A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics & Chemistry. It was found that all students passed in at least one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, at most 11 students passed in both Mathematics and Physics, at most 15 students passed in both Physics and Chemistry, at most 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is
Numerical Answer Type:
Enter a numerical valueAnswer: 10 to 10+4 marks
Solution & Explanation
Related Formula
n(M P C) = n(M) + n(P) + n(C) - n(M P) - n(P C) - n(M C) + n(M P C)$$n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)$$
Core Logic
Set Operations diagram for Q21 - JEE Main 2024 Morning
Let x$x$ be the number of students who passed in all three subjects, so n(M P C) = x$n(M \cap P \cap C) = x$.
We are given:
n(M) = 20, n(P) = 25, n(C) = 16$n(M) = 20, n(P) = 25, n(C) = 16$n(M P C) = 40$n(M \cup P \cup C) = 40$
Constraints:
n(M P) ≤ 11$n(M \cap P) \le 11$n(P C) ≤ 15$n(P \cap C) \le 15$n(M C) ≤ 15$n(M \cap C) \le 15$
If we let n(M P) = 11$n(M \cap P) = 11$, then the region representing only M P$M \cap P$ is 11 - x$11 - x$. For this to be non-negative, x ≤ 11$x \le 11$.
Step 1: Testing x = 11
Set Operations diagram for Q21 - JEE Main 2024 Morning
Assume x = 11$x = 11$.
To maximize, let's fix n(M P) = 11$n(M \cap P) = 11$. This leaves 0$0$ students in (M P) - (M P C)$(M \cap P) - (M \cap P \cap C)$.
Let n(M C) = 11 + z$n(M \cap C) = 11 + z$ and n(P C) = 11 + y$n(P \cap C) = 11 + y$.
Constraints limit these subsets:
11 + z ≤ 15 ⇒ z ≤ 4$11 + z \le 15 \Rightarrow z \le 4$11 + y ≤ 15 ⇒ y ≤ 4$11 + y \le 15 \Rightarrow y \le 4$
Filling the Venn diagram nodes:
Only M = 20 - 11 - z - 0 = 9 - z$M = 20 - 11 - z - 0 = 9 - z$
Only P = 25 - 11 - y - 0 = 14 - y$P = 25 - 11 - y - 0 = 14 - y$
Only C = 16 - 11 - y - z = 5 - y - z$C = 16 - 11 - y - z = 5 - y - z$
Summing all disjoint areas:
Total = (9 - z) + 0 + (14 - y) + z + 11 + y + (5 - y - z) = 40$= (9 - z) + 0 + (14 - y) + z + 11 + y + (5 - y - z) = 40$39 - y - z = 40 ⇒ y + z = -1$39 - y - z = 40 \Rightarrow y + z = -1$
Since subsets cannot be negative, x=11$x=11$ is impossible.
Step 2: Testing x = 10
Set Operations diagram for Q21 - JEE Main 2024 Morning
Assume x = 10$x = 10$.
Let's construct a valid distribution with x = 10$x = 10$.
We can choose intersection values:
n(M P) = 10$n(M \cap P) = 10$, so only M P$M \cap P$ is 0$0$.
n(M C) = 11$n(M \cap C) = 11$, so only M C$M \cap C$ is 1$1$.
n(P C) = 10$n(P \cap C) = 10$, so only P C$P \cap C$ is 0$0$.
This gives:
Only M = 20 - 10 - 0 - 1 = 9$M = 20 - 10 - 0 - 1 = 9$
Only P = 25 - 10 - 0 - 0 = 15$P = 25 - 10 - 0 - 0 = 15$
Only C = 16 - 10 - 1 - 0 = 5$C = 16 - 10 - 1 - 0 = 5$
Summing up = 9 + 15 + 5 + 0 + 1 + 0 + 10 = 40$9 + 15 + 5 + 0 + 1 + 0 + 10 = 40$.
All conditions are satisfied perfectly.
Hence, the maximum number is 10.
Pattern Recognition
In multi-constraint Venn diagram maximization, test the theoretical upper bound sequentially downward until a non-negative configuration for all disjoint subsets is achieved.
Keywords:#maximum number of students passed#JEE Main 2024 Morning Q21#Sets JEE Main 2024#Set Operations JEE Main 2024
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Q54jee_main_2025_07_april_eveningRange of Rational Functions
If the range of the function f(x) = (5 - x)/(x² - 3x + 2)$f(x) = \frac{5 - x}{x^2 - 3x + 2}$, x ≠ 1, 2$x \neq 1, 2$, is (-∞ , α ] [ β , ∞)$(-\infty , \alpha ] \cup [ \beta , \infty)$, then α² +β²$\alpha^2 +\beta^2$ is equal to :
A.190$190$
B.192$192$
C.188$188$
D.194$194$
Solution
Related Formula
For a quadratic equation Ax² + Bx + C = 0$Ax^2 + Bx + C = 0$ to yield real roots, its discriminant must satisfy:
D = B² - 4AC ≥ 0$$D = B^2 - 4AC \ge 0$$
Core Logic
Set y = (5 - x)/(x² - 3x + 2)$y = \frac{5 - x}{x^2 - 3x + 2}$:
Case I: If y = 0$y = 0$, the equation simplifies to x - 5 = 0 x = 5$x - 5 = 0 \implies x = 5$, which is a valid part of the domain. Thus, 0$0$ belongs to the range.
Case II: If y ≠ 0$y \neq 0$, for x$x$ to be real, D ≥ 0$D \ge 0$:
For rational expressions of the form LinearQuadratic$\frac{\text{Linear}}{\text{Quadratic}}$, converting to a quadratic in x$x$ and forcing D ≥ 0$D \ge 0$ establishes the range boundaries elegantly.
