inequity 54

magical cook · Posted: Sun Dec 23, 2007 11:42 am Post subject: inequity 54

If n denotes a numner 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..

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

my way was...
n^2 , 1/100
-1/10 < n < 1/10
since n is negative
-1/10 < n
n needs to be reciprocal
-10 < n

but which is greater than -10 and which is not answer (not even an option)

Could anyone pls take a look at my procedure and give me advise where I got it wrong...

gabriel · Posted: Sun Dec 23, 2007 12:34 pm Post subject: Re: inequity 54

magical cook · Posted: Sun Dec 23, 2007 2:08 pm Post subject:

ahhh, my bad.. Embarassed

thank you !

Bschool08

magical cook, can you confirm that A is the right answer? i'm not seeing how -10<1/n is actually n<-10---i'm missing something here? if n is actually negative shouldn't taking it over to the other side yield n>10 making the answer E? if you can confirm would appreciate it - thanks

magical cook · Posted: Sun Dec 23, 2007 7:56 pm Post subject:

Hi, the answer is A (this is gmatprep question, so OA should be correct Wink

)

Bschool08 · Posted: Sun Dec 23, 2007 9:21 pm Post subject:

ok- thanks for confirming

samirpandeyit62 · Posted: Sun Dec 23, 2007 9:32 pm Post subject:

n^2 < 1/100

i.e n <1> -1/10

since n is to the left of 0, hence n > -1/10

i.e n > -1/10

i.e 10 > -1/n

or 1/n (reciprocal of n)< -10

A