General chapter practice

Correct answer: B. 3

Explanation: To determine membership, solve the defining equation rather than guessing from the options. Starting with x + 2 = 5, subtract 2 from both sides: x = 5 − 2 = 3. Since 3 is a natural number, it satisfies both conditions and belongs to Y. Testing the other choices in the equation gives values different from 5, so option B is the only correct answer.

Correct answer: A. Z = {x : x is a multiple of 5 and 5 ≤ x ≤ 20}

Explanation: A correct set-builder description must generate every listed element and no extra element. Each member of Z is a multiple of 5, and the inclusive bounds 5 ≤ x ≤ 20 restrict the multiples to 5, 10, 15, and 20 exactly. Option B also includes 1, 2, 4, and 20; option C includes many even numbers; option D gives primes. Therefore option A is correct.

Correct answer: A. Positive factors of 8

Explanation: The governing concept is factor identification. A positive factor of 8 divides 8 exactly, leaving remainder zero. Checking the listed values, 8 ÷ 1 = 8, 8 ÷ 2 = 4, 8 ÷ 4 = 2, and 8 ÷ 8 = 1. These are all the positive factors of 8. Multiples of 8 continue as 8, 16, 24, and so on, so option A is correct.

Correct answer: A. A ⊆ B

Explanation: The governing concept is subset notation. A ⊆ B means that every element of A must also belong to B. The elements 1, 2, and 3 all occur in B, so A is a subset of B. Conversely, B is not a subset of A because 4 is missing from A. The sets are also unequal, and 4 ∉ A. Therefore option A is correct.

Correct answer: B. 3

Explanation: The governing concept is the cardinality of a set. The notation n(A) denotes the number of distinct elements in set A; it does not denote the largest element, the sum of the elements, or a numerical operation on them. In A = {2, 4, 6}, the distinct members are 2, 4, and 6, so the set contains exactly three elements. Therefore n(A) = 3 and option B is correct. Option A counts too few elements, option C adds an element that is not present, and option D mistakes the largest listed member, 6, for the number of members. If an element were repeated in a set, it would still be counted only once, but no repetition occurs here. Directly counting the displayed distinct members gives the answer.

Correct answer: A. 3

Explanation: The governing set principle is that repeated listings do not create new elements. Although five entries are written, the distinct elements of B are only 2, 4, and 6. We can rewrite the set without duplicates as B = {2, 4, 6}. Therefore its cardinality is n(B) = 3. The value 5 counts written entries, not distinct set elements, so option A is correct.

Correct answer: C. ∅

Explanation: The governing concept is the definition of an empty set: it is a set containing no elements. The symbol ∅ is specially used for this set. In contrast, 0 is a number, while {0} is a one-element set whose only member is 0. Similarly, {1} contains one element, namely 1. Therefore, option C alone represents the empty set; the braces in B and D show that those sets are not empty.

Correct answer: A. 0

Explanation: The governing concept is cardinality, written n(C), which means the number of distinct elements in set C. The empty set C = ∅ has no elements at all, so counting its elements gives n(C) = 0. The symbol ∅ in option C denotes the set itself, not the numerical size of that set. Option B would describe a singleton, and option D is incorrect because the cardinality of an empty set is well-defined.

Correct answer: A. D = {M, A, T, H, E, I, C, S}

Explanation: The governing concept is roster notation: a set is written by listing its distinct elements, and repetition does not create a new element. MATHEMATICS contains M, A, T, H, E, I, C, and S. Although M, A, and T occur more than once in the word, each is listed only once in a set. Thus option A is correct; B incorrectly repeats letters, C lists vowels only, and D omits several valid letters.

Correct answer: D. 6

Explanation: The governing idea is interpreting a set-builder condition with a strict inequality. Taking natural numbers as 1, 2, 3, and so on, the condition x < 6 produces E = {1, 2, 3, 4, 5}. Since the sign is strictly less than, 6 is excluded rather than included. Therefore 1, 3, and 5 belong to E, while option D, 6, does not.

Correct answer: A. F = {x : x is a whole number and x < 5}

Explanation: The governing concept is set-builder notation: the rule must generate every member of F and no other member. Whole numbers are 0, 1, 2, 3, 4, and so on; imposing x < 5 leaves exactly {0, 1, 2, 3, 4}. Under the usual school convention used here, natural numbers begin at 1, so B omits 0. C gives larger integers and D gives only even numbers. Hence A is correct.

Correct answer: A. It is a finite set

Explanation: The governing concept is classification of sets by size and clarity of definition. The condition “month having 31 days” is objective and identifies exactly seven names: January, March, May, July, August, October, and December. The set therefore has a limited number of clearly specified elements, so it is finite. It is neither empty nor infinite, and the rule is well-defined. Thus option A is correct.

