For how many integer values of x, is |x – 3| + |x + 1| + |x|

Question

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0

(B) 2

(C) 4

(D) 6

(E) Infinite

broall · Accepted Answer

If $x<-1$ we have $ (3-x)+(-x-1) + (-x) = 2-3x<10 \implies 3x>-8 \implies x>-\frac{8}{3}$. Hence $-\frac{8}{3}-6$. Hence $-1 \leq x < 0$ If $0 \leq x < 3$ we have $(3-x)+(x+1)+x=4+x<10 \implies x<6$. Hence $0 \leq x < 3$ If $x \geq 3$ we have $(x-3)+(x+1)+x=3x-2 <10 \implies 3x<12 \implies x<4$. Hence $3 \leq x <4$ Combine all possible cases, we have $-\frac{8}{3} \leq x <4$. Now $x$ could recieve 6 integer values $\{-2; -1; 0; 1; 2; 3\}$ The answer is D

ehsan090 · Answer

Easy one
first take positive
x-3+x+1+x<10
3x-2<10
3x<12
x<4
Now
Take mods negative
-x+3-x-1-x<10
-3x+2<10
-3x<8
x>-2.8===-2
mean
-2.8<x<4
-2,-1,0,1,2(total 6 integers)

sobby · Answer

I guess plunging in value will be simplest way to solve this....
0,1,2,3,-1,-2

vityakim@gmail.com · Answer

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite

My 2 cents.
The word "integer" is important.

So there are below points to consider.
You need at least able to come up with the below inequality.

x>=3
0<=x<3
-1<=x<0
-1>x

At this point, just plug in number.
-1>x, plug in -2 (u will see that anything less than -2 doesn't work)
-1<=x<0, plug in -1
0<=x<3, plug in 0,1,2
x>=3, plug in 3 (u will see that anything bigger than 3 doesn't work)
Hence, D.

Tan2017 · Answer

Firstly while looking at the question we can eliminate Answer choice E as this question consists of an absolute sum with a range of less than value (10 in this case). Thus there are finite set of possibilities.

After this we can determine the critical points -1,0,3 and check for the no of integer values that satisfy the equation

Now note all the 3 critical points satisfy the equation hence we have 3 values to start with.

Range 1: x<-1---------Only -2 satisfies this. Any value less than -2 gives a sum of greater than 10
Range 2: -1<x<0------ No Integer Values
Range 3: 0<x<3 ------ Both 1 and 2 Satisfy the inequality 
Range 4: x>3-------Since even 4 doesn't satisfy, therefore no value greater than 4 would. Hence no values

Therefore a total of 6 integer values satisfy the inequality- (-2,-1,0,1,2,3)

EMPOWERgmatRichC · Answer

Hi All,

The prompt limits us the INTEGER values for X - and since we're adding 3 absolute value totals together, there can't be that many sums that are LESS than 10. This is confirmed by the answer choices (that are focused on relatively small numbers), so it's likely that we should be able to just 'brute force' this question and find all of the solutions...

|X - 3| + |X + 1| + |X| < 10

Let's start with the easiest number and work our way up...

IF... X=0, then the sum = 4 
IF... X=1, then the sum = 5 
IF... X=2, then the sum = 6 
IF... X=3, then the sum = 7 
IF... X=4, then the sum = 10, but THAT is too big

We can't forget about NEGATIVE numbers though... 
IF... X = -1, then the sum = 5 
IF... X = -2, then the sum = 8

At this point, we have 6 possibilities, and it makes no sense that there would be an 'infinite' number of solutions, so we can stop working. If you want to go one more step though, then you can...

IF... X = -3, then the sum = 11, and THAT is too big.

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

Kinshook · Answer

Asked: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10? |x – 3| = Distance of x from 3 |x + 1| = Distance of x from -1 |x| = Distance of x from 0 Sum of above distances < 10 At x=-1 ; |x – 3| + |x + 1| + |x| = 4+0+1=5 < 10 OK At x=-2 ; |x – 3| + |x + 1| + |x| = 5+1+2=8 < 10 OK At x=-3 ; |x – 3| + |x + 1| + |x| = 6+2+3=11 > 10 NOT OK If x<-2 ; |x – 3| + |x + 1| + |x| = > 10 NOT OK At x=0 ; |x – 3| + |x + 1| + |x| = 3+1+0=4 < 10 OK At x=1 ; |x – 3| + |x + 1| + |x| = 2+2+1=5 < 10 OK At x=2 ; |x – 3| + |x + 1| + |x| = 1+3+2=6 < 10 OK At x=3 ; |x – 3| + |x + 1| + |x| = 0+4+3=7 < 10 OK At x=4 ; |x – 3| + |x + 1| + |x| = 1+5+4=10 = 10 NOT OK If x>3; |x – 3| + |x + 1| + |x| = >= 10 NOT OK For x ={-2,-1,0,1,2,3} above conditions are valid. IMO D

Doer01 · Answer

Hi Bunuel , I have gone through GMAT Club math book to understand how three point system works. Thanks for putting it together. It sometimes still confuses me when it comes to putting greater than and greater than or equal to signs. For eg: In this question, we have -1, 0, and 3 as transition points. So would the ranges be: x<-1 -1<=x<0 0<=0<3 x>3 OR x<=-1 -1<=x<=0 0<=x<=3 x>=3 and why? I'd appreciate your help.

