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

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Joined: 02 Sep 2009
Posts: 62469
How many integers of x are there so that x + 3 < 1 < 2x + 9 ?  [#permalink]

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25 Nov 2019, 22:12
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25% (medium)

Question Stats:

68% (01:12) correct 32% (01:44) wrong based on 65 sessions

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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

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

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25 Nov 2019, 22:33
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.

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

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25 Nov 2019, 22:39
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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25 Nov 2019, 23:28
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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25 Nov 2019, 23:32
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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26 Nov 2019, 02:39
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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26 Nov 2019, 05:01
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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26 Nov 2019, 14:14
1
x +3 < 1 —> x <—2
2x+ 9 > 1 —> x >—4

—> x=—3
There is only one integer.

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

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28 Nov 2019, 16:37
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.

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Re: How many integers of x are there so that x + 3 < 1 < 2x + 9 ?   [#permalink] 28 Nov 2019, 16:37
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