There are three different cable channels namely Ahead, Luck and Bang. In a survey it was
found that 85% of viewers respond to Bang, 20% to Luck, and 30% to Ahead. 20% of viewers
respond to exactly two channels and 5% to none.
1) What percentage of the viewers responded to all three?
(A) 10
(B) 12
(C) 14
(D) None of these
2) Assuming 20% respond to Ahead and Bang, and 16% respond to Bang and Luck,
what is the percentage of viewers who watch only Luck?
(A) 20
(B) 10
(C) 16
(D) None of these
Vein Diagram-how to attack these questions..
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 154
- Joined: Tue Aug 26, 2008 12:59 pm
- Location: Canada
- Thanked: 4 times
You can a. draw or venn diagram or b. use formulas or c. use a tablerahulg83 wrote:There are three different cable channels namely Ahead, Luck and Bang. In a survey it was
found that 85% of viewers respond to Bang, 20% to Luck, and 30% to Ahead. 20% of viewers
respond to exactly two channels and 5% to none.
1) What percentage of the viewers responded to all three?
(A) 10
(B) 12
(C) 14
(D) None of these
2) Assuming 20% respond to Ahead and Bang, and 16% respond to Bang and Luck,
what is the percentage of viewers who watch only Luck?
(A) 20
(B) 10
(C) 16
(D) None of these
Using a table is best if you have two options like black and white etc
Otherwise use a. or b like for this question.
I'll solve 1) here. Look at the Venn diagram attached.
Both 20% is in orange.
The total viewers is 100.
Make sure you account for the total viewers and take out the double counts (intersections).
The total viewers are (A U B U L) - 20 - all3 + 5
85 + 20 +30 -20 - all3 + 5 = 100%
120 - all3 = 100
all3 = 20
I'll go with A) for 1.
Someone else can solve 2.
- Attachments
-
- dmateer25
- Community Manager
- Posts: 1049
- Joined: Sun Apr 06, 2008 5:15 pm
- Location: Pittsburgh, PA
- Thanked: 113 times
- Followed by:27 members
- GMAT Score:710
niraj_a wrote:ppl,
for 1, if all3 = 20, then how is A i.e. 10% the answer? shouldn't it be 20%?
Let total = 100
100 = 85 + 20 + 30 + 5 – 20 – 2(x)
100 = 120 -2x
2x = 20
x = 10
-
- Master | Next Rank: 500 Posts
- Posts: 154
- Joined: Tue Aug 26, 2008 12:59 pm
- Location: Canada
- Thanked: 4 times
-
- Senior | Next Rank: 100 Posts
- Posts: 71
- Joined: Sat Sep 20, 2008 5:48 am
- Thanked: 5 times
-
- Master | Next Rank: 500 Posts
- Posts: 154
- Joined: Tue Aug 26, 2008 12:59 pm
- Location: Canada
- Thanked: 4 times
I think the answer is D. none because of 20.
Can someone please post OA or confirm without reasonable doubt!
Can someone please post OA or confirm without reasonable doubt!
-
- Master | Next Rank: 500 Posts
- Posts: 154
- Joined: Tue Aug 26, 2008 12:59 pm
- Location: Canada
- Thanked: 4 times
Yes hes got it right!dmateer25 wrote:niraj_a wrote:ppl,
for 1, if all3 = 20, then how is A i.e. 10% the answer? shouldn't it be 20%?
Let total = 100
100 = 85 + 20 + 30 + 5 – 20 – 2(x)
100 = 120 -2x
2x = 20
x = 10
-
- Senior | Next Rank: 100 Posts
- Posts: 62
- Joined: Thu Oct 23, 2008 9:20 am
- adilka
- Senior | Next Rank: 100 Posts
- Posts: 79
- Joined: Thu Oct 23, 2008 9:28 am
- Location: Canada
- Thanked: 1 times
- GMAT Score:700
Answer 1: Here's my logic based on this diagram:
A+B+C is 30+85+20 = 135
But this is double counting all the orange areas twice - X (once in A and once in L), Y (once in B and once in A) and Z (once in L and once in B) so we have to substract the X+Y+Z once to eliminate the double conting.
W is counted 3 times so we have to take out 2 of these to eliminate the "triple" couting.
Since 5% of the population don't watch any channels at all total is
A+B+C - (X+Y+Z) - 2W = 100-5
X+Y+Z = 20 (given) hence
30+85+20 + 20 - 2w = 100-95
W = 10!
A+B+C is 30+85+20 = 135
But this is double counting all the orange areas twice - X (once in A and once in L), Y (once in B and once in A) and Z (once in L and once in B) so we have to substract the X+Y+Z once to eliminate the double conting.
W is counted 3 times so we have to take out 2 of these to eliminate the "triple" couting.
Since 5% of the population don't watch any channels at all total is
A+B+C - (X+Y+Z) - 2W = 100-5
X+Y+Z = 20 (given) hence
30+85+20 + 20 - 2w = 100-95
W = 10!
- Attachments
-
-
- Senior | Next Rank: 100 Posts
- Posts: 62
- Joined: Thu Oct 23, 2008 9:20 am
let n(A)=a,n(b)=b,n(c)=c
n(a intersaction b intersaction c intersaction)=h
then n(a intersaction b)=e+h,n(a intersaction c)=f+h,n(b intersaction c)=g+h
100-95=a+b+c-[(e+h)+(f+h)+(g+h)]+h
95=(a+b+c)-[(e+f+g)+3h]+h
95=(85+30+20)-[(20)+3h]+h
95=135-20-2h
2h=20
h=10
n(a intersaction b intersaction c intersaction)=h
then n(a intersaction b)=e+h,n(a intersaction c)=f+h,n(b intersaction c)=g+h
100-95=a+b+c-[(e+h)+(f+h)+(g+h)]+h
95=(a+b+c)-[(e+f+g)+3h]+h
95=(85+30+20)-[(20)+3h]+h
95=135-20-2h
2h=20
h=10
- Attachments
-