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Question (1):
List the elements of the following sets.
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Answer:
i) Omit o and u from the vowels a, e, i, o, u to get {a,e,i}
ii) There are exactly two such names {Karnataka, Kerala}
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Question (2):
List the elements of the following sets and specify whether they are finite or infinite:
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Answer:
(i) Since there is no natural number lying between 2 and 3, the set is a null set i.e. f. This is a finite set.
(i) Since there are infinite number of real number between 4 and 1, the set is an infinite set.
(iii) Since there are infinite number of multiplies of 4 i.e. {4,8,12,16....}, the set is an infinite set.
(iv) Although the set is finite at any given time, it would be almost impossible to list its elements.
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Question (3):
Write down the following statements using set notation.
(i) x is an element of set A
(ii) y is not an element of set
(iii) A is a subset of B
(iv) A is a proper subset of B
(v) A contains B
(vi) A is superset of B
(vii) Neither A is a subset of B, nor B is a subset of A.
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Answer:
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Question (4):
Express the following sets in set builder form.
(i) Set A consists of the letters p, q, r, s and t
(ii) Set C = {1,3,5,7,....}
(iii) Set D = {8}
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Answer:
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Question (5):
If A = {2,3}, B = {3,4,5}, C = {3,4,6}, find
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Answer:
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Question (6):
A = {a, b, c, e}; B = {b, c, d, e, f}; C = {d, e, f, g}
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Answer:
(i)
(ii)
(iii)
(iv)
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Question (7):
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Answer:
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Question (8):
Compute
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Answer:
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Question (9):
Let A = {1, 2, 3}, B = {2, 3, 4, 5} and C = {3, 4, 5, 6} be three subsets of the universal set U = {1, 2, 3, 4, 5, 6, 7}. Verify the following identities
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Answer:
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Question (10):
For any two sets A and B, prove that A' - B' = B - A.
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Answer:
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Question (11):
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Answer:
From (i) and (ii),
Similarly,
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Question (12):
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Answer:
(Bl contains all the elements of A and some other elements also.)
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Question (13):
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Answer:
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Question (14):
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Answer:
I. Proof:
II. Proof:
 .........(i)
 ........ (ii)
From (i) and (ii),
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Question (15):
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Answer:
Let
 ... (i)
Let,
 ... (ii)
From (i) and (ii), we get
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Question (16):
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Answer:
Let us assume,
and let
But x  B and x  B cannot hold good simulataneously. This is a contradiction. Hence our assumption is wrong and the theorem is true.
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Question (17):
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Answer:
We shall prove this in two parts,
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Question (18):
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Answer:
Thus verified.
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Question (19):
Each person in a group of 50 can speak either English or Hindi or both. If 35 persons can speak English and 25 can speak both; find the number of those who speak Hindi only.
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Answer:
Let
n (E) = number of persons who can speak English
n (H) = number of persons who can speak Hindi
To find:
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Question (20):
There are 64 persons in a group. Each of them eats either apple or plantain or both. Those who eat apples are 36 in number. The number of those who eat both apples and plantain is 21. How many eat plantains only?
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Answer:
Let n(A) = number of persons who eat apples
n(P) = number of persons who eat plantains
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Question (21):
In a class of 80 boys, there are 60 boys who play chess and 35 play table tennis. Use Venn diagrams to show how many boys play both the games? How many play chess only and how many play tennis only?
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Answer:
Let n(C) = number of boys of who play chess
n(T) = number of boys who play table tennis
Let x = number of boys who play both the games
From the above diagram it is clear that,
Those who play chess only = 60 - x = 60 - 15 = 45
Those who play table tennis only = 35 - 15 = 20
To find:
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Question (22):
At dinner, seven people ordered the special of the day. Three people order from the list of the menu, one person orders only salad. Find the number 'm' (dinners ordered by the people).
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Answer:
The sets are disjoint. (assuming no one orders more than one dinner) m = 7 + 3 + 1 = 10, the total number of dinners ordered.
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Question (23):
A survey shows that 71% of Indians like to watch cricket, whereas 64% like to watch hockey. What percentage of Indians like to watch both cricket and hockey? (Assuming that every indian watches at least one of these games)
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Answer:
Let n(C) = percentage of Indians who watch cricket and
n (H) = percentage of Indians who watch hockey.
Hence 35% of Indians like to watch both cricket and hockey.
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Question (24):
In a group of 80 persons, 30 drink Fanta but not Limca and 41 drink Fanta.
1. How many drink Fanta and Limca both?
2. How many drink Limca, but not Fanta?
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Answer:
Let n(F) = number who drink Fanta
n(L) = number who drink Limca
Now,
9 persons drink Fanta and Limca both.
