Practice Questions for CSIR NET Group Theory : Sets and Relations I

1. Let \(D\) be the set of tuples \((w_1, \ldots, w_{10})\), where \(w_i \in \{1, 2, 3\}\), \(1 \leq i \leq 10\), and \(w_i + w_{i+1}\) is an even number for each \(i\) with \(1 \leq i \leq 9\). Then the number of elements in \(D\) is

(a.) \(2^{11} + 1\)
(b.) \(2^{10} + 1\)
(c.) \(3^{10} + 1\)
(d.) \(3^{11} + 1\)

Explanation for Question 1:
We have the conditions: \(\omega_{i}+w_{i+1}\) is even \(\omega_{i} \in \{1, 2, 3\}\) For different values of \(\omega_{i}\), we get the following possibilities:
1. \(\omega_{i}=1 \Rightarrow \omega_{i+1}=3\) or \(\omega_{i+1}=1\)
2. \(\omega_{i}=2 \Rightarrow \omega_{i+1}=2\)
3. \(\omega_{i}=3 \Rightarrow \omega_{i+1}=1\) or \(\omega_{i+1}=3\)
Now, let's consider the cases:
Case I - \(\omega_{1}=1\)
Number of possibilities \(=2^{9}\)
Case II - \(\omega_{1}=2\)
Number of possibilities \(=1\)
Case III - \(\omega_{1}=3\)
Number of possibilities \(=2^{9}\)
Therefore, the total number of possibilities \(=2^{10}+1\).

2. The number of surjective maps from a set of 4 elements to a set of 3 elements is

(a.) 36
(b.) 64
(c.) 69
(d.) 81

Explanation for Question 2:
The number of surjective maps \(=3^{4}-\sum_{r=1}^{3-1}(-1)^{r}(3-r)^{4} = 36\).

3. If \(n\) is a positive integer such that the sum of all positive integers \(a\) satisfying \(1 \leq a \leq n\) and \(\operatorname{GCD}(a, n)=1\) is equal to \(240n\), then the number of summands, namely, \(\varphi(n)\), is

(a.) 120
(b.) 124
(c.) 240
(d.) 480

Explanation for Question 3:
We have the equation \(n \times \phi(n) = \sum \phi(n)\).

4. An ice cream shop sells ice creams in five possible flavors: Vanilla, Chocolate, Strawberry, Mango, and Pineapple. How many combinations of three-scoop cones are possible? [Note: The repetition of flavors is allowed, but the order in which the flavors are chosen does not matter.]

(a.) 10
(b.) 20
(c.) 35
(d.) 243

Explanation for Question 4:
Let \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\) be flavors of ice creams.
Case I - No repartition:
Number of possibilities \(=^{5}C_{3} = 10\)
Case II - Flavor is repeated twice:
Number of possibilities \(=^{5}C_{2} \times 2 = 20\)
Case III - All same flavors:
Number of possibilities \(=5\)

5. We are given a class consisting of 4 boys and 4 girls. A committee that consists of a President, a Vice-President, and a Secretary is to be chosen among the 8 students of the class. Let \(a\) denote the number of ways of choosing the committee in such a way that the committee has at least one boy and at least one girl. Let \(b\) denote the number of ways of choosing the committee in such a way that the number of girls is greater than or equal to that of the boys. Then

(a.) \(a = 288\)
(b.) \(b = 168\)
(c.) \(a = 144\)
(d.) \(b = 192\)

Explanation for Question 5:
Among 4 boys \(b_1, b_2, b_3, b_4\) and 4 girls \(g_1, g_2, g_3, g_4\), we choose a committee of 3 people.
Case I - Committee has at least 1 boy and at least 1 girl:
Number of ways to select 1 boy and 2 girls = \({4 \choose 2} \times 3! = 144\)
Number of ways to select 1 girl and 2 boys = \({4 \choose 2} \times 3! = 144\)
Therefore, \(a = 288\).
Case II - Number of girls is greater than or equal to the number of boys:
Number of ways to choose 2 girls and 1 boy = 144
Number of ways to choose 3 girls = \({4 \choose 3} \times 3! = 24
Therefore, \(b = 168\).

