is (x-3)^2, which is always positie irrespective of what value x takes.


The point is that \(\sqrt{x^2}=|x|\). Next, \(|x| = x\), when \(x \geq 0\) and \(|x| = -x\), when \(x \leq 0\).

We have \(\sqrt{x^2 - 6x + 9}=\sqrt{(x-3)^2}\). Thus according to the above property \(\sqrt{x^2 - 6x + 9}=\sqrt{(x-3)^2}=|x-3|\). Next, since \(x \le 2\), then \(x - 3 < 0\) and therefore \(|x-3| = -(x-3) = 3-x\).

Notice here that since \(x \le 2\), then 3 - x will be positive, so the condition that the square root must give a non-negative result is satisfied.
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New post 04 Jul 2017, 21:48
The question says " each expression under the square root is greater than or equal to 0"
Hence, It comes out that x is less than or equal to 2 by second expression and for the first expression why at all we are thinking it to be square of (x-3) as this will result in values greater or equal to 3 not valid as it is less than 2. but if we take it as square of (3-x) then it will less than 3 satisfying outcome from other equation.

Therefore, the shorter method of solving will be as follows.

Step 1. from second expression find x is less than or equal to 2
Step 2. solve first expression as sqaure of (3-x) not square of (x-3) { as this gives x is greater or equal to 3} and then you get the answer.

It avoids complexity of modulus.

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New post 24 Oct 2017, 23:53
In the GMAT Club math book:
"When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
That is, sqrt(25) = 5, NOT +-5.

So, why sqrt(x^2) must be |x|, not just +x?

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Terabyte wrote:
In the GMAT Club math book:
"When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
That is, sqrt(25) = 5, NOT +-5.

So, why sqrt(x^2) must be |x|, not just +x?


\(\sqrt{x^2}=|x|\):[/b]

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Hope it's clear.
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New post 25 Oct 2017, 00:05
Bunuel wrote:

\(\sqrt{x^2}=|x|\):[/b]

The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Hope it's clear.


Yes, it's clear. Thank you very much, Bunuel. Learning a lot from your posts. Aiming for Q50

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Re: M08-16   [#permalink] 25 Oct 2017, 00:05

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