GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 26 Jun 2019, 16:03

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

M08-16

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Intern
Intern
avatar
B
Joined: 04 May 2017
Posts: 1
Re: M08-16  [#permalink]

Show Tags

New post 01 Nov 2017, 08:23
1
I think this is a high-quality question and I agree with explanation.
Manager
Manager
avatar
G
Joined: 01 Feb 2017
Posts: 227
Re: M08-16  [#permalink]

Show Tags

New post 09 Nov 2017, 01:51
2
Using the plug-in method is a faster way to get correct answer.

As we know that expression under square root is <=0, x must be <=2.

Plug in any value of x that is <=2, say x=1, in given expression​and each answer choice.

Only Choice A will match.

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 24 Jun 2013
Posts: 1
GMAT ToolKit User Reviews Badge
Re: M08-16  [#permalink]

Show Tags

New post 23 Nov 2017, 04:28
Bunuel wrote:
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. \(x + \sqrt{x - 2}\)



Hie Bunuel,

Can you please provide link to similar kind of question
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55801
Re: M08-16  [#permalink]

Show Tags

New post 23 Nov 2017, 05:28
prk0789 wrote:
Bunuel wrote:
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. \(x + \sqrt{x - 2}\)



Hie Bunuel,

Can you please provide link to similar kind of question



8. Exponents and Roots of Numbers



_________________
Intern
Intern
avatar
Joined: 17 Feb 2018
Posts: 1
Re M08-16  [#permalink]

Show Tags

New post 20 Feb 2018, 02:00
1
I think this is a high-quality question and I agree with explanation.
Manager
Manager
avatar
G
Joined: 26 Dec 2017
Posts: 156
Reviews Badge
Re: M08-16  [#permalink]

Show Tags

New post 02 Jun 2018, 06:27
Here as per condition we can get both x<=2 and x>=3
wondering how only x<=2 condition was considered initially
_________________
--If you like my post pls give kudos
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55801
Re: M08-16  [#permalink]

Show Tags

New post 02 Jun 2018, 09:27
Manager
Manager
User avatar
S
Joined: 26 Feb 2018
Posts: 51
Location: India
GMAT 1: 640 Q45 V34
GPA: 3.9
WE: Web Development (Computer Software)
M08-16  [#permalink]

Show Tags

New post Updated on: 22 Jun 2018, 23:02
Hey Bunuel!

I'm a bit confused over here. I do understand that \(\sqrt{x^2} = |x|\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).

So the entity, \(|x-3| ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?

Let me know where am I going wrong.

Thanks a lot
PrayasT

Originally posted by PrayasT on 22 Jun 2018, 14:46.
Last edited by PrayasT on 22 Jun 2018, 23:02, edited 1 time in total.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55801
Re: M08-16  [#permalink]

Show Tags

New post 22 Jun 2018, 22:34
PrayasT wrote:
Hey Bunnel!

I'm a bit confused over here. I do understand that \(\sqrt{x^2} = |x|\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).

So the entity, \(|x-3| ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?

Let me know where am I going wrong.

Thanks a lot
PrayasT


Absolute value is always non-negative so \(|x-3| ≥ 0\) is true for any value of x. Or in other way, since the square of a number is non-negative then \(\sqrt{(x-3)^2}\) is defined for any value of x.

We get the range of x from \(\sqrt{2 - x}\). The square root to be defined the expression under the square root must be greater than or equal to 0 --> \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\).
_________________
Manager
Manager
User avatar
S
Joined: 26 Feb 2018
Posts: 51
Location: India
GMAT 1: 640 Q45 V34
GPA: 3.9
WE: Web Development (Computer Software)
Re: M08-16  [#permalink]

Show Tags

New post 22 Jun 2018, 23:05
Bunuel wrote:
PrayasT wrote:
Hey Bunnel!

I'm a bit confused over here. I do understand that \(\sqrt{x^2} = |x|\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).

So the entity, \(|x-3| ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?

Let me know where am I going wrong.

Thanks a lot
PrayasT


Absolute value is always non-negative so \(|x-3| ≥ 0\) is true for any value of x. Or in other way, since the square of a number is non-negative then \(\sqrt{(x-3)^2}\) is defined for any value of x.

We get the range of x from \(\sqrt{2 - x}\). The square root to be defined the expression under the square root must be greater than or equal to 0 --> \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\).



That makes perfect sense! Thanks a lot!

Regards,
PrayasT
Intern
Intern
avatar
B
Joined: 08 Aug 2017
Posts: 31
Re: M08-16  [#permalink]

Show Tags

New post 20 Jul 2018, 21:36
I solved in the following manner:

Once you realize x<=2, plug in x=2 in the prompt, you get 0.
Now check answer choices by plugging in x=2 to get the value 0.

Only answer choice A fits.
Intern
Intern
avatar
B
Joined: 26 Dec 2016
Posts: 34
Location: Canada
Concentration: Finance, Entrepreneurship
GMAT 1: 710 Q45 V41
GMAT 2: 610 Q41 V33
GMAT 3: 710 Q47 V40
GPA: 3.5
Reviews Badge
Re: M08-16  [#permalink]

Show Tags

New post 12 Aug 2018, 11:25
What a great question.
_________________
GMAT 1: 610
GMAT 2: 710
GMAT 3: 710
Intern
Intern
avatar
B
Joined: 08 Jan 2017
Posts: 17
Location: India
GMAT 1: 630 Q45 V31
GMAT 2: 670 Q48 V34
GPA: 4
WE: Engineering (Computer Software)
Re: M08-16  [#permalink]

Show Tags

New post 17 Nov 2018, 04:52
Hi Bunuel,

In ques - Link: https://gmatclub.com/forum/d01-183474.html#p1571995

You have made a point that:
"When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root.

That is:
√9=3, NOT +3 or -3;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3."


Why same is not valid here?
Why are we considering the value of sqrt as positive and negative?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55801
Re: M08-16  [#permalink]

Show Tags

New post 17 Nov 2018, 08:14
SSwami wrote:
Hi Bunuel,

In ques - Link: https://gmatclub.com/forum/d01-183474.html#p1571995

You have made a point that:
"When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root.

That is:
√9=3, NOT +3 or -3;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3."


Why same is not valid here?
Why are we considering the value of sqrt as positive and negative?


Where do you see a contradiction? When we have that \(\sqrt{x^2}=|x|\), still the square root is giving a non-negative result because the absolute value of a number is always non-negative. Please read the previous pages, I think several posts there explain this clearly.
_________________
Intern
Intern
avatar
B
Joined: 18 Dec 2018
Posts: 3
Re M08-16  [#permalink]

Show Tags

New post 23 Dec 2018, 08:58
Why is the expression >=0 but becomes <=0 in the explanation? If according to the setting of the question, shouldn't the answer be D?
Intern
Intern
avatar
B
Joined: 18 Dec 2018
Posts: 3
Re M08-16  [#permalink]

Show Tags

New post 23 Dec 2018, 11:43
Because we know X<=2, -(X-3) instead of (X-3).
But how do we make sure if -(X-3) satisfy X<=2?
Intern
Intern
avatar
B
Joined: 22 Oct 2018
Posts: 4
GMAT 1: 770 Q50 V45
Premium Member
Re M08-16  [#permalink]

Show Tags

New post 07 Jan 2019, 21:10
Fab Question!
GMAT Club Bot
Re M08-16   [#permalink] 07 Jan 2019, 21:10

Go to page   Previous    1   2   [ 37 posts ] 

Display posts from previous: Sort by

M08-16

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel



Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne