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Manager  S
Joined: 29 Jan 2019
Posts: 81
Location: India
GPA: 4
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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2
ydmuley wrote:
I feel quick way to solve the problem is as per below.

$$|x – 3| + |x + 1| + |x| < 10$$

$$-10 < x - 3 + x + 1 + x < 10$$

$$-10 < 3x - 2 < 10$$

$$-8 < 3x < 12$$

$$-8/3 < x < 4$$

$$-2.6 < x < 4$$

Now, lets write down the integer values which fall in this range.

$$-2, -1, 0, 1, 2, 3$$

So in all there are 6 integer values, which satisfy the above equation.

Hence, Answer is D = 6

CAN WE RELY ON THIS METHOD FOR SUCH QUESTIONS?
CEO  V
Joined: 03 Jun 2019
Posts: 3360
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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Bunuel 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

$$Asked:$$ 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
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
CEO  V
Joined: 03 Jun 2019
Posts: 3360
Location: India
GMAT 1: 690 Q50 V34 WE: Engineering (Transportation)
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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rupeshnverbal wrote:
ydmuley wrote:
I feel quick way to solve the problem is as per below.

$$|x – 3| + |x + 1| + |x| < 10$$

$$-10 < x - 3 + x + 1 + x < 10$$

$$-10 < 3x - 2 < 10$$

$$-8 < 3x < 12$$

$$-8/3 < x < 4$$

$$-2.6 < x < 4$$

Now, lets write down the integer values which fall in this range.

$$-2, -1, 0, 1, 2, 3$$

So in all there are 6 integer values, which satisfy the above equation.

Hence, Answer is D = 6

CAN WE RELY ON THIS METHOD FOR SUCH QUESTIONS?

It is better to take sum of distances approach.
The question is
Where x can lie on the number line such that sum of distances of x from 0, -1 and 3 is less than 10.
Plot the number line and then find x values satisfying this condition.
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
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Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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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
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Vaibhav

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Senior Manager  G
Joined: 19 Sep 2017
Posts: 250
Location: United Kingdom
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For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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Bunuel 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

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?

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Cheers!!
Math Expert V
Joined: 02 Sep 2009
Posts: 65807
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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Doer01 wrote:
Bunuel 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

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?

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 $$x \geq 3$$

...
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Intern  B
Joined: 15 Aug 2014
Posts: 29
GMAT 1: 630 Q46 V31 Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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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.
Manager  B
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Schools: IIMA PGPX"20 (A)
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Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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

Only -3 doesnt satisfy the equation ,
Intern  B
Joined: 20 Mar 2019
Posts: 23
For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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@s wrote:
Tanvi94 wrote:
ydmuley wrote:
I feel quick way to solve the problem is as per below.

$$|x – 3| + |x + 1| + |x| < 10$$

$$-10 < x - 3 + x + 1 + x < 10$$

$$-10 < 3x - 2 < 10$$

$$-8 < 3x < 12$$

$$-8/3 < x < 4$$

$$-2.6 < x < 4$$

Now, lets write down the integer values which fall in this range.

$$-2, -1, 0, 1, 2, 3$$

So in all there are 6 integer values, which satisfy the above equation.

Hence, Answer is D = 6

Can someone confirm if this approach can be used to solve the above?

It works and its really quick. wow. 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
Manager  G
Joined: 27 Feb 2019
Posts: 156
GMAT 1: 720 Q48 V41
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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For GMAT online, if i get this question, all i would do is plug in values....
Manager  B
Joined: 17 Sep 2019
Posts: 121
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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chetan2u
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. Math Expert V
Joined: 02 Aug 2009
Posts: 8794
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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GMAT0010 wrote:
chetan2u
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. 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
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Manager  G
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Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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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.
Manager  B
Joined: 17 Sep 2019
Posts: 121
Re: For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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chetan2u wrote:
GMAT0010 wrote:
chetan2u
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. 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

Thanks, I understood much better now. Intern  B
Joined: 22 Jun 2020
Posts: 34
For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?  [#permalink]

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ydmuley wrote:
I feel quick way to solve the problem is as per below.

$$|x – 3| + |x + 1| + |x| < 10$$

$$-10 < x - 3 + x + 1 + x < 10$$

$$-10 < 3x - 2 < 10$$

$$-8 < 3x < 12$$

$$-8/3 < x < 4$$

$$-2.6 < x < 4$$

Now, lets write down the integer values which fall in this range.

$$-2, -1, 0, 1, 2, 3$$

So in all there are 6 integer values, which satisfy the above equation.

Hence, Answer is D = 6

Wow, thank you so much for this! Literally none of the other answers made much sense, and were all so convoluted. For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?   [#permalink] 13 Jul 2020, 16:11

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