Selina solutions

Selina solutions

Grade 7

Set Concepts (Some Simple Divisions by Vedic Method) | Exercise 13(D)

Question 10

Q10)Let n(A) = 31, n(B) = 20 and n(A \cap B) = 6, find:

(i) n(A – B)

(ii) n(B – A)

(iii) n(A \cup B)

Solution :

(i) Given:n(A) = 31, n(B) = 20 and n(A \cap B) = 6

We know that,

n(A – B) = n(A) – n(A \cap B)

n(A – B) = 31 -6 = 25

(ii) Given : n(A) = 31, n(B) = 20 and n(A \cap B) = 6

We know that,

n(B – A) = n(B) – n(A \cap B)

n(B – A) = 20 – 6 = 14

(iii) Given : n(A) = 31, n(B) = 20 and n(A \cap B) = 6

We know that,

n(A \cup B) = n(A) + n(B) – n(A \cap B)

n(A\cup B) = 31 + 20 - 6 = 45

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subject-cta

Question 10

Q10)Let n(A) = 31, n(B) = 20 and n(A \cap B) = 6, find:

(i) n(A – B)

(ii) n(B – A)

(iii) n(A \cup B)

Solution :

(i) Given:n(A) = 31, n(B) = 20 and n(A \cap B) = 6

We know that,

n(A – B) = n(A) – n(A \cap B)

n(A – B) = 31 -6 = 25

(ii) Given : n(A) = 31, n(B) = 20 and n(A \cap B) = 6

We know that,

n(B – A) = n(B) – n(A \cap B)

n(B – A) = 20 – 6 = 14

(iii) Given : n(A) = 31, n(B) = 20 and n(A \cap B) = 6

We know that,

n(A \cup B) = n(A) + n(B) – n(A \cap B)

n(A\cup B) = 31 + 20 - 6 = 45

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

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Book a free class now
subject-cta
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