Bunuel · Answer

You are right, the transition points are -1, 0 and 3. You should include each in either of the ranges with = sign and it does not matter in which you include. Fpr example, the ranges you consider could be:

x < -1 ,  -1 \leq x \leq 0 ,  0 < x < 3  and  x \geq 3

x \leq -1 ,  -1 < x < 0 ,  0 \leq x \leq 3  and  x > 3

x \leq -1 ,  -1 < x \leq 0 ,  0 < x < 3  and  x \geq 3

...

Manku · Answer

The best approach, perhaps not very efficient, is the one using critical points (CP).

As per the equation, we have following CPs.
-1, 0, and 3

Lets start by identifying possible integer values in these ranges.

1. when x< -1

|x-3| becomes -(x-3)
|x+1| becomes -(x+1)
|x| becomes -x

So, -x+3-x-1-x < 10
gives us x> [-8][/3]
combining with the constraint we get the following range : [-8][/3]<x<-1
which gives us 1 possible integer value = -2

2. when -1< x < 0

|x-3| becomes -(x-3)
|x+1| becomes (x+1)
|x| becomes -x

So, -x+3 + x+1 - x < 10
gives us x > -6
combining with the constraint we get the following range : -1< x < 0
which gives us 2 possible integer value = -1 and 0

3. when 0< x < 3
|x-3| becomes -(x-3)
|x+1| becomes (x+1)
|x| becomes x

So, -x+3 + x + 1 + x < 10
gives us x < 6
combining with the constraint we get the following range : 0< x < 3
which gives us 2 possible integer value = 1 and 2

4. when x> 3
|x-3| becomes (x-3)
|x+1| becomes (x+1)
|x| becomes x

So, x-3 + x+ 1 +x < 10
gives us x< 4
combining with the constraint we get the following range : 3< x < 4
which gives us 1 possible integer value = 1

Total possible integer solution for x : 6

Hope it helps.

GMAT0010 · Answer

chetan2u
I read your method for 2 or more mods.
I tried to use it here, but I feel lost.
I took |x| on the RHS and squared on both sides.
I finally got 3x^2-8x-96 < -20|x| . From here I reasoned that x can't take positive values for the equation to hold. But if can't hold when x = -4.
Please help me. How can I better approach this?

chetan2u · Answer

You can square only when both sides are positive. Here 10-|x| can be negative for all x>10. If I were to answer I’ll divide this question in two parts. 1) when x>0 The x in all three mods will get added, 3 will get subtracted and 1 will get added when we open mod. So 3x-3+1=3x-2<10....3x<12......x<4 2) when x<0, say x=-a The x in all three mods will get added, 3 will also get added and 1 will be subtracted. So -3a-3+1=-3a-2, but this will also be in mod. So |-3a-2|<10.....3a+2<10....a<8/3 Multiply by negative sign So -a>-8/3.....x>-8/3 So -8/3

GMATGuruNY · Answer

|a| = the distance between a and 0.
|a-b| = the distance between a and b.
|a+b| = the distance between a and -b.

Question stem, rephrased:
For how many integer values of x is (distance between x and 3) + (distance between x and -1) + (distance between x and 0) < 10?

For the sum of the three distances to be less than 10, x must be close to the outermost values (-1 and 3).
If we test integer values close to -1 and 3, we find that only the following satisfy |x - 3| + |x + 1| + |x| < 10:
-2, -1, 0, 1, 2, 3.

D

yashikaaggarwal · Answer

Took me 2.15 minutes without pen and paper.
So the solution I got is 
When ever we have integer value in this type.
Always put X = 0 (to know whether we have solution or not)
|X-3|+|x+1|+|X| < 10
Put X = 0
4 < 10 (so we do have solution chances)

Put X = -1 (I took it -1, you can do for +1 one too)
5 < 10

Put X = -2
8 < 10
Now we know going beyond -2 is useless.

X = 1 
5 < 10

X = 2
6 < 10

X = 3
7 < 10

Going beyond is useless.

So we are left with 6 values.
The answers stated above are also good. 
Whatever answer suits you best apply it.

Posted from my mobile device

soniasw16 · Answer

Is there a faster, less error prone approach to such questions?

GMATCoachBen · Answer

soniasw16

GMATGuruNY  posted an excellent solution above that's logical and shorter. It can often be helpful to think in terms of  distances  when we're dealing with absolute values. For example, |x - 3| translates to "the  distance  between x and 3".

I've pasted it below:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

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For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

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