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VP
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Set Theory Made Easy [#permalink]
 23 Nov 2007, 14:25
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See paper attached - enjoy
Please report any comments or errors.
Thanks
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CEO
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among 200 ppl, 56% like strawberry, 44% like apple, and 40% like raspberry. IF 30% of people like both strawbery and apple, what is the LARGEST possible number of ppl who like raspberry but do not like either strawberry or apple?
20
60
80
86
92
is this one considered set theory? can u solve it too?
thanks
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VP
Joined: 08 Jun 2005
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bmwhype2 wrote: among 200 ppl, 56% like strawberry, 44% like apple, and 40% like raspberry. IF 30% of people like both strawbery and apple, what is the LARGEST possible number of ppl who like raspberry but do not like either strawberry or apple?
20
60
80
86
92
is this one considered set theory? can u solve it too?
thanks
the answer is (B) - see my paper
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SVP
Joined: 28 Dec 2005
Posts: 1612
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KillerSquirrel wrote: bmwhype2 wrote: among 200 ppl, 56% like strawberry, 44% like apple, and 40% like raspberry. IF 30% of people like both strawbery and apple, what is the LARGEST possible number of ppl who like raspberry but do not like either strawberry or apple?
20
60
80
86
92
is this one considered set theory? can u solve it too?
thanks the answer is (B) - see my paper still having a bit of difficulty ... can you show us ?
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CEO
Joined: 29 Aug 2007
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1
This post received
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bmwhype2 wrote: among 200 ppl, 56% like strawberry, 44% like apple, and 40% like raspberry. IF 30% of people like both strawbery and apple, what is the LARGEST possible number of ppl who like raspberry but do not like either strawberry or apple?
20
60
80
86
92
is this one considered set theory? can u solve it too?
thanks
total = S + A + R - (SA + SR + AR) - 2(SAR)
200 = S + A + R - (SA + SR + AR) - 2(SAR)
only S = S - SA - SR - SAR
only A = A - SA - AR - SAR
only R = R - SR - AR - SAR
200 = only S + only A + only R + (SA + SR + AR) + 2(SAR)
200 = (S - SA - SR - SAR) + (A - SA - AR - SAR) + (R - SR - AR - SAR) + (SA + SR + AR) + 2(SAR)
to maximize only R = (R - SR - AR - SAR), we need to minimize SR, AR and SAR. so lets make it 0:
200 = S + A + R - (SA + SR + AR) - 2(SAR)
200 = 112 + 88 + R - (60 + 0 + 0) - 2(0)
so R = 60
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VP
Joined: 08 Jun 2005
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Among 200 people, 56% like strawberry, 44% like apple, and 40% like raspberry. If 30% of the people like strawbery and apple, what is the greatest possible number of people who like raspberry but do not like either strawberry or apple?
(A) 20
(B) 60
(C) 80
(D) 86
(E) 92
Total = Set1 + Set2 + Set3 + Neither - Both - (All three X 2)
200 = 112 + 88 + 80 + 0 – (60+X) - 0*2
X = 20
This is the additional number of people who like both.
80 – 20 = 60
People who like only strawberry minus the people who like strawberry and something else.
The answer is (B)
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Senior Manager
Joined: 06 Aug 2007
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bmwhype2 wrote: among 200 ppl, 56% like strawberry, 44% like apple, and 40% like raspberry. IF 30% of people like both strawbery and apple, what is the LARGEST possible number of ppl who like raspberry but do not like either strawberry or apple?
20
60
80
86
92
is this one considered set theory? can u solve it too?
thanks
This one is 60 ......and would tell us that 0 ppl like all three.
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SVP
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how do you know that the amount that like neither is 0, and that the number who like all three are 0 ?
or, are you setting them to 0 to find out the MAX number who like two of the three ?
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VP
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pmenon wrote: how do you know that the amount that like neither is 0, and that the number who like all three are 0 ?
or, are you setting them to 0 to find out the MAX number who like two of the three ?
Yes - you are correct
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CEO
Joined: 21 Jan 2007
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KillerSquirrel wrote: Among 200 people, 56% like strawberry, 44% like apple, and 40% like raspberry. If 30% of the people like strawbery and apple, what is the greatest possible number of people who like raspberry but do not like either strawberry or apple?
(A) 20
(B) 60
(C) 80
(D) 86
(E) 92
Total = Set1 + Set2 + Set3 + Neither - Both - (All three X 2)
200 = 112 + 88 + 80 + 0 – (60+X) - 0*2
X = 20
This is the additional number of people who like both.
80 – 20 = 60
People who like only strawberry minus the people who like strawberry and something else.
The answer is (B)
KS, you are a true contributor to the Quant forum.
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CEO
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This is how I did it.