\ 39 people drink Limca but not Fanta.
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Question (25):
Out of 120 students who secured first class marks in Mathematics or English, 70 obtained first class in Mathematics and 31 in English and Mathematics. How many students secured first class marks in English only?
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Answer:
Number of people who secured first class marks in English only is
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Question (26):
In a group of 1500 people, 950 can speak Hindi and 650 can speak both Hindi and Kannada. How many can speak Hindi only? How many can speak Kannada only?
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Answer:
Let
Number who can speak Hindi only=
Number who can speak Kannada only=
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Question (27):
A class has 175 students. The following is the description of students studying one or more of the following subjects in this class. Mathematics 100, Physics 70, Chemistry 46, Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students are enrolled in Mathematics alone, Physics alone and Chemistry alone?
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Answer:
Let M = Set of students who study Mathematics
P = Set of students who study Physics
C = Set of students who study Chemistry
Let us denote the number of students in each bounded region by the letter a, b, c, d, e, f, g as shown in the figure.
Solving these equations we get,
Number of students who have not offered any of these three subject = 175 - 153 = 22
Students enrolled in Mathematics only = e = 60
Students enrolled in Chemistry only = g = 13
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Question (28):
Represent the following using Venn diagram
A = {2, 3, 4, 5} B= {2, 3}
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Answer:
A is a subset of B.
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Question (29):
Represent the following using Venn diagrams:
(i) 
(ii) 
(iii) A - B and B - A
(iv) Al
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Answer:
(iii) A - B and B - A
(iv) Al
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Question (30):
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Answer:
Furthermore,
Hence from (1) and (2),
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Question (31):
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Answer:
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Question (32):
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Answer:
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Question (33):
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Answer:
Let
 
The statement in the second pair of brackets is impossible. Thus, we arrive at a contradiction. The assumption is false and hence the theorem is true.
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Question (34):
Using Euclid's division algorithm, find the HCF of the following numbers. 624, 936, 264
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Answer:
Using Euclid's division algorithm
HCF (936, 624)
936= 624 x 1 + 312
624 = 312 x 2 + 0
HCF (936, 624) = 312
HCF (312, 264)
312 = 264 x 1 + 48
264 = 48 x 5 + 24
48 = 24 x 2+ 0
HCF (312, 264) = 24
Using long division
i) 624, 936, 264
HCF (936, 624) = HCF (624, 312) = 312
Now take 312 and 264
 HCF of 624, 936 and 264 is 24.
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Question (35):
Using Euclid's division algorithm, find the HCF of the following numbers. 2103, 9945, 9216
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Answer:
Hence HCF of 9945, 9216 and 2103 is 3.
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Question (36):
Using Euclid's division algorithm, find the HCF of the following numbers. 13620, 221, 6810
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Answer:
HCF (6810, 221) = 1
HCF (13620, 6810, 221) = 1 which shows that 6810 and 221 are co-prime.
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Question (37):
Using Euclid's division algorithm, find the HCF of the following numbers. 594, 1848, 792.
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Answer:
594, 1848, 792
HCF (1848, 792)
1848 = 792 x 2 + 264
792 = 264 x 3 + 0
HCF (1848, 792) = 264
HCF (594, 264)
594 = 264 x 2 + 66
264 = 66 x 4 + 0
HCF (594, 264) = 66
HCF (594, 1848, 792) = 66
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Question (38):
Find the HCF and LCM of 144 and 180 by the prime factor method. [2 Marks]
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Answer:
144 2 x 2 x 2 x 2 x 3 x 3
180 = 2 x 2 x 3 x 3 x 5
HCF (144, 180) = 2 x 2 x 3 x 3 = 36
LCM (144, 180) = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720
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Question (39):
Write the factor tree for the following numbers and write them as the product of primes
i) 81 ii) 32 iii) 45 iv) 120 v) 200 vi) 100
[each 2 Marks]
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Answer:
i) 81 = 3 x 3 x 3 x 3 = 34
ii) 32 =2 x 2 x 2 x 2 x 2 = 25
iii) 45 = 3 x 3 x 5 = 32 x 5
iv) 120 = 2 x 2 x 2 x 3 x 5 = 23 x 3 x 5
v) 200 = 2 x 2 x 2 x 5 x 5 = 23 x 52
vi) 100 =2 x 2 x5 x 5 =22 x 52
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Question (40):
Given HCF (117, 221) =13. Find LCM (117, 221) [2 Marks]
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Answer:
HCF (117, 221) = 13
We know that the product of HCF and LCM of 2 numbers is equal to the product of the two numbers.
