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Chapter 6

Set Concepts — Assertion-Reason Type Questions

Class - 7 Concise Mathematics Selina



Assertion-Reason Type Questions

Question 17

Assertion (A) : Let A = {x | x is a composite number less than 4}. Then set A is an empty set.

Reason (R) : {0} is not an empty set.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

The smallest composite number is 4, so there is no composite number less than 4. Hence A = { }, which is an empty set.

Thus, assertion (A) is true.

{0} is a set with one element 0, so it is not an empty set.

Thus, reason (R) is true.

Hence, Option 3 is the correct option.

Question 18

Assertion (A) : Let A = {1, 3, 5} and B = {5, 1, 3} then A is subset of B, B is also a subset of A.

Reason (R) : If B is a subset of A, then each element of set A is present in set B.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

A = {1, 3, 5} and B = {5, 1, 3} = {1, 3, 5}. Since both sets have the same elements, A ⊆ B and B ⊆ A.

Thus, assertion (A) is true.

If B is a subset of A, then each element of B is present in A, not each element of A present in B.

Thus, reason (R) is false.

Hence, Option 1 is the correct option.

Question 19

Assertion (A) : Let A = {x | x is prime number} and B = {x | x is composite number}. Then sets A and B are not disjoint sets.

Reason (R) : Two non-empty sets, A and B are said to be disjoint, if they do not have any element in common.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

No number is both prime and composite, so sets A and B have no element in common, which means they are disjoint sets. Hence the statement that they are not disjoint is false.

Thus, assertion (A) is false.

Two non-empty sets are said to be disjoint if they do not have any element in common, which is the correct definition.

Thus, reason (R) is true.

Hence, Option 2 is the correct option.

Question 20

Assertion (A) : Let A = {a, b, c, d} and B = {a, b, e, c, f, d} then A is a proper subset of B.

Reason (R) : No set is a proper subset of itself.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

A = {a, b, c, d} and B = {a, b, c, d, e, f}. Every element of A is in B and A ≠ B, since B has the extra elements e and f. So A is a proper subset of B.

Thus, assertion (A) is true.

No set is a proper subset of itself, which is a correct statement.

Thus, reason (R) is true.

Hence, Option 3 is the correct option.

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