Previous year questions: Sets

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Every question below comes from an official source: six from the JEE Main 2026 papers released by the National Testing Agency and four from the NCERT Exemplar for Class 11 Mathematics. The source and the marks appear in the heading. Try each question before reading the worked answer, and note that the JEE items use the same set language as the class notes, so the route is always to identify the set, then count or simplify.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The set A contains the first 101 terms of an arithmetic progression that starts at 1 and rises by 5 each time. The set B contains the first 71 terms of an arithmetic progression that starts at 9 and rises by 7 each time. Find how many members of A intersection B are divisible by 3.

Answer: Write the two progressions as numbers congruent to 1 modulo 5 and to 9 modulo 7. A common member has the form 9 + 7k, and the condition 9 + 7k congruent to 1 modulo 5 reduces to 2k congruent to 2, so k is congruent to 1 modulo 5. The common members are therefore 16, 51, 86 and so on, an arithmetic progression with difference 35. The largest term of B is 9 plus 70 times 7, that is 499, while A ends at 501, so the limit is 499. Solving 16 + 35m at most 499 gives m at most 13, so there are 14 common terms, for m from 0 to 13. A common term is divisible by 3 when 1 + 2m is 0 modulo 3, which gives m congruent to 1 modulo 3. The values m equal to 1, 4, 7, 10 and 13 give five terms. The official key records 5.

Q2 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Find the number of elements of the set S of ordered pairs (r, k), with k an integer, for which 36C(r+1) equals 6 times 35Cr divided by (k squared minus 3).

Answer: Simplify the binomial ratio first. The quotient 36C(r+1) divided by 35Cr equals 36 over (r + 1), because the two factorials cancel except for the factor 36 and the denominator r + 1. The equation becomes 36 over (r+1) equal to 6 over (k squared minus 3). Cross multiplying gives 6(r + 1) equal to 36(k squared minus 3), so r equals 6k squared minus 19. The binomial coefficient 35Cr is defined only for r between 0 and 35, so check integer squares. For k squared equal to 4, r is 5; for k squared equal to 9, r is 35; smaller squares give negative r and larger squares exceed 35. Each of k equal to plus or minus 2 and plus or minus 3 gives one pair, so S has four elements.

Q3 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Let S be the set of x in the closed interval from minus pi to pi for which sin x times (sin x + cos x) equals a, where a is an integer. Find the number of elements in S.

Answer: Simplify the expression first. The product expands to sin squared x plus sin x cos x, which equals half of (1 plus sin 2x minus cos 2x). The combination sin 2x minus cos 2x equals root 2 times sin(2x minus pi by 4), so the whole expression lies between (1 minus root 2) by 2 and (1 plus root 2) by 2. The only integers in that range are 0 and 1. For a equal to 0, the equation is sin(2x minus pi by 4) equal to minus 1 by root 2, and counting solutions as the angle runs over an interval of length 4 pi gives five values. For a equal to 1, the same sine equals 1 by root 2, which gives four values. The two groups do not overlap, so S has 9 elements, matching the official key.

Q4 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

Take A equal to the set of integers from 2 to 6. Define a relation R on the product A cross A by saying that (x, y) is related to (z, w) when x divides z and y is at most w. Find the number of elements in R.

Answer: The relation compares the two pairs position by position, so count the choices separately for each position. For the first entries, count pairs (x, z) from A cross A with x dividing z. The divisor 2 works with z equal to 2, 4 and 6, giving three; 3 works with 3 and 6, giving two; each of 4, 5 and 6 works only with itself, giving one each. The first position therefore allows 3 + 2 + 1 + 1 + 1 = 8 pairs. For the second entries, count ordered pairs (y, w) with y at most w, which is the number of pairs with repetition allowed, 15. Multiplying the independent counts gives 8 times 15 = 120 elements. The official answer is 120.

Q5 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Let A be the set containing 1, 4 and 7, and B the set containing 2, 3 and 8. In the product (A cross B) cross (A cross B), keep the pairs of ordered pairs for which a1 + b2 divides a2 + b1. Find the number of elements kept.

