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

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Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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New post 28 Apr 2018, 23:14
3
13
00:00
A
B
C
D
E

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  95% (hard)

Question Stats:

47% (02:41) correct 53% (02:28) wrong based on 139 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.
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Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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New post 29 Apr 2018, 10:12
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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.
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Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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New post 22 May 2018, 08:08
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.

VeritasPrepKarishma / mikemcgarry can you please advise?
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Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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New post 15 Aug 2018, 06:10
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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New post Updated on: 30 Aug 2018, 08:43
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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.

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, 01:47.
Last edited by pratik2018 on 30 Aug 2018, 08:43, edited 1 time in total.
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Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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New post 30 Aug 2018, 01:49
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
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Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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New post 30 Aug 2018, 02:01
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.
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Re: Out of 80 people, some drink tea, some drink coffee, while some drink  [#permalink]

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Re: Out of 80 people, some drink tea, some drink coffee, while some drink   [#permalink] 30 Sep 2019, 10:19
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