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# Set Theory Made Easy

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Director
Joined: 08 Jun 2005
Posts: 887
Set Theory Made Easy  [#permalink]

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23 Nov 2007, 13:25
9
8
See paper attached - enjoy

Please report any comments or errors.

Thanks

Attachments

Set Theory.pdf [52.27 KiB]
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SVP
Joined: 21 Jan 2007
Posts: 1882
Location: New York City

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23 Nov 2007, 13:48
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
Director
Joined: 08 Jun 2005
Posts: 887

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23 Nov 2007, 13:53
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

VP
Joined: 28 Dec 2005
Posts: 1076

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23 Nov 2007, 19:25
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 ?
SVP
Joined: 29 Aug 2007
Posts: 1766

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23 Nov 2007, 20:35
1
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
Director
Joined: 08 Jun 2005
Posts: 887

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23 Nov 2007, 21:25
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)
Manager
Joined: 06 Aug 2007
Posts: 152

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24 Nov 2007, 17:30
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.
VP
Joined: 28 Dec 2005
Posts: 1076

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24 Nov 2007, 20:17
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 ?
Director
Joined: 08 Jun 2005
Posts: 887

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24 Nov 2007, 20:24
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

SVP
Joined: 21 Jan 2007
Posts: 1882
Location: New York City

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27 Nov 2007, 07:13
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.
SVP
Joined: 21 Jan 2007
Posts: 1882
Location: New York City

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27 Nov 2007, 07:27
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
Director
Joined: 21 Jul 2006
Posts: 971

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27 Nov 2007, 07:58
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?
Director
Joined: 21 Jul 2006
Posts: 971

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27 Nov 2007, 08:16
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?
Senior Manager
Joined: 09 Oct 2007
Posts: 335

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27 Nov 2007, 09:37
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.
Director
Joined: 08 Jun 2005
Posts: 887

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27 Nov 2007, 20:33
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.

Director
Joined: 08 Jun 2005
Posts: 887

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27 Nov 2007, 20:37
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.

Director
Joined: 08 Jun 2005
Posts: 887

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27 Nov 2007, 20:41
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!

Intern
Joined: 25 Apr 2012
Posts: 2
Re: Set Theory Made Easy  [#permalink]

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06 May 2012, 14:03
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
Intern
Joined: 25 Apr 2012
Posts: 2
Re: Set Theory Made Easy  [#permalink]

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06 May 2012, 14:12
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
Non-Human User
Joined: 09 Sep 2013
Posts: 14464
Re: Set Theory Made Easy  [#permalink]

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21 Feb 2020, 08:06
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Re: Set Theory Made Easy   [#permalink] 21 Feb 2020, 08:06
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# Set Theory Made Easy

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