Answer: Write one ordered pair as (a1, b1) and the other as (a2, b2). The divisor is d equal to a1 + b2, whose possible values from the two sets are 3, 4, 6, 7, 9, 10, 12 and 15, with 9 arising twice. For each divisor value, count the pairs (a2, b1) for which a2 + b1 is a multiple of d, since a2 and b1 range over A and B. For d equal to 3 the multiples 3, 6, 9, 12 and 15 give six choices; for d equal to 4 the choices are 4 and 12, giving two; for d equal to 6 the choices are 6 and 12, giving two; for d equal to 7, 10, 12 and 15 a single multiple comes from the sets. The divisor 9 appears twice, and each appearance allows the two representations of 9, so it contributes four. Adding 6 + 2 + 2 + 1 + 1 + 1 + 4 gives 18 elements. The official answer is 18.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

On the set containing minus 2, minus 1, 0, 1 and 2, define a relation R by putting (a, b) in R exactly when 1 + ab is positive. Two statements are made: the number of elements in R is 17, and R is an equivalence relation. Decide which statement is true.

Answer: Count first. The condition fails only when ab is at most minus 1. Every pair with a zero in it satisfies the condition, giving 9 pairs inside a 5 by 5 array. Pairs with both entries of the same sign also satisfy it, and there are 4 positive by positive pairs plus 4 negative by negative pairs, adding 8 more. Pairs such as (1, minus 1) fail because the product is minus 1, and pairs with product minus 2 or minus 4 fail as well. The total is 17, so the first statement is true.

The second statement fails because the relation is not transitive. The pair (2, 0) is in R and the pair (0, minus 1) is in R, but (2, minus 1) is not in R since 1 plus minus 2 is negative. So the relation is reflexive and symmetric but not transitive, and only the first statement is true. The official key records the first statement only.

Q7 (NCERT Exemplar, 3 marks)

Write the following sets in roster form: A is the set of real x with 2x + 11 = 15, B is the set of real x with x squared equal to x, and C is the set of positive factors of a prime number p.

Answer: Solve each condition. For A, the equation gives 2x equal to 4, so x is 2 and A is the singleton set containing 2. For B, x squared equal to x rearranges to x(x minus 1) equal to 0, so the real solutions are 0 and 1, and B is the set containing 0 and 1. For C, a prime number has exactly two positive factors, 1 and itself, so C is the set containing 1 and p. Source: NCERT Exemplar Class 11 Mathematics, Chapter 1, Exercise 1.3, Short Answer Type, Q1.

Q8 (NCERT Exemplar, 3 marks)

With L equal to the set of 1, 2, 3 and 4, M equal to the set of 3, 4, 5 and 6, and N equal to the set of 1, 3 and 5, verify that L minus (M union N) equals (L minus M) intersection (L minus N).

Answer: Compute the left side. M union N is the set of 1, 3, 4, 5 and 6, and removing those elements from L leaves the set containing only 2. Compute the right side. L minus M is the set of 1 and 2, and L minus N is the set of 2 and 4. Their intersection is the set containing 2. Both sides equal the set containing 2, so the identity is verified. The identity is a special case of De Morgan's law written with differences, which is why the two expressions agree. Source: NCERT Exemplar Class 11 Mathematics, Chapter 1, Exercise 1.3, Short Answer Type, Q5.

Q9 (NCERT Exemplar, 3 marks)

Out of 100 students, 15 passed English, 12 passed Mathematics and 8 passed Science. Also 6 passed English and Mathematics, 7 passed Mathematics and Science, 4 passed English and Science, and 4 passed all three. Find the number who passed English and Mathematics but not Science, the number who passed Mathematics and Science but not English, the number who passed Mathematics only, and the number who passed more than one subject.

Answer: Subtract the triple overlap from each pairwise figure to get the exclusive overlaps. English and Mathematics but not Science is 6 minus 4, that is 2. Mathematics and Science but not English is 7 minus 4, that is 3. Mathematics only is found by removing from 12 the two pairwise overlaps that involve Mathematics and adding back the triple overlap once, which gives 12 minus 6 minus 7 plus 4, that is 3. The number passing more than one subject is the sum of the three pairwise figures minus twice the triple figure, since the triple students were counted twice too many times: 6 + 7 + 4 minus 8 equals 9. Source: NCERT Exemplar Class 11 Mathematics, Chapter 1, Exercise 1.3, Long Answer Type, Q24.

Q10 (NCERT Exemplar, 3 marks)

In a class of 60 students, 25 play cricket, 20 play tennis, and 10 play both games. Find the number of students who play neither game.

Answer: The number who play at least one game is the union count, 25 plus 20 minus 10, that is 35. The class has 60 students, so the number outside the union is 60 minus 35, that is 25. The same result follows from a Venn diagram: write 10 in the overlap, 15 in the cricket only region, 10 in the tennis only region, and the remaining region holds 25. Source: NCERT Exemplar Class 11 Mathematics, Chapter 1, Exercise 1.3, Long Answer Type, Q25.