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by rahul.s » Mon Feb 08, 2010 9:54 pm
i'm having some difficulty with this problem:

the 38 movies in a video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. however, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy and 4 are both drama and comedy. how many action-drama-comedies are there?

OA is 2

can't seem to figure it out. i tried it algebraically as well as plugging in numbers, but can't seem to get it.
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by papgust » Mon Feb 08, 2010 10:08 pm
This is the formula for using it algebraically.

True # of objects = (total # Action) + (total # Drama) + (total #Comedy) - (# only AD) - (# only DC) - (# only AC) - 2(# ADC)

Let x be the ALL three categories.

Total #Action = 10
Total #Drama = 20
Total #comedy = 18

A n D = 5
#only A n D = 5-x (We need to eliminate double count by subtracting all 3 category number i.e. intersection of all 3 sets)

A n C = 3
#only A n C = 3-x

D n C = 4
#only D n C = 4-x

Total movies (A U D U C) = 38

Now, apply everything to the formula above

38 = 10+20+18 - [(5-x)+(3-x)+(4-x)] - 2*x
38 = 48 - [12-3x] - 2*x
x = 2
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by shashank.ism » Tue Feb 09, 2010 12:02 am
rahul.s wrote:i'm having some difficulty with this problem:

the 38 movies in a video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. however, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy and 4 are both drama and comedy. how many action-drama-comedies are there?

OA is 2

can't seem to figure it out. i tried it algebraically as well as plugging in numbers, but can't seem to get it.

Ans: Let N(A) be the no. of action movies , N(D) the no. of drama movies. and N(C) be no. of comedy movies.
N(A) = 10, N(B) = 20, N(C) =18, and N(A or B or C) = 38
so N(A and D) = 5, N(A and C) = 3, N(D and C) = 4.

N(A or B or C) = N(A) + N(B) + N(C) - N(A and D) - N(A and C) - N(D and C) + 2 N(A and B and C)
[we subtract the common part i.e. {N(A and D) + N(A and C) + N(D and C)} but in this way we subtracted N(A and B and C) three times so lastly we add it again 2 times.]
--> 38 = 10 + 20 + 18 - 5 - 3 -4 + 2N(A and B and C) --> N(A and B and C) = (12 -10)/2 = 1
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by shashank.ism » Tue Feb 09, 2010 12:15 am
papgust wrote:This is the formula for using it algebraically.

True # of objects = (total # Action) + (total # Drama) + (total #Comedy) - (# only AD) - (# only DC) - (# only AC) - 2(# ADC)

Let x be the ALL three categories.

Total #Action = 10
Total #Drama = 20
Total #comedy = 18

A n D = 5
#only A n D = 5-x (We need to eliminate double count by subtracting all 3 category number i.e. intersection of all 3 sets)

A n C = 3
#only A n C = 3-x

D n C = 4
#only D n C = 4-x

Total movies (A U D U C) = 38

Now, apply everything to the formula above

38 = 10+20+18 - [(5-x)+(3-x)+(4-x)] - 2*x
38 = 48 - [12-3x] - 2*x
x = 2
Papgust you took the right approach but in mid way u got out of track.
Now if you are taking only
A n D = 5
#only A n D and not including AnDnC = 5-x (We need to eliminate double count by subtracting all 3 category number i.e. intersection of all 3 sets)

A n C = 3
#only A n C and not including AnDnC= 3-x

D n C = 4
#only D n C and not including AnDnC = 4-x


then why are you again subtracting 2x ( I don't know for what reason, instead you will have to add x which is not included because u have subtracted it in each of them)
so eq. would be like this.
38 = 10+20+18 - [(5-x)+(3-x)+(4-x)] + x

you can take other approach too like what I have taken.
38 = 10+20+18 - [(5)+(3)+(4)] - 2*x ( this 2x is subtracted because it comes thrice in 5,3,4 and we need it only once.)
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by thephoenix » Tue Feb 09, 2010 12:17 am
the best way to do is by venn diag

draw three intersecting venn dig and assume the portion common to all 3 as x
now the portion common to A and D will be 5-x ( as bth action and drama is 5)
similarly portion common to Cand D will be 4-x ( as bth action and drama is 4)
and portion common to A and C will be 3-x ( as bth action and drama is 3)

now the rest of the portion of A which not common to any one is= 10-x-(5-x)-(3-x)=2+x
similarly for C=11+x
for D=11+x

now tot=2+x+11+x+11+x+(5-x)+(4-x)+(3-x)+x=36+x=38--->x=2

most of the things which i have written can be done by looking at circles

i wish i cud draw the circles here but there are no such tools

suggestion for moderators pls include such option to draw and to write inside the fig.. it will help
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by papgust » Tue Feb 09, 2010 12:41 am
shashank.ism wrote: Papgust you took the right approach but in mid way u got out of track.
Now if you are taking only
A n D = 5
#only A n D and not including AnDnC = 5-x (We need to eliminate double count by subtracting all 3 category number i.e. intersection of all 3 sets)

A n C = 3
#only A n C and not including AnDnC= 3-x

D n C = 4
#only D n C and not including AnDnC = 4-x


then why are you again subtracting 2x ( I don't know for what reason, instead you will have to add x which is not included because u have subtracted it in each of them)
so eq. would be like this.
38 = 10+20+18 - [(5-x)+(3-x)+(4-x)] + x

you can take other approach too like what I have taken.
38 = 10+20+18 - [(5)+(3)+(4)] - 2*x ( this 2x is subtracted because it comes thrice in 5,3,4 and we need it only once.)
Hey shashank,

We are subtracting a 'x' from the doubles portion because when it's already counted in set A and set B. So we need to subtract an x from doubles.

