Sets Test

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Sets Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

If out of 150 students who read at least one newspaper The Times of India, The Hindustan Times and The Hindu. There are 65 who read The Times of India, 41 who read The Hindu and 50 who read The Hindustan Times. What is the maximum possible number of students who read all the three newspaper?

A
$$7$$

B
$$42$$

C
$$3$$

D
Cannot be determined

Solution

From venn diagram

$$a+b+c=150$$

$$a+2b+3c=156$$

Hence $$b+2c=6$$

To maximise c we take minimum value of b that is 0.

Hence $$ c=3$$
Question 2
1 / -0

Directions For Questions

In a survey of 25 students, it was found that 15 had taken mathematics, 12 had taken physics and 11 had taken chemistry, 5 had taken mathematics and chemistry, 9 had taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken all the three subjects.

...view full instructions

Find the number of students that had taken only mathematics

A
2

B
3

C
4

D
5

Solution

Let M=Set of students who took Mathematics
C=Set of students who took Chemistry
P=Set of student who took Physics
$$\therefore n(M)=15,n(P)=12,n(C)=11,n(M\cap C)=5,n(M\cap P)=9,n(P\cap C)=4,n(M\cap P \cap C)=3$$
No.of students that had taken only mathematics
$$\Rightarrow n(M-(P\cup C))=n(M)-n(M \cap(P\cup C))$$
$$= n(M)-n((M\cap P)\cup(M\cap C))$$
$$= n(M)-[n(M\cap P)+n(M\cap C)-n((M\cap P)\cap(M\cap C))]$$
$$= n(M)-n(M\cap P)-n(M\cap C)+n(M\cap P\cap C)$$
$$= 15-9-5+3=4$$
Question 3
1 / -0

Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey ; 80 played cricket and basketball and 40 played cricket and hockey 24 player all the three games. The number of boys who did not play any game is

A
128

B
216

C
240

D
160

Solution

Total no. of boys n(X)=800
Let n(C)=boys who played cricket
n(H)=boys who played hockey
n(B)=boys who played basket ball
$$\therefore n(C)=224,n(H)=240,n(B)=336,n(B\cap H)=64,n(C\cap B)=80,n(C\cap H)=40,n(C\cap H\cap B)=24$$

$$\therefore n(C\cup H\cup B)=n(C)+n(H)+n(B)-n(B\cap H)-n(C\cap B)-n(B\cap H)+n(C\cap H\cap B)$$
$$\Rightarrow 224+240+336-64-80-40+24=640$$
The number of boys who did not play any game=$$n(X)-n(C\cup H\cup B)$$
$$\Rightarrow 800-640=160$$
Question 4
1 / -0

Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in $$A\cup B$$ ?

A
$$9$$

B
$$6$$

C
$$3$$

D
$$18$$

Solution

Let $$A$$ be the left circle and $$B$$ be the right circle.There are $$3$$ elements in $$A$$ and $$6$$ elements in $$B$$.
The union of $$A$$ and $$B$$ contains elements that are in either circle.Thus,the union of $$A$$ and $$B$$ will be all of the elements in $$A$$ along with all of the element $$B$$.However,you have to subtract the elements that are in the overlapping area because you are counting twice.
If $$A$$ and $$B$$ don't overlap at all,then the union will ontain $$9$$ elements.If $$A$$ is completely inside $$B$$ then the union will only contain $$6$$ elements,which is the minimum no. of elements in the union of $$A$$ and $$B$$.
let $$A=1,2, B=2,3$$
$$\therefore A\cup B=1,2,3$$ which is $$3$$ elements.
$$\therefore A $$ has $$2$$ elements ,$$B$$ has $$2$$ elements,and there is $$1$$ element overlapping.
$$\therefore 2+2-1=3!=3\times 2\times 1=6$$
Question 5
1 / -0

How many subsets can be formed from the set $$\left \{p, q, r\right \}$$ is?

A
6

B
7

C
8

D
9

Solution

$$\left\{p,q,r \right\}$$ It has 3 elemente
$$\therefore 2^n=2^3=8 Subset$$
Question 6
1 / -0

If $$A=\left \{1, 2, 3, .....9\right \}$$ and $$B=\left \{2, 3, 4, 5, 7, 8\right \}$$, then A-B is given by

A
$$\left \{1, 6, 7, 8\right \}$$

B
$$\left \{1, 6, 9\right \}$$

C
$$\left \{1, 9\right \}$$

D
$$\left \{6, 9\right \}$$

Solution

$$A=\left\{1,2,3,4,5,6,7,8,9 \right\}$$
$$B=\left\{2,3,4,5,7,8 \right\}$$
A-B is a set of all those elements of A which are not in B.
$$\therefore A-B=\left\{1,6,9 \right\}$$
Question 7
1 / -0

