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If n denotes a number to the left of 0 on the number line

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reciprocals and negatives  [#permalink]

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New post Updated on: 25 Nov 2011, 22:59
4
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


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
flip signs as n is negative
1/10n > 1
note 1/n * -1/10 = +1/10n
now multiply both sides by 10

10 * 1/10n < 1 * 10
Don't flip signs
1/n < 10

Answer is A! Please explain your answers.

Originally posted by study on 25 Nov 2011, 02:02.
Last edited by study on 25 Nov 2011, 22:59, edited 3 times in total.
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Re: reciprocals and negatives  [#permalink]

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New post 25 Nov 2011, 14:51
Thanks, Ian. I edited the post. That was a typo. Would you please look at the post now and correct my mistake. Thanks.
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Re: reciprocals and negatives  [#permalink]

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New post 26 Nov 2011, 07:14
study wrote:
Thanks, Ian. I edited the post. That was a typo. Would you please look at the post now and correct my mistake. Thanks.


In your edited post, the negative sign vanished - it still needs to be there. You have:

-1/10 < n

Now if we multiply by 1/n on both sides, we must reverse the inequality, since 1/n is negative:

-1/10n > 1

Now we can multiply by 10 on both sides:

-1/n > 10

Now we can multiply by -1 on both sides, reversing the inequality since we are multiplying by a negative:

1/n < -10

Note that you can also do this problem very quickly by finding any suitable number for n (say -1/100) and working out the reciprocal of n.
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Re: reciprocals and negatives  [#permalink]

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New post 26 Nov 2011, 18:22
study wrote:

Yes, the negative sign vanished, cuz negative * negative = positive. So a negative 1/n * negative 1/10 = + 1/10n
And that is exactly the part I don't understand. Why would you retain a negative after multiplying a negative number by another negative number? Would you please explain
To illustrate
-2 * -5 = 10. Not -10
so why would -1/n * -1/10 = -1/10n?



If you multiply, say, -1 by x, the result is -x. It makes no difference if x is positive or negative. If x is negative, then -x is a *positive* number, even if it might look negative because of the negative sign in front. There is a second negative sign 'hidden' inside of 'x'.

That's the issue with the step you took in your edited post. When you multiply 1/n by -1/10, the result is *always* equal to -1/10n. It makes no difference at all if n is positive or negative. If n is negative, then -1/10n is a positive number, because you have two negatives in the fraction, one in the numerator and one in the denominator (since n is negative).

I'd strongly suggest you review this part of algebra, because it's fundamental in many GMAT questions.
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new  [#permalink]

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New post 07 Mar 2012, 05:14
can you explain why are you flipping sign?

n> - 1/10.

why 1/n<-10 ?

if n=4 for example 4>-1/10 and 1/4>-10. why to flip sign?
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New post 07 Mar 2012, 06:23
akakhidze wrote:
can you explain why are you flipping sign?

n> - 1/10.

why 1/n<-10 ?

if n=4 for example 4>-1/10 and 1/4>-10. why to flip sign?


Welcome to GMAT Club. Below is an answer to your question.

Since n denotes a number to the left of 0 on the number then n is negative, so it can not be 4 as you assumed.

Next, when you multiply (or divide) an inequality by a negative number you must flip the sign of the inequality.

So, for \(n>-\frac{1}{10}\) --> multiply by negative -10 and flip the sign: \(-10n<1\) --> divide by negative \(n\) and flip the sign again: \(-10>\frac{1}{n}\).

For a complete solution refer to the posts above, for example: if-n-denotes-a-number-to-the-left-of-0-on-the-number-line-91659.html#p667838

Hope it helps.
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 17 Jun 2013, 05:51
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 22 Jun 2013, 07:20
Hi Bunuel
Can it be right if go like the following way?

given (-n)^2<1/100
=>n^2<1/00
=>n<1/10

Now Q ask us to find reciprocal of n
So take the reciprocal on both side simply we get
1/n<10 Ans
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 07 Nov 2013, 02:03
prasannajeet wrote:
Hi Bunuel
Can it be right if go like the following way?

given (-n)^2<1/100
=>n^2<1/00
=>n<1/10

Now Q ask us to find reciprocal of n
So take the reciprocal on both side simply we get
1/n<10 Ans


No.
Given: n < 0
\(n^2 < (1/100)\)
Note that you will not use -n because n is negative. n already includes the negative sign.
When you take square root, you get |n| < 1/10 (and not n < 1/10). This means -1/10 < n < 1/10. But since n < 0, -1/10 < n < 0.

We need to find the value of reciprocal i.e. 1/n.

n > -1/10
1/n < -10 Note that in an inequality, if both sides of the inequality have the same sign (positive or negative), the sign of inequality (<, >) flips when you take the reciprocal. Here both sides are negative so sign flips.
(Or do what Bunuel did: multiply by -10/n.)
So reciprocal is less than -10.

Instead, I would do this question by thinking of some numbers and figuring out the logic - discussed here: if-n-denotes-a-number-to-the-left-of-0-on-the-number-line-91659.html#p811517
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 03 Oct 2014, 19:39
Bunuel wrote:
hardnstrong wrote:
i dont get your point bangolarian
we have n^2 < 100 so -1/10 < n < 1/10
how you got n < - 1/10

Can somebody please explain if n > -1/10. does that mean its reciprocal is 1/n < -10 . Do we change less than and greater than sign with reciprocal of any interger? :?:


Refer to my previous post:

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 \(n<0\) expression \(-\frac{10}{n}>0\), 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 \(\frac{1}{n}<-10<0\). OR \(\frac{1}{n}<-10\).

Answer: A.

Hope it helps.







Hi Bunnel ,

I am weak with inequalities and flipping of signs , where could i get theory for it / additional similar practice questions ?

