Sets Test

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Sets Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$A \cap B = \phi$$, then $$A \Delta B$$ =

A
$$A \cap B$$

B
$$A \cup B$$

C
$$A - B$$

D
$$B - A$$

Solution

We know that $$A\Delta B=A\cup B-A\cap B $$

Given $$A\cap B=\phi $$

$$\therefore A\Delta B=A\cup B-\phi=A\cup B $$

Hence, the answer is $$A\cup B$$.
Question 2
1 / -0

There are 19 hockey players in a club. On a particular day 14 were wearing the prescribed hockey shirts, while 11 were wearing the prescribed hockey pants. None of then was without hockey pant or hockey shirt. How many of them were in complete hockey uniform?

A
8

B
6

C
9

D
7

Solution

We can look at it in 2 ways

First by set theory

$$n(A\cap B) = n(A) + n(B)-n(A\cup B)$$

$$=14+11-19=6$$

Qualitatively, we know that 14 people are wearing prescribed hockey shirts,which leaves us with 5 players who must be wearing hockey pants. So out of 11 players who are wearing hockey pants, 5 are not wearing hockey shirts while the other 6 are in complete uniform.
Question 3
1 / -0

In a class, 20 opted for Physics, 17 for Maths, 5 for both and 10 for other subjects. The class contains how many students?

A
$$35$$

B
$$42$$

C
$$52$$

D
$$60$$

Solution

By set theory

$$n(P\cup M) = n(P) + n(M)-n(P\cap M)$$

$$ =20+17-5=32$$

So total no. of students $$= 32+10 =42$$

$$32$$ opted for at least one subject from Physics and maths while $$10$$ opted for other.
Question 4
1 / -0

In a community of 175 persons, 40 read the Times, 50 read the Samachar and 100 do not read any. How many persons read both the papers?

A
10

B
15

C
20

D
25

Solution

Since 100 do not read any

$$n(T\cup S)=175-100=75$$

By set theory

$$n(T\cap S) = n(T) + n(S)-n(T\cup S)$$

$$=40+50-75 = 15$$
Question 5
1 / -0

In a party, 70 guests were to be served tea or coffee after dinner. There were 52 guests who preferred tea while 37 preferred coffee. Each of the guests liked one or the other beverage. How many guests liked both tea and coffee?

A
15

B
18

C
19

D
33

Solution

By set theory

$$n(T\cap C) = n(T) + n(C)-n(T\cup C)$$

$$=52+37-70 = 19$$
Question 6
1 / -0

In a group of $$15, 7$$ have studied German, $$8$$ have studied French, and $$3$$ have not studied either. How many of these have studied both German and French?

A
$$0$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Since $$3$$ have neither studied German nor French

$$n(G\cup F) = 15-3 = 12$$

By set theory

$$n(G\cap F) = n(G) + n(F)-n(G\cup F)$$

$$=7+8-12 = 3$$
Question 7
1 / -0

In a certain group of 75 students, 16 students are taking physics, geography and English; 24 students are taking physics and geography, 30 students are taking physics and English; and 22 students are taking geography and English. However, 7 students are taking only physics, 10 students are taking only geography and 5 students are taking only English. How many of these students are taking physics?

A
n(P) = 45

B
n(P) = 35

C
n(P) = 65

D
n(P) = 55

Solution

7 students are taking Physics only.

So

$$7=n(P)-n(P\cap G)-n(P\cap E) +n(P\cap G \cap E)$$

$$\Rightarrow n(P)=7$$ $$+ n (P\cap G)+n(P\cap E) -n(P\cap G \cap E)$$

So $$n(P)=7+24+30-16=45$$
Question 8
1 / -0

One Hundred Twenty-five $$(125)$$ aliens descended on a set of film as Extra Terrestrial Beings. $$40$$ had two noses, $$30$$ had three legs, $$20$$ had four ears, $$10$$ had two noses and three legs, $$12$$ had three legs and four ears, $$5$$ had two noses and four ears and $$3$$ had all the three unusual features. How many were there without any of these unusual features?

A
$$5$$

B
$$35$$

C
$$80$$

D
None of these

Solution

Let $$A=$$ set of aliens having two noses
$$B=$$ set of aliens having three legs
$$C=$$ set of aliens having four ears

$$n(A)=40,n(B)=30,n(C)=20,n(A\cap B)=10,n(B\cap C)=12,n(C\cap A)=5,n(A\cap B\cap C)=3$$

$$n(A\cup B\cup C)=n(A)+n(B)+n(C)-(n(A\cap B)+n(B\cap C)+n(C\cap A))+n(A\cap B\cap C)$$
$$n(A\cup B\cup C)=40+30+20-(10+12+5)+3$$
$$n(A\cup B\cup C)=40+30+20-(10+12+5)+3$$
$$n(A\cup B\cup C)=66$$
Aliens having any of the deformities $$=66$$
So, aliens without any deformities $$=125-66=59$$
Question 9
1 / -0

In a class of 80 children, 35% children can play only cricket, 45% children can play only table-tennis and the remaining children can play both the games. In all, how many children can play cricket?

A
$$55$$

B
$$44$$

C
$$36$$

D
$$28$$

Solution

Clearly $$35\%$$ children can play cricket. Also $$20\%$$ can play both.

So $$55\%$$ children can play cricket

Total no. of kids = $$0.55\times 80 =44$$
Question 10
1 / -0

In a survey of brand preference for toothpastes, $$82$$ of the population (number of people covered for the survey is $$100$$) liked at least one of the brands: I, II and III. $$40$$ of those liked brand I, $$25$$ liked brand II and $$35$$ liked brand III. If $$8$$ of those asked, showed liking for all the three brands, then what percentage of those liked more than one of the three brands?

A
$$13$$

B
$$10$$

C
$$8$$

D
$$5$$

Solution

By set theory

$$n(I\cup II \cup III) = n(I) + n(II)+n(III)-n(I\cap II)-n(II\cap III)-n(I\cap III) +n(I\cap II\cap III)$$

$$82=40+25+35-x+8$$

$$\Rightarrow x = 26$$

But

$$n(I\cap II)+n(II\cap III)+n(I\cap III)$$ counts $$n(I\cap II\cap III)$$ thrice whereas it needs to be counted only once.

Hence answer $$=26-2\times8=10$$

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Sets Test - 23

Result

Sets Test - 23

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SHARING IS CARING

Question 1

$$A \cap B$$

$$A \cup B$$

$$A - B$$

$$B - A$$

Solution

Question 2

8

6

9

7

Solution

Question 3

$$35$$

$$42$$

$$52$$

$$60$$

Solution

Question 4

10

15

20

25

Solution

Question 5

15

18

19

33

Solution

Question 6

$$0$$

$$3$$

$$4$$

$$5$$

Solution

Question 7

n(P) = 45

n(P) = 35

n(P) = 65

n(P) = 55

Solution

Question 8

$$5$$

$$35$$

$$80$$

None of these

Solution

Question 9

$$55$$

$$44$$

$$36$$

$$28$$

Solution

Question 10

$$13$$

$$10$$

$$8$$

$$5$$

Solution

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