JEE Main Maths questions with solutions

Q1. Let A be the set {(n, 2n): n ∈ N} and let B be the set {(2n, 3n): n ∈ N}. What is the intersection A ∩ B?

1. {(n, 6n): n ∈ N}
2. {(2n, 6n): n ∈ N}
3. {(n, 3n): n ∈ N}
4. ∅

Answer: ∅

The sets A and B consist of pairs where the first elements are defined differently, leading to no common pairs. Specifically, the first element of A is n while in B it is 2n, meaning there are no values of n that satisfy both conditions simultaneously.

Q2. Let set A contain 3 elements and set B contain 6 elements. Then the cardinality of their union must satisfy

1. 3 ≤ n(A ∪ B) ≤ 6
2. 3 ≤ n(A ∪ B) ≤ 9
3. 6 ≤ n(A ∪ B) ≤ 9
4. 0 ≤ n(A ∪ B) ≤ 9

Answer: 6 ≤ n(A ∪ B) ≤ 9

The union of sets A and B can have a minimum of 6 elements if all elements of A are also in B, and a maximum of 9 elements if all elements are distinct. Therefore, the cardinality of the union must fall within the range of 6 to 9.

Q3. Given the sets A = {1, 2, 5} and B = {3, 4, 5, 9}, what is A ∩ B?

1. {1, 2, 5, 9}
2. {1, 2, 3, 4, 9}
3. {1, 2, 3, 4, 5, 9}
4. None of these

Answer: None of these

The intersection contains only elements common to both sets. A = {1,2,5} and B = {3,4,5,9} share only 5, so A intersect B = {5}. Since {5} is not among the options, the answer is 'None of these'.

Q4. At a conference with 100 attendees, 29 are Indian women and 23 are Indian men. Among the Indian attendees, 4 are doctors, and 24 are either men or doctors. If there are no foreign doctors, how many foreigners and how many women doctors are present at the conference?

1. 48, 1
2. 34, 3
3. 46, 4
4. 42, 2

Answer: 48, 1

Indians = 29 + 23 = 52, so foreigners = 100 - 52 = 48 (all doctors are Indian). For Indians, |men or doctors| = 24 = 23 + 4 - (male doctors), giving 3 male doctors, so women doctors = 4 - 3 = 1. Answer: 48 foreigners and 1 woman doctor.

Q5. Let X and Y be two non-empty sets, and let A be a non-empty set such that X ∩ A = Y ∩ A = A and X ∪ A = Y ∪ A. Which of the following must be true?

1. X is a proper subset of Y
2. Y is a proper subset of X
3. X and Y are equal
4. X and Y have no common elements

Answer: X and Y are equal

X cap A = A means A is a subset of X, so X cup A = X. Similarly Y cup A = Y. The given X cup A = Y cup A then becomes X = Y, so the two sets must be equal.

Q6. If A and B are non-empty sets with A containing B, then which of the following is true?

1. B' − A' = A − B
2. B' − A' = B − A
3. A' − B' = A − B
4. A' ∩ B' = B − A

Answer: B' − A' = A − B

With B contained in A: B' - A' = B' intersect A = A intersect B' = A - B. Since B is a subset of A, B - A is empty, so option 'B'-A' = B-A' is false. The correct identity is B' - A' = A - B.

Q7. In a city with 10,000 households, it is observed that 40% of the households subscribe to newspaper A, 20% subscribe to newspaper B, and 10% subscribe to newspaper C. Also, 5% subscribe to both A and B, 3% subscribe to both B and C, and 4% subscribe to both A and C. If 2% of the households subscribe to all three newspapers, then

1. 3,300 households subscribe to A only
2. 1,400 households subscribe to B only
3. 4,000 households subscribe to none of A, B and C
4. All of the above statements are true

Answer: All of the above statements are true

With N=10000: A-only = A-AB-AC+ABC = 4000-500-400+200 = 3300; B-only = 2000-500-300+200 = 1400; union = 4000+2000+1000-500-300-400+200 = 6000, so none = 10000-6000 = 4000. All three statements hold, so the correct choice is 'All of the above are true'.

Q8. In a conflict, 70% of the fighters lost one eye, 80% lost an ear, 75% lost an arm, and 85% lost a leg. If x% of them lost all four of these parts, what is the least possible value of x?