6. In a group of 265 persons, 200 like singing, 110 like dancing, and 55 like painting. If 60 persons like both singing and dancing, 30 like both singing and painting, and 10 like all three activities, then the number of persons who like only dancing and painting is

(a.) 10
(b.) 20
(c.) 30
(d.) 40

Explanation for Question 6:
Total number of people = 265.
Number of people who like to:
Sing = 200
Dance = 110
Paint = 55
Sing and Paint = 30
Sing and Dance = 60
Dance and Paint = \(x\)
Sing, Dance, and Paint = 10
Number of people who only like to dance and sing is 50
Dance and Paint = \(x - 10\)
Sing and Paint = 20
Total = 120 + 20 + 50 + 10 + 60 - x + x - 10 + 35 - x = 265.
Therefore, \(x = 20\).

7. The last two digits of \(7^{81}\) are

(a.) 07
(b.) 17
(c.) 37
(d.) 47

Explanation for Question 7:
\[ \begin{align*} & p^{\phi(m)} \equiv 1(\bmod m) \\ & \Rightarrow 7^{40} \equiv 1(\bmod 100) \\ & \Rightarrow 7^{80} \equiv 01(\bmod 100) \\ & \Rightarrow 7^{81} = 07(\bmod 100) \end{align*} \]

8. The number of positive divisors of 50,000 is

(a.) 20
(b.) 30
(c.) 40
(d.) 50

Explanation for Question 8:
\[ \begin{align*} & 50000 = 2^4 \times 5^5 \\ & \Rightarrow \tau(50000) = 5 \times 6 = 30 \end{align*} \]

9. The last digit of \((38)^{2011}\) is

(a.) 6
(b.) 2
(c.) 4
(d.) 8

Explanation for Question 9:
Observe that \(38(\bmod 10) \equiv 8(\bmod 10)\) and \((8)^{4k+1} \equiv 8(\bmod 10)\), \((8)^{4k+2} \equiv 4(\bmod 10)\), \((8)^{4k+3} \equiv 2(\bmod 10)\). \[ \begin{align*} & 2011 = 4 \times 52 + 3 \\ & \Rightarrow (38)^{2011} \equiv 2(\bmod 10) \end{align*} \]

10. The number of multiples of \(10^{44}\) that divide \(10^{33}\) is

(a.) 11
(b.) 12
(c.) 121
(d.) 144

Explanation for Question 10:
\(10^{44} \times 10^{11} = 10^{55}\) and the number of factors of \(10^{11} = \tau\left(2^{11} \times 5^{11}\right) = 144\).

11. The number of elements in the set \(\{m: 1 \leq m \leq 100, \text{m and 100 are relatively prime}\}\) is

(a.) 100
(b.) 250
(c.) 300
(d.) 400

Explanation for Question 11:
\(\phi(1000) = \phi\left(2^3\right) \times \phi\left(5^3\right) = 4 \times 100 = 400\).

12. The unit digit of \(2^{100}\) is given by

(a.) 2
(b.) 4
(c.) 6
(d.) 8

Explanation for Question 12:
\[ \begin{align*} & 2^{4k+1} \equiv 2(\bmod 10) \\ & 2^{4k+1} \equiv 4(\bmod 10) \\ & 2^{4k+3} \equiv 8(\bmod 10) \\ & 2^{4k} \equiv 6(\bmod 10) \end{align*} \]
\[ \begin{align*} & 100 = 4 \times 25 \\ & \Rightarrow 2^{100} \equiv 6(\bmod 10) \end{align*} \]

13. Let \(U(n)\) be the set of all positive integers less than \(n\) and relatively prime to \(n\). For \(n=248\), the number of elements in \(U(n)\) is

(a.) 60
(b.) 120
(c.) 180
(d.) 240

Explanation for Question 13:
\(\phi(n) = \phi(248) = \phi\left(2^3 \times 31\right) = \phi\left(2^3\right) \cdot \phi(31) = 4 \times 30 = 120\).

