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by fercho81 » Sun Sep 06, 2009 10:52 am
If n denotes a number to the left of zero on the number line such that the square of n is less than 1/100, then the reciprocal of n must be:

1)less than -10
2)between -1 and -1/10
3)between -1/10 and 0
4)between 0 and 1/10
5)greater than 10

Thanks in advance for your answers. The answer is B.
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by krumpy » Sun Sep 06, 2009 1:42 pm
Are you sure it's B? Wouldn't the square of the reciprocal of -1 be 1, which is greater than 1/100? I would think it would be A:

n<-1/10 => n^2<1/100
so the reciprocal is less than -10
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by tom4lax » Sun Sep 06, 2009 3:34 pm
If we pick n = -1/100, then n^2 is 1/10,000, which satisfies the condition.

Reciprocal of n is -100.

Answer is A?
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It is E.

by enniguy » Sun Sep 06, 2009 5:00 pm
Darn. I didn't read the question properly! Yes it is B. <Edited my previous Post>

If n^2 < 1/100 then,

-1/10 < n < 1/10. Calculating the reciprocal,
the equation will be,
-10 < 1/n > 10. How? Consider a similar equation:

-1/2 < 1/4 < 1/2. Here n = 1/4. Calculate the reciprocal:
-2 < 4 > 2.

But, n is less than zero and so 1/n will also be less than 0. So the equation becomes,
-10 < 1/n (Reciprocal of n) < 0.

Also, n value will is between 0 and -1/10. So any value between 0 and -1 will have a reciprocal with value less than -1. Hence,

-10 < 1/n < -1.
Last edited by enniguy on Mon Sep 07, 2009 3:20 am, edited 3 times in total.
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by bharathh » Sun Sep 06, 2009 9:23 pm
Question states that n is left of zero on the number line. So n is negative. D and E are ruled out

Next part states that its square is less than 1/100...
1/400 is less than 1/100. Its negative root would be -1/20

Reciprocal of -1/20 is -20 which is less than -10

Answer has to be A
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