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SubjectsMathematics
समुच्चय और उनके निरूपण
In this Class 10 Mathematics topic from the chapter “Sets,” students learn how to describe and represent a collection of well-defined objects using clear mathematical language. They explore common forms such as descriptive statements, roster or tabular notation, and set-builder notation, while identifying elements and understanding the symbols used for membership and non-membership. The topic builds accuracy in reading, writing, comparing, and interpreting sets, providing a foundation for later ideas involving relationships and operations on sets.
TOPIC PRACTICE
Up to 25 questions from this page. Select your focus, then start.
The listed elements are exactly the first five positive multiples of 4. Substituting n = 1, 2, 3, 4, and 5 into 4n gives 4, 8, 12, 16, and 20, with no extra elements. Option B produces every positive even number up to 20, option C describes infinitely many multiples of 4, and option D includes every natural number from 4 through 20. Therefore, option A is the only exact set-builder representation.
A set records distinct objects, so a repeated letter is written only once. The word “mathematics” contains the letters m, a, t, h, e, m, a, t, i, c, and s. Removing repetitions leaves {m, a, t, h, e, i, c, s}. Option B repeats letters and therefore is not a proper roster form of a set, option C omits letters, and option D lists vowels rather than all distinct letters. Thus A is correct.
The elements 7, 14, 21, 28, and 35 are the first five positive multiples of 7. Writing an element as 7n and allowing n to take the values 1 through 5 produces exactly those five numbers. Option B includes every natural number up to 35, option C gives multiples of 5, and option D gives the first five square numbers. Therefore, the correct description is option A.
The positive factors of 15 are 1, 3, 5, and 15, while the positive factors of 25 are 1, 5, and 25. The numbers occurring in both lists are 1 and 5. Therefore, the set of common factors is C = {1, 5}. A number must divide both 15 and 25 exactly to belong to C.
The inequalities are strict, so neither 50 nor 70 can be included. The multiples of 5 between these endpoints are 55, 60, and 65. Each is a natural number and is divisible by 5, so all three satisfy the definition. Therefore, the roster form is I = {55, 60, 65}, which is option A. The other options incorrectly include one or both boundary values.
The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Applying x > 5 removes 1, 2, 3, and 5. The remaining divisors are 6, 10, 15, and 30, so J = {6, 10, 15, 30}. Option A is correct. Option B wrongly includes 5, while C omits 6 and D ignores the inequality condition.
The absolute-value condition |x| ≤ 2 means that x is at a distance of at most 2 from zero. Among the integers, the values satisfying this are −2, −1, 0, 1, and 2. Both endpoints are included because the inequality is less than or equal to, and zero is also an integer satisfying the condition. Therefore, option A gives the correct roster form.
Every member of P is a positive multiple of 3, so it can be written as 3n, where n is natural. The first term 3 corresponds to n = 1, and the last term 30 corresponds to n = 10 because 30 = 3 × 10. Thus n ranges from 1 through 10, giving P = {3n : n ∈ N, 1 ≤ n ≤ 10}. Option D stops at 27 and omits 30.
We inspect the natural numbers strictly between 10 and 25 and retain only primes. The prime numbers in that interval are 11, 13, 17, 19, and 23. Numbers such as 15 and 21 are composite, so they cannot be included. The endpoints 10 and 25 are excluded by the strict inequalities; 25 is also composite. Hence option A is the correct roster form.
A divisor of 48 is a natural number that divides 48 without leaving a remainder. The relevant positive divisors are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. The condition x < 10 removes 12 and every larger divisor. Therefore the required roster form is T = {1, 2, 3, 4, 6, 8}. Option A includes every divisor below 10, while the other options either omit a valid divisor or include 12.
To determine the members of U, solve the defining equation x² − 1 = 0. Using the difference of squares, x² − 1 = (x − 1)(x + 1). Hence either x − 1 = 0, giving x = 1, or x + 1 = 0, giving x = −1. Both values belong to the integers, so both must be included. Therefore U = {-1, 1}, making option A correct. Zero does not satisfy the equation because 0² − 1 = −1.
The positive divisors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100. The phrase two-digit means that the number must be at least 10 and at most 99. Filtering the divisor list with this digit condition leaves 10, 20, 25, and 50. The number 100 is a divisor but has three digits, while 1, 2, 4, and 5 have one digit. Thus V = {10, 20, 25, 50}, so option A is correct.