Thanks and Regards ,
Sheldon Rodrigues
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 04 Oct 2014, 03:22
shelrod007 wrote:
Bunuel wrote:
hardnstrong wrote:
i dont get your point bangolarian
we have n^2 < 100 so -1/10 < n < 1/10
how you got n < - 1/10

Can somebody please explain if n > -1/10. does that mean its reciprocal is 1/n < -10 . Do we change less than and greater than sign with reciprocal of any interger? :?:


Refer to my previous post:

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 \(n<0\) expression \(-\frac{10}{n}>0\), 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 \(\frac{1}{n}<-10<0\). OR \(\frac{1}{n}<-10\).

Answer: A.

Hope it helps.







Hi Bunnel ,

I am weak with inequalities and flipping of signs , where could i get theory for it / additional similar practice questions ?

Thanks and Regards ,
Sheldon Rodrigues


Theory on Inequalities
Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html
Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 14 Feb 2016, 09:25
These types of problems can be easier to do visually :

Step 1: range of n^2 < 1 /100 | --------- 0 <--------.01 -----|

Step 2: range of \sqrt{n^2} = \sqrt{1/100} (question states only to the left of 0) : | ---------(-1/10)-------->0 ---------|

* note to see if the range of n is going towards 0 or away test a number of n^2 and see what the sq root gives you i.e. \sqrt{1/10000} = +/- \frac{1}{100} which is smaller.

Step 3: inverse of the range n | <--------- (-10) ---- 0 ----------|
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 28 Mar 2017, 10:57
I approached it this way:
n is number to the left from 0
n^2<1/100
question asks 1/n-?
so from the data we know that -1/10<n<1/10
let's flip so -10>1/n>10
As we know that n is to the left from 0,
then 1/n<-10
Answer is A
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 08 Jul 2017, 20:13
Bunuel wrote:
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

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.


Bunuel, why do you Multiply \(-\frac{1}{10}<n<0\) by \(-\frac{10}{n}\)?

Can you not just take the reciprocal of \(-\frac{1}{10}<n<0\) to give you: \(-10<\frac{1}{n}<0\)?
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 09 Jul 2017, 02:28
LakerFan24 wrote:
Bunuel wrote:
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

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.


Bunuel, why do you Multiply \(-\frac{1}{10}<n<0\) by \(-\frac{10}{n}\)?

Can you not just take the reciprocal of \(-\frac{1}{10}<n<0\) to give you: \(-10<\frac{1}{n}<0\)?


As you can see you are not getting correct inequality, so you cannot do that way.

-10 < -5 < -1 but -1/10 < -1/5 < -1 is not correct.
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 10 Jul 2017, 19:39
Math mods and others - Please check my approach:

Given: n is less than 0, implying n is negative.

Also given: \(n^2\) < \(\frac{1}{100}\)

Square rooting both sides:

n< -\(\frac{1}{10}\) AND n < \(\frac{1}{10}\)

Reciprocating both the inequalities,
\(\frac{1}{n}\) <-10 AND \(\frac{1}{n}\) <10

Since, \(\frac{1}{n}\) MUST be less than -10 and 10 at all times, the option that always satisfies this is A.

Is this a right approach? I always falter when removing the squares on both sides of the equation/inequality and I want to make sure my approach is correct so that I can register it in my noggin.
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 26 Aug 2017, 08:41
experts, please correct me if i'm wrong:

isn't it easy to just take a glance here and eliminate A/C?
- You're told the SQUARE of n = fraction meaning n = fraction
--> The reciprocal of n must therefore b an integer. Elim B, C, D straight away

Between A & E, you know that n = negative, so its reciprocal cannot be positive. Elim E.

Ans: A
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 17 Dec 2017, 07:37
A

n= 1/x
(1/x)^2 < 1/100
LHS positive since x is negative and square is positive and RHS is positive

cross multiply
x^2 >100
(x-10)(x+10)>0

(-infinity,-10) U ( 10,+infinity)
positive part is not included as x is negative since n is negative ( left of zero 0) so reciprocal of negative cant be positive

answer is x or (1/n) <-10
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If n denotes a number to the left of 0 on the number line  [#permalink]

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New post 06 Jan 2018, 10:41
AKProdigy87 wrote:
The answer is A.

We are given two pieces of information:

1) \(n^2 < \frac{1}{100}\)

2) \(n < 0\)

SO:

\(n^2 < \frac{1}{100}\)

\(|n| < \frac{1}{10}\)

BUT n < 0 SO:

\(n > -\frac{1}{10}\)

To determine the conditions on the reciprocal of n:

\(\frac{1}{n} < -\frac{10}{1}\)

\(\frac{1}{n} < -10\)


Hello Bunuel, could help me to clear my doubts :-)

according to general rule when variable is under radical sign we square both sides -- >

for example if \(\sqrt{x}\) = 4 then we need to squae both sides of equation

so \(\sqrt{(x)}\) = \((4)^2\)

---> x= 16

now when it comes to the solution above from this equation \(n^2 < \frac{1}{100}\) we get this \(|n| < \frac{1}{10}\) - why ? should not we square both sides as I did in my example ? :?

Why are we applying this formula \(\sqrt{x^2}\) = \(|x|\) if this doesnt look like \(n^2 < \frac{1}{100}\) ? <-- (here we dont have radical sign) why 100 in denominator is reduced by 10 ? shouldn't 100 be multiplied by itself as in my example ? in my example the number 4 turns into 16 and here it get reduced...so I am confused

Also ss there a difference between \(\sqrt{x^2}\) and \(\sqrt{x}\) ?

many thanks ! :)

perhaps you friends can help :? :-) chetan2u , niks18
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