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If n denotes a number to the left of 0 on the number line such that 24 Nov 2018, 15:05
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A video explanation can be found here:
https://www.youtube.com/watch?v=lG5iq9xcsws

First note that if n^2 < 1/100, the range of values of n would be

-1/10 < n < 1/10

But that range is further restricted – since we’re told n is to the left of 0, then

-1/10 < n < 0

(Note that C is a trap answer)

Thinking about reciprocals, we move from negative fractions to negative whole numbers. (The correct answer may become easier to visualize if you draw out the number line.)

Answer A
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Gladiator59 chetan2u Bunuel VeritasKarishma
Hello could you provide your assistance regarding my question? It has to do with inequalities.

lets say we have -1<10n (n<0) when we divide by n I know that it should be like this -1/n>10 and then 1/n < -10
But what I want to ask is when we divide by n, which we know is a negative number, why can't we proceed like this, -1<10n (n<0) 1/n>10 since -1/(a negative number) = positive number
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UNSTOPPABLE12
Gladiator59 chetan2u Bunuel VeritasKarishma
Hello could you provide your assistance regarding my question? It has to do with inequalities.

lets say we have -1<10n (n<0) when we divide by n I know that it should be like this -1/n>10 and then 1/n < -10
But what I want to ask is when we divide by n, which we know is a negative number, why can't we proceed like this, -1<10n (n<0) 1/n>10 since -1/(a negative number) = positive number
Show more

If we know -1 < 10n
(n is a negative number so the right hand side is negative too. n could take values such as -1/20, -1/50 etc)

To process it easily, we can just divide both sides by 10 (n gets separated out)
-1/10 < n
Since n < 0, we get -1/10 < n < 0

If instead, we divide both sides by n, we get
-1/n > 10
How did you get 1/n > 10?
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Gladiator59 chetan2u Bunuel VeritasKarishma
Hello could you provide your assistance regarding my question? It has to do with inequalities.

lets say we have -1<10n (n<0) when we divide by n I know that it should be like this -1/n>10 and then 1/n < -10
But what I want to ask is when we divide by n, which we know is a negative number, why can't we proceed like this, -1<10n (n<0) 1/n>10 since -1/(a negative number) = positive number

If we know -1 < 10n
(n is a negative number so the right hand side is negative too. n could take values such as -1/20, -1/50 etc)

To process it easily, we can just divide both sides by 10 (n gets separated out)
-1/10 < n
Since n < 0, we get -1/10 < n < 0

If instead, we divide both sides by n, we get
-1/n > 10
How did you get 1/n > 10?
Show more

Well, to be frank with you I was really confused between handling inequalities when there is a variable and when we have a specific number what I did was to consider n<0 but then imagined in my head that n is eg.-2 so i both changed the sign of inequality and the sign of the fraction to positive, which is wrong so instead of just dividing -1<10n by n(n<0) and getting -1/n>10 I would divide by n and simultaneously think that n is a number like eg.-2 so I would also change the sign of -1/n to 1/n which is wrong . thank you VeritasKarishma for your reply.
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Gladiator59 chetan2u Bunuel VeritasKarishma
Hello could you provide your assistance regarding my question? It has to do with inequalities.

lets say we have -1<10n (n<0) when we divide by n I know that it should be like this -1/n>10 and then 1/n < -10
But what I want to ask is when we divide by n, which we know is a negative number, why can't we proceed like this, -1<10n (n<0) 1/n>10 since -1/(a negative number) = positive number

If we know -1 < 10n
(n is a negative number so the right hand side is negative too. n could take values such as -1/20, -1/50 etc)

To process it easily, we can just divide both sides by 10 (n gets separated out)
-1/10 < n
Since n < 0, we get -1/10 < n < 0

If instead, we divide both sides by n, we get
-1/n > 10
How did you get 1/n > 10?

Well, to be frank with you I was really confused between handling inequalities when there is a variable and when we have a specific number what I did was to consider n<0 but then imagined in my head that n is eg.-2 so i both changed the sign of inequality and the sign of the fraction to positive, which is wrong so instead of just dividing -1<10n by n(n<0) and getting -1/n>10 I would divide by n and simultaneously think that n is a number like eg.-2 so I would also change the sign of -1/n to 1/n which is wrong . thank you VeritasKarishma for your reply.
Show more

Yes, you don't have to change the sign of the fraction. Note that -1/n is a positive number because n is negative. When you make it 1/n, you are making it negative again.
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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.
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Why exactly do you multiply by -10/n? What is the reasoning behind this decision. Thanks.
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Bunuel
hardnstrong

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.

Why exactly do you multiply by -10/n? What is the reasoning behind this decision. Thanks.
Show more

The questions asks: ...the reciprocal of n must be. So, we multiply by -10/n to get the the reciprocal of n (1/n) in the inequality and get the answer.

Hope it's clear.
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Video solution from Quant Reasoning starts at 0:28
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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hey Bunuel,

can you please elaborate why we did not flip the sign after multiplying by-10/n?


