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

 It is currently 14 Oct 2019, 08:39 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

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.  If 4<=x<=6 and 2<=y<=3, then the minimum possible value of

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

Hide Tags

GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

1
1
11 00:00

Difficulty:   35% (medium)

Question Stats: 71% (01:52) correct 29% (01:51) wrong based on 221 sessions

HideShow timer Statistics

GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

$$?\,\, = \,\,\min \,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,\, = \,\,\min \,\left| {{y^2} - {x^2}} \right|$$

$$2 \le y \le 3\,\,\,\,\, \Rightarrow \,\,\,\,\,4 \le {y^2} \le 9$$

$$4 \le x \le 6\,\,\,\,\, \Rightarrow \,\,\,\,\,16 \le {x^2} \le 36\,\,\,\,\, \Rightarrow \,\,\,\,\, - 36 \le - {x^2} \le - 16$$

$$\left. \matrix{ 4 \le {y^2} \le 9 \hfill \cr - 36 \le - {x^2} \le - 16\,\,\, \hfill \cr} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\, - 32 \le {y^2} - {x^2} \le - 7\,\,\,\,\, \Rightarrow \,\,\,\,\,7 \le \left| {{y^2} - {x^2}} \right| \le 32\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 7$$

The correct answer is (D).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
VP  D
Joined: 31 Oct 2013
Posts: 1472
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

$$4\leq x \leq6$$

$$2\leq y \leq3$$

Given,

$$|y^2 - x^2|$$

but we are asked to find out the minimum value of |y^2 - x^2| .

we must work with corner values / extremes values of x and y.

$$2^2 - 4^2$$ = 4 - 16 = - 12 = | -12 | = 12

$$2^2 - 6^2$$= 4 - 36 = -32 = | -32| = 32

$$3^2 - 4^2$$= 9 - 16 = -7 = | - 7| = 7.

$$3^2 - 6^2$$= 9 - 36 = -27 = | -27| = 27.

D is the correct answer.
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2974
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

1
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

For minimum Possible value of ∣(y−x)(y+x)∣ the absolute values of (y-x) and (y+x) must be MINIMUM

Minimum ABSOLUTE Value of y-x = l3-4l = l-1l = 1

Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7

i.e. Minimum value of l(y-x)(y+x)l = l1*7l = 7

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Director  G
Joined: 09 Mar 2018
Posts: 997
Location: India
Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

So the given expression can be written as

$$|y^2 - x^2|$$

Only case possible is when we maximize y = 4 and minimize x = 3

|9-16|

7
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Manager  B
Joined: 02 Nov 2018
Posts: 50
Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

GMATinsight wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

For minimum Possible value of ∣(y−x)(y+x)∣ the absolute values of (y-x) and (y+x) must be MINIMUM

Minimum ABSOLUTE Value of y-x = l3-4l = l-1l = 1

Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7

i.e. Minimum value of l(y-x)(y+x)l = l1*7l = 7

Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries..
Intern  Joined: 16 May 2018
Posts: 4
Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries.
Board of Directors D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4749
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Values of x can be 4 , 5 & 6 ; Values of y can be 2 , 3

SO, the Possible min value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$

Will be $$\,\left| {\left( {3 - 4} \right)\left( {4 + 3} \right)} \right|\,$$ = 1*7 = 7 , Answer must be (D)
_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Manager  B
Joined: 02 Nov 2018
Posts: 50
Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of  [#permalink]

Show Tags

jamalabdullah100 wrote:
GMATinsight wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 11)

If $$\,4 \le x \le 6\,$$ and $$\,2 \le y \le 3$$ , the minimum possible value of $$\,\left| {\left( {y - x} \right)\left( {y + x} \right)} \right|\,$$ is:

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

For minimum Possible value of ∣(y−x)(y+x)∣ the absolute values of (y-x) and (y+x) must be MINIMUM

Minimum ABSOLUTE Value of y-x = l3-4l = l-1l = 1

Minimum ABSOLUTE Value of y+x = l3+4l = l7l = 7

i.e. Minimum value of l(y-x)(y+x)l = l1*7l = 7

Why can't the Minimum ABSOLUTE Value of y+x = 2+4 = 6? Both y=2 and x=4 are included in the boundaries..

Can someone help with the above please? It doesn't make sense to me.. Re: If 4<=x<=6 and 2<=y<=3, then the minimum possible value of   [#permalink] 23 Sep 2019, 05:06
Display posts from previous: Sort by

If 4<=x<=6 and 2<=y<=3, then the minimum possible value of

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  