1. 10
2. 12
3. 15
4. None of these

Answer: 10

The minimum percentage losing all four = 100 - [(100-70)+(100-80)+(100-75)+(100-85)] = 100 - (30+20+25+15) = 100 - 90 = 10. So the least value of x is 10.

Q9. Given a universal set U with n(U) = 700, n(A) = 200, n(B) = 300, and n(A ∩ B) = 100, what is the value of n(A' ∩ B')?

1. 400
2. 600
3. 300
4. None of these

Answer: 300

By De Morgan, A' n B' = (A u B)'. n(A u B) = 200 + 300 - 100 = 400, so n(A' n B') = 700 - 400 = 300.

Q10. Consider the following statements: Statement-1: If B = U A, then n(B) = n(U) − n(A), where U denotes the universal set. Statement-2: For any three sets A, B and C, if C = A − B, then n(C) = n(A) − n(B). Which of the following is correct?

1. Statement-1 is true, and Statement-2 correctly explains Statement-1.
2. Statement-1 is true, but Statement-2 does not correctly explain Statement-1.
3. Statement-1 is false, while Statement-2 is true.
4. Statement-1 is true, while Statement-2 is false.

Answer: Statement-1 is true, while Statement-2 is false.

If B is the complement of A then n(B) = n(U) - n(A), so Statement-1 is true. But for C = A - B, n(A-B) = n(A) - n(A intersect B), which equals n(A) - n(B) only when B is a subset of A; in general it is false. Hence Statement-1 true, Statement-2 false.

Q11. In a class containing 80 students, labeled 1 to 80, every student with an odd number chooses Cricket, every student whose number is a multiple of 5 chooses Football, and every student whose number is a multiple of 7 chooses Hockey. How many students choose none of these three games?

1. 13
2. 24
3. 28
4. 52

Answer: 28

Odd numbers (Cricket), multiples of 5 (Football) and multiples of 7 (Hockey) are chosen. A student picks none only if the number is even and not a multiple of 5 or 7. Counting 1-80 gives 28 such students.

Q12. A group of 60 students contains 23 who play hockey, 15 who play basketball, and 20 who play cricket. Also, 7 students play both hockey and basketball, 5 play both cricket and basketball, 4 play both hockey and cricket, and 15 students do not play any of these three games. Which statement is true?

1. 4 students play hockey, basketball, and cricket
2. 20 students play hockey but not cricket
3. 1 student plays hockey and cricket but not basketball
4. All of the above are correct

Answer: 1 student plays hockey and cricket but not basketball

Union = 60 - 15 = 45. By inclusion-exclusion 45 = 23+15+20 -7-5-4 + n(all), giving n(all)=3. Then hockey-and-cricket-only = 4 - 3 = 1, so exactly 1 student plays hockey and cricket but not basketball.

Q13. Let A denote the set of all divisors of 15, B the set of prime numbers less than 10, and C the set of even numbers less than 9. Then the set (A ∪ C) ∩ B is

1. {1, 3, 5}
2. {1, 2, 3}
3. {2, 3, 5}
4. {2, 5}

Answer: {2, 3, 5}

The correct option includes the elements that are both in the union of the divisors of 15 and the even numbers less than 9, and also in the set of prime numbers less than 10. The divisors of 15 are {1, 3, 5, 15}, and the even numbers less than 9 are {0, 2, 4, 6, 8}. The union of these sets is {1, 2, 3, 4, 5, 6, 8, 15}, and the intersection with the prime numbers less than 10, which are {2, 3, 5, 7}, results in {2, 3, 5}.

Q14. Two finite sets contain m and n elements, respectively. The first set has 112 more subsets than the second set. What are the values of m and n?

1. 4, 7
2. 7, 4
3. 4, 4
4. 7, 7

Answer: 7, 4

The number of subsets of a set with m elements is given by 2^m, and for a set with n elements, it is 2ⁿ. The equation 2^m - 2ⁿ = 112 holds true for m = 7 and n = 4, as it results in 128 - 16 = 112.

Q15. In a school, 30 students study both Mathematics and Chemistry. This common group is 10% of all Mathematics enrolments and 12% of all Chemistry enrolments. What is the total number of students who study at least one of these two subjects?

