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Math Expert V
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What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 64% (01:55) correct 36% (01:58) wrong based on 469 sessions

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What is the minimum value of |x-4| + |x+3| + |x-5| ?
A. -3
B. 3
C. 5
D. 7
E. 8

Kudos for correct solution

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Math Expert V
Joined: 02 Aug 2009
Posts: 7953
Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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4
8
chetan2u wrote:
What is the minimum value of |x-4| + |x+3| + |x-5| ?
A. -3
B. 3
C. 5
D. 7
E. 8

OA after three days

Good Explanation Engr2012.

The Q tests us on the understanding of Property of a Modulus..

CONCEPT:

-
when we have two modulus, the value will be the minimum between the two Critical Points and will increase on either side of CP.

the two extremeties are -3 and 5, so |x+3| + |x-5| will remain constant, 3+5=8, within the range from -3 to 5..
so we have to keep x-4 as minimum to have the minimum value for |x-4| + |x+3| + |x-5| ..
x-4 will be 0 at x=4..
so our min value will OCCUR at x=4 and will be 8
E

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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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4
7
chetan2u wrote:
What is the minimum value of |x-4| + |x+3| + |x-5| ?
A. -3
B. 3
C. 5
D. 7
E. 8

OA after three days

The ranges that this question needs to be looked at are:

1. x<-3
2. -3 $$\leq$$ x < 4
3. 4 $$\leq$$ x < 5
4. x $$\geq$$ 5

Taking each one of them one by one:

1. x<-3 , the given expression will become, f(x) = |x-4| + |x+3| + |x-5| = -(x-4)-(x+3)-(x-5) = -3x+6 . Slope of the line f(x) = -3x+6 is <0 and thus this line will not have a defined minimum (= -inf !)

2. In the range, -3 $$\leq$$ x < 4, the equation takes the form, f(x) = 12-x, again negative slope, potential of a minimum value in the given range. The minimum value will be at x=-3 ---> 12+3=15. Not in the options, move on.

3. 4 $$\leq$$ x < 5 , equation becomes f(x) = x+4, positive slope ---> minimum value will be when x=4 ---> 4+4 =8 .

4. x $$\geq$$ 5, equation becomes, f(x)= 3x-6, positive slope and as such the minimum will be at x=5 , giving you f(x)=9 as the minimum value.

Hence E, 8 is the correct answer.

Method 2:

|x-a| is the distance of x from a.

Thus f(x) = |x-4| + |x+3| + |x-5| = distance of x from 4 + distance of x from -3 + distance of x from 5.

The ranges remain the same as the ones mentioned above, but now let us assume some values in the given ranges, (ignore the first one as it should be pretty obvious that this range will not give a minimum value!)

Ranges # 2 and 3, -3 $$\leq$$ x < 4, take x=-2, you get f(x) =13, move to x=0, you get f(x) = 12, the values are decreasing, good. Now take x= 3, you get f(x) = 9, for x=4, you get f(x)=8, when you take x=5, f(x) = 9, starts to increase again.

Thus 8 , E is the correct answer.

This method though depends on the fact that the options given are integers and the nature of the values (decreasing to increasing) does not vary inconsistently. If the options would have been decimals or the nature of the values would have been varying, then this method will not be a recommended course of action.

Hope this helps.

FYI, the graph of the given functions look like this:

Attachment: 2016-03-12_23-27-34.jpg [ 58.43 KiB | Viewed 9808 times ]
General Discussion
Marshall & McDonough Moderator D
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Posts: 1686
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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5
2
A cannot be the answer as all the three terms are in modulus and hence the answer will be non negative.

|x-4| >= 0 --> Minimum occurs at x = 4
|x+3| >= 0 --> Minimum occurs at x = -3
|x-5| >= 0 --> Minimum occurs at x = 5

x = -3 --> Result = 7 + 0 + 8 = 15. Also any negative value will push the combined value of |x-4| + |x-5| to a value > 9.
x = 4 --> Result = 0 + 7 + 1 = 8
x = 5 --> Result = 1 + 8 + 0 = 9
x = 3 --> Result = 1 + 6 + 2 = 9

So minimum value of the expression occurs at x = 4 and the resultant value = 8

Manager  Joined: 17 Oct 2015
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What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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1
1
Hello,

We can start here by defining the Critical point limits for potential sign changes at may occur.
1. x<-3:

x-4=-ve, x+3=-ve and x-5=-ve==>AFTER OPENING MODs==> -x+4-x-3-x+5==>-3x+6==>infintely positive.

2. -3=<x<0

x-4=-ve, x+3=+ ve and x-5=-ve==> After mod. are opened==> -x+4+x+3-x+5==>12-x==> min. poss value would be if x=-3===> 15.

