Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 29 Jan 2019
Posts: 81
Location: India
GPA: 4
WE: Business Development (Computer Software)

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
16 Aug 2019, 23:04
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
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]
Show Tags
16 Aug 2019, 23:17
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
_________________
Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com



CEO
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]
Show Tags
16 Aug 2019, 23:20
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.
_________________
Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com



Manager
Joined: 19 Nov 2017
Posts: 249
Location: India
GPA: 4

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
20 Aug 2019, 21:45
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
_________________
Vaibhav Sky is the limit. 800 is the limit.
~GMAC



Senior Manager
Joined: 19 Sep 2017
Posts: 250
Location: United Kingdom
GPA: 3.9
WE: Account Management (Other)

For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
07 Jan 2020, 03:07
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? I'd appreciate your help.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 65807

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
07 Jan 2020, 03:17
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? 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 \(x \geq 3\) ...
_________________



Intern
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]
Show Tags
19 Apr 2020, 07:09
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
x3 becomes (x3) x+1 becomes (x+1) x becomes x
So, x+3x1x < 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
x3 becomes (x3) 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 x3 becomes (x3) 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 x3 becomes (x3) x+1 becomes (x+1) x becomes x
So, x3 + 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
Joined: 02 Mar 2020
Posts: 62
Location: India

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
19 Apr 2020, 11:40
Hint : I also took values starting from 3 to 3
Only 3 doesnt satisfy the equation ,



Intern
Joined: 20 Mar 2019
Posts: 23

For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
30 Apr 2020, 07:22
@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
Joined: 27 Feb 2019
Posts: 156

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
19 May 2020, 12:39
For GMAT online, if i get this question, all i would do is plug in values....



Manager
Joined: 17 Sep 2019
Posts: 121

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
05 Jul 2020, 10:54
chetan2uI 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^28x96 < 20x . 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?



Math Expert
Joined: 02 Aug 2009
Posts: 8794

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
05 Jul 2020, 18:32
GMAT0010 wrote: chetan2uI 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^28x96 < 20x . 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 10x 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 3x3+1=3x2<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 3a3+1=3a2, but this will also be in mod. So 3a2<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
_________________



Manager
Joined: 05 Jan 2020
Posts: 142

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
05 Jul 2020, 18:55
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
Joined: 17 Sep 2019
Posts: 121

Re: For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
05 Jul 2020, 21:13
chetan2u wrote: GMAT0010 wrote: chetan2uI 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^28x96 < 20x . 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 10x 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 3x3+1=3x2<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 3a3+1=3a2, but this will also be in mod. So 3a2<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
Joined: 22 Jun 2020
Posts: 34

For how many integer values of x, is x – 3 + x + 1 + x < 10?
[#permalink]
Show Tags
13 Jul 2020, 16:11
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



Go to page
Previous
1 2
[ 35 posts ]