1. 520
2. 490
3. 560
4. 480

Answer: 520

The total number of Mathematics students can be calculated as 30 is 10% of the total, leading to 300 Mathematics students. Similarly, for Chemistry, 30 is 12% of the total, resulting in 250 Chemistry students. Using the principle of inclusion-exclusion, the total number of students studying at least one subject is 300 + 250 - 30, which equals 520.

Q16. Given two sets A and B with n(A) = 1000 and n(B) = 500, and with n(A ∩ B) at least 1, let n(A ∪ B) = p. Which range must p satisfy?

1. 500 ≤ p ≤ 1000
2. 1001 ≤ p ≤ 1498
3. 1000 ≤ p ≤ 1498
4. 1000 ≤ p ≤ 1499

Answer: 1000 ≤ p ≤ 1499

n(AuB) = 1000 + 500 - n(AnB). Since n(AnB) can be from 1 (max union) up to 500 (min union, B subset of A), p ranges from 1000 to 1499, i.e. 1000 <= p <= 1499.

Q17. How many ordered pairs {a, b} with integers a and b satisfy 2a² + 3b² = 35, where Z denotes the set of all integers?

1. 2
2. 4
3. 8
4. 12

Answer: 8

Testing b^2 in {0,1,4,9}: 3b^2 = 3 leaves 2a^2 = 32 -> a = +/-4, and 3b^2 = 27 leaves 2a^2 = 8 -> a = +/-2. So (a,b) = (+/-4,+/-1) and (+/-2,+/-3), giving 8 ordered integer pairs.

Q18. Let A, B and C be finite sets with n(A)=10, n(B)=15, n(C)=20, n(A∩B)=8 and n(B∩C)=9. Which of the following can be the value of n(A∪B∪C)?

1. 26
2. 27
3. 28
4. Any one of 26, 27, or 28 may occur

Answer: Any one of 26, 27, or 28 may occur

n(AUBUC) = 45 - 8 - 9 - n(A∩C) + n(A∩B∩C). Since n(A∩C) and n(A∩B∩C) are not fixed, testing feasible (nonnegative-region) values yields totals of 26, 27 and 28. Hence any one of 26, 27 or 28 may occur.

Q19. In a city with 10,000 households, it is observed that 40% of the households purchase newspaper A, 20% purchase newspaper B, and 10% purchase newspaper C. Also, 5% purchase both A and B, 3% purchase both B and C, and 4% purchase both A and C. If 2% of the households subscribe to all three newspapers, how many households buy only newspaper A?

1. 3100
2. 3300
3. 2900
4. 1400

Answer: 3300

To find the number of households that buy only newspaper A, we subtract those who buy A and B, A and C, and all three newspapers from the total purchasing A. This calculation shows that 3300 households exclusively purchase newspaper A.

Q20. Consider the following assertions: Assertion 1: If A ∪ B = A ∪ C and A ∩ B = A ∩ C, then B = C. Assertion 2: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Choose the correct statement about these assertions.

1. Assertion 1 is true and Assertion 2 is true; Assertion 2 correctly explains Assertion 1.
2. Assertion 1 is true and Assertion 2 is true; Assertion 2 does not correctly explain Assertion 1.
3. Assertion 1 is false and Assertion 2 is true.
4. Assertion 1 is true and Assertion 2 is false.

Answer: Assertion 1 is true and Assertion 2 is true; Assertion 2 does not correctly explain Assertion 1.

Assertion 1 is a valid theorem: if A∪B=A∪C and A∩B=A∩C then B=C. Assertion 2 is the distributive law A∪(B∩C)=(A∪B)∩(A∪C), also true. Both are true, but the distributive law does not directly explain the cancellation result, so 'both true, A2 does not explain A1' is correct.

Q21. Given f(x)=(x(x-p))/(q-p)+(x(x-q))/(p-q), where p≠ q, determine the value of f(p)+f(q).

1. f(p-q)
2. f(p+q)
3. f(p(p+q))
4. f(q(p-q))

Answer: f(p+q)

Combining the terms, f(x) = x(x-p)/(q-p) + x(x-q)/(p-q) simplifies to f(x) = x. Hence f(p)+f(q) = p+q = f(p+q).