The integers satisfying −3 ≤ x ≤ 3 are −3, −2, −1, 0, 1, 2, and 3. The square of an integer is even exactly when the integer itself is even, because an even number has an even factor and an odd number has an odd square. Among the listed integers, the even values are −2, 0, and 2. Therefore W = {-2, 0, 2}, so option A is correct.
The prime numbers less than 10 are 2, 3, 5, and 7, because each has exactly two positive factors: 1 and itself. Thus the description “x is a prime number less than 10” produces precisely X = {2, 3, 5, 7}. Option B is incomplete because it excludes 2, the only even prime number. Factors of 10 and composite numbers less than 10 produce different sets, so options C and D are incorrect.
The positive multiples of 6 are 6, 12, 18, 24, 30, 36, 42, and so on. The condition x ≤ 36 stops the list at 36, so 42 and all later multiples are excluded. The roster therefore contains exactly 6, 12, 18, 24, 30, and 36. Zero is not listed because the question is using the positive natural multiples intended by the options; 1 is not a multiple of 6. Hence option A is correct.
List the positive divisors of 18 first: 1, 2, 3, 6, 9, and 18. The set requires divisors that are also even. From this list, 2, 6, and 18 are even, whereas 1, 3, and 9 are odd. Therefore the elements satisfying both conditions are C = {2, 6, 18}. Number 4 is not a divisor of 18, so option C is wrong, and option D omits the valid divisor 2. Hence option A is correct.
To find G, first list the positive divisors of 42: 1, 2, 3, 6, 7, 14, 21, and 42. From this list, identify the numbers that are prime. The numbers 2, 3, and 7 are prime because each has exactly two positive factors, 1 and itself. The number 1 is neither prime nor composite, while 6, 14, 21, and 42 are composite. Therefore, the set of prime divisors of 42 is G = {2, 3, 7}, so option A is correct.
To determine B, test positive natural numbers against x² < 20. We get 1² = 1, 2² = 4, 3² = 9, and 4² = 16, all below 20. However, 5² = 25, so 5 and every larger positive natural number are excluded. Hence B = {1, 2, 3, 4}, exactly the same set as A. Therefore A = B, making option A correct; B is neither empty nor infinite.
The digits of 2026, in order, are 2, 0, 2, and 6. However, a set records whether an element is present, not how many times it occurs. The digit 2 appears twice but is listed only once in the set. Therefore the distinct-digit set is A = {0, 2, 6}. Option B is correct; options A and D incorrectly preserve repetition, while C omits the digit 0.
A prime number is greater than 1 and has exactly two positive divisors: 1 and itself. Checking the natural numbers below 10 gives 2, 3, 5, and 7 as primes. The number 1 is excluded because it has only one positive divisor, and 4, 6, and 8 are composite. Therefore the correct roster, or listing, form is P = {2, 3, 5, 7}.
The governing concept is converting set-builder notation into roster form. The symbol ℤ restricts x to integers, and the inclusive inequality −2 ≤ x ≤ 2 includes every integer from −2 through 2. Listing them gives −2, −1, 0, 1, and 2, so the set has five elements and option A is correct. Option B lists only the endpoints, option C omits both endpoints, and option D wrongly treats the set as empty.
The interval [2, 5] is a closed interval because square brackets are used at both ends. This means that every real number from 2 through 5 is included, including both endpoints 2 and 5. Among the choices, 2 is included, whereas 1, 6, and 5.5 lie outside the interval. Thus option A is correct.
An open interval (2, 5) contains every real number strictly greater than 2 and strictly less than 5. The round brackets exclude both endpoints. Thus 3, 4, and 4.9 lie inside the interval, but 2 is not included. Therefore, option A is correct. Notice that 5 would also be excluded, but it is not among the listed choices.
In interval notation, a square bracket includes the endpoint and a parenthesis excludes it. The left bracket in [2, 5) includes 2, so x may equal 2. The right parenthesis excludes 5, so x must be less than 5. Therefore the equivalent inequality is 2 ≤ x < 5, making option A correct.
In interval notation, a round bracket excludes an endpoint and a square bracket includes it. The interval (2, 5] therefore excludes 2, giving x > 2, and includes 5, giving x ≤ 5. Combining both conditions produces 2 < x ≤ 5, which is option A. Options B, C, and D either include 2, exclude 5, or do both incorrectly.
QUIZ COMPLETE
