Simple way to solve Veritas absolute question

Mo2men wrote:For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0

(B) 2

(C) 4

(D) 6

(E) Infinite

Take a quick look at the answer choices. There can't possibly be an infinite number of integer values that would make that expression less than 10. So all we have to do is show that there are more than 4 possibilities, and the answer will have to be D. (And if there are 4 or fewer, it won't take that long to test those.)

If x = 0, we get |0 – 3| + |0 + 1| + |0| = 3 + 1 + 0 = 4. Less than 10. so this works
If x = 1, we get |1 – 3| + |1 + 1| + |1| = 2 + 2 + 1 = 5. Less than 10. so this works
If x = 2, we get |2 – 3| + |2 + 1| + |2| = 1 + 3 + 2 = 6. Less than 10. so this works
If x = 3, we get |3– 3| + |3 + 1| + |3| = 0 + 4 + 3 = 7. Less than 10. so this works
If x = 4, we get |4 – 3| + |4 + 1| + |4| = 1 + 5 + 4 = 10. Not less than 10

We've got four values that work. Let's try a negative

If x = -1, we get |-1 – 3| + |-1 + 1| + |-1| = 4 + 0 + 1 = 5. Less than 10. so this works
So there's more than four possibilities. That means the answer must be D

And if we want to be thorough
If x = -2, we get |-2 – 3| + |-2 + 1| + |-2| = 5 + 1 + 2 = 8. Less than 10. so this works, and our six values are x = -2, -1, 0, 1, 2, or 3

Mo2men wrote:For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0

(B) 2

(C) 4

(D) 6

(E) Infinite

|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.

The correct answer is D.

For a more algebraic approach to a similar problem, check my post here:
https://www.beatthegmat.com/absolute-val ... 17161.html

Hi Experts ,

Please check and advise.

Case a -

|x - 3| + |x + 1| + |x| < 10

3x<12

x<4

case b -

|x - 3| + |x + 1| + |x| < 10

3x<-8

x<-8/3, but this can not be possible, because we have to tell no. of integers right ?

So what will be the next?

Please explain.

rsarashi wrote:Hi Experts ,

Please check and advise.

Case a -

|x - 3| + |x + 1| + |x| < 10

3x<12

x<4

case b -

|x - 3| + |x + 1| + |x| < 10

3x<-8

x<-8/3, but this can not be possible, because we have to tell no. of integers right ?

So what will be the next?

Please explain.

First, it seems as though you combined all the expressions into one. |x - 3| + |x + 1| + |x| is not the same as |3x - 2.|

An additional, broader point: you forgot to flip the sign in the second case. A radically simplified version of this problem would give us

|3x - 2| < 10

If that expression is less than 10 units from 0, then we know it's either less than 10 or greater than -10.
1) 3x - 2 < 10 ---> 3x < 12 ---> x < 4 (You did this correctly
2) 3x - 2 > -10 --> 3x > -8 --> x > (-8/3)

Together: (-8/3) < x < 4
The integers in that range: -2, -1, 0, 1, 2, 3. There are 6 of them.

DavidG@VeritasPrep wrote:

rsarashi wrote:Hi Experts ,

Please check and advise.

Case a -

|x - 3| + |x + 1| + |x| < 10

3x<12

x<4

case b -

|x - 3| + |x + 1| + |x| < 10

3x<-8

x<-8/3, but this can not be possible, because we have to tell no. of integers right ?

So what will be the next?

Please explain.

You forgot to flip the sign in the second case. A radically simplified version of this problem would give us

|3x - 2| < 10

If that expression is less than 10 units from 0, then we know it's either less than 10 or greater than -10.
1) 3x - 2 < 10 ---> 3x < 12 ---> x < 4 (You did this correctly
2) 3x - 2 > -10 --> 3x > -8 --> x > (-8/3)

Together: (-8/3) < x < 4
The integers in that range: -2, -1, 0, 1, 2, 3. There are 6 of them.

Dear David,

I have 2 questions based on the solution above.

1- How come we added the 3 terms although they are all inside modulus? what is the rule or restrictions?

2- If the question is |x - 3| - |x + 1| - |x|<10 , can I solve it using the same above? if yes, should it be |-4-x|<10

thanks

Mo2men wrote:

DavidG@VeritasPrep wrote:

rsarashi wrote:Hi Experts ,

Please check and advise.

Case a -

|x - 3| + |x + 1| + |x| < 10

3x<12

x<4

case b -

|x - 3| + |x + 1| + |x| < 10

3x<-8

x<-8/3, but this can not be possible, because we have to tell no. of integers right ?

So what will be the next?

Please explain.

You forgot to flip the sign in the second case. A radically simplified version of this problem would give us

|3x - 2| < 10

If that expression is less than 10 units from 0, then we know it's either less than 10 or greater than -10.
1) 3x - 2 < 10 ---> 3x < 12 ---> x < 4 (You did this correctly
2) 3x - 2 > -10 --> 3x > -8 --> x > (-8/3)

Together: (-8/3) < x < 4
The integers in that range: -2, -1, 0, 1, 2, 3. There are 6 of them.

Dear David,

I have 2 questions based on the solution above.

1- How come we added the 3 terms although they are all inside modulus? what is the rule or restrictions?

2- If the question is |x - 3| - |x + 1| - |x|<10 , can I solve it using the same above? if yes, should it be |-4-x|<10

thanks

Excellent questions. I should have been clearer - I was trying to illustrate a general point about absolute value by taking a different, simpler version of the prompt.

To be clear: it is NOT a valid mathematical move to dump everything inside the absolute brackets.

You can see this with simple numbers. If x = 2, then |x - 3| + |x + 1| + |x| = 1 + 3 + 2 = 6

But if x =2, then |3x - 2| = 4.