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How many integers of x are there so that x + 3 < 1 < 2x + 9 ?

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How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 25 Nov 2019, 23:12
1
4
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A
B
C
D
E

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Question Stats:

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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 25 Nov 2019, 23:33
1
How many integers of x are there so that x+3<1<2x+9 ?

(A) Four
(B) Three
(C) Two
(D) One
(E) None

x+3 < 1 & 1 < 2x+9
x < -2 & x > -4

Only -3 satisfies both.

Answer D.
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 25 Nov 2019, 23:39
1
x+3 < 1 < 2x+9
--> x + 3 < 1 and 2x + 9 > 1
--> x < 1 - 3 & 2x > -8
--> x < -2 & x > - 4
--> -4 < x < -2

So, only possible value of x = -3 --> One possible value

IMO Option D
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 26 Nov 2019, 00:28
1
x+3<1<2x+9

From x+3<1, x<-2
From 2x+9>1, x>-4

so x=-3

OA:D
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 26 Nov 2019, 00:32
1
the given relation x+3<1<2x+9
is only valid for x= -3
IMO D ; one is correct

How many integers of x are there so that x+3<1<2x+9 ?

(A) Four
(B) Three
(C) Two
(D) One
(E) None
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 26 Nov 2019, 03:39
1
Quote:
How many integers of x are there so that x+3<1<2x+9?

(A) Four
(B) Three
(C) Two
(D) One
(E) None


\(x+3<1<2x+9\)
\(x+3<1…x<-2\)
\(1<2x+9…-8<2x…-4<x\)
\(-4<x<-2:x=-3\)

Ans (D)
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 26 Nov 2019, 06:01
1
x+3<1<2x+9
This is the same as x+3<1 and 1<2x+9
x+3<1 means x<-2
and 1<2x+9 means -4<x
we now have -4<x<-2
-3 is the only integer that falls within the given range.

The answer is therefore option D.
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 26 Nov 2019, 15:14
1
x +3 < 1 —> x <—2
2x+ 9 > 1 —> x >—4

—> x=—3
There is only one integer.

The answer is D

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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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New post 28 Nov 2019, 17:37
Bunuel wrote:

Competition Mode Question



How many integers of x are there so that \(x + 3 < 1 < 2x + 9\) ?

(A) Four
(B) Three
(C) Two
(D) One
(E) None


Since x + 3 < 1, then x < -2.

Since 1 < 2x + 9, then -8 < 2x, and therefore, -4 < x.

We see that -4 < x < -2. Since x is an integer, x can be only -3. Therefore, there is only one solution for x.

Answer: D
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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?   [#permalink] 28 Nov 2019, 17:37
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