NIMCET 2021 Previous Year Questions (PYQs) – Page 7 of 12
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Solution
Solution
Given: Values \(0,1,2,\dots,50\) with frequencies \(1,\binom{50}{1},\binom{50}{2},\dots,\binom{50}{50}\) (i.e., \(\binom{50}{k}\) for value \(k\)).
Total frequency: $$\sum_{k=0}^{50}\binom{50}{k}=2^{50}.$$
Sum of (value × frequency): $$\sum_{k=0}^{50} k\binom{50}{k}=50\cdot 2^{49}\quad\text{(identity: }\sum k\binom{n}{k}=n2^{\,n-1}\text{)}.$$
Arithmetic Mean (AM): $$\bar{x}=\frac{\sum k\binom{50}{k}}{\sum \binom{50}{k}} =\frac{50\cdot 2^{49}}{2^{50}}=\frac{50}{2}=25.$$
Answer: \(25\).
Solution
Given: \(A=\{1,2,3,4\}\), \(B=\{3,4,5\}\)
\(A\cup B=\{1,2,3,4,5\}\Rightarrow |A\cup B|=5\)
\(A\cap B=\{3,4\}\Rightarrow |A\cap B|=2\)
\(A\triangle B=(A\cup B)\setminus(A\cap B)=\{1,2,5\}\Rightarrow |A\triangle B|=3\)
Size of Cartesian product: \(|(A\cup B)\times(A\cap B)\times(A\triangle B)| = 5\times 2\times 3 = \mathbf{30}\).
Solution
Solution
Let \(|S|=m\).
Count incidences via the \(A_i\): There are 30 sets each of size 5, so total memberships \(=30\times 5=150\). Each element of \(S\) lies in exactly 10 of the \(A_i\), so also \(= m\times 10\). Hence \(m=\dfrac{150}{10}=15\).
Count incidences via the \(B_j\): There are \(n\) sets each of size 3, so total memberships \(=n\times 3\). Each element of \(S\) lies in exactly 9 of the \(B_j\), so also \(= m\times 9 = 15\times 9=135\).
Thus \(n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.\)
Solution
Solution
Solution
Solution
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Commented Dec 24, 2021
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