14. Two finite sets have \(m\) and \(n\) elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. Then the values of \(m\) and \(n\) are respectively:

(a.) 8, 5
(b.) 6, 3
(c.) 4, 1
(d.) None of these

Explanation for Question 14:
If \(|A| = m\), then \(|P(A)| = 2^m\). Given \(2^m - 2^n = 56\), it implies \(2^n(2^{m-n} - 1) = 56\). Therefore, \(n = 3, m = 6\).

15. Let \(A\) be a set containing \(n\) elements, then its power set \(P(A)\) contains:

(a.) \(n\) elements
(b.) \(2^n\) elements
(c.) \(n^2\) elements
(d.) None of these

Explanation for Question 15:
If \(|A| = n\), then \(|P(A)| = 2^n\).

16. 20 teachers of a school either teach mathematics or physics. 12 of them teach mathematics while 4 teach both the subjects. Then the number of teachers teaching physics is

(a.) 12
(b.) 8
(c.) 16
(d.) None of these

Explanation for Question 16:
There are 12 teachers teaching math, \(x\) teachers teaching physics, and 4 teachers teaching both. Therefore, the number of teachers teaching only math is \(12 - 4 = 8\), and the number of teachers teaching only physics is \(x - 4\). The total number of teachers is \(8 + x - 4 + 4 = 20\), which implies \(x = 12\).

17. Let \(A=\{1,2,3,4\}\) and let \(R=\{(2,2),(3,3),(4,4),(1,2)\}\) be a relation in \(A\). Then \(R\) is

(a.) Reflexive
(b.) Symmetric
(c.) Transitive
(d.) None of these

Explanation for Question 17:
\((1,1) \notin R\), so \(R\) is not reflexive.
\((1,2) \in R\), but \((2,1) \notin R\), so \(R\) is not symmetric.
To check transitivity, you need to verify that \((a,b) \in R\) and \((b,c) \in R\) imply \((a,c) \in R\). The transitive property is not mentioned in the solution provided. To complete the explanation, you should check if \((a,b) \in R\) and \((b,c) \in R\) indeed imply \((a,c) \in R).

18. Let \(A=\{1,2,3,4\}\) and \(R\) be a relation in \(A\)

\(R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,1),(1,3)\}\). Then \(R\) is

(a.) Reflexive
(b.) Symmetric
(c.) Transitive
(d.) An equivalence relation

Explanation for Question 18:
\((1,1), (2,2), (3,3), (4,4) \in R\), so \(R\) is reflexive.
\(R\) is also symmetric.
However, \((2,1), (1,3) \in R\), but \((2,3) \notin R\), so \(R\) is not transitive.

19. A survey shows that \(63\%\) of the Americans like cheese whereas \(76\%\) like apples. If \(x\%\) of the Americans like both cheese and apples, then

(a.) \(x = 39\)
(b.) \(x = 63\)
(c.) \(39 \leq x \leq 63\)
(d.) None of these

Explanation for Question 19:
Solve the equation \(63 + 76 - x = 100\) to find \(x\): \(x = 139 - 100 = 39\).

20. A set contains \(2n+1\) elements. The number of subsets of this set containing more than \(n\) elements is equal to

(a.) \(2^{n-1}\)
(b.) \(2^n\)
(c.) \(2^{n+1}\)
(d.) \(2^{2n}\)

Explanation for Question 20:
a) If \(n = 0\), then \(2n + 1 = 1\). The number of subsets with cardinality greater than 0 is 1, which is \(2^0\). So, options a and c are false.
b) If \(n = 1\), then \(2n + 1 = 3\). The number of subsets with cardinality greater than 1 is 4, which is \(2^2\). So, option b is false.