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vipulgoyal
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If x is not equal to 1, is x^2/(x-1) greater than x?
(1) x is not an integer.
(2) x is positive.
I think OE needs another view, OE as follows..
Is \frac{x^2}{x-1}>x --> is \frac{x^2}{x-1}-x>0? --> is \frac{x^2-x^2+x}{x-1}>0 --> is \frac{x}{x-1}>0 --> is x<0 or x>1?
(1) x is not an integer --> clearly insufficient: for example if x=1.5 the the answer will be YES but if x=0.5 then the answer will be NO.
(2) x is positive. Also not sufficient.
(1)+(2) The values of x from (1) also satisfy (2) (x can be positive fraction from the range (0,1) or some non integer more than 1) thus even taken together statements are not sufficient.
OA Answer: E.
As B says x is +ve , which means range, from 0 to positive "excluding 0"
from a and b , range becomes 0<x<1 "not including 0 and 1"
in that case ans should be C
(1) x is not an integer.
(2) x is positive.
I think OE needs another view, OE as follows..
Is \frac{x^2}{x-1}>x --> is \frac{x^2}{x-1}-x>0? --> is \frac{x^2-x^2+x}{x-1}>0 --> is \frac{x}{x-1}>0 --> is x<0 or x>1?
(1) x is not an integer --> clearly insufficient: for example if x=1.5 the the answer will be YES but if x=0.5 then the answer will be NO.
(2) x is positive. Also not sufficient.
(1)+(2) The values of x from (1) also satisfy (2) (x can be positive fraction from the range (0,1) or some non integer more than 1) thus even taken together statements are not sufficient.
OA Answer: E.
As B says x is +ve , which means range, from 0 to positive "excluding 0"
from a and b , range becomes 0<x<1 "not including 0 and 1"
in that case ans should be C
