Roster Form and Classification of Numbers

Correct answer: A. Cubes of natural numbers less than 30

Explanation: The governing concept is describing a finite set by identifying the common property shared by every element. We have 1 = 1³, 8 = 2³, and 27 = 3³, so all three numbers are cubes of natural numbers. Each is also less than 30. The next cube of a natural number is 4³ = 64, which is not less than 30; therefore the description is complete and does not omit another qualifying number. Option A is correct. Option B is false because 8 is not a perfect square. Option C is false because 1 and 27 are not prime. Option D is false because 8 and 27 do not divide 30 exactly.

Correct answer: C. T = {1, 2}

Explanation: The governing concept is interpreting a set-builder condition together with the convention stated in the question that natural numbers start at 1. We list the natural numbers and retain those satisfying x ≤ 2. The resulting elements are 1 and 2, so T = {1, 2}. The inclusive symbol ≤ is important because it includes the endpoint 2. Hence option C is correct. The set is not empty because it contains two elements, not a singleton because a singleton has exactly one element, and not infinite because its members stop at 2. If a different convention included zero, the set could be written {0, 1, 2}, but under the convention used here the given answer is {1, 2}.

Correct answer: A. Cubes of natural numbers less than 70

Explanation: The governing concept is classifying a set by a common numerical form and verifying the stated bound. We have 1 = 1³, 8 = 2³, 27 = 3³, and 64 = 4³, so every element is the cube of a natural number. Each value is less than 70. The next natural-number cube is 5³ = 125, which exceeds 70, so no additional cube should be included under this description. Therefore option A is correct. Option B is false because 8, 27, and 64 are not perfect squares. Option C is false because several listed values do not divide 70 exactly. Option D is false because 8, 27, and 64 are composite, while 1 is neither prime nor composite.

Correct answer: A. Empty set

Explanation: First solve the defining equation: x + 1 = 1. Subtracting 1 from both sides gives x = 0. Under the convention used in this question, natural numbers begin with 1, so 0 is not an allowed natural number. Consequently, no natural number satisfies the complete condition defining R. A set with no elements is called the empty set and is written as ∅. It is not a singleton, because a singleton has exactly one member. It is not infinite and cannot have two elements either. Therefore R = ∅, making option A the unique correct answer. The conclusion depends on applying both the equation and the stated domain of natural numbers.

Correct answer: B. 5

Explanation: The governing concept is converting a bounded inequality into a finite roster and then counting its distinct members. The lower condition 3 ≤ x includes 3 because equality is allowed. The upper condition x < 8 excludes 8 because the inequality is strict. The natural numbers satisfying both conditions are B = {3, 4, 5, 6, 7}. Counting this roster gives five elements, so option B is correct. Option A misses one value, usually 7 or 3. Option C incorrectly includes 8, and option D confuses the upper boundary with the number of members. The symbols ≤ and < must therefore be interpreted before counting.

Correct answer: A. F = {4, 6, 8, 9}

Explanation: The governing concept is classification of digits followed by roster-form representation. A composite number is a positive integer greater than 1 that has more than two positive factors. Considering the digits from 0 through 9, the composite digits are 4, 6, 8, and 9. For example, 4 has factors 1, 2, 4; 6 has 1, 2, 3, 6; 8 has 1, 2, 4, 8; and 9 has 1, 3, 9. The digits 2, 3, 5, and 7 are prime, while 1 is neither prime nor composite. Zero is also neither a composite digit nor a prime digit. Thus the exact roster form is {4, 6, 8, 9}, making option A correct. B lists primes, C incorrectly includes 1, and D includes 0 and the prime digit 2.