Elitmus
Exam
Numerical Ability
Percentage
In an opinion poll 78% of those asked were in the favour of atleast one of the three sports persons to be included in the commonwealth organizing commitee Saina, Dhoni and Anand. 50% of those asked favoured Saina, 30% favoured Dhoni 20% favored Anand.If 5% of those asked favoured all of them,what percentage of those asked favoured atleast one person ??
Options
A)61
B)56
C)51
D)22
Read Solution (Total 12)
-
- Let x% be in favour of A and B
y% be in favour of B and C
z % be in favour of A and C
=> 78 - 5 - (x + y + z) = (50-5-x-z) + (30-5-x-y) + (20-5-y-z)
=> (x + y + z) = 12%
=> (78 - (x+y+z)-5) = 61% - 11 years agoHelpfull: Yes(35) No(4)
- comes out to be 61 for exactly one.
- 11 years agoHelpfull: Yes(7) No(0)
- opinion poll 78%
let X% for A&B
Y% for B&C
Z% for C&A
78-5-(X+Y+Z)=(50-5-X-Z) + (30-5-x-y) + (20-5-y-z)
=> 73 -X-Y-Z=100-15-2X-2Y-2Z
=> (X+Y+Z)=12%
78-5-12=61% ans - 9 years agoHelpfull: Yes(7) No(0)
- 61 for exactly one person
- 11 years agoHelpfull: Yes(3) No(1)
- let total members be 100
78% of 100 is 78
raina=50%of 78=39
dhoni=30%of 78~23
anand=20%of 78~16
all 5%of 78 all=3.4~3
use venn diagram dude
so atleast one person=only raina+only dhoni+only anand+raina&&dhoni&&anand
=raina-all+dhoni-all+anand-all+all
=39-3+23-3+16-3
=36+20+13=69
69*100/78=88.8 - 11 years agoHelpfull: Yes(2) No(13)
- 61%
5% were in favour of all
Let x% be in favour of A and B
y% be in favour of B and C
z % be in favour of A and C
=> 78 + 5 - (x + y + z) = (50+5-x-z) + (30+5-x-y) + (20+5-y-z)
=> (x + y + z) = 32%
=> (78 - (x+y+z)+5) = 51% favoured exactly one. - 9 years agoHelpfull: Yes(1) No(0)
- In the last line it is exactly one person by this the answer is 56
- 11 years agoHelpfull: Yes(0) No(11)
- will u pls explain the answer....mr. sumit
- 11 years agoHelpfull: Yes(0) No(3)
- @sumit pls explain d answer?
- 11 years agoHelpfull: Yes(0) No(1)
- could anyone tell me the another way of solving this question?
I am not getting this ways - 11 years agoHelpfull: Yes(0) No(1)
- Let x% be in favour of A and B
y% be in favour of B and C
z % be in favour of A and C
=> 78 - 5 - (x + y + z) = (50-5-x-z) + (30-5-x-y) + (20-5-y-z)
=> (x + y + z) = 12%
=> (78 - (x+y+z)-5) = 61%
Read more at http://www.m4maths.com/18476-In-an-opinion-poll-78-of-those-asked-were-in-the-favour-of-atleast-one-of-the-three-sports-persons.html#2cH2IsUPbhIZsrWs.99 - 9 years agoHelpfull: Yes(0) No(0)
- Abxjdjdheisisudh
- 2 years agoHelpfull: Yes(0) No(0)
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