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How many possible integer values are there for x if 4x  3 [#permalink]
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Updated on: 20 Oct 2013, 04:03
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How many possible integer values are there for x if 4x  3 < 6 ? A. One B. Two C. Three D. Four E. Five What I did here was to separate two scenarios: a. 4x – 3 < 6 and b. 4x – 3 >  6
I simplified x in both scenarios that I ended up in a range of 2.25 < x < .75 Therefore, my answer is 3 integers (2, 1, 0)
I would like to know if this method is correct and if I should use such method for problems that involve absolute value and inequality.
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Originally posted by cpcalanoc on 19 Dec 2004, 18:33.
Last edited by Bunuel on 20 Oct 2013, 04:03, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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u can do this way. Directly substitute the value of x by trial and error method . Start from 0 and move on. u will get 0,1 and 2 as it asked for positive integers



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Cpcalanoc you choose the right way BUT from 4x3<6 => 4x<9 =>
x<2.25
the same mistake in the other
So 0.75 < x < 2.25
then the answer (0, 1, 2)
to Rakesh1239 your answer is right but where you see the word positive?



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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20 Nov 2013, 05:25
cpcalanoc wrote: How many possible integer values are there for x if 4x  3 < 6 ? A. One B. Two C. Three D. Four E. Five What I did here was to separate two scenarios: a. 4x – 3 < 6 and b. 4x – 3 >  6
I simplified x in both scenarios that I ended up in a range of 2.25 < x < .75 Therefore, my answer is 3 integers (2, 1, 0)
I would like to know if this method is correct and if I should use such method for problems that involve absolute value and inequality. You can try algebraically, If x>=0, then you will have the range 0<=x<9/4 If x<0, you get x>3/4 but this solution is not valid so x must vbe x>=0. Hence, you have 3 integer solutions from the above inequality: 0, 1 and 2. Hope this helps Cheers J



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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26 Mar 2014, 09:06
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[quote="cpcalanoc"]How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five
Solution: 4x3 < 6
let 4x=a therefore we have a3 < 6 ==> read this as origin is at +3 and we have to move +6 to the right and 6 to the left
(the less than sign represents that the a must be within boundaries )
(36)3(3+6)
now, we have 3<a<9
but a =4x ==> 3<4x<9
dividing all values by +4 we have 0.75 <x < 2.24
Now question says Integer values (not rational ) therefore we have 0,1,2
Hence 3



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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11 Apr 2014, 07:29
My Strategy , X is an integer ( + ,  , 0 ) Principles : Integers ,Module and Inequality Plug In Method Start with 0 , 1 , 1 , 2 as values By Plugging in 4x  3 < 6 Is Yes for 0 , 1, 2 But not with 1 So 3 Values After seeing Bunnel solution we can use Module Basics with a Range in Number line 6<4x  3< 6 , 3<4x<9 , \frac {3}{4} <x< \frac{9}{4}, So  0,75<x< 2.25 So Range from 0 , 1 , 2 If any wrong or any Suggestions in my Strategy pls Correct it
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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11 Apr 2014, 12:12
kanusha wrote: My Strategy , X is an integer ( + ,  , 0 ) Principles : Integers ,Module and Inequality Plug In Method Start with 0 , 1 , 1 , 2 as values By Plugging in 4x  3 < 6 Is Yes for 0 , 1, 2 But not with 1 So 3 Values After seeing Bunnel solution we can use Module Basics with a Range in Number line 6<4x  3< 6 , 3<4x<9 , \frac {3}{4} <x< \frac{9}{4}, So  0,75<x< 2.25 So Range from 0 , 1 , 2 If any wrong or any Suggestions in my Strategy pls Correct it your strategy is good but suppose if you have lot of integers, how many will you keep plugging? The second method is better as you quickly know the range and based on conceptual understanding. given 4x  3 can take both positive and negative values. It can lie between 6 and 0 or 0 and +6.



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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27 Apr 2014, 09:58
Bunuel wrote: How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five
\(4x  3 < 6\);
Get rid of the modulus: \(6<4x3< 6\);
Add 3 to all three parts: \(3<4x< 9\);
Divide by 4: \(\frac{3}{4}<x< \frac{9}{4}\) > \(0.75<x< 2.25\).
x can take following integer values: 0, 1, and 2.
Answer: C. @Bunuel  The question stems talks about how many positive integer values of x. IF we are talking about positive integer then it should be only {1,2} and not {0). Am I missing anything here?



