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reciprocals and negatives
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Updated on: 25 Nov 2011, 22:59
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
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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
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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
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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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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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Re: If n denotes a number to the left of 0 on the number line
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17 Jun 2013, 05:51



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Re: If n denotes a number to the left of 0 on the number line
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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
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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: ifndenotesanumbertotheleftof0onthenumberline91659.html#p811517
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Re: If n denotes a number to the left of 0 on the number line
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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
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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 InequalitiesSolving Quadratic Inequalities  Graphic Approach: solvingquadraticinequalitiesgraphicapproach170528.htmlInequality tips: tipsandhintsforspecificquanttopicswithexamples172096.html#p1379270inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.htmleverythingislessthanzero108884.htmlgraphicapproachtoproblemswithinequalities68037.html
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Re: If n denotes a number to the left of 0 on the number line
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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
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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
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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
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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
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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
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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
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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 (x10)(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
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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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