# PE 5 - Venn Diagram | LR - Venn Diagram

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**Answer the next 5 questions based on the information given below.**

300 students of CAT4ALL institute wrote CAT exam in 2023. Two-thirds of the students could not clear cutoff for any of the three sections in the exam. For every two students who cleared cutoff for section only section A, one more cleared the cutoff of only section B. For every five students who could clear cutoff only section C, three less could clear cutoff of only section B. The number of students who could clear cutoff of both B and C was 30 and the ratio of students who could clear cutoff of both A and C to those who could clear cutoff of both A and B was 5 : 7.

**PE 5 - Venn Diagram | LR - Venn Diagram**

The number of students who could clear cutoff of exactly two sections was 30 and the ratio of the number of students who could clear cutoff of both C and B but not A to the number of students who could clear cutoff of both A and C but not B is 2 : 1. How many students could clear cutoff of all three sections?

Answer: 20

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**Explanation** :

Two-thirds of the students who could not clear cutoff of any of the three sections = (2/3) × 300 = 200.

∴ Students who cleared cutoff of at least one section = 300 – 200 = 100.

For every two students who could clear cutoff of section A, one more could clear cutoff of section B.

∴ Only A : Only B = 2 : (2 + 1) = 2 : 3

Similarly, Only C : Only B = 5 : (5 − 3) = 5 : 2

∴ Only A : Only B : Only C = 4 : 6 : 15

Given in the question: a + b + c = 30 and c : b = 2 : 1.

∴ c = 2b and a = 30 – 3b and t = 30 – 2b

Now, the ratio of students who could clear cutoff of both A and C to those who could clear cutoff of both A and B was 5 : 7

⇒ (b+t)/(a+t) = 5/7

⇒ (30-b)/(60-5b) = 5/7

⇒ 210 – 7b = 300 – 25b

⇒ 18b = 90

⇒ b = 5

∴ t = 30 – 2b = 20

Hence, 20.

Workspace:

**PE 5 - Venn Diagram | LR - Venn Diagram**

What is the maximum number of students who could clear cutoff of only section C?

Answer: 30

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**Explanation** :

Consider the solution to first question of this set.

Total students who could clear cutoff of at least one section = 100 = 4x + 6x + 15x + a + b + c + t

⇒ 100 = 25x + a + b + 30

⇒ 25x = 70 – (a + b)

Since a, b ≥ 0

Maximum value of x can be 2.

∴ Maximum number of students who cleared cutoff of only section C = 15x = 30.

Hence, 30.

Workspace:

**PE 5 - Venn Diagram | LR - Venn Diagram**

How many students could clear cutoff of exactly one section? (Refer to the data from the first question of the set, if required.)

- (a)
40

- (b)
50

- (c)
25

- (d)
None of these

Answer: Option B

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**Explanation** :

Consider the solution to first question of this set.

100 = 25x + 5 + 15 + 10 + 20

⇒ x = 2

Students who could clear the cutoff of only one section = 8 + 12 + 30 = 50.

Hence, option (b).

Workspace:

**PE 5 - Venn Diagram | LR - Venn Diagram**

After rechecking 10 students from only C shifted to only A. 5 students each from only A and only B shifted to only C. What is the ratio of students who could clear cutoff for section A, B and C respectively. [Assume data from the first question.]

- (a)
52 : 53 : 65

- (b)
48 : 57 : 65

- (c)
53 : 52 : 65

- (d)
None of these

Answer: Option C

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**Explanation** :

Consider the solution to previous question of this set.

After reshuffling

After reshuffling students who cleared cutoff for section

A = 13 + 5 + 15 + 20 = 53

B = 7 + 15 + 20 + 10 = 52

C = 30 + 10 + 20 + 5 = 65

∴ A : B : C = 53 : 52 : 65

Hence, option (c).

Workspace:

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