Regarding the triples, remember that we are not taking (only #A), (only #B), and (only #C). We are taking (total #A), (total #B), and (total #C). It's already counted in sets A, B and C. So we are deducting the triples portion twice.
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by shashank.ism » Tue Feb 09, 2010 1:48 am
papgust wrote: Hey shashank,

We are subtracting a 'x' from the doubles portion because when it's already counted in set A and set B. So we need to subtract an x from doubles.

Regarding the triples, remember that we are not taking (only #A), (only #B), and (only #C). We are taking (total #A), (total #B), and (total #C). It's already counted in sets A, B and C. So we are deducting the triples portion twice.

Papgust I think u r having correct equation..
Last edited by shashank.ism on Tue Feb 09, 2010 1:59 am, edited 1 time in total.
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by shashank.ism » Tue Feb 09, 2010 1:51 am
This is the diagram which I have short out, so that this problem could be understood easily...
You can short out this from the folowing venn diagram.
Image
Now we can see
blue = 10 - (5-x) - (3-x) - x
red = 20 - (5-x) - (4-x) - x
green = 18 - (3-x) -(4-x) -x
light blue = 5-x
yellow = 3-x
purple = 4-x
white = x
add all these
10 - (5-x) - (3-x) - x + 20 - (5-x) - (4-x) - x + 18 - (3-x) -(4-x) -x + 5-x + 3-x + 4-x + x
= 10+20+18-(5-x)-(4-x)-(3-x)+x -3x
--> 38 = 48 - 12 + x --> x=2 Ans.
Last edited by shashank.ism on Tue Feb 09, 2010 2:09 am, edited 1 time in total.
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by papgust » Tue Feb 09, 2010 2:05 am
shashank.ism wrote: Papgust I think u r having correct equation..
I was about to reply to your post. but suddenly you edited the post. I hope its clear now. :)
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by shashank.ism » Tue Feb 09, 2010 2:12 am
papgust wrote:
shashank.ism wrote: Papgust I think u r having correct equation..
I was about to reply to your post. but suddenly you edited the post. I hope its clear now. :)
yeah I just missed out the 3x which is also added in 1st three values. Solving this type of eq. after very long time.. Now all the concepts got fresh when I made that colorful venn diagram. :)
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by ajith » Tue Feb 16, 2010 1:58 am
rahul.s wrote:i'm having some difficulty with this problem:

the 38 movies in a video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. however, some movies are classified under more than one category: 5 are both action and drama, 3 are both action and comedy and 4 are both drama and comedy. how many action-drama-comedies are there?

OA is 2

can't seem to figure it out. i tried it algebraically as well as plugging in numbers, but can't seem to get it.
n(A U B U C) = n(A) + n(B) + n(C) - [n(A n B) + n(B n C) + n(C n A) ] + n(A n B n C)
38 = 10+20+18 - 5 -3 -4 + x
38= 36+x
x=2
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by aslan » Sun Nov 28, 2010 2:37 pm
I guess papgust approach is the most efficient here.

Although the Venn diagram is easier to conceptualize, but it does take long and there are more chance of errors in the + and - that a person will make.Makes work more messy in a 2 min time.
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by danielphonics » Mon Nov 14, 2011 4:33 pm
shashank.ism wrote:
papgust wrote:This is the formula for using it algebraically.

True # of objects = (total # Action) + (total # Drama) + (total #Comedy) - (# only AD) - (# only DC) - (# only AC) - 2(# ADC)

Let x be the ALL three categories.

Total #Action = 10
Total #Drama = 20
Total #comedy = 18

A n D = 5
#only A n D = 5-x (We need to eliminate double count by subtracting all 3 category number i.e. intersection of all 3 sets)

A n C = 3
#only A n C = 3-x

D n C = 4
#only D n C = 4-x

Total movies (A U D U C) = 38

Now, apply everything to the formula above

38 = 10+20+18 - [(5-x)+(3-x)+(4-x)] - 2*x
38 = 48 - [12-3x] - 2*x
x = 2
Papgust you took the right approach but in mid way u got out of track.
Now if you are taking only
A n D = 5
#only A n D and not including AnDnC = 5-x (We need to eliminate double count by subtracting all 3 category number i.e. intersection of all 3 sets)

A n C = 3
#only A n C and not including AnDnC= 3-x

D n C = 4
#only D n C and not including AnDnC = 4-x


then why are you again subtracting 2x ( I don't know for what reason, instead you will have to add x which is not included because u have subtracted it in each of them)
so eq. would be like this.
38 = 10+20+18 - [(5-x)+(3-x)+(4-x)] + x

you can take other approach too like what I have taken.
38 = 10+20+18 - [(5)+(3)+(4)] - 2*x ( this 2x is subtracted because it comes thrice in 5,3,4 and we need it only once.)

... I tried this approach but.... it doesn't work, that equation yields : I get 38 = 36- 2x, and therefore x= (-1).

Or .. am I working a bit too late?
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