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If n denotes a number to the left of 0 on the number line [#permalink]
29 Dec 2009, 07:50

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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.

Re: Inequalities and reciprocals [#permalink]
21 Mar 2010, 10:24

nifoui wrote:

If n denotes a number to the left of the 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

now lets say p = 1/n There can be two equations n < 0 => 1/p <0 since 1 cannot be negative this implies p< 0 and n^2 < 1/100 => 1/p^2 < 1/100 =>p^2>100 => (p -10)(p+10) > 0 => either p <-10 or p >10 since p < 0 from first equation we can deduce that p < -10 hence answer is A _________________

Re: Inequalities and reciprocals [#permalink]
21 Mar 2010, 10:28

nifoui wrote:

If n denotes a number to the left of the 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

The number of left of the 0 on the number line. We know that n is negative. The square of n is less than 1/100 so, -1/10 < n < 0. Choose a number between -1/10 and 0, for example -1/20. The reciprocal of that is -20. Thus the answer is A. _________________

Re: Inequalities and reciprocals [#permalink]
21 Mar 2010, 17:32

A good approach for these kind of problems is always a Trail&Error/Substitution.

In this case since the square must be less than \frac{1}{100} consider any number in denominator more than 100. lets say 121 that will give us \frac{1}{121}.

Since n is -ve the square root of this will be \frac{-1}{11}. and the reciprocal of this number is -11.

Now Go back to the options and compare the one that satisfy the answer.

If there is only one option then that's your answer. If there are two try to eliminate one by reason.(can also try a different fraction.)

Re: PS Number line [#permalink]
23 Mar 2010, 05:41

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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.

Re: PS Number line [#permalink]
24 Mar 2010, 00:07

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.

Bunuel, Thanks so much for your explanation

As i am weak in this type of problems so just need clarification on 1 more thing 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.

because n < 0 and from above we know n >-1/10 so instead of multiplying by -10/n can we straight so it like this -> if n>-1/10 , therefore 10>-1/n which means -10>1/n

Re: PS Number line [#permalink]
12 May 2010, 09:55

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Bunuel wrote:

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.

Answer: A.

Perhaps it's easier to remember this regarding switching of the signs...

When a and b are both positive or both negative:

* If a < b then 1/a > 1/b * If a > b then 1/a < 1/b

However when either a or b is negative (but not both) the direction stays the same:

* If a < b then 1/a < 1/b * If a > b then 1/a > 1/b

Take the above rules into consideration.... n^2 < 1/100

|n| < 1/10

but as n<0, therefore

-n < 1/10 or we can write as n > -1/10

( think of the above mentioned rules regarding the signs of the inequality, and as we know that n is also negative and right side of the inequality is also negative, thus taking the reciprocal will invert the sign of the inequality )

Re: if n denotes a number to the left of 0 [#permalink]
02 Nov 2010, 05:17

Expert's post

1

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anilnandyala wrote:

if n denotes a number to the left of 0 on the number line such that the square of n is less then 1/100 then reciprocal of n must be? a less then -10 b B/w -1 & -1/10 c B/w -1/10 & 0 d B/w 0 & 1/10 e greater then 10

Since the number is to the left of 0, it must be a negative number. The squares pf -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). _________________

Re: Number properties [#permalink]
21 Aug 2011, 06:49

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If n denotes a number to the left of 0 - means that n is negative square of n is less than 1/100 - means n is less than -1/10 then the reciprocal of n must be 1/(1/10) = less than -10 _________________

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Re: Gmatprep number line PS [#permalink]
03 Sep 2011, 00:39

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800_gal 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

kkalyan wrote:

so when n^2 < 1/100 then n<-1/10 It means when we take square root on both side we have to change the sign? Can anyone explain the logic?

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

n^2<\frac{1}{(10)^2}

Taking square root both sides:

\sqrt{n^2}<\sqrt{\frac{1}{(10)^2}}

\hspace{3} \sqrt{n^2}=|n|

\therefore \hspace{3} |n|<\frac{1}{10}

|x|<y is equivalent to -y<x<+y

Similarly, -\frac{1}{10}<n<\frac{1}{10}

From the stem we know, n<0

So, -\frac{1}{10}<n<0

Write those separately: -\frac{1}{10}<n Flip RHS and LHS and change the symbol to ">" n>-\frac{1}{10} Multiply both sides by -10: -10*n<1 Divide each side by n(n is -ve so signs flip again) -10>\frac{1}{n}

Flip RHS and LHS and change the symbol to "<" \frac{1}{n}<-10--------1

Also, n<0-----------2

In 1 and 2; 1 is more limiting \frac{1}{n}<-10

Ans: "A"

************************************************************** Note: Few of these steps could be avoided, but I just wanted to make clear how the inequality symbols and polarity signs change. _________________

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