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In a room containing $28$ people, there are $18$ people who speak English, $15$, people who speak Hindi and $22$ people who speak Kannada. $9$ persons speak both English and Hindi, $11$ persons speak both Hindi and Kannada whereas $13$ persons speak both Kannada and English. How many speak all three languages?

  1. $9$
  2. $8$
  3. $7$
  4. $6$
asked in Set Theory & Algebra by Veteran (59.7k points)
edited by | 964 views
0
the should mention that all the people know speaking some language else we can not assume the sample space

2 Answers

+13 votes
Best answer
Apply set formula of $A$ union $B$ union $C$ ....
$28 = (18 + 15 + 22) - (9 + 11 + 13) + x$
$28 = 55 - 33 + x$
$x = 6$
answered by Veteran (55.6k points)
edited by
+2

your answer is correct but solution is not 100% correct  

please refer Arjun's answer to https://gateoverflow.in/2314/gate1993_17

point is that it is not mentioned anywhere that all 28 people must speak atleast one of three language.

hence answer could be either 0 ,1 ,2 ,3 ,4 ,5 ,6 , but as in option only 6 is given , hence answer is 6

+9

mehul vaidya

u are absolutely correct. Since the question is not giving any information about the people who don't speak any of the languages, we should not neglect it.

| E | = 18, | H | =15 , |K| =22 , |E∩ H| =9 , |E∩K| = 11 , |H∩K|, |E∩H∩K| =X

Let the people who don't speak any of the languages is Y;

According to the principle of Mutual Exclusion:

|E∪ H∪ K|  = 18 + 15+ 22 -(9+11+13) + X

28 - Y = 22 + X

X + Y = 6 ----------(1)

There are 7 possible pairs of values which satisfy this equation here;

(6,0),(5,1)(4,2)(3,3)

Since we have to follow the options, we will take pair(6,0) indicates X =6 otherwise X can be 5, 4,3,2,1,0

0 votes

 In question given that,

n(EUHUK) =28 ,n(E∩H) =9 ,n(E∩K) =11 ,n(H∩K) =13 ,n(E)=18 ,n(H)=15 , n(K)=22

n(EUHUK) =  n(E) + n(H)+ n(K) - [ n(E∩H) + n(E∩K) + n(H∩K) ] + n(E∩H∩k)

28 = 18 +15 +22 -[9+11+13] + n(E∩H∩k)

n(E∩H∩k) = 6

Option (D)6 ,is correct.

answered by Loyal (7.2k points)
Answer:

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