Correct answer: A. H = {1, 2, 3, 4, 6, 12}

Explanation: The governing concept is membership by divisibility. A natural number is a factor of 12 when it divides 12 exactly, leaving remainder zero. The positive divisors are found from the factor pairs 1×12, 2×6, and 3×4, giving H = {1, 2, 3, 4, 6, 12}. Option B is mostly a list of even numbers, C contains multiples of 12, and D contains odd numbers, so only A is complete and correct.

Correct answer: A. I = {−3, −2, −1}

Explanation: The governing concept is combining an integer classification with a strict inequality. Negative integers are less than zero, and the integers greater than −4 are −3, −2, −1, 0, 1, and so on; restricting them to negative integers leaves only −3, −2, and −1. The strict sign x > −4 excludes −4, while 0 is not negative. Therefore option A is the correct roster form.

Correct answer: A. T={x: x is a prime number less than 10}

Explanation: The governing idea is set-builder description: a condition must produce every member of the given set and no extra member. The prime numbers less than 10 are exactly 2, 3, 5, and 7, so option A matches T. Odd numbers would also include 1 and 9; even numbers would include 2, 4, 6, and 8; divisors of 10 are 1, 2, 5, and 10. Therefore A is unambiguous.

Correct answer: D. 2

Explanation: The governing concept is interpreting strict and inclusive inequalities in a set definition. The condition 2<x excludes 2, while x≤6 includes 6. Because x is natural, the resulting set is U={3,4,5,6}. Thus 3, 4, and 6 are members, but 2 is not. Option D is therefore the only element outside U; confusing < with ≤ would lead to a wrong choice.

Correct answer: A. V={1,2,3,6}

Explanation: A factor of 6 is a natural number that divides 6 exactly, leaving remainder zero. The factor pairs are 1×6 and 2×3, so the positive natural factors are 1, 2, 3, and 6. Zero cannot be a factor because division by zero is undefined, and 12 is not a divisor of 6. Therefore option A is the complete roster form of V.

Correct answer: A. C={1,3,5,7}

Explanation: The governing process is substitution in a set-builder rule. Whole numbers less than 4 are n=0,1,2,3. Substituting into x=2n+1 gives x=1,3,5,7 respectively. Hence C={1,3,5,7}, which is option A. Option B incorrectly starts with n=1; option C ignores the rule, and option D includes 0, which the formula never produces for these values.

Correct answer: B. 5

Explanation: The governing concept is counting integers in a half-open interval. Since -3<x, the integer -3 is excluded; since x≤2, the endpoint 2 is included. Listing the permitted integers gives -2,-1,0,1,2. There are five distinct values, so the cardinality of D is 5. Therefore option B is correct; counts of 4, 6, or 7 result from mishandling an endpoint or adding an outside integer.

Correct answer: A. E={6,12,18,24}

Explanation: A roster form lists every element explicitly. The positive multiples of 6 below 30 are obtained as 6×1=6, 6×2=12, 6×3=18, and 6×4=24. The next multiple, 30, is excluded because the condition says less than 30, and 0 is not positive. Thus E={6,12,18,24}, so option A is correct.

Correct answer: A. G={1,10}

Explanation: Using nonnegative integer exponents, the relevant powers are 10⁰=1, 10¹=10, and 10²=100. The condition x<100 includes 1 and 10 but excludes 100, so G={1,10}. Zero is not the value of 10⁰; it is the exponent in that expression. Therefore option A is correct, while the other choices include the boundary or an invalid value.

Correct answer: C. 5∈H

Explanation: The governing concept is membership under inclusive bounds. Because 1≤x≤4 includes both endpoints, the set is H={1,2,3,4}. Consequently, 1∈H, 3∈H, and 4∈H are true statements. The number 5 lies beyond the upper limit, so 5∈H is false. Hence option C is the required answer; treating the question as asking for a true statement would reverse the task.

Correct answer: A. I={1,3,9}

Explanation: The governing concept is identifying positive divisors and writing them in roster form. The factor pairs of 9 are 1×9 and 3×3, so its positive factors are 1, 3, and 9. Six is not a factor because 9÷6 is not an integer, and zero cannot be a factor because division by zero is undefined. Therefore option A gives the complete set.

Correct answer: B. J={3,4,5,6,7,8}

Explanation: The governing idea is converting strict inequalities into a roster form. Greater than 2 means x>2, so 2 is excluded; less than 9 means x<9, so 9 is also excluded. The natural numbers between these boundaries are 3,4,5,6,7,8. Thus option B contains exactly the required members. The other options include one or both excluded endpoints.

Correct answer: D. -4

Explanation: A constant term is a term that contains no variable. In 7a - 4, the term 7a contains the variable a, whereas -4 contains no variable. Therefore, -4 is the constant term. The option 7 is only the coefficient of a in 7a; it is not a separate term in the expression. Exam tip: To identify the constant term, choose the term that has no variable.