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25 Nov 2005, 10:56
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Is sqrt(x) a prime number?
1. l3x-7l=2x+2
2. x^2=9
1. l3x-7l=2x+2
2. x^2=9
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25 Nov 2005, 11:01
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From A) x=9 and x=1 then yes sqrtx is a prime
From B) x=3, x=-3 for radical sign x must be positive so x=-3 is out and again yes sqrtx is not a prime
so i would pick D)
my mistake 1 is not a prime so A) is not suff
the ans should be B)
From B) x=3, x=-3 for radical sign x must be positive so x=-3 is out and again yes sqrtx is not a prime
so i would pick D)
my mistake 1 is not a prime so A) is not suff
the ans should be B)
25 Nov 2005, 11:15
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we are looing for sqrt(x)
from statement 1,
l3x-7l=2x+2
so either 3x-7 =2x+2 or -3x+7=2x+2
so x = 9 or x =1
so sqrt(x) = +3 , -3 , +1 or -1
+3 , -3 are prime numbers but +1 or -1 are not prime numbers
so statement 1 is not sufficient.
from statement 2,
x^2 = 9
so x= 3 or -3
sqrt(3) = 1.6..... which is not a prime
sqrt(-3) is imaginary number. so again this statement is not sufficient.
again combining both statement doesn't help us.
so my answer is E.
from statement 1,
l3x-7l=2x+2
so either 3x-7 =2x+2 or -3x+7=2x+2
so x = 9 or x =1
so sqrt(x) = +3 , -3 , +1 or -1
+3 , -3 are prime numbers but +1 or -1 are not prime numbers
so statement 1 is not sufficient.
from statement 2,
x^2 = 9
so x= 3 or -3
sqrt(3) = 1.6..... which is not a prime
sqrt(-3) is imaginary number. so again this statement is not sufficient.
again combining both statement doesn't help us.
so my answer is E.