3. 0=<x<4:

x-4 -ve, x+3 +ve, and x-5 -ve==>-x+12==>min poss. value would be when x=0==>min. value of function in this limit=12.

4. 4=<x<5:
x-4 +ve, x+3 +ve, x-5 -ve.===>x+4 (+ve) and minimum poss. value as per this limit=8.

5. x>=5:
x-4 +ve, x+3 +ve and x+5 +ve.===>3x+6==> infinitely positive.

Hence minimum poss. value of function overall=8=Option E.

I understand i shouldn't have put 0 in between for extra limit. but personally i'm kind of scared of zero. it sometimes leads to disasters when ignored.

Also, do let me know Engr2012 and chetan2u how much time a person should take solving this. i took 4:56 minutes.
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What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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debbiem wrote:
Hello,

We can start here by defining the Critical point limits for potential sign changes at may occur.
1. x<-3:

x-4=-ve, x+3=-ve and x-5=-ve==>AFTER OPENING MODs==> -x+4-x-3-x+5==>-3x+6==>infintely positive.

2. -3=<x<0

x-4=-ve, x+3=+ ve and x-5=-ve==> After mod. are opened==> -x+4+x+3-x+5==>12-x==> min. poss value would be if x=-3===> 15.

3. 0=<x<4:

x-4 -ve, x+3 +ve, and x-5 -ve==>-x+12==>min poss. value would be when x=0==>min. value of function in this limit=12.

4. 4=<x<5:
x-4 +ve, x+3 +ve, x-5 -ve.===>x+4 (+ve) and minimum poss. value as per this limit=8.

5. x>=5:
x-4 +ve, x+3 +ve and x+5 +ve.===>3x+6==> infinitely positive.

Hence minimum poss. value of function overall=8=Option E.

I understand i shouldn't have put 0 in between for extra limit. but personally i'm kind of scared of zero. it sometimes leads to disasters when ignored.

Also, do let me know Engr2012 and chetan2u how much time a person should take solving this. i took 4:56 minutes.

3 mods questions are rare in GMAT (didnt see any across my 3 attempts) and as such if they do come in the GMAT, try to remember what chetan2u has mentioned above that minimum value will be between the 2 extreme critical points. But your detailed method is correct and should have taken you 2-3 minutes.

In this case, x=0 should not be taken into account as it is NOT a critical point. If you had a mod |x| in the given expression, then yes, x=0 would have become a critical point.

Additionally, for absolute value questions, try to plot a rough graph of what it should look like and it will help you in honing onto the range that you need to worry about. As shown in the graph in my post, the only value that you were supposed to worry about would have been in the range $$4 \leq x < 5$$. This can reduce the time taken to solve this question to less than 2 minutes.

Hope this helps.
Manager  Joined: 17 Oct 2015
Posts: 101
Location: India
Concentration: Finance
Schools: ISB '18 (S)
GMAT 1: 690 Q47 V37 GMAT 2: 700 Q44 V41 WE: Corporate Finance (Commercial Banking)
Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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Engr2012 wrote:
debbiem wrote:
Hello,

We can start here by defining the Critical point limits for potential sign changes at may occur.
1. x<-3:

x-4=-ve, x+3=-ve and x-5=-ve==>AFTER OPENING MODs==> -x+4-x-3-x+5==>-3x+6==>infintely positive.

2. -3=<x<0

x-4=-ve, x+3=+ ve and x-5=-ve==> After mod. are opened==> -x+4+x+3-x+5==>12-x==> min. poss value would be if x=-3===> 15.

3. 0=<x<4:

x-4 -ve, x+3 +ve, and x-5 -ve==>-x+12==>min poss. value would be when x=0==>min. value of function in this limit=12.

4. 4=<x<5:
x-4 +ve, x+3 +ve, x-5 -ve.===>x+4 (+ve) and minimum poss. value as per this limit=8.

5. x>=5:
x-4 +ve, x+3 +ve and x+5 +ve.===>3x+6==> infinitely positive.

Hence minimum poss. value of function overall=8=Option E.

I understand i shouldn't have put 0 in between for extra limit. but personally i'm kind of scared of zero. it sometimes leads to disasters when ignored.

Also, do let me know Engr2012 and chetan2u how much time a person should take solving this. i took 4:56 minutes.

3 mods questions are rare in GMAT (didnt see any across my 3 attempts) and as such if they do come in the GMAT, try to remember what chetan2u has mentioned above that minimum value will be between the 2 extreme critical points. But your detailed method is correct and should have taken you 2-3 minutes.

In this case, x=0 should not be taken into account as it is NOT a critical point. If you had a mod |x| in the given expression, then yes, x=0 would have become a critical point.

Additionally, for absolute value questions, try to plot a rough graph of what it should look like and it will help you in honing onto the range that you need to worry about. As shown in the graph in my post, the only value that you were supposed to worry about would have been in the range $$4 \leq x < 5$$. This can reduce the time taken to solve this question to less than 2 minutes.

