Venn Diagrams MCQ Quiz - Objective Question with Answer for Venn Diagrams - Download Free PDF

Given :

Total candidates = 500

Failed in English only = 30

Failed in Hindi only = 75

Failed in Mathematics only = 50

Failed in English and Hindi = 15

Failed in Hindi and Mathematics = 17

Failed in Mathematics and English = 17

Failed in all three = 5

Concept used:

Candidates who failed in at least one subject are counted by summing those who failed in only one subject, those who failed in exactly two subjects, and those who failed in all three. However, remember that those who failed in exactly two subjects and all three subjects get counted extra times in the "only" categories, so we adjust for this by adding back those numbers.

Calculations :

So, the calculation goes as follows:

Total failed = (30 + 75 + 50) + (15 + 17 + 17) - 2 x 5

The subtraction by 2 x 5 is essential because the 5 candidates who failed in all three were counted three times in the single failures and once more in the double failures, requiring us to subtract them twice to adjust for the overcounting.

Calculating the number:

Total failed = 155 + 49 - 10 = 194

Therefore, the number of candidates passing in  subjects is the total number of students minus the number who failed any subject:

Passed in all subjects = Total candidates - Total failed = 500 - 194 = 306

Passed in two or more subjects =  306 + candidates failed in only one subject = 306 + 30 + 75 + 50 = 461

So, the correct answer is option 1) 461.

Given : 

Total candidates appeared in an examination = 500

30 candidates failed in English only

75 failed in Hindi only

50 failed in mathematics only and 15 failed in both English and Hindi

17 failed in both Hindi and Mathematics 

17 failed in both Mathematics and English 

5 failed in all three tests

Concept used :

Venn diagram concept

Calculations :

Candidates failed in atleast one subject

⇒ Candidates failed in one subject + candidates failed on two subject + candidates failed in three subject

Candidates failed in english = 30

Candidates failed in Hindi = 75

Candidates failed in mathematics = 50

Candidates failed in english and hindi only = 10

Candidates failed in english and mathematics only = 12

Candidates failed in Hindi and mathematics only = 12

Candidates failed in Hindi, english and mathematics = 5

Candidates failed in atleast one subject = 30 + 75 + 50 + 10 + 12 + 12 + 5

⇒ 194

Percentage of candidates failed in atleast one subject

⇒ (Candidates failed in atleast one subject/Total candidates) × 100

⇒ (194/500) × 100

⇒ 38.8 %

∴ Percentage of candidates failed in atleast one subject is 38.8%

Given : 

Total candidates who appeared in an examination = 500

30 candidates failed in English only

75 failed in Hindi only

50 failed in mathematics only

15 failed in both English and Hindi

17 failed in both Hindi and Mathematics 

17 failed in both Mathematics and English 

5 failed in all three tests

Concept used :

Venn diagram concept

Calculations :

Candidates failed in only english = 30

Candidates failed in only hindi = 75

Candidates failed in only mathematics = 50

Candidates failed in only one subject = 30 + 75 + 50

⇒ 155

The percentage of candidates who failed in only one subject

⇒ (Candidates failed in only one subject/total candidates) × 100

⇒ (155/500) × 100

⇒ 31%

∴ The percentage of candidates who failed in only one subject is 31%.

Given : 

Total candidates appeared in an examination = 500

30 candidates failed in English only

75 failed in Hindi only

50 failed in mathematics only and 

15 failed in both English and Hindi

17 failed in both Hindi and Mathematics 

17 failed in both Mathematics and English 

5 failed in all three tests

Concept used :

Venn diagram concept

Calculations :

Candidates failed in at least 2 subjects = candidates failed in 2 subjects + candidates failed in 3 subjects

Candidates failed in only English and mathematics = 12

Candidates failed in only English and Hindi = 10

Candidates failed in only Hindi and mathematics = 12

Candidates failed in Hindi, English, and mathematics = 5

Candidates failed in at least 2 subjects = 12 + 10 + 12 + 5 = 39

Percentage of candidates who failed in at least two subjects

⇒ (candidates failed in at least 2 subjects/total candidates) × 100

⇒ (39/500) × 100 = 39/5

⇒ 7.8 %

∴ The percentage of candidates who failed in at least two subjects is 7.8%.

Concept Used:

Venn diagram representation of set operations:

• A': Complement of A (everything outside A)
• B': Complement of B (everything outside B)
• A ∩ B: Intersection of A and B (overlapping region)
• A ∪ B: Union of A and B (everything in A or B or both)

Calculation:

The shaded region covers everything outside both circles A and B.

Let's analyze each option:

1. A' ∩ B: This represents everything outside A that is also inside B. This would be only the part of B that doesn't overlap with A.
2. A ∩ B: This represents the intersection of A and B (the overlapping region).
3. (A ∩ B)': This represents everything outside the intersection of A and B. This would be everything except the overlapping region.
4. A' ∩ B': This represents everything outside A that is also outside B. This is exactly what the shaded region shows.

