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"

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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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~fluke

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