Problem Solving

Can somebody help solve this one?




Thanks a lot :roll:

A

x<0, and (x^2)<(1/100), thus x<-(1/10), reciprocal of this is -10

x<-10

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

If n = -1/11 then this squared would be 1/121 which would meet our condition. The reciprocal of -1/11 is -11 or greater than -10.

A.

square of n <1/100

-1/10<n<+1/10 and n is left of 0 implies

-1/10 <n<0

reciprocate it
-10>1/n>1/0

1/n is always less than -10

Answer A

bstalling wrote:A

x<0, and (x^2)<(1/100), thus x<-(1/10), reciprocal of this is -10

x<-10

x<-1/10 is wrong it has to be x>-1/10.


x^2 < (1/100) ==> -1/10<x
==> -10 >1/x
==> 1/x<-10

Thanks
raama