Sets
25 JEE Maths previous year questions on Sets — options free on every question; 2 include the answer & explanation free, the rest unlock with PYQ Pass.
Let \(A={1,2,3,\ldots \ldots .,10}\) and \(B={\frac{m}{n}:m,n\in A,m Then \(n(B)\) is equal to
31
\(\text{for}m=1\Rightarrow \frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots \ldots \ldots \frac{1}{10}\to 9\\ \text{for}m=2\Rightarrow \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}\to 4\\ \text{for}m=3\Rightarrow \frac{3}{4},\frac{3}{5},\frac{3}{7},\frac{3}{8},\frac{3}{10}\to 5\\ \text{for}m=4\Rightarrow \frac{4}{5},\frac{4}{7},\frac{4}{9}\to 3\\ \text{for}m=5\Rightarrow \frac{5}{6},\frac{5}{7},\frac{5}{8},\frac{5}{9},\to 4\\ \text{for}m=6\Rightarrow \frac{6}{7}\to 1\\ \text{for}m=7\Rightarrow \frac{7}{8},\frac{7}{9},\frac{7}{10},\to 3\\ \text{for}m=8\Rightarrow \frac{8}{9}\to 1\\ \text{for}m=9\Rightarrow \frac{9}{10}\to 1\\ \text{then}n(B)=31\)
Let \(A={(\alpha ,\beta )\in \mathrm{R}\times \mathrm{R}:|\alpha -1|\leq 4\text{ and }|\beta -5|\leq 6}\) and \(B=\left\{(\alpha ,\beta )\in \mathrm{R}\times \mathrm{R}:16(\alpha -2{)}^{2}+9(\beta -6{)}^{2}\leq 144\right\}\). Then
\(\mathrm{B}\subset \mathrm{A}\)
Set A:
\(A=\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}:|\alpha-1| \leq 4\) and \(|\beta-5| \leq 6\}\)
This represents a rectangle with:
\(−3\leq \alpha \leq 5\) (from \(\alpha −1\in [−4,4]\))
\(−1\leq \beta \leq 11\) (from \(\beta −5\in [−6,6]\))
Set B:
\(B=\left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}: 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}\)
This is an ellipse centered at \((2,6)\).
Standard form: \(\frac{(\alpha-2)^2}{9}+\frac{(\beta-6)^2}{16} \leq 1\)
So, it has:
Horizontal semi-axis \(= 3\)
Vertical semi-axis \(= 4\)
Hence, the ellipse spans:
\(\alpha \in [2−3,2+3]=[−1,5]\)
\(\beta \in [6−4,6+4]=[2,10]\)
Now compare sets:
Rectangle A: \(\alpha \in [−3,5],\beta \in [−1,11]\)
Ellipse B: completely lies within bounds of A
So, B is a subset of A, but A is not a subset of B (e.g., point \((−3,−1)\) is in A but not in B)
Let \(A\) and \(B\) be two finite sets with \(m\) and \(n\) elements, respectively. The total number of subsets of the set \(A\) is \(56\) more than the total number of subsets of \(B\). Then the distance of the point \(P(m, n)\) from the point \(Q(-2,-3)\) is:
[JEE Main 2024, 27 Jan (Shift 2)]
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Let \(A\) and \(B\) be two finite sets with \(m\) and \(n\) elements, respectively. The total number of subsets of the set \(A\) is \(56\) more than the total number of subsets of \(B\). Then the distance of the point \(P(m, n)\) from the point \(Q(-2,-3)\) is:
[JEE Main 2024, 27 Jan (Shift 2)]
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Let \(A={1,2,3,\ldots \ldots .,10}\) and \(B=\left\{\frac{m}{n}: m, n \in A, m [JEE Main 2025, 22 Jan (Shift 1)]
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Let \(S =\left\{x \in R :(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}\). Then the number of elements in \(S\) is:
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In a statistical investigation of 1003 families of Calcutta, it was found that 63 families have neither a radio nor a TV, 794 families have a radio, and 187 have a TV. The number of families having both a radio and a TV is:
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Let \(\mathrm{A}={1,6,11,16,\ldots }\) and \(\mathrm{B}={9,16,23,30,\ldots }\) be the sets consisting of the first \(2025\) terms of two arithmetic progressions. Then \(n(A\cup B)\) is
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Let \(\mathrm{A}={1,6,11,16,\ldots }\) and \(\mathrm{B}={9,16,23,30,\ldots }\) be the sets consisting of the first \(2025\) terms of two arithmetic progressions. Then \(n(A\cup B)\) is
