Set Theory Test 14

Result

Set Theory Test 14

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In a group, if 60% of people drink tea and 70% drink coffee. What is the maximum possible percentage of people drinking either tea or coffee but not both?

A
$$100\%$$

B
$$70\%$$

C
$$30\%$$

D
$$10\%$$

Solution

To find maximum possible percentage of people drinking either coffee or tea, we can assume everyone drinks at least either of the options.

Hence

$$a+b=100$$

$$a+2b=60+70=130$$

$$b=30$$

hence $$a=70$$ which is the required area of the venn diagram.
Question 2
1 / -0

If A and B are two disjoint sets and N is universal set, then $$\displaystyle A^{\circ}\cup \left [ \left ( A\cup B \right )\cap B^{\circ} \right ]$$ is

A
$$\displaystyle \phi $$

B
$$A$$

C
$$B$$

D
$$N$$

Solution

In the Venn Diagram, we can see that $$ (A \cup B) \cap B' $$ covers the pink coloured region inside $$N$$ except $$B$$.
And $$ A' $$ is the region which is shown using striking lines , that is the region outside $$A$$
And if we perform $$ A' \cup (A \cup B) \cap B' $$, it will cover the pink region as well s the striken region, which is nothing but the complete set $$N$$.
Question 3
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 4
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 5
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 6
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 7
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 8
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 9
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 10
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 $$A$$
$$A=(1,2,5,6,7,11,12,15,16,17,21,22,25,26,27,31,32,35,36,37)$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Set Theory Test 14

Result

Set Theory Test 14

Score

-

Rank

-

SHARING IS CARING

Question 1

$$100\%$$

$$70\%$$

$$30\%$$

$$10\%$$

Solution

Question 2

$$\displaystyle \phi $$

$$A$$

$$B$$

$$N$$

Solution

Question 3

$$7$$

$$42$$

$$3$$

Cannot be determined

Solution

Question 4

2

3

4

5

Solution

Question 5

128

216

240

160

Solution

Question 6

$$9$$

$$6$$

$$3$$

$$18$$

Solution

Question 7

6

7

8

9

Solution

Question 8

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

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

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

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

Solution

Question 9

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 10

$$10$$

$$13$$

$$17$$

$$20$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.