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# If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)

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Veritas Prep GMAT Instructor
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If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]

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29 Oct 2010, 11:18
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If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?

(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7

(Still high on mods! Next week, will make questions on some other topic.)
[Reveal] Spoiler: OA

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by Bunuel on 06 Jul 2013, 02:33, edited 1 time in total. Added the OA. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2783 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: Question of the Day - II [#permalink] ### Show Tags 29 Oct 2010, 18:51 7 This post received KUDOS VeritasPrepKarishma wrote: Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)? (A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7 (Still high on mods! Next week, will make questions on some other topic.) put x=0 we get f(x) = 5..rule out D and E put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E Now the answer is either 3 or 4. Reason for above checking of values: for every value of x> 3 the f(x) is quite big because of 4x-1 for every value of x < -1 the f(x) if bigger than 3 and 4. for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3 => only 4 is the probable answer. Thus we only need to check x>=-1 and x<= 1/4 f(x) in this domain is = 1-4x + 3-x + x+1 = 5-4x => for f(x) to be minimum the x should be +ve => for x = 1/4 , f(x) = 4. This can be solved using graph as well by plotting the f(x) in different domains. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html Retired Moderator Joined: 02 Sep 2010 Posts: 803 Location: London Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 01:59 10 This post received KUDOS 1 This post was BOOKMARKED Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x) Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3 g(x) is minimum at x=(1/4) where it is equal to 0. Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point f(1/4)=g(1/4)+h(1/4)=0+4 Hence answer is 4 _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 08:53 45 This post received KUDOS Expert's post 13 This post was BOOKMARKED That's right! I can count on you guys to always give the correct answer and some innovative ways of solving... Let me give you the method I would use to solve it too... Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend... So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4 So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible? Attachment: Ques.jpg [ 4.56 KiB | Viewed 14954 times ] Can I say that minimum total distance will be covered from point 1/4? Attachment: Ques1.jpg [ 3.43 KiB | Viewed 14937 times ] The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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30 Oct 2010, 08:57
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In other words, as my mentor says, assume there is one guy on point 3, one on point -1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x - 3) to f(x)...
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 11 Jul 2010 Posts: 224 Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 09:29 would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 16:17 3 This post received KUDOS Expert's post 3 This post was BOOKMARKED gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment: Ques.jpg [ 5.2 KiB | Viewed 14965 times ] Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6| f(x) = |2x - 3| + |4x + 7| etc _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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09 Nov 2010, 23:23
Great Method Karishma !!

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Re: Question of the Day - II [#permalink]

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14 Jun 2011, 02:14
for x = 0, f(x) = 5. POE options C,D and E.
for x = 1/4, f(x) = 4.
B it is.
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Re: Question of the Day - II [#permalink]

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14 Jun 2011, 03:47
Very good question thanks for posting
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Re: Question of the Day - II [#permalink]

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15 Mar 2012, 11:07
VeritasPrepKarishma wrote:
gmat1011 wrote:

The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance

Yes, that is right! The answer still remains 1/4.
Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4.
Attachment:
Ques.jpg

Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6|
f(x) = |2x - 3| + |4x + 7| etc

Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.
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Re: Question of the Day - II [#permalink]

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15 Mar 2012, 11:24
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Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.

Forget these numbers. Think logically.

My house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles)

Now say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance.
If there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 15 Mar 2012, 11:33 5 This post received KUDOS Expert's post 2 This post was BOOKMARKED ficklehead wrote: In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? Please correct me if I am wrong. x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together. If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11. To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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15 Mar 2012, 14:48
VeritasPrepKarishma wrote:
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.

x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance.
Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together.
If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet.
If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11.

To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11

Thanks Karishma for this detailed explanation.
I got it now.
I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.

To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
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Re: Question of the Day - II [#permalink]

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15 Mar 2012, 22:03
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I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.

To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?

Look at the diagram below. Make a number line in such questions.
Attachment:

Ques3.jpg [ 5.1 KiB | Viewed 13919 times ]

For x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between -1 and 1 and they will cover a total distance of 11 only. So x can take any value -1 < x < 1 and the value of the expression will be 11.

|2x-3|+|4x+7| = 2|x-3/2| + 4|x+7/4|

Attachment:

Ques4.jpg [ 5.35 KiB | Viewed 13934 times ]

There are 4 people at -7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4
For these people to meet covering the minimum distance, the 2 people X and Y should travel to point -7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is -7/4.
Check by putting x = -7/4. You get |2x-3|+|4x+7| = 13/2

Also see that when you put x = 0 or 3/2 etc, the value of the expression is higher.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 09:01 Thanks a lot, Karishma. Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 19:48 I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ? Manager Joined: 28 Jul 2011 Posts: 238 Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 14:33 I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 21:07 kuttingchai wrote: I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Put x = 1/4 and the minimum value you will get is 4. How did you get x = 1/2? I would suggest you to check out one of the approaches mentioned above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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20 Mar 2012, 21:17
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I am wondering how can this method be used in questions where there are negative between terms :
Ex: minimum value of : |x+6|-|x-1| ?

You can do it with a negative sign too.
You want to find the minimum value of (distance from -6) - (distance from 1)

Make a number line with -6 and 1 on it.
(-6)..........................(1)

Think of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .

What if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7

If you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.
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Re: Question of the Day - II   [#permalink] 20 Mar 2012, 21:17

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