The shaded region represents A' ∩ B'.

Hence option 4 is correct.

Given:

Percent of readers who read all the three = 10% 

Percent of readers who read C = 45% 

Percent of readers who read at least two = 35% 

Percent of readers who read A and B = 15% 

Calculations:

Let the percent of readers who read only A and C = a%

Let the percent of readers who read only B and C = b%

Percent of readers who read only C  = c%

According to the question

Percent of readers who read C = 45%

⇒ a% + 10% + b% + c% = 45%

⇒ a% + b% + c% = 35%      ----(1)

Percent of readers who read at least two = 35% 

⇒ a% + 10% + 5% + b% = 35% 

⇒ a% + b% = 20%      ----(2)

Put the value of a% + b% from equation (2) to equation (1)

⇒ 20% + c% = 35% 

⇒ c% = 15% 

∴ Percent of readers who read only C is 15%

Calculation:

Given: n(Z) = 90, 

From the above diagram, we get

n(X) = a + 46, n(Y) = b + 51, n(Z) = 90 (given)

n(X ∩ Y) = 16 + 18 = 34, 

n(Y ∩ Z)  = 17 + 18 =35,

n(X ∩ Z) = 12 + 18 = 30, 

n(X ∩ Y ∩ Z) = 18

Now the required values

n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) 

⇒ (a + 46) + (b + 51) + (90) - 34 - 35 - 30 + 18

⇒  a + b + 106

Hence, option (4) is correct. 

Concept:  

• Let a population be which consists of n number of individuals.
• Let x number of individuals in the population which belongs to category 1.
• Let y number of individuals in the population which belongs to category 2.
• Number of individuals in the population which belong categories 1 and 2 = x + y - n

Calculation:

Given: 

• 63% of soldiers lost one eye, 82% lost one ear, 68% lost one arm, 91% lost one leg, and, x% lost all four organs.
• Soldiers who lost one eye and one ear = (63 + 82 – 100) % = 45%
• Soldiers who lost one eye, one ear and one arm = (45 + 68 – 100) % = 13%
• Soldiers who lost one eye, one ear, one arm and one leg = (13 + 91 – 100) % = 4% = x%
• So, the minimum value of x is 4.
• Hence, the correct answer is option 1.

Calculation:

Total number of students in class = 60

Students who like music = A + B = 45     …. (1)

Students who like dancing = B + C = 50     …. (2)

Students who like nothing =5

Students who like both music and dancing = B

We know that A + B + C + 5 = 60

From equation 1^st,

⇒ 45 + C + 5 = 60

⇒ C = 10

Put the value of C in equation 2^nd, we get

⇒ B = 40

Hence, 40 students like both music and dancing.

Calculation:

Let A = Students who play cricket, B = Students who play football

⇒ n (A) = 50% and n (B) = 40%

⇒ n (A ∩ B) = 10%

⇒ n (A ∪ B) = n (A) + n (B) - n (A ∩ B) = 50 + 40 - 10 = 80%

As we know that (A ∪ B) ’ = A’ ∩ B’

Calculation:

Let, x number of persons taking milk only.

According to the question

60 + 15 + (x - 5) + (50 - x) + x = 150

120 + x = 150

x = 150 - 120 = 30

∴ The required value is 30.

Number of students passed in at least one subject = (The total number of students - The number of students who failed in all subjects)

⇒ 450 - 60

⇒ 390

∴ The number of students who passed in at least one subject is 390.

Given:

No of people speaking Hindi =  n(H) = 50

No of people speaking Tamil = n(T) = 20 

No of people speaking both Hindi and Tamil = n(H & T) = 10 

Formula used:

n(A or B) = n(A) + n(B) - n(A & B)

Calculation:

n(H or T) = 50 + 20 - 10 = 60

∴ No of people speaking Hindi or Tamil = 60

Concept:

In a Venn diagram circles are used, that overlap or don't overlap to show the differences and commonalities among things or groups of things.

Calculation:

This problem can be solved using the Venn diagram

so, (A - B) ∪ (B - A) ∪ (A ∩ B) = A ∪ B.

Hence, the correct answer is A ∪ B.

Calculation:

Given: n(Z) = 90 and the number of elements in Y and Z are in the ratio 4 : 5

To find: b = ?

n(Y) = 16 + 18 + 17 + b

n(Z) = 12 + 18 + 17 + c = 90 

\(\rm \frac{n(y)}{n(z)}=\frac{16+18+17+b}{12+18+17+c}=\frac{4}{5}\\ \Rightarrow \frac{51+b}{90}=\frac{4}{5}\\ \Rightarrow (51+b)\times 5=360 \\\Rightarrow255+5b=360\\ \Rightarrow5b=105\\ \Rightarrow b=21\)

Hence, option (3) is correct.