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# How many integer values are there for x such that 1 < 3x + 5 < 17?

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How many integer values are there for x such that 1 < 3x + 5 < 17?  [#permalink]

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20 Dec 2015, 04:16
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Difficulty:

35% (medium)

Question Stats:

67% (00:43) correct 33% (00:40) wrong based on 213 sessions

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How many integer values are there for x such that 1 < 3x + 5 < 17?

A. Two
B. Three
C. Four
D. Five
E. Six

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Re: How many integer values are there for x such that 1 < 3x + 5 < 17?  [#permalink]

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20 Dec 2015, 04:42
1
1 < 3x + 5 < 17
=> -4 < 3x < 12
=> -4/3 < x < 4
x can take integer values -1,0 , 1 , 2 , 3

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Re: How many integer values are there for x such that 1 < 3x + 5 < 17?  [#permalink]

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08 May 2016, 13:30
The inequality given is 1 < 3x+5 < 17
it can further reduced to -4 < 3x < 12
finally -4/3 < x < 4
So can only take 5 integer values i.e. -1,0,1,2,3
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Re: How many integer values are there for x such that 1 < 3x + 5 < 17?  [#permalink]

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09 May 2016, 08:28
Bunuel wrote:
How many integer values are there for x such that 1 < 3x + 5 < 17?

A. Two
B. Three
C. Four
D. Five
E. Six

$$1 < 3x + 5 < 17$$

=> $$1 -5 < 3x< 17 - 5$$

=> $$-4 < 3x< 12$$

=> $$\frac{-4}{3} < x< 4$$

=> $$-1\frac{1}{3} < x< 4$$

Possible value of x is {-1 0 , 1 , 2 , 3 }

So, Answer will be (D) 4
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Re: How many integer values are there for x such that 1 < 3x + 5 < 17?  [#permalink]

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09 May 2016, 09:27
1<3x+5<17
or -4<3x<12

x can have -1,0,1,2,3- i.e 5 values to be >-4 and <12
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Re: How many integer values are there for x such that 1 < 3x + 5 < 17?  [#permalink]

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04 Nov 2018, 17:03
Bunuel wrote:
How many integer values are there for x such that 1 < 3x + 5 < 17?

A. Two
B. Three
C. Four
D. Five
E. Six

Solving the inequality, we have:

1 < 3x + 5 < 17

-4 < 3x < 12

-4/3 < x < 4

So there are 5 integers between -4/3 and 4, namely, -1, 0, 1, 2 and 3.

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Re: How many integer values are there for x such that 1 < 3x + 5 < 17? &nbs [#permalink] 04 Nov 2018, 17:03
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# How many integer values are there for x such that 1 < 3x + 5 < 17?

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