Q22. A real-valued function f(x) obeys the relation f(x-y)=f(x)f(y)-f(a-x)f(a+y), where a is a fixed constant, and f(0)=1. Then the value of f(2a-x) is

1. -f(x)
2. f(x)
3. f(a)+f(a-x)
4. f(-x)

Answer: -f(x)

The given functional equation can be manipulated by substituting specific values for x and y, revealing that f(2a-x) is indeed the negative of f(x). This relationship holds due to the symmetry and properties of the function as defined by the equation.

Q23. The set of all real values of x for which the function f(x)=(3)/(4-x²)+log₁₀(x³-x) is defined is:

1. (-1,0)∪(1,2)∪(2,∞)
2. (a,2)
3. (-1,0)∪(a,2)
4. (1,2)∪(2,∞)

Answer: (-1,0)∪(1,2)∪(2,∞)

The term 3/(4-x^2) requires x != +/-2, and log10(x^3-x) requires x^3-x = x(x-1)(x+1) > 0, i.e. x in (-1,0) U (1,inf). Removing x = 2 gives the domain (-1,0) U (1,2) U (2,inf).

Q24. Let A = {1, 2, 3, 4, 5} and B = {2, 3, 6, 7}. Find the number of elements in the set (A × B) ∩ B × (B × A).

1. 18
2. 6
3. 4
4. 0

Answer: 4

(A×B)∩(B×A) consists of pairs (x,y) with x,y in both A and B, i.e. in A∩B = {2,3}. So the count is |A∩B|^2 = 2^2 = 4.

Q25. A relation R is defined on the set of integers Z by the condition that (x,y) belongs to R exactly when x² + y² = 9. Which statement below is false?

1. R = {(0,3), (0,-3), (3,0), (-3,0)}
2. The domain of R is {-3, 0, 3}
3. The range of R is {-3, 0, 3}
4. None of these

Answer: None of these

Integer solutions of x^2+y^2=9 are (0,+/-3) and (+/-3,0). The relation set, its domain {-3,0,3}, and its range {-3,0,3} are all stated correctly. Since none of statements (a)-(c) is false, the false-statement answer is 'None of these'.

Q26. If f(x)=√(1+x²), then which of the following statements is true?

1. f(xy)=f(x) f(y)
2. f(xy)≥ f(x) f(y)
3. f(xy)≤ f(x) f(y)
4. None of these

Answer: f(xy)≤ f(x) f(y)

f(x)f(y)=sqrt((1+x^2)(1+y^2))=sqrt(1+x^2+y^2+x^2y^2), while f(xy)=sqrt(1+x^2y^2). Since x^2+y^2>=0, we always have f(xy) <= f(x)f(y).

Q27. For the function f(x)=√(x-√(1-x²)), what is its domain?

1. [-1,-1/√(2)]∪[1/√(2),1]
2. [-1,1]
3. (-∞,-1/2]∪[1/√(2),+∞)
4. [1/√(2),1]

Answer: [1/√(2),1]

Require 1-x^2 >= 0 so x in [-1,1], and x - sqrt(1-x^2) >= 0 means x >= sqrt(1-x^2), which forces x >= 0 and x^2 >= 1-x^2, i.e. x >= 1/sqrt(2). Combining gives the domain [1/sqrt(2), 1].

Q28. Find the period of the function |sin³(x/2)| + |cos⁵(x/5)|.

1. 2π
2. 10π
3. 8π
4. 5π

Answer: 10π

The period of the function is determined by the individual periods of its components. The period of | sin³(x/2) is 4 and the period of | cos⁵(x/5) is 10, making the least common multiple 10.

Q29. Given that the number of elements in set A is 4 and in set B is 3, and the Cartesian product A × B × C has 24 elements, what is the number of elements in set C?

1. 288
2. 1
3. 12
4. 2

Answer: 2

|A x B x C| = 4 * 3 * |C| = 24, so |C| = 24/12 = 2.

Q30. Let S = {1, 2, 3, 4, 5}. If R = {(x, y): x + y < 6}, then how many ordered pairs are in R?

1. 8
2. 10
3. 6
4. 5

Answer: 10

For x+y<6 with x,y in {1..5}: x=1 gives y=1..4 (4), x=2 gives y=1..3 (3), x=3 gives y=1,2 (2), x=4 gives y=1 (1), x=5 gives none. Total = 4+3+2+1 = 10.