100% = A% only + B% only + C% only - Both - 2*ALL + Neither
100 = 56 + 44 + Raspberry - [AB + AC + BC ] - 2*0 + 0
100 = 100 + R - [30 + 0 + 0] + 0
100 = 70 + R
R = 30
30% of 200 = 60
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SVP
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In a consumer survey, 85% of those surveyed like at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% like product 3. If 5% of the people in the survey like all 3 of products, what % of the survey participants likes more than one of the three products?
A) 5
B) 10
C) 15
D) 20
E) 25
I understand how you solved this first problem from your attachment. However, I don't understand why you added Both + Neither together. The question clearly asks for the % participants who likes more than one of the three products. Those who are neither means that they don't like any, so how can you include such people to those who like more than one product? 
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There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?
a) 94
b) 115
c) 132
d) 136
e) 140
in the problem just before this one, you mentioned that when you have 3 sets in which you also have a number that is included in all the 3 sets, and another number that is included in the 2 sets, you subtract the number that is included in the 3 sets from the number that is included in the 2 sets. how come you didn't do the same with this problem? you considered 32 as it is without subtracting 4 from it in order to end up with 28 as the number for the doubles. would you explain?
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Senior Manager
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tarek99 wrote: There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?
a) 94
b) 115
c) 132
d) 136
e) 140
in the problem just before this one, you mentioned that when you have 3 sets in which you also have a number that is included in all the 3 sets, and another number that is included in the 2 sets, you subtract the number that is included in the 3 sets from the number that is included in the 2 sets. how come you didn't do the same with this problem? you considered 32 as it is without subtracting 4 from it in order to end up with 28 as the number for the doubles. would you explain?
Tarek, they key on this is the wording of the problem. Note that it says exactly 2 sports. That means that 4 play 3 sports, 32 play only 2 sports, so you don;t subtract 32-4.
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VP
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asdert wrote: tarek99 wrote: There are 230 students. 80 play football, 42 play soccer and 12 play rugby. 32 play exactly 2 sports and 4 play all three. How many students play none?
a) 94
b) 115
c) 132
d) 136
e) 140
in the problem just before this one, you mentioned that when you have 3 sets in which you also have a number that is included in all the 3 sets, and another number that is included in the 2 sets, you subtract the number that is included in the 3 sets from the number that is included in the 2 sets. how come you didn't do the same with this problem? you considered 32 as it is without subtracting 4 from it in order to end up with 28 as the number for the doubles. would you explain? Tarek, they key on this is the wording of the problem. Note that it says exactly 2 sports. That means that 4 play 3 sports, 32 play only 2 sports, so you don;t subtract 32-4. Good answer - asdert - It's all in the wording - you should know what you are looking for.
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VP
Joined: 08 Jun 2005
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bmwhype2 wrote: among 200 ppl, 56% like strawberry, 44% like apple, and 40% like raspberry. IF 30% of people like both strawbery and apple, what is the LARGEST possible number of ppl who like raspberry but do not like either strawberry or apple?
20
60
80
86
92
is this one considered set theory? can u solve it too?
thanks
You have the same problem solved in my paper - the wording is different the numbers are the same.
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VP
Joined: 08 Jun 2005
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tarek99 wrote: In a consumer survey, 85% of those surveyed like at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% like product 3. If 5% of the people in the survey like all 3 of products, what % of the survey participants likes more than one of the three products? A) 5 B) 10 C) 15 D) 20 E) 25 I understand how you solved this first problem from your attachment. However, I don't understand why you added Both + Neither together. The question clearly asks for the % participants who likes more than one of the three products. Those who are neither means that they don't like any, so how can you include such people to those who like more than one product? Thanks for the heads up! - This is a misprint, should be "Both + All three"
I will correct and repost it 
Thanks!
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Re: Set Theory Made Easy [#permalink]
06 May 2012, 15:03
This is a relatively easy question from OG 11 diagnostic test which I got incorrect . A marketing firm determined that, of 200 households surveyed, 80 used neither Brand A nor Brand B soap, 60 used only Brand A soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap? (A) 15 (B) 20 (C) 30 (D) 40 (E) 45 As per the set theory attachment at the beginning of the thread Total = A + B + Neither - Both 200 = 60 + 3x + 80 - x 60 = 2x therefore x =30. OG answers this as A = 15 . Can someone kindly correct my understanding. Thanks in Advance
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Re: Set Theory Made Easy [#permalink]
06 May 2012, 15:12
OK i guess I did figure it out myself.I should have read this more carefully.
The crux lies in the word 'A only' and 'B only' where as the formula is for set A
Therefore the equation is
200 = (60 + x ) + (3x + x ) + 80 - x
200 = 140 + 4x
60 = 4x
x =15
Therefore answer A
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Re: Set Theory Made Easy
[#permalink]
06 May 2012, 15:12
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