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

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Manager
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Updated on: 25 Nov 2011, 21:59
4

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

Originally posted by study on 25 Nov 2011, 01:02.
Last edited by study on 25 Nov 2011, 21:59, edited 3 times in total.
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25 Nov 2011, 04:41
study wrote:

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

First, that is not what E says (E says that 1/n is greater than positive 10, not negative 10), so that might have suggested that your answer wasn't quite right. I've highlighted your mistake in red. When you multiply on both sides of the inequality by 1/n, you must reverse the inequality, because 1/n is negative.
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25 Nov 2011, 13:51
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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26 Nov 2011, 06:14
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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26 Nov 2011, 17:22
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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07 Mar 2012, 04:14
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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07 Mar 2012, 05:23
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?

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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17 Jun 2013, 04:51
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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22 Jun 2013, 06:20
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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07 Nov 2013, 01:03
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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03 Oct 2014, 18:39
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 $$\frac{1}{n}<-10$$.

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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04 Oct 2014, 02:22
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 $$\frac{1}{n}<-10$$.

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
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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14 Feb 2016, 08:25
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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28 Mar 2017, 09:57
I approached it this way:
n is number to the left from 0
n^2<1/100
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
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Re: If n denotes a number to the left of 0 on the number line  [#permalink]

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08 Jul 2017, 19:13
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 $$\frac{1}{n}<-10<0$$.

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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09 Jul 2017, 01:28
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 $$\frac{1}{n}<-10<0$$.

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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10 Jul 2017, 18:39
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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26 Aug 2017, 07:41
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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17 Dec 2017, 06:37
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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06 Jan 2018, 09:41
AKProdigy87 wrote:

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
If n denotes a number to the left of 0 on the number line &nbs [#permalink] 06 Jan 2018, 09:41

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