Latest Elitmus Aptitude Question SOLUTION: 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

• Placement Papers
• /
• Elitmus

Elitmus Question Papers:

(#M40019212) ELITMUS question Keep an EYE

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

Asked In Elitmus Prakash Gupta (10 years ago)

Unsolved Read Solution (7)

Is this Puzzle helpful?   (6)   (1) Submit Your Solution

Read Solution (7) :

IFFAT IRAM  (10 years ago)

f% likes only facebook
e% likes only elitmus
g% likes both
1% likes none
so,
99% likes either facebook or elitmus or both
=> f+e+g=99
also, e=f+2
if f is even , then e is even
or if f is odd, then e is odd
in any case sum of f and e would be even
so, f+e+g=99
=> even + g=odd
g has to be odd to satisfy the above relation
ans) g is odd

Is this solution Helpfull? Yes (55) | No (2)  

venkatesh  (10 years ago)

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

Is this solution Helpfull? Yes (7) | No (1)  

TEJ PRAKASH TEWARI  (10 years ago)

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.

Is this solution Helpfull? Yes (5) | No (2)  

PREM KUMAR  (10 years ago)

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;

Is this solution Helpfull? Yes (3) | No   

Mohit Jain  (10 years ago)

Prakash Gupta, Do you have elitmus paper photos, if yes then please mail me..
mohitjain012@gmail.com , Thank you in advance.

Is this solution Helpfull? Yes (3) | No   

surabhi sugandha  (10 years ago)

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

Is this solution Helpfull? Yes (2) | No (2)  

Anubhav Guleria  (10 years ago)

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

Is this solution Helpfull? Yes (1) | No   

Please Login to Read Solution.