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reciprocals and negatives

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reciprocals and negatives [#permalink]

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New post 24 Nov 2011, 21:51
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

80% (00:07) correct 20% (03:52) wrong based on 7 sessions

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Can someone please explain this:

If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and -1/10
E. greater than 10

I get ans E

n^2 < 1/100
-1/10 < n < 1/10
n is negative so forget right side of equation
multiply both sides by 1/n
1/n * -1/10 < n * 1/n
-1/10n < 1
now multiply both sides by -10
-10 * -1/10n < 1 * -10
Flip signs
1/n > -10

Answer E!

But answer is A! Please explain your answers.


OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-n-denotes ... 91659.html

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Re: reciprocals and negatives [#permalink]

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New post 24 Nov 2011, 22:27
Thanks, in that case it goes:

n^2 < 1/100
-1/10 < n < 1/10
n is negative so forget right side of equation
multiply both sides by 1/n
since n is negative flip signs
1/n * -1/10 > n * 1/n

now the left side becomes positive as -1/n * -1/10 > 1
ie 1/10n > 1


now multiply both sides by 10
since 10 is now positive, do not flip signs
1/n > 10

still get Ans E

Subhashghosh or anyone else would you please explain now Thanks.


subhashghosh wrote:
study wrote:
Can someone please explain this:

If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and -1/10
E. greater than 10

I get ans E

I'm not solving this, just trying to give a pointer.

n^2 < 1/100
-1/10 < n < 1/10
n is negative so forget right side of equation
multiply both sides by 1/n
1/n * -1/10 < n * 1/n

-1/10n < 1
if n is negative, 1/n is also the same.

now multiply both sides by -10
-10 * -1/10n < 1 * -10
Flip signs
1/n > -10

Answer E!

But answer is A! Please explain your answers.

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Re: reciprocals and negatives [#permalink]

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New post 25 Nov 2011, 22:00
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I looked at it in this way. |n| would definitely be less than 1/10 (e.g. 1/11) or else n^2 would be greater than 1/100 (e.g 1/9 * 1/9 or 1/81)

But as n is left to side of 0 in number line , so n<-1/10 , e.g. n=-1/11 or -1/12 or so on.

Thus inverting n , we get -11 or -12 or so on. Thus the reciprocal of n must definitely be less than -10 . So option (A) is the correct answer.
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Re: reciprocals and negatives [#permalink]

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New post 28 Nov 2011, 01:26
study wrote:
Can someone please explain this:

n^2 < 1/100
-1/10 < n < 1/10
n is negative so forget right side of equation
multiply both sides by 1/n
1/n * -1/10 < n * 1/n
-1/10n < 1
now multiply both sides by -10
-10 * -1/10n < 1 * -10
Flip signs
1/n > -10

Answer E!

But answer is A! Please explain your answers.


You made the mistake here ... when you know that n<0 , then 1/n will also be less than 0. So while multiplying by 1/n , you have to change the sign of the inequality.
i.e -1/10n > 1 or (Multilying both sides by -10) , -1/10n * -10 < 1 * -10 or n<-10 .. Thus answer (A).
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Re: reciprocals and negatives [#permalink]

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New post 28 Nov 2011, 23:27
thisiszico2006 wrote:
study wrote:
Can someone please explain this:

n^2 < 1/100
-1/10 < n < 1/10
n is negative so forget right side of equation
multiply both sides by 1/n
1/n * -1/10 < n * 1/n
-1/10n < 1
now multiply both sides by -10
-10 * -1/10n < 1 * -10
Flip signs
1/n > -10

Answer E!

But answer is A! Please explain your answers.


You made the mistake here ... when you know that n<0 , then 1/n will also be less than 0. So while multiplying by 1/n , you have to change the sign of the inequality.
i.e -1/10n > 1 or (Multilying both sides by -10) , -1/10n * -10 < 1 * -10 or n<-10 .. Thus answer (A).



Thanks I got this qs wrong while solving- ur ex makes it easy

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Re: reciprocals and negatives [#permalink]

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New post 03 Dec 2011, 00:05
This is my approach:

Since the number is to the left of 0, it must be a negative number.
The square of -1/10 will be 1/100. If the square has to be less than 1/100, the number,n, must be something like -1/15 to get 1/225 etc. So numbers lying between -1/10 and 0 will have squares less than 1/100.
But C is not the answer since we need to find reciprocal of n. Reciprocal of -1/10 is -10, of -1/15 is -15. Therefore, we see that numbers less than -10 are reciprocals of n and hence our answer is (A).
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Re: reciprocals and negatives [#permalink]

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New post 10 Dec 2011, 01:05
Just pick a number. I select \(\frac{-1}{100}\), which is negative and its square is \(\frac{1}{10000}\) and less than \(\frac{1}{100}\)

its reciprocal is: \(-100\)

Answer A

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Re: reciprocals and negatives [#permalink]

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New post 21 Aug 2017, 02:15
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Expert's post
study wrote:
Can someone please explain this:

If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and -1/10
E. greater than 10

I get ans E

n^2 < 1/100
-1/10 < n < 1/10
n is negative so forget right side of equation
multiply both sides by 1/n
1/n * -1/10 < n * 1/n
-1/10n < 1
now multiply both sides by -10
-10 * -1/10n < 1 * -10
Flip signs
1/n > -10

Answer E!

But answer is A! Please explain your answers.


We have \(n<0\) and \(n^2<\frac{1}{100}\)

\(n^2<\frac{1}{100}\) --> \(-\frac{1}{10}<n<\frac{1}{10}\), but as \(n<0\) --> \(-\frac{1}{10}<n<0\).

Multiply the inequality by \(-\frac{10}{n}\), (note as \(n<0\), then \(-\frac{10}{n}>0\), and we don't have to switch signs) --> \((-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})\) --> so finally we'll get \(\frac{1}{n}<-10<0\).

Answer: A.

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Re: reciprocals and negatives   [#permalink] 21 Aug 2017, 02:15
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