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# Out of 80 people, some drink tea, some drink coffee, while some drink

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DS Forum Moderator
Joined: 21 Aug 2013
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Location: India
Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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28 Apr 2018, 22:14
3
11
00:00

Difficulty:

95% (hard)

Question Stats:

46% (02:43) correct 54% (02:36) wrong based on 157 sessions

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Out of 80 people, some drink tea, some drink coffee, while some drink both. There might also be some who drink neither tea nor coffee. How many drink only one drink, i.e., either only tea or only coffee?

(1) 60% of those who drink tea, also drink coffee.

(2) 60% of those who do not drink tea, also do not drink coffee.
Manager
Joined: 22 Jun 2017
Posts: 71
Location: Brazil
GMAT 1: 600 Q48 V25
GPA: 3.5
WE: Engineering (Energy and Utilities)
Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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29 Apr 2018, 09:12
2
1
Option C.

(1) Not sufficient.

(2) Not sufficient.

(1) + (2) Sufficient:

x = just tea
y = tea and coffee
z = just coffee
w = neither tea nor coffee

x+y+z+w = 80

> from (1):

(x+y) * 0,6 = y

6x = 4y

y = $$\frac{6x}{4}$$

> from (2):

(z+w) * 0,6 = w

6z = 4w

w = $$\frac{6z}{4}$$

> (1) + (2):

x + $$\frac{6x}{4}$$ + z + $$\frac{6z}{4}$$ = 80

x+z = 32.
Intern
Joined: 12 Jul 2017
Posts: 33
Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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22 May 2018, 07:08
I think the answer is E.
People who only drink tea= a
people who only drink coffee=b
people who drink both=c
people who drink neither=x
a+b+c+x=80 (from question stem) - eq1
1) Insufficient as it leads to c=0.6*(a+c) or c=(3/2)*a (eq2)
2) Insufficient as it leads to (b+x)(0.6)=a+x or x=(3/2)*b-(5/2)*a (eq3)

1) and 2) we have 3 equations and 4 unknowns
it will lead to: a+b+(3/2)*a+(3/2)*b-(5/2)*a=80 => (5/2)*b=80; b=32
however it gives no info about a+b.

Manager
Joined: 01 Jan 2018
Posts: 69
Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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15 Aug 2018, 05:10
Let a= Who drink tea only
b= Who drink coffee only
c = Who drink both
d = Who drink none

Now, as per St. 1 c=0.6*(a+c)=>6a=4c Not sufficient as we want only sum of only a and b in answer.
From St. 2 (b+d)*0.6=a+d=>6b=10a+4d
Not sufficient for reason as in St. 1

Combining the two we have below equations -
6b=10a+4d...(1)
6a=4c...(2)
and a+b+c+d=80...(3)

We can substitute for d from eq 1 and for c from eq 2 to get the answer. We can choose choice C as the answer.We don't need to calculate till the end.
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Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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Updated on: 30 Aug 2018, 07:43
1
Out of 80 people, some drink tea, some drink coffee, while some drink both. There might also be some who drink neither tea nor coffee. How many drink only one drink, i.e., either only tea or only coffee?

(1) 60% of those who drink tea, also drink coffee.

(2) 60% of those who do not drink tea, also do not drink coffee.

Using complement signs here as ' in the following descriptions. Please correct me if I am wrong.

1. From 60% of n(T) also drink coffee so : n(TnC) = 60 % of n(T)
then remaining 40 % of n(T) don't drink coffee so n(only T) = 40% of n(T)
inufficient.

2. 60% of n(T') also doesnt drink coffee so n(C') which is n(TuC)' complement= 60% of n(T')
hence 40% of n(T') drink coffee so n(only C) = 40% of n(T')

Doesnt say anything how much percentage that reflects to.

1&2 combined. We have all four components of the overlapping sets :
n(TnC) = 60% of n(T)
n(TuC)' = 60% of n(T')

n(TnC) = 60% of n(T)
n(TnC) = 6/10 * { n(only T) + n(TnC)}
10*n(TnC) = 6*n(only T) + 6 * n(TnC)
4*n(TnC) = 6*n(only T)
n(TnC) = 6/4 * n(only T) ---------(a)

n(TuC)' = 60% of n(T')
n(TuC)'= 60/100 { n(only C) + n(TuC)'}
n(TuC)' = 6/10 * { n(only C) + n(TuC)'}
10*n(TuC)' = 6*n(only C) + 6 * n(TuC)'
4*n(TuC)' = 6*n(only C)
n(TuC)' = 6/4 * n(only C) --------(b)

Total = 80
n(only T) + n(only C) + n(Tnc) + n(Tuc)' = 80
Substituting values from (a) and (b)
6/4* n(only T) + n(only T) + n(only C) + 6/4* n(only C) = 80
n(only T) + n(only C) = 32
Cannot solve with to find n(T) and n(T') as only known is n(T) + n(T') = 80

Originally posted by pratik2018 on 30 Aug 2018, 00:47.
Last edited by pratik2018 on 30 Aug 2018, 07:43, edited 1 time in total.
Manager
Joined: 10 Apr 2018
Posts: 108
GPA: 3.56
WE: Engineering (Computer Software)
Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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30 Aug 2018, 00:49
amanvermagmat wrote:
Out of 80 people, some drink tea, some drink coffee, while some drink both. There might also be some who drink neither tea nor coffee. How many drink only one drink, i.e., either only tea or only coffee?

(1) 60% of those who drink tea, also drink coffee.

(2) 60% of those who do not drink tea, also do not drink coffee.

Hi AmanVarma,
It was a real exercise for my brain.
Can you please mention the source of the question.

Thanks
The Graceful
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Every EXPERT was a beginner once...
Don't look at the clock. Do what it does, keep going
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To achieve great things, two things are needed:a plan and not quite enough time - Leonard Bernstein.

Manager
Joined: 05 Nov 2015
Posts: 62
Location: India
Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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30 Aug 2018, 01:01
Total 80
OT - unknown ...?(**60% of those who drink Tea also drink coffee) -- can not be 60% of 80
OC - unknowns
Both -- ** (can be calculated but we dont have OT)
None -- 80 - OT (60% of those who do onot drink Tea also do not drink coffee)
Hence E

FillFM wrote:
Option C.

(1) Not sufficient.

(2) Not sufficient.

(1) + (2) Sufficient:

x = just tea
y = tea and coffee
z = just coffee
w = neither tea nor coffee

x+y+z+w = 80

> from (1):

(x+y) * 0,6 = y

6x = 4y

y = $$\frac{6x}{4}$$

> from (2):

(z+w) * 0,6 = w

6z = 4w

w = $$\frac{6z}{4}$$

> (1) + (2):

x + $$\frac{6x}{4}$$ + z + $$\frac{6z}{4}$$ = 80

x+z = 32.
Re: Out of 80 people, some drink tea, some drink coffee, while some drink &nbs [#permalink] 30 Aug 2018, 01:01
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