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

Best way to solve this question would be to plug in integers into the equation.
0,1, -1, -2, 2 etc. The answer cannot be infinite so it has to be 6.
Hence, D

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.

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 \(x \geq 3\)

...

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.

Hint : I also took values starting from -3 to 3

Only -3 doesnt satisfy the equation ,

This approach is not correct because it gets rid of the absolute values without making any consideration on the signs.
Indeed in this case he is assuming x to be greater than 3 (otherwise there would be at least one sign change).
The most efficient way to solve this is plugging in numbers, as explained above from other users

For GMAT online, if i get this question, all i would do is plug in values....

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?

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<x<4....

x=-2,-1,0,1,2,3

The transition points are -1, 0, and 3.
At x = 3, value = 7
At x = -1, value = 5

For every increment of 1 unit on the number line (for x>3 and x<-2) the total value will increase by 3.
=> at x = 4, value = 7+3 = 10 (discard as value should be less than 10)
=> at x = -2, value = 5+3 = 8
=> at x = -3, value = 8+3 = 11 (discard as value should be less than 10)

The final range of x is -2 to 3. => 6 integer values.

Thanks, I understood much better now.

Wow, thank you so much for this! Literally none of the other answers made much sense, and were all so convoluted.

Hi ydmuley

How and from where did you learn this concept?

Please refer...

I am really having hard time understanding absolute value, absolute value with inequalities and the abssolute valie with two or more variables.

Thank you.

Posted from my mobile device

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

(A) 0

(B) 2

(C) 4

(D) 6

(E) Infinite

Positive:
|x – 3| + |x + 1| + |x| < 10
x -3 + x + 1 + x <10
x < 4

Negative:
|x – 3| + |x + 1| + |x| < 10
-x + 3 - x - 1 - x < 10
x > -8/3

-8/3 < x < 4

x = -2, -1, 0, 1, 2, 3 All work

Answer is D.

Is it correct to have the inequality sign change from \(-8/3\)< to \(-8/3\)≤ ?
No difference for this question, but if it were -9/3

|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.

Bunuel
Thanks for posting this question. always some variations to learn.
I went through some sections of the gmat club book as well.
What would be your approach? the inequalities question are really case by case and could have quite a bit of a variation.
there are a few solutions here but I am not sure if the logic behind is universal for all cases. for example,
Negative:
|x – 3| + |x + 1| + |x| < 10
-x + 3 - x - 1 - x < 10
x > -8/3
this is likely a viable approach by breaking up the possible magnitude in the positive spectrum and negative spectrum on the number line.
However, my approach after reading the gmat club mathbook would have convinced me to find the relationship of the three absolute values by setting them equal to each other e.g., |x-3|=x and evaluate at when x is positive/negative...which seems a little time consuming, many ways to Rome I guess?
GMATGuruNY
thanks to your explanation I was able to understand the use of transition points a little better.