GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 13 Jul 2020, 07:44

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

If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +

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

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 65238
If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 26 May 2020, 01:42
14
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

17% (02:05) correct 83% (01:31) wrong based on 78 sessions

HideShow timer Statistics

If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5

Project PS Butler


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS



Are You Up For the Challenge: 700 Level Questions

_________________
Senior Manager
Senior Manager
User avatar
G
Joined: 05 Aug 2019
Posts: 301
Location: India
Concentration: Leadership, Technology
GMAT 1: 600 Q50 V22
GPA: 4
CAT Tests
If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post Updated on: 28 May 2020, 21:34
2
f(x)=|x|+|x+3|+|x+6|+|x+9|+|x+12|+|x+15|+|x+18|+|x+21|
For the equation to be minimum, individual items should be minimized.
As all are absolute values, They can occur only on the inflection points.
As there are 8 terms in AP, it should be at mean point = average of 4th and 5th term
Avg(9,12) = 10.5
But as x is an integer, x should be near the average point
x = -11, -10
So 2 values, B

Correction - I should have checked the expression value before jumping to answer, Apparently expression value on x =-9, -10,-11,-12 remains the same.
Hence 4 values

Originally posted by sambitspm on 26 May 2020, 02:13.
Last edited by sambitspm on 28 May 2020, 21:34, edited 1 time in total.
Intern
Intern
avatar
B
Joined: 07 Oct 2016
Posts: 4
Re: If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 26 May 2020, 02:30
x= 0 -> 0 + 3 + 6 + 9 + 12 + 15+ 18 + 21
x = - 3 -> 3 + 0 + 3 + 6 + 9 + 12 + 15 + 18
x = - 6 -> 6 + 3 + 0 + 3 + 6 + 9 + 12 + 15
x = - 9 -> 9 + 6 + 3 + 0 + 3 + 6 + 9 + 12
x= - 12 -> 12 + 9 + 6 + 3 + 0 + 3 + 6 + 9
x= - 15 -> 15 + 12 + 9 + 6 + 3 + 0 + 3 + 6
x= - 18 -> 18 + 15 + 12 + 9 + 6 + 3 + 0 + 3
x= = -21 -> 21 + 18 + 15 + 12 + 9 + 6 + 3 + 0

-> 2 values when x = - 9 and x = -12 so that the value of f(x) is minimum
GMAT Club Legend
GMAT Club Legend
User avatar
V
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 4361
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Reviews Badge
If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 26 May 2020, 02:34
1
1
Bunuel wrote:
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions


Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum


there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D
Attachments

Screenshot 2020-05-26 at 6.14.29 PM.png
Screenshot 2020-05-26 at 6.14.29 PM.png [ 128.15 KiB | Viewed 782 times ]


_________________
Prepare with PERFECTION to claim Q≥50 and V≥40 !!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

Most affordable l Comprehensive l 2000+ Qn ALL with Video explanations l LINK: Courses and Pricing
Click for FREE Demo on VERBAL & QUANT
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
FREE GMAT Resource: 22 FREE (FULL LENGTH) GMAT CATs LINKS l NEW OG QUANT 50 Qn+VIDEO Sol.
Manager
Manager
avatar
S
Joined: 05 Jan 2020
Posts: 129
Re: If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 26 May 2020, 07:41
1
1
Please excuse as I could not present the number line screenshot.

Representing the numbers on number line will provide a visual and easily comprehensible solution.
The terminal points are 0 and -21.
Let's consider the extreme terms only. Any value of x that lies between these two values (inclusive) will provide us the minimum result of the modulus operands.

Since the sequence has even number of terms, therefore, the minimum value will occur for all values of x between the two mid terms (inclusive). So min value will occur at -9, -10, -11, -12.

Note:
1) First arrange the terms of a modulus function in ascending order.
2) If the number of terms in the function is odd, then the min value will occur at the mid term. If the number of terms is even, then the min value will occur for all the values between the two mid terms.
Manager
Manager
avatar
G
Joined: 12 Mar 2019
Posts: 168
Re: If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 27 May 2020, 23:27
GMATinsight wrote:
Bunuel wrote:
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions


Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum


there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D


In actual exam there would be enough time to do math for all values. My question is why answer is 4 values and not 2, If 10.5 is midterm, can't we only consider -10 and -11 as values. why we included -9 and -12 . Appreciate reply. Thanks
Manager
Manager
avatar
S
Joined: 05 Jan 2020
Posts: 129
Re: If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 28 May 2020, 08:41
1
rishab0507 wrote:
GMATinsight wrote:
Bunuel wrote:
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions


Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum


there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D


In actual exam there would be enough time to do math for all values. My question is why answer is 4 values and not 2, If 10.5 is midterm, can't we only consider -10 and -11 as values. why we included -9 and -12 . Appreciate reply. Thanks


Hi Rishab,

The question deals with modulus operands. That basically is the distance (absolute value) of a point with reference to another point. It will always be positive.
Now coming to your question, consider any point between -9 and -12. The sum of the distances (absolute values) from -9 to x and x to -12 will always remain the same.

To simplify it further, consider x=-10. Distance from -9 to -10, d1=1. Distance from -10 to -12, d2=2. Sum of d1+d2=3.
If x=-9, d1=0 and d2=3. Again, sum=3. You can see that the absolute value will always be 3 for any point between -9 and -12 inclusive. Any value of x beyond these points will increase the absolute value and thus, we won't get the minimum value.

Num of integer values from -9 to -12 is 4. Ans: 4

Kudos if it helps!
Manager
Manager
avatar
G
Joined: 12 Mar 2019
Posts: 168
Re: If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +  [#permalink]

Show Tags

New post 28 May 2020, 12:26
Lipun wrote:
rishab0507 wrote:
GMATinsight wrote:
Bunuel wrote:
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions


Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum


there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D


In actual exam there would be enough time to do math for all values. My question is why answer is 4 values and not 2, If 10.5 is midterm, can't we only consider -10 and -11 as values. why we included -9 and -12 . Appreciate reply. Thanks


Hi Rishab,

The question deals with modulus operands. That basically is the distance (absolute value) of a point with reference to another point. It will always be positive.
Now coming to your question, consider any point between -9 and -12. The sum of the distances (absolute values) from -9 to x and x to -12 will always remain the same.

To simplify it further, consider x=-10. Distance from -9 to -10, d1=1. Distance from -10 to -12, d2=2. Sum of d1+d2=3.
If x=-9, d1=0 and d2=3. Again, sum=3. You can see that the absolute value will always be 3 for any point between -9 and -12 inclusive. Any value of x beyond these points will increase the absolute value and thus, we won't get the minimum value.

Num of integer values from -9 to -12 is 4. Ans: 4

Kudos if it helps!


Kudos to u. I am finding hard to understand modulus question, It seems your concepts are rock solid in this. Can you suggest me from where to study concept of modulus, mostly 700 level ques of gmat come from this section and i am struggling, Have you found any good material on this forum or anywhere you can suggest.
thanks
GMAT Club Bot
Re: If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +   [#permalink] 28 May 2020, 12:26

If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x +

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





cron

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