GATE CSE 1998 | Question: 2.4

Question

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?

• A. $9$
• B. $8$
• C. $7$
• D. $6$

Comments

@HitechGa No this is wrong! 

"So we assume that cardinality of the union of the sets is equal to the number of people i.e. each people speaks at least one of the three languages." 

The union cannot be assumed to be 28 unless the problem explicitly states that every person speaks at least one language. Even if pairwise intersections are given, the cardinality of E intersection H intersection K may still vary, so the union is not automatically fixed. 

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Note : From mathematical standpoint, $|E \cup H \cup K|$ is not automatically equal to 28 until /unless the problem explicitly states that every person in the room speaks at least one of the three languages.

Credit : Grok (mentor of Go classes) :

Generally when the union is not given and exact ans is there for the ques we assume total as 28 only.Several places and books follow the same only.

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Generalised Solution

Let $E$, $H$, and $K$ represent the sets of people who speak English, Hindi, and Kannada, respectively.

We are given:

• $|E| = 18, \quad |H| = 15, \quad |K| = 22$

• $|E \cap H| = 9, \quad |H \cap K| = 11, \quad |K \cap E| = 13$

• Total people in the room ($|U|$) = $28$

Let the number of people who speak all three languages be $|E \cap H \cap K| = x$.

Using the Principle of Inclusion-Exclusion (PIE):

$$|E \cup H \cup K| = |E| + |H| + |K| - (|E \cap H| + |H \cap K| + |K \cap E|) + |E \cap H \cap K|$$

Substitute the given values into the equation:

$$|E \cup H \cup K| = 18 + 15 + 22 - (9 + 11 + 13) + x$$

$$|E \cup H \cup K| = 55 - 33 + x$$

$$|E \cup H \cup K| = 22 + x$$

Finding the Bounds for $x$

Note : A Venn diagram is valid only if every disjoint region has a real, possible number of elements in it.so  the number of people must be >=0

 

Let's look at the region representing people who speak Hindi only:

$$\text{Hindi only} = |H| - |E \cap H| - |H \cap K| + |E \cap H \cap K|$$

$$\text{Hindi only} = 15 - 9 - 11 + x = x - 5$$

Since the number of people cannot be negative:

$$x - 5 \ge 0 \implies x \ge 5$$

We also know that the total union cannot exceed the total number of people in the room ($|U| = 28$):

$$|E \cup H \cup K| \le 28 \implies 22 + x \le 28 \implies x \le 6$$

Combining these bounds, we get:

$$5 \le x \le 6$$

So, Mathematically, $|E \cup H \cup K|$ can be either 27 or 28, which yields two valid scenarios.

 

Answer 1

Apply set formula of $A$ union $B$ union $C$
$28 = (18 + 15 + 22) - (9 + 11 + 13) + x$
$28 = 55 - 33 + x$
$x = 6$

Correct Answer: $D$

Comments

@mehul vaidya

did not get you bro

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@mehul vaidyadid not get you bro

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akash.dinkar12 X cannot be less than 5 ...

To know why it is not less than 5 , I have given my own answer plz check and correct me if I am wrong 

Click on link : https://gateoverflow.in/1676/gate-cse-1998-question-2-4?show=532987#c532987

Note :  A Venn diagram is valid only if every disjoint region has a real, possible number of elements in it.so  the number of people must be >=0

Answer 2

Using Venn diagram

Comments

nice explanation

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Best answer ISTG!. Thanks.

Answer 3

 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.

Answer 4

Let A be the people who speaks English.
Let B be the people who speaks Hindi.
Let C be the people who speaks Kannada.
(A∪B∪C) = A + B + C - (A∩B) - (B∩C) - (C∩A) + (A∩B∩C)
28 = 18 + 15 + 22 - 9 - 11 - 13 + (A∩B∩C)
∴ (A∩B∩C) = 6

Comments

by venn diagram

Total – None = a + b = c – (ab +bc + ca) + (abc)

28 – none = 18 + 15 + 22 -(9+11+13) + (abc)

(abc)- none = 6

Answer 5

Apply Inclusion Exclusion principle

people who speak English = A

people who speak Hindi = B

people who speak Kannada = C

APPLY FORMULA=

n(AUBUC) =  n(A) + n(B)+ n(C) - [ n(A∩B) + n(B∩C) + n(A∩C) ] + n(A∩B∩C)

28 = 18 + 15 + 22 - 9 - 11 - 13 + n(A∩B∩C)

∴ (A∩B∩C) = 6

 

Answer 6

Just because the options are given everybody is using the the total number of people as the triple union of english Hindi and Kannada speaking people! This is wrong. One must include a less than or equal to sign as there is nothing explicitly mentioned!

 

But wait,