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

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

From \(n^2<\frac{1}{100}\), it follows that \(-\frac{1}{10}<n<\frac{1}{10}\), but as \(n<0\), we have \(-\frac{1}{10}<n<0\).

Multiply the inequality by \(-\frac{10}{n}\) (note: as \(n<0\), then \(-\frac{10}{n}>0\), and we don't have to switch signs). This leads to \((-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})\) , which simplifies to \(\frac{1}{n}<-10<0\).

Answer: A.
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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:

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

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

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 )

1/n < -10

The answer is (A).


n^2 < 1/100

therefore n < -1/10 or n > 1/10

n > 1/10 means that 10 > 1/n (not one of the answers)

n < -1/10 means that 1/n < -10 which is (A)

Given n^ 2 < 1/100 and n is negative.

n = -1/11 or n = -1/13.

So reciprocal of n = -11 or n = -13.

I wud go with option A

Step 1 : n<0 and n^2 <1/100 so choose n=-1/11

Step 2: Reciprocal of chosen n=-11

Step 3: Choose A

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.)

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

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

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.

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.

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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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?

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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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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We are given that n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100. This means n is negative, and it is between 0 and -1/10 since (-1/10)^2 = 1/100. Thus, n could be equal to values such as -1/11, -1/12, -1/13, etc. Notice that when we square these values, they are all less than 1/100.

Let’s take the reciprocal of any of these values listed. The reciprocal of -1/11 is -11. Similarly, the reciprocal of -1/12 is -12 and the reciprocal of -1/13 is -13. Since these reciprocals are all less than -10, answer choice A is correct.

Answer: A
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