General chapter practice

The governing concept is conversion from set-builder form to roster form by listing every value that satisfies the stated condition. Here 3x≤15. Since 3 is positive, division by 3 preserves the inequality direction and gives x≤5. The question states x∈N, and the convention used here is N={1,2,3,...}; therefore the eligible natural numbers are 1, 2, 3, 4, and 5. The equality sign is important, so x=5 must be included. Thus C={1,2,3,4,5}, making option B correct. Option A incorrectly includes 0, option C omits 5, and option D lists multiples of 3 rather than values of x.

The governing idea is counting positive multiples of a fixed number subject to an upper bound. The positive multiples of 7 are 7×1, 7×2, 7×3, and so on. We need 7k<50 with k a positive integer. Dividing by 7 gives k<50/7≈7.14, so k can be 1,2,3,4,5,6, or 7. The corresponding elements are 7, 14, 21, 28, 35, 42, and 49, which makes seven distinct elements. The next multiple, 7×8=56, is already greater than 50 and is excluded. Therefore option B, 7, is correct. Zero is not positive, and the strict phrase “less than 50” would exclude 50 even if it were a multiple.

Each listed element is a square of a consecutive natural number: 1=1², 4=2², 9=3², 16=4², and 25=5². Therefore the rule x=n² with 1≤n≤5 produces exactly the given set, so option A is correct. Option B gives even numbers, C gives 4 through 8, and D gives multiples of 3.

The notation F={0} places the number 0 inside braces, so F contains exactly one element. By definition, any set with one element is a singleton set; therefore option B is correct. The empty set is written as ∅ or {}, and it contains no elements. Thus {0} is not empty and must not be confused with ∅.

A common divisor must divide both numbers exactly. The divisors of 18 are 1, 2, 3, 6, 9, and 18. Testing these against 24 leaves 1, 2, 3, and 6, because each divides 24 without remainder. Number 9 does not divide 24, and 4 does not divide 18. Therefore G={1,2,3,6}, so option A is correct.