[JEE Main 2025, 4 Apr (Shift 1)]
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Let \(A=\{n \in[100,700] \cap N: n\) is neither a multiple of 3 nor a multiple of 4\(\}\). Then the number of elements in \(A\) is
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\(\text{ Let }A={n\in [100,700]\cap N:n\text{ is neither a multiple of }3\text{ nor a multiple of }4}\text{.}\)
then the number of elements in A is
[JEE Main 2024, 6 Apr (Shift 1)]
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Let \(A={1,2,3,\ldots \ldots .,10}\) and \(B={\frac{m}{n}:m,n\in A,m Then \(n(B)\) is equal to
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Let \(A={1,2,3,\ldots \ldots .,10}\) and \(B={\frac{m}{n}:m,n\in A,m Then \(n(B)\) is equal to [JEE Main 2025]
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A set A has 3 elements and another set B has 6 elements. Then
[JEE Main 2023]
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In a class of 140 students numbered 1 to 140 , all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is:
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Let \( S=\{1,2,3, \ldots, 100\} \). The number of non-empty subsets \( A \) of \( S \) such that the product of elements in \( A \) is even, is
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If \(A=\{x \in R:|x|<2\}\) and \(B=\{x \in R:|x-2| \geq 3\}\) : then:
[JEE Main 2020, 9 Jan (Shift 2)]
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If \(A = \{ x \in R:|x - 2| > 1\}\), \(B=\left\{x\in R:\sqrt{{x}^{2}-3}>1\right\}\) and \(\ C = \{ x \in R:|x - 4| \geq 2\}\) and \(Z\) is the set of all integers, then the number of subsets of the set \((A \cap B \cap C)^{c} \cap Z\) is
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An organization awarded 48 medals in event \(A\), 25 in event \(B\) and 18 in event \(C\). If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
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Let \( \bigcup_{i=1}^{50} X_{i}=\bigcup_{i=1}^{n} Y_{i}=T \), where each \( X_{i} \) contains \(10\) elements and each \( Y_{i} \) contains \(5\) elements. If each element of the set \( T \) is an element of exactly \(20\) of sets \( X_{i}'s \) and exactly 6 of sets \( Y_{i}'s \) then \( n \) is equal to
[JEE Main 2020, 4 Sep (Shift 2)]
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Let \(\overset{50}{\underset{i=1}{\cup }}{X}_{i}=\overset{50}{\underset{i=1}{\cup }}{Y}_{i}=T\) where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi's and exactly 6 of sets Yi's, then n is equal to
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A survey shows that \(63 \%\) of the people in a city read newspaper \(A\) whereas \(76 \%\) read newspaper \(B\). If \(x\%\) of the people read both the newspapers, then a possible value of \(x\) can be:
[JEE Main 2020, 4 Sep (Shift 1)]
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Let Z, be the set of all integers,
\(A=\left\{\left(x,y\right)\in Z\times Z:{\left(x-2\right)}^{2}+{y}^{2}\leq 4\right\}\\ B=\left\{\left(x,y\right)\in Z\times Z:{x}^{2}+{y}^{2}\leq 4\right\}\\ C=\left\{\left(x,y\right)\in Z\times Z:{\left(x-2\right)}^{2}+{(y-2)}^{2}\leq 4\right\}\)
If the total number of relations from \(A\cap B\) to \(A\cap C\) is 2p, then the value of p is:
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A survey shows that \(73 \%\) of the persons working in an office like coffee, whereas \(65 \%\) like tea. If \(x\) denotes the percentage of them, who like both coffee and tea, then \(x\) cannot be:
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Let the number of elements in sets A and B be five and two respectively. Then the number of subsets of A × B each having at least 3 and at most 6 elements is:
[JEE Main 2023, 08 Apr (Shift 1)]
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