Hope this helps.

Thanks! will def try graphs Manager  B
Joined: 03 Oct 2013
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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I tried using the extreme points and a couple of integers around them to see the trend.

The three points I used are x = -3, 4 and 5

With x = -3 the value is 15
With x = 4 the value is 8
With x = 5 the value is 9

because of the mod (x+3) term any value greater than 5 will lead to a sum greater than 8.

I think the minimum value is 8.
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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Vyshak wrote:
A cannot be the answer as all the three terms are in modulus and hence the answer will be non negative.

|x-4| >= 0 --> Minimum occurs at x = 4
|x+3| >= 0 --> Minimum occurs at x = -3
|x-5| >= 0 --> Minimum occurs at x = 5

x = -3 --> Result = 7 + 0 + 8 = 15. Also any negative value will push the combined value of |x-4| + |x-5| to a value > 9.
x = 4 --> Result = 0 + 7 + 1 = 8
x = 5 --> Result = 1 + 8 + 0 = 9
x = 3 --> Result = 1 + 6 + 2 = 9

So minimum value of the expression occurs at x = 4 and the resultant value = 8

I tried the same way! One question, from where did you get x=3?? (x=-3 or 4 or 5 sounds good though!) Why did you try with x=3?

Intern  B
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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The question can be solved by plotting the origin of the three modulus sign on the number line.
Please note |x-4| actually means distance from the point 4. So, 4 is origin.
Similarly, |x+3| denotes distance from -3.

So, in the question we need to find a point whose sum of distances from 4 , -3 and 5 is minimum.

Point -3,4 and 5 on the number line and try to figure out a point whose sum is minimum.

Please note it will lie always in the middle point when odd number of origins are there.
So, the minimum will lie at x=4.

Put x=4 in the equation, you will 8 as the answer
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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chetan2u wrote:
What is the minimum value of |x-4| + |x+3| + |x-5| ?
A. -3
B. 3
C. 5
D. 7
E. 8

Kudos for correct solution

value of a mod is min when it is equal to 0

when x = 4
0 + 7 + 1 = 8

when x= -3
7 + 0 + 8 = 15

when x = 5
1 + 8 + 0 =9

so minimum = 8
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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chetan2u wrote:
chetan2u wrote:
What is the minimum value of |x-4| + |x+3| + |x-5| ?
A. -3
B. 3
C. 5
D. 7
E. 8

OA after three days

Good Explanation Engr2012.

The Q tests us on the understanding of Property of a Modulus..

CONCEPT:

-
when we have two modulus, the value will be the minimum between the two Critical Points and will increase on either side of CP.

the two extremeties are -3 and 5, so |x+3| + |x-5| will remain constant, 3+5=8, within the range from -3 to 5..
so we have to keep x-4 as minimum to have the minimum value for |x-4| + |x+3| + |x-5| ..
x-4 will be 0 at x=4..
so our min value will OCCUR at x=4 and will be 8
E
Well i was just wondering whether in only this case , mod of (x+3) and mod of (x-5) will remain constant for any value of x(of course i tried only integer value) , or is it general that between two extremities the sum of mod will be always constant for any value of x in between. Well , ofcourse you said that the value will be minimum between the C.P
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Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?  [#permalink]

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brains wrote:
chetan2u wrote:
chetan2u wrote:
What is the minimum value of |x-4| + |x+3| + |x-5| ?
A. -3
B. 3
C. 5
D. 7
E. 8

OA after three days

Good Explanation Engr2012.

The Q tests us on the understanding of Property of a Modulus..

CONCEPT:

-
when we have two modulus, the value will be the minimum between the two Critical Points and will increase on either side of CP.

the two extremeties are -3 and 5, so |x+3| + |x-5| will remain constant, 3+5=8, within the range from -3 to 5..
so we have to keep x-4 as minimum to have the minimum value for |x-4| + |x+3| + |x-5| ..
x-4 will be 0 at x=4..
so our min value will OCCUR at x=4 and will be 8
E
Well i was just wondering whether in only this case , mod of (x+3) and mod of (x-5) will remain constant for any value of x(of course i tried only integer value) , or is it general that between two extremities the sum of mod will be always constant for any value of x in between. Well , ofcourse you said that the value will be minimum between the C.P

Yes it is for all values, even fraction..
Reason :- you are adding in one and equivalent is getting subtracted from other..

|x+3|+|x-5|....
x as 0...3+5=8
x as 1 |4|+|-4|=8
x as 1/2...|3+1/2|+|1/2-5|=|7/2+9/2=16/2=8
_________________ Re: What is the minimum value of |x-4| + |x+3| + |x-5| ?   [#permalink] 10 Sep 2018, 05:56
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