CORRECTION there are two communities facebook and elitmus f alies likes only facebook and e alies like only elitmus g

ans : g = odd

from given data e% + f% - g% = 99% --> (1)
since 1% donot like either e or f
and given that f% = e% -2%
so from (1) e% + e% - g% - 2% = 99%
2e% - g% = 101%
g% = 2e% - 101% = even - odd = odd
11 years agoHelpfull: Yes(7) No(1)
Easiest one Ans. b)g is odd
Explanation:suppose there are total 100 people
now f% people like facebook i.e. n(F)=f
e% people like elitmus i.e. n(E)=e=f+2
g% people like both i.e. n(F^E)=g
1% people dont like anything i.e. n!(F^E)=1
n(F U E)=n(F)+n(E)-n(F^E)
100-1=99=f+f+2-g
2f-g=97
Since 2f will always produce even number
So g should be always odd.
11 years agoHelpfull: Yes(5) No(2)
Ans: g is odd
Explanation:

Since
a 'union' b=a+b-a 'intersection' b
100-1=f+e-g;
science e=f+2;
97=2f-g;

2f is alway even.
so g should be odd to equate with left side;
11 years agoHelpfull: Yes(3) No(0)
Prakash Gupta, Do you have elitmus paper photos, if yes then please mail me..
mohitjain012@gmail.com , Thank you in advance.
11 years agoHelpfull: Yes(3) No(0)
g is odd...
f% likes only facebook
e% likes only elitmus
g% likes both
1% likes none
so...f+e-g=(facebook U elitmus)
2+2f-g=(facebook U elitmus) (because e-f=2)
2+2f-g=(1-l)
even-g=odd (because 2+2f is even and 1-l is odd)
so...g has to be odd
11 years agoHelpfull: Yes(2) No(2)
b
Explaination:
clearly f +g +e =99
and since e is 2 more than f then either both odd or both even....in any case sum of e and f will be even ,but final sum(i.e. 99) is odd so g has to be odd
11 years agoHelpfull: Yes(1) No(0)

CORRECTION..!!!!
there are two communities facebook and elitmus.f% alies likes only facebook and e% alies like only elitmus.g% like both of them.1% do not like neither facebook nor elitmus. e is 2 more than f?
a) g is even.
b) g is odd
c) g can be odd or even
d) can't determined