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M07-31

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M07-31  [#permalink]

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New post 16 Sep 2014, 00:35
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A
B
C
D
E

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  35% (medium)

Question Stats:

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

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Re M07-31  [#permalink]

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New post 16 Sep 2014, 00:35
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Re: M07-31  [#permalink]

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New post 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.


Answer: B




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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Re: M07-31  [#permalink]

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New post 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.


Answer: B




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.....
check on your range x<-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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Re M07-31  [#permalink]

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New post 25 Jul 2016, 10:38
I think this is a high-quality question and I agree with explanation.
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Re M07-31  [#permalink]

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New post 05 Sep 2017, 20:42
I think this is a high-quality question and I agree with explanation.
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Re: M07-31  [#permalink]

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New post 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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New post 03 Dec 2017, 20:49
adkikani wrote:
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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Re: M07-31  [#permalink]

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New post 03 Dec 2017, 21:33
Bunuel

Thanks for your two cents

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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New post 03 Dec 2017, 21:37
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Re: M07-31  [#permalink]

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New post 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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