HCF x LCM = Product of the 2 numbers.
 13 x LCM =117 x 221
LCM =
= 1989
Therefore, LCM (117, 221) = 1989
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Question (41):
Find the HCF and LCM of 1152 and 1664 by using the fundamental theorem of arithmetic. [2 Marks]
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Answer:
Steps:
1. Find the prime factors of 1152 and 1664
2. Find the common factors and write their products as HCF.
3. Find LCM by dividing the product of the 2 numbers by their HCF.
Prime factors of 1152 = 27 x 32
Prime factors of 1664 = 27 x 13
HCF (1152, 1664) = 27 = 128
LCM (1152, 1664) = 
= 1152 x 13
= 14976
Therefore,
HCF (1152, 1664) =128
LCM (1152, 1664) = 149 |
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Question (42):
Check whether 12n can end with the digit '0' for any  (n belongs to N means n is a natural number).
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Answer:
If the number 12n ends with '0', then it should be divisible by a prime number 5, which means that '5' is a prime factor of 12n.
This is not possible because, the prime factors of 12 are only 2 and 3. (12n) = (2 x 2 x 3)n. Here we do not have 5 as a factor. Hence by the principle of fundamental theorem of arithmetic, we have no other prime in the factorization of 12n other than 2 and 3. So we can say that there is no natural number n for which 12n ends with the digit zero. |
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Question (43):
Find why 5 x 11 x 19 x 23 + 23 is a composite number.
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Answer:
23 is a common factor. So the given number can be written as
23 (5 x 11 x 19) + 23
= 23 (1045) + 23
= 23 (1045 +1)
= 23 x 1046
= 2 x 23 x 528
This is the product of prime numbers. Therefore the given number is a composite number.
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Question (44):
Prove that  is an irrational number by contradiction method. [3 Marks]
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Answer:
Let assume that  is a rational number.
Write  (p and q are co-prime, they are integers and  )
Squaring both sides
p2 = 3q2 or 3q2 = p2
==> 3 divides p2 and 3 divides p …(1)
Let p = 3x
p2 = 9x2
So, 3q2 = 9x2
q2 = 3x2
==> 3 divides q2 and 3 divides q …(2)
From (1) and (2), we get 3 is a common factor of p and q. This contradicts the assumption that p and q are co-prime.
Therefore  is not a rational number.
Hence is an irrational numbe |
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Question (45):
Prove that is an irrational number by representing it in decimal form. [2 Marks]
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Answer:
Prove that is irrational
 is a non-terminating and non-repeating decimal.
 is irrational.
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Question (46):
Prove that is an irrational number by contradiction method. [3 Marks]
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Answer:
Prove that is irrational by contradiction method.
Let us assume that is rational number.
 (p and q are integers, q 0, p and q have no common factor)
Squaring both the sides
Let p = 5y ==> p2 = 25y2
5q2 = 25y2, q2 = 5y2
==> 5 divides q2
==> 5 divides q
From (1) and (2) we get 5 is a common factor of p and q.
This contradicts the assumption that p and q have no common factor.
 is an irrational number |
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Question (47):
Prove that  is an irrational number by contradiction method.
[3 Marks]
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Answer:
Prove that  is irrational by contradiction method.
Let us assume that  is a rational number.
 (p and q are integers,  p and q are co-prime)
Squaring both the sides,
p2 = 7q2 7 divides p2and 7 divides p …(1)
Let p = 7x
 4.9x2 = 7q2 7x2 = q2
 7 divides q2
 7 divides q …(2)
From (1) and (2) we get '7' is a common factor of p and q, which is contradicting our assumption that p and q are co-prime.
 is an irrational number.
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Question (48):
Show that  is an irrational number by contradiction method.
[2 Marks] |
Answer:
Show that  is an irrational number.
Let us assume that  a rational number
Let 
Here y is rational, 3 is rational
 is rational which implies  is rational. But we know that  is irrational. So our assumption is wrong.
 is an irrational number.
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Question (49):
Without actual division, say whether the following are terminating decimal or non terminating repeating decimals.
(i) , (ii)  , (iii)  , (iv)  , (v) 
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Answer:
(i)  . Here factors of 80 are in the form 24 x 5. (2n 5m). So it is terminating decimal.
(ii)  which when simplifies becomes  which is a terminating decimal.
(iii)  ; factors of 250 = 2 x 53 ….  Terminating decimal
(iv)  ; factors of 17, 1 and 17 are non terminating repeating decimal
(v)  ; factors of 15 = 3 x 5: '3' is a factor, so it is non terminating repeating decimal.
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