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For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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09 Jan 2017, 07:01
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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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09 Jan 2017, 07:30
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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 If \(x<1\) we have \( (3x)+(x1) + (x) = 23x<10 \implies 3x>8 \implies x>\frac{8}{3}\). Hence \(\frac{8}{3}<x<1\) If \(1 \leq x< 0\) we have \((3x)+(x+1) +(x) = 4x < 10 \implies x>6\). Hence \(1 \leq x < 0\) If \(0 \leq x < 3\) we have \((3x)+(x+1)+x=4+x<10 \implies x<6\). Hence \(0 \leq x < 3\) If \(x \geq 3\) we have \((x3)+(x+1)+x=3x2 <10 \implies 3x<12 \implies x<4\). 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
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For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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29 Jan 2017, 09:30
Hi,
I'm just writing my approach.(tricky)
just a quick look at answers we can understand E could not be right, there is no way that sum of 3 positive large integer be less that 10 and if it infinite all of large integers must be solutions for example 1000 or1000 (consider that we have always finite number of integers if the solution is not infinity). for other choices we have 4 modulus x<1 , x>3, 1<x<0 & 0<x<3 . we want just integers so there is no int between 0 and 1 (second inequality) and there is just 1&2 as ints between 0 and 3. up to know we should check if (1, 0,1,2,3) is the solution and it is easy to just calculate in mind that all of them are solutions. therefore the answer must be D.



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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29 Jan 2017, 11:52
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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 I guess plunging in value will be simplest way to solve this.... 0,1,2,3,1,2



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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16 May 2017, 04:57
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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.



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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18 May 2017, 12:48
vityakim@gmail.com 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
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. Can you please elaborate your approach a bit?



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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18 May 2017, 18:11
My way is probably not the best way but this is how my mind works.
So, there are 3 points to consider. 1, 0 and 3.
If you don't get how I got these numbers, please refer to the free GMAT Club Math Book.
x<1 1<=x<0 0<=x<3 x>=3
At this point, there is an algebra way but I 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 and it works. 0<=x<3, plug in 0,1,2 and they work. x>=3, plug in 3 (u will see that anything bigger than 3 doesn't work) Hence, D.



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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18 May 2017, 19:29
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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<1Only 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>3Since 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)



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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19 May 2017, 10:51
vityakim@gmail.com wrote: My way is probably not the best way but this is how my mind works.
So, there are 3 points to consider. 1, 0 and 3.
If you don't get how I got these numbers, please refer to the free GMAT Club Math Book.
x<1 1<=x<0 0<=x<3 x>=3
At this point, there is an algebra way but I 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 and it works. 0<=x<3, plug in 0,1,2 and they work. x>=3, plug in 3 (u will see that anything bigger than 3 doesn't work) Hence, D. Thanks. Got it now.



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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19 Jun 2017, 05:46
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Could someone please tell me on what basis we have classified these ranges?
Range 1: x<1Only 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>3Since even 4 doesn't satisfy, therefore no value greater than 4 would. Hence no values



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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19 Jun 2017, 10:20
swamygali wrote: Could someone please tell me on what basis we have classified these ranges?
Range 1: x<1Only 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>3Since even 4 doesn't satisfy, therefore no value greater than 4 would. Hence no values There is a method called critical point for these kind of questions.Through that method we get these ranges.Below URL can help.Also refer to gmat club math book. https://gmatclub.com/forum/absolutemod ... l#p1622372



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For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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21 Jun 2017, 23:48
swamygali wrote: Could someone please tell me on what basis we have classified these ranges?
Range 1: x<1Only 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>3Since even 4 doesn't satisfy, therefore no value greater than 4 would. Hence no values Realize that the equation x – 3 + x + 1 + x need to be less than 10, so we can set the min. and max. value : 1. First, lets work with the positive integers. Here, X+X+1 be the max. value, hence the deciding factor. The max. value will be 4. Because, at X=4, X+X+1 will be 9 and since X3 cannot be negative, we can set the max. value of X set at 4. 2. X cannot be less than 5 : With 5, X3 = 8, and since, x + 1 + x is always positive, and the min. positive integer is 1, so 5 is the min. value. Now, Just plug in the integers between 4 and 5, inclusive, and you are done.  You can also work with the answer choices. Plug 0 and 1000  one will work and other won't. Eliminate A and E. Plug 1  work Plug 2  work. Eliminate A,B and E. At this point, we have 3 values that satisfy the eq. So, we must decide between 4 and 6. We must try to find 2 integers that work. If 2 integers will work, then we can eliminate C and select D. Plug (1)  work. Plug (2)  work. Now, you don't need to know what is the sixth value of X  just select D and move on. Cheers !!
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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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23 Jun 2017, 13:56
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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
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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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07 Jul 2017, 00:21
hi, I did apply the counting method : by algebraic solving we know 3<x<4 Only values that would fit is 3,2,1,0,1,2 so answer is 6 (by counting) But is this method correct?



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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13 Jul 2017, 11:14
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Easy one first take positive x3+x+1+x<10 3x2<10 3x<12 x<4 Now Take mods negative x+3x1x<10 3x+2<10 3x<8 x>2.8===2 mean 2.8<x<4 2,1,0,1,2(total 6 integers)
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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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13 Jul 2017, 11:15
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Easy one first take positive x3+x+1+x<10 3x2<10 3x<12 x<4 Now Take mods negative x+3x1x<10 3x+2<10 3x<8 x>2.8===2 mean 2.8<x<4 2,1,0,1,2(total 6 integers)
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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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20 Jan 2018, 15:51
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: GMAT assassins aren't born, they're made, Rich
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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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19 Feb 2018, 11:50
Would it be possible to square this equation and solve it?



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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06 Mar 2018, 22:48
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?



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Re: For how many integer values of x, is x – 3 + x + 1 + x < 10? [#permalink]
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13 May 2018, 08:11
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.




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