Thanks in advance.
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kadamhari825
hey Bunuel,

can you please elaborate why we did not flip the sign after multiplying by-10/n?


Thanks in advance.
Show more

I think this is explained in the solution, no?

"As \(n<0\), then \(-\frac{10}{n}>0\), and we don't have to switch signs".
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topmbaseeker
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
Show more

STRATEGY: As with all GMAT Problem Solving questions, we should immediately ask ourselves, Can I use the answer choices to my advantage?
In this case, the answer is a resounding yes, because we can just find a value of n that satisfies the given information, and then take the reciprocal of n.
From here, we should give ourselves about 20 seconds to identify a faster approach.
In this case, we can probably perform some algebra, but that will likely take longer than testing a value

We're told that the square of n is less than 1/100
In other words: n² < 1/100
So, for example, it could be the case that n = 1/20, since (1/20)² = 1/400, and 1/400 < 1/100

Important: The question also tells us that n denotes a number to the left of 0 on the number line
In other words, n is NEGATIVE, which means n = 1/20 doesn't satisfy this condition.
However, n = -1/20 meets all of the criteria since -1/20 is negative, and (-1/20)² = 1/400, and 1/400 < 1/100

If n = -1/20, then the reciprocal of n = -20/1 = -20
Only answer choice A works.
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If n denotes a number to the left of 0 on the number line such that 01 Dec 2022, 21:20
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Asked: 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

n < 0
n^2 < 1/100
n > -1/10
1/n < -10

IMO A
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Bunuel
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.
Show more

Hi Bunuel, I don't understand what you did after figuring out that -1/10 < n < 0. I would think that I only need to get the reciprocal so I would do -10 < 1/n < 0.

Do signs change?
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Bunuel
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.

Hi Bunuel, I don't understand what you did after figuring out that -1/10 < n < 0. I would think that I only need to get the reciprocal so I would do -10 < 1/n < 0.

Do signs change?
Show more


This is not correct.

We have \(-\frac{1}{10}<n<0\).

Say n = -1/100, so 1/n = -100: \(-\frac{1}{10}<-\frac{1}{100}<0\). Would it be correct to write -10 < -100 < 0? No.

We multiply \(-\frac{1}{10}<n<0\) by -10/n to get the the reciprocal of n (1/n) in the inequality and get the answer.

Hope it's clear.
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Kindly explain how come square of a number is less than ZERO.....
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deepak.brijendra
Kindly explain how come square of a number is less than ZERO.....
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The square of a number is always greater than or equal to 0. Can you specify which post you are referring to where it suggests that the square of a number is less than 0?
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Bunuel
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.

Hi Bunuel, I don't understand what you did after figuring out that -1/10 < n < 0. I would think that I only need to get the reciprocal so I would do -10 < 1/n < 0.

Do signs change?


This is not correct.

We have \(-\frac{1}{10}<n<0\).

Say n = -1/100, so 1/n = -100: \(-\frac{1}{10}<-\frac{1}{100}<0\). Would it be correct to write -10 < -100 < 0? No.

We multiply \(-\frac{1}{10}<n<0\) by -10/n to get the the reciprocal of n (1/n) in the inequality and get the answer.

Hope it's clear.
Show more

Hi Bunuel!

Thank you for your response. If you multiple n by -10/n, wouldn't it just become -10?
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Hi Bunuel, I don't understand what you did after figuring out that -1/10 < n < 0. I would think that I only need to get the reciprocal so I would do -10 < 1/n < 0.

Do signs change?


This is not correct.

We have \(-\frac{1}{10}<n<0\).

Say n = -1/100, so 1/n = -100: \(-\frac{1}{10}<-\frac{1}{100}<0\). Would it be correct to write -10 < -100 < 0? No.

We multiply \(-\frac{1}{10}<n<0\) by -10/n to get the the reciprocal of n (1/n) in the inequality and get the answer.

Hope it's clear.

Hi Bunuel!

Thank you for your response. If you multiple n by -10/n, wouldn't it just become -10?
Show more

Yes:

    \((-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})\)

    \(\frac{1}{n}<-10<0\).
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Hello Bunuel, thank you for your continued support. I’m curious about how you realized that multiplying by -10/n would give you the reciprocal of n. Is this a math trick or rule, or did you notice that this algebraic manipulation would leave 1/n in the inequality? Thank you!
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thegmatmaster
Hello Bunuel, thank you for your continued support. I’m curious about how you realized that multiplying by -10/n would give you the reciprocal of n. Is this a math trick or rule, or did you notice that this algebraic manipulation would leave 1/n in the inequality? Thank you!
Show more

We have \(-\frac{1}{10}<n<0\) and want to evaluate 1/n. How can we do this? If we multiply by -10/n, we'll get 1/n on the left side, and n is eliminated in the middle. So, it's a simple algebraic trick.
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