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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28 Apr 2014, 01:34
chanakya84 wrote: Bunuel wrote: How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five
\(4x  3 < 6\);
Get rid of the modulus: \(6<4x3< 6\);
Add 3 to all three parts: \(3<4x< 9\);
Divide by 4: \(\frac{3}{4}<x< \frac{9}{4}\) > \(0.75<x< 2.25\).
x can take following integer values: 0, 1, and 2.
Answer: C. @Bunuel  The question stems talks about how many positive integer values of x. IF we are talking about positive integer then it should be only {1,2} and not {0). Am I missing anything here? The question does not specify that the integers must be positive: "how many possible integer values are there for x..." Hope it's clear.
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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28 Apr 2014, 06:46
Bunuel wrote: chanakya84 wrote: Bunuel wrote: How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five
\(4x  3 < 6\);
Get rid of the modulus: \(6<4x3< 6\);
Add 3 to all three parts: \(3<4x< 9\);
Divide by 4: \(\frac{3}{4}<x< \frac{9}{4}\) > \(0.75<x< 2.25\).
x can take following integer values: 0, 1, and 2.
Answer: C. @Bunuel  The question stems talks about how many positive integer values of x. IF we are talking about positive integer then it should be only {1,2} and not {0). Am I missing anything here? The question does not specify that the integers must be positive: "how many possible integer values are there for x..." Hope it's clear. Ahh..my mistake.... Read the question wrong.



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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31 Oct 2016, 13:59
You can figure this out by taking the equation and solving:
SCENARIO 1: (4x  3) < 6 4x<9 x<(9/4)
SCENARIO 2: (4x  3) > 6 4x>3 x>(3/4)
So (0.75)<x<2.25 > The only integers within this range are 0,1,2 > Hence, 3 is the correct answer (C)



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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11 Sep 2017, 21:33
Hi, I think answer will be B. We can get 1,2 .zero is neither positive or negative and question is asking for positive values only.
PS Any number to the left of zero on a number line; can be integer or noninteger. Note: The number zero is neither positive nor negative
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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11 Sep 2017, 21:36



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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11 Sep 2017, 21:41
Thanks,I got it I didn't read the question properly
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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11 Sep 2017, 22:58
4x  3 < 6 remove the modulus Case 1) 4x3 <6 4x<9 x<9/4 x<2.25
Case 2) (4x6)<6 4x3>6 4x>3 x>3/4 x>0.75 Hence 0.75 < x < 2.25
Therefore only possible values for x are 0,1,2.
Hence Option C.
Kudos if it helps.



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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25 Nov 2017, 03:30
Bunuel wrote: How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five
\(4x  3 < 6\);
Get rid of the modulus: \(6<4x3< 6\);
Add 3 to all three parts: \(3<4x< 9\);
Divide by 4: \(\frac{3}{4}<x< \frac{9}{4}\) > \(0.75<x< 2.25\).
x can take following integer values: 0, 1, and 2.
Answer: C. Hi! Why is zero seen as an integer here? I understand 1,2 but zero isn't counted as an integer. Please correct my understanding. Thank you!



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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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25 Nov 2017, 04:11
Madhavi1990 wrote: Bunuel wrote: How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five
\(4x  3 < 6\);
Get rid of the modulus: \(6<4x3< 6\);
Add 3 to all three parts: \(3<4x< 9\);
Divide by 4: \(\frac{3}{4}<x< \frac{9}{4}\) > \(0.75<x< 2.25\).
x can take following integer values: 0, 1, and 2.
Answer: C. Hi! Why is zero seen as an integer here? I understand 1,2 but zero isn't counted as an integer. Please correct my understanding. Thank you! ZERO:1. 0 is an integer.2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself. Brush up fundamentals before attempting questions: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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12 Mar 2018, 06:07
cpcalanoc wrote: How many possible integer values are there for x if 4x  3 < 6 ? A. One B. Two C. Three D. Four E. Five What I did here was to separate two scenarios: a. 4x – 3 < 6 and b. 4x – 3 >  6
I simplified x in both scenarios that I ended up in a range of 2.25 < x < .75 Therefore, my answer is 3 integers (2, 1, 0)
I would like to know if this method is correct and if I should use such method for problems that involve absolute value and inequality. Question: Number of possible values of x which are integer? Given: 4x  3 < 6 6+3 < 4x < 6+3 3/4 < x < 9/4 0.75 < x < 2.25 values are 0, 1, 2 (Three integer values) (C)
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Re: How many possible integer values are there for x if 4x  3 [#permalink]
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13 Mar 2018, 16:54
cpcalanoc wrote: How many possible integer values are there for x if 4x  3 < 6 ?
A. One B. Two C. Three D. Four E. Five We can first solve when the expression (4x  3) is positive: 4x  3 < 6 4x < 9 x < 9/4 = 2 ¼ Now we can solve when the expression (4x  3) is negative: (4x  3) < 6 4x + 3 < 6 4x < 3 x > 3/4 Thus: 3/4 < x < 2 1/4 So x can be any of 3 integer values: 0, 1, or 2. Answer: C
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