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# M07-31

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Math Expert
Joined: 02 Sep 2009
Posts: 58381

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16 Sep 2014, 00:35
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Difficulty:

35% (medium)

Question Stats:

76% (01:01) correct 24% (01:11) wrong based on 41 sessions

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If $$x^2-100 \lt 300$$ then $$x$$ can take how many integer values?

A. 42
B. 39
C. 38
D. 37
E. 19

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Joined: 02 Sep 2009
Posts: 58381

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16 Sep 2014, 00:35
2
1
Official Solution:

If $$x^2-100 \lt 300$$ then $$x$$ can take how many integer values?

A. 42
B. 39
C. 38
D. 37
E. 19

Given: $$x^2-100 \lt 300$$;

$$x^2 \lt 400$$;

$$|x| \lt 20$$;

$$-20 \lt x \lt 20$$, so $$x$$ can take 39 integer values from -19 to 19, inclusive.

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Joined: 20 Jul 2012
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11 May 2016, 23:36
Bunuel wrote:
Official Solution:

If $$x^2-100 \lt 300$$ then $$x$$ can take how many integer values?

A. 42
B. 39
C. 38
D. 37
E. 19

Given: $$x^2-100 \lt 300$$;

$$x^2 \lt 400$$;

$$|x| \lt 20$$;

$$-20 \lt x \lt 20$$, so $$x$$ can take 39 integer values from -19 to 19, inclusive.

One question here :-
can the above equation not be written as (x-20)(x+20) < 0
in that case the rang of x differs i.e x <20 or x <-20

I know i am missing some concept here (or worst some mistake) as x taking a value less than -20, just does not satisfy the give equation
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12 May 2016, 00:09
grsm wrote:
Bunuel wrote:
Official Solution:

If $$x^2-100 \lt 300$$ then $$x$$ can take how many integer values?

A. 42
B. 39
C. 38
D. 37
E. 19

Given: $$x^2-100 \lt 300$$;

$$x^2 \lt 400$$;

$$|x| \lt 20$$;

$$-20 \lt x \lt 20$$, so $$x$$ can take 39 integer values from -19 to 19, inclusive.

One question here :-
can the above equation not be written as (x-20)(x+20) < 0
in that case the rang of x differs i.e x <20 or x <-20

I know i am missing some concept here (or worst some mistake) as x taking a value less than -20, just does not satisfy the give equation

Hi
you are correct in equation, BUT wrong in inference from it..
(x-20)(x+20)<0...
this means that x-20 and x+20 have to be of opposite sign...
this is possible in range from -20 to +20.....
put x as -21..
so (-21-20)(-21+20) = -41*-1= 41 which is >0 so range does not hold
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25 Jul 2016, 10:38
I think this is a high-quality question and I agree with explanation.
Current Student
Joined: 23 Jul 2015
Posts: 143

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05 Sep 2017, 20:42
I think this is a high-quality question and I agree with explanation.
IIMA, IIMC School Moderator
Joined: 04 Sep 2016
Posts: 1366
Location: India
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03 Dec 2017, 15:36
niks18

Quote:
$$x^2 \lt 400$$;

$$|x| \lt 20$$;

I understood that $$\sqrt{}x^2 = |x|$$ but
$$\sqrt{400}$$ can take both values as -20 and +20
Why did we consider only positive values?
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Math Expert
Joined: 02 Sep 2009
Posts: 58381

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03 Dec 2017, 20:49
niks18

Quote:
$$x^2 \lt 400$$;

$$|x| \lt 20$$;

I understood that $$\sqrt{}x^2 = |x|$$ but
$$\sqrt{400}$$ can take both values as -20 and +20
Why did we consider only positive values?

Not clear what are you implying there.

$$\sqrt{400}=20$$ ONLY, not +20 and -20. In contrast x^2 = 400 has TWO solutions, x = 20 and x = -20.

In the question above we DO consider the possibility of x being negative:
$$x^2 \lt 400$$;

$$|x| \lt 20$$

$$x^2 \lt 400$$;

$$-20 < x <20$$
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03 Dec 2017, 21:33
Bunuel

Quote:
Not clear what are you implying there.

$$\sqrt{400}=20$$ ONLY, not +20 and -20. In contrast x^2 = 400 has TWO solutions, x = 20 and x = -20.

Is this because GMAT does not test imaginary roots ie complex number of the form $$\sqrt{-i}$$
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It's the journey that brings us happiness not the destination.

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Math Expert
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03 Dec 2017, 21:37
Bunuel

Quote:
Not clear what are you implying there.

$$\sqrt{400}=20$$ ONLY, not +20 and -20. In contrast x^2 = 400 has TWO solutions, x = 20 and x = -20.

Is this because GMAT does not test imaginary roots ie complex number of the form $$\sqrt{-i}$$

GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers.
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18 Oct 2018, 04:54
x^2−100<300
x^2 < 400
-20<x<20
so number of integers for x apart from 20 and -20 = 39
"B"
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Re: M07-31   [#permalink] 18 Oct 2018, 04:54
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# M07-31

Moderators: chetan2u, Bunuel