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

Question

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.

FillFM · Answer

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.

afa13 · Answer

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?

dvishal387 · Answer

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.

pratik2018 · Answer

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

TheGraceful · Answer

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

Thanks
The Graceful

s8kadian · Answer

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

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

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