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Q52jee_main_2025_24_jan_eveningOne-One and Onto Functions
The function f:(-∞,∞)arrow(-∞,1)$f:(-\infty,\infty)\rightarrow(-\infty,1)$, defined by f(x)= 2x-2-x2x+2-x$f(x)=\frac{2^{x}-2^{-x}}{2^{x}+2^{-x}}$ is:
A.One-one but not onto$\text{One-one but not onto}$
B.Onto but not one-one$\text{Onto but not one-one}$
C.Both one-one and onto$\text{Both one-one and onto}$
D.Neither one-one nor onto$\text{Neither one-one nor onto}$
Solution
Related Formula
A function is one-one if its derivative is strictly monotonic (always positive or always negative) across its domain. It is onto if its range equals its co-domain.
Core Logic
Rewrite the function by multiplying the numerator and denominator by 2^x$2^x$:
Thus, the range of the function is (-1, 1)$(-1, 1)$. Since the given co-domain is (-∞, 1)$(-\infty, 1)$ and Range ≠ Co-domain$\text{Range} \neq \text{Co-domain}$ , the function is not onto.
Pattern Recognition
The expression given is a shifted form of the hyperbolic tangent function (x ln 2)$\tanh(x \ln 2)$. Hyperbolic tangent always maps to (-1, 1)$(-1, 1)$, making its restriction against (-∞, 1)$(-\infty, 1)$ non-surjective (not onto).
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Q58jee_main_2025_24_jan_eveningLinear Programming and Inequalities in Two Variables
Let the points ((11)/(2),α)$(\frac{11}{2},\alpha)$ lie on or inside the $\triangle$ with sides x+y=11$x+y=11$, x+2y=16$x+2y=16$ and 2x+3y=29$2x+3y=29$ Then the product of the smallest and the largest values of a is equal to:
A.22$22$
B.44$44$
C.33$33$
D.55$55$
Solution
Related Formula
For a vertical line x = x₀$x = x_0$ crossing a bounded region, the valid coordinates of y$y$ sit between the boundary lines intersecting that specific line vertical plane.
Core Logic
The point given is fixed at x = (11)/(2) = 5.5$x = \frac{11}{2} = 5.5$. We evaluate the values of y$y$ along this vertical line segment across each boundary edge.
Linear Programming region graph for Q58 - JEE Main 2025 Evening
Step 1: Evaluate Intersections
Substitute x = (11)/(2)$x = \frac{11}{2}$ into the three linear constraints:
From x + y = 11$x + y = 11$:
(11)/(2) + y = 11 ⇒ y = 11 - 5.5 = 5.5$$\frac{11}{2} + y = 11 \Rightarrow y = 11 - 5.5 = 5.5$$
Checking the internal region of the $\triangle$ bounded by these lines, the valid range for α$\alpha$ along the section line is delimited by y = 5.5$y = 5.5$ and y = 6$y = 6$:
Instead of drawing full coordinate diagrams or computing all three vertex points, evaluating values directly at the fixed coordinate constraint x = 5.5$x = 5.5$ saves time during multi-line area problems.
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Q71jee_main_2025_24_jan_eveningNumber of Functions under Constraints
Number of functions f:1,2, ,100arrow0,1$f:\{1,2,\dots,100\}\rightarrow\{0,1\}$, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to \_\_\_\_.
Numerical Answer.Answer: 392
Solution
Related Formula
Fundamental Counting Principle: If an operation can be performed in n₁$n_1$ ways, followed by a second operation in n₂$n_2$ ways, the total configurations equal n₁ × n₂$n_1 \times n_2$.
Core Logic
The domain set contains integers from 1 to 100. We divide the mapping requirements across distinct subsets of this domain.
Function mapping grid for Q71 - JEE Main 2025 Evening
Step 1: Choose the single element from 1, 2, , 98$\{1, 2, \dots, 98\}$
We must assign the image value 1 to exactly one positive integer from the subset 1, 2, , 98$\{1, 2, \dots, 98\}$. The number of ways to pick this single element is :
981 = 98 ways$$\binom{98}{1} = 98 \text{ ways}$$
Step 2: Mapping remaining elements
The remaining 97 elements in the 1, 2, , 98$\{1, 2, \dots, 98\}$ subset cannot map to 1, so they must map to 0. This leaves exactly 1 choice per remaining element.
For the final two elements in the domain, 99 and 100, there are no structural constraints:
Separate domains tightly into restricted blocks vs completely free components. Realizing that elements 99 and 100 behave independently with full co-domain targets leaves a clear product formulation.
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Qjee_main_2025_24_jan_morningSymmetric Property of Functions
Let f(x) = 2x + 2 + 1622x + 1 + 2x + 4 + 32$f(x) = \frac{2^{x + 2} + 16}{2^{2x + 1} + 2^{x + 4} + 32}$. Then the value of 8(f((1)/(15)) + f((2)/(15)) + … + f((59)/(15)))$8\left(f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)\right)$ is equal to:
A.118$118$
B.92$92$
C.102$102$
D.108$108$
Solution
Related Formula
Many finite fractional sum questions involving functional terms rely on identifying an underlying symmetric summation invariant, typically of the form f(x) + f(k-x) = constant$f(x) + f(k-x) = \text{constant}$.
Core Logic
First simplify the expression for f(x)$f(x)$ algebraically:
Whenever a symmetric set of arguments is presented inside a summation matching (k)/(n) + (N-k)/(n) = constant$\frac{k}{n} + \frac{N-k}{n} = \text{constant}$, look for an algebraic reduction of f(x)$f(x)$ that yields a uniform constant sum for symmetric pairs.
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More Sets Questions — jee_main_2024_30_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.