The set of real numbers r satisfying
$$\displaystyle \frac{3 r^2- 8r+5}{4r^2-3r+7}>0$$ is

A
the set of all real numbers

B
the set of all positive real numbers

C
the set of all real numbers strictly between 1 and 5/3

D
the set of all real numbers which are either less than 1 or greater than 5/3

Solution

$$\displaystyle \frac{3.r^2 - 8.r + 5 }{4. r^2 - 3.r + 7} > 0$$
$$\Rightarrow 3r^2 -3r -5r + 5 > 0$$ and $$4r^2 - 3r + 7 > 0 $$ as D < 0
$$\Rightarrow 3r(r-1) -5(r -1) > 0$$
$$\Rightarrow (3r - 5) (r - 1) > 0$$
$$(3r - 5) (r - 1) > 0$$
$$r \varepsilon (- \infty, 1) \cup (5/3, \infty)$$
Question 8
1 / -0

Let $$ S=\left\{1,2,3,.....40\right\} $$ and let $$A$$ be a subset of $$S$$ such that no two elements in $$A$$ have their sum divisible by $$5$$ What is the maximum number of elements possible in $$A$$?

A
$$10$$

B
$$13$$

C
$$17$$

D
$$20$$

Solution

There are $$20$$ maximum number of elements possible in $$A$$
$$A=(1,2,5,6,7,11,12,15,16,17,21,22,25,26,27,31,32,35,36,37)$$
Question 9
1 / -0

Look at the set of numbers below.
Set : $$\left \{6, 12, 30, 48\right\}$$
Which statement about all the numbers in this set is NOT true?

A
They are all multiples of $$3$$

B
They are all even numbers

C
They are all factors of $$48$$

D
They are all divisible by $$2$$

Solution

Here above statement C is not true because 30 is not factors of 48.
So, Answer (C) They are all factors of 48
Question 10
1 / -0

In a class of 55 students, the number of students studying different subjects are 23 in Mathematics and 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects, The number of students who have taken exactly one subject is

A
20

B
24

C
23

D
22

Solution

Total no. of student=55
Let n(M)=student who studying mathematics
n(C)=student who studying chemistry
n(P)=student who studying Physics
$$\therefore n(M)=23,n(P)=24,n(C)=19,n(M\cap P)=12,n(M\cap C)=9,n(P\cap C)=7,n(M\cap P\cap C)=4$$
Number of student who studying mathematics but not physic and chemistry
$$\Rightarrow n(M)-[(n(M\cap C)+n(M\cap P)]+n(M\cap P\cap C)$$
$$\Rightarrow 23-[9+12]+4$$
$$\Rightarrow 23-21+4=6$$

Number of student who studying chemistry but not physic and matehematics
$$\Rightarrow n(C)-[(n(M\cap C)+n(P\cap C)]+n(M\cap P\cap C)$$
$$\Rightarrow 19-[9+7]+4$$
$$\Rightarrow 19-16+4=7$$

Number of student who studying physics but not mathematics and chemistry
$$\Rightarrow n(P)-[(n(M\cap P)+n(P\cap C)]+n(M\cap P\cap C)$$
$$\Rightarrow 24-[12+7]+4$$
$$\Rightarrow 24-19+4=9$$

$$\therefore$$no. of student studying exactly one subject$$=6+7+9=22$$

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Re-Attempt Weekly Quiz Competition

Sets Test - 28

Result

Sets Test - 28

Score

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Rank

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SHARING IS CARING

Question 1

$$7$$

$$42$$

$$3$$

Cannot be determined

Solution

Question 2

2

3

4

5

Solution

Question 3

128

216

240

160

Solution

Question 4

$$9$$

$$6$$

$$3$$

$$18$$

Solution

Question 5

6

7

8

9

Solution

Question 6

$$\left \{1, 6, 7, 8\right \}$$

$$\left \{1, 6, 9\right \}$$

$$\left \{1, 9\right \}$$

$$\left \{6, 9\right \}$$

Solution

Question 7

the set of all real numbers

the set of all positive real numbers

the set of all real numbers strictly between 1 and 5/3

the set of all real numbers which are either less than 1 or greater than 5/3

Solution

Question 8

$$10$$

$$13$$

$$17$$

$$20$$

Solution

Question 9

They are all multiples of $$3$$

They are all even numbers

They are all factors of $$48$$

They are all divisible by $$2$$

Solution

Question 10

20

24

23

22

Solution

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