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Sets Practice Questions - DPP for JEE
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Page 1
1. (a) From Venn-Euler’s Diagram.
2. (d) A = {(n, 2n) : n ? N} and B = {(2n, 3n)}: n ? N
Listing few members of each set
A = {(1, 2), (2, 4), (3, 6),....}
B = {(2, 3), (4, 6), (6, 9)......}
There is no member common to both these sets, hence.
A n B = f
3. (a) bN = {bx : x ?N}
cN = {cx : x ?N}
? bN n cN = {x : x is multiple of b and c both}
= { x: x is multiple of l.c.m. of b and c }
= { x : x is multiple of b c}
[given b and c are relatively prime ? l.c.m. of b andc = bc]
? bN n cN = {bc x : x ?N} = dN (Given)
? d = bc.
4. (d)
We have to find
Now
=
Page 2
1. (a) From Venn-Euler’s Diagram.
2. (d) A = {(n, 2n) : n ? N} and B = {(2n, 3n)}: n ? N
Listing few members of each set
A = {(1, 2), (2, 4), (3, 6),....}
B = {(2, 3), (4, 6), (6, 9)......}
There is no member common to both these sets, hence.
A n B = f
3. (a) bN = {bx : x ?N}
cN = {cx : x ?N}
? bN n cN = {x : x is multiple of b and c both}
= { x: x is multiple of l.c.m. of b and c }
= { x : x is multiple of b c}
[given b and c are relatively prime ? l.c.m. of b andc = bc]
? bN n cN = {bc x : x ?N} = dN (Given)
? d = bc.
4. (d)
We have to find
Now
=
=
=
= 23 – 12 – 9 + 4 = 27 – 21 = 6
= 24 – 12 – 7 + 4 = 9
= 19 – 7 – 9 + 4 = 23 – 16 = 7
5. (c) We have
min n (A ? B) = max {n(A), n (B)} = max {3, 6} = 6
max n (A ? B) = n (A) + n (B) = 9
? 6 = n (A ? B) = 9
6. (a) See the following Venn diagram
n (I) = 29 + 23 = 52
n (F) = 100 – 52 = 48
?
?
7. (c) Suppose a ? X and a?A
? a? X ? A ? a? Y ? A
? a? Y and a?A ( Q X ?A = Y ?A)
? a?YnA ? YnA is non-empty
This contradicts that YnA = f
So, X = Y
Page 3
1. (a) From Venn-Euler’s Diagram.
2. (d) A = {(n, 2n) : n ? N} and B = {(2n, 3n)}: n ? N
Listing few members of each set
A = {(1, 2), (2, 4), (3, 6),....}
B = {(2, 3), (4, 6), (6, 9)......}
There is no member common to both these sets, hence.
A n B = f
3. (a) bN = {bx : x ?N}
cN = {cx : x ?N}
? bN n cN = {x : x is multiple of b and c both}
= { x: x is multiple of l.c.m. of b and c }
= { x : x is multiple of b c}
[given b and c are relatively prime ? l.c.m. of b andc = bc]
? bN n cN = {bc x : x ?N} = dN (Given)
? d = bc.
4. (d)
We have to find
Now
=
=
=
= 23 – 12 – 9 + 4 = 27 – 21 = 6
= 24 – 12 – 7 + 4 = 9
= 19 – 7 – 9 + 4 = 23 – 16 = 7
5. (c) We have
min n (A ? B) = max {n(A), n (B)} = max {3, 6} = 6
max n (A ? B) = n (A) + n (B) = 9
? 6 = n (A ? B) = 9
6. (a) See the following Venn diagram
n (I) = 29 + 23 = 52
n (F) = 100 – 52 = 48
?
?
7. (c) Suppose a ? X and a?A
? a? X ? A ? a? Y ? A
? a? Y and a?A ( Q X ?A = Y ?A)
? a?YnA ? YnA is non-empty
This contradicts that YnA = f
So, X = Y
8. (d) (a)
? A – B = A n B' ...(i)
x ?A and x ?B'
? x ?A' and x ?B' ? x?B' and x ?A'
? x?B'–A'
? A – B = B' – A' ...(ii)
Clearly (a) is not correct. Also from (i) (c) is not correct.
Next let x ?A – (A – B)
? x?A and x?A – B
? x?A and [x ?A or x ?B]
A – (A – B) = A n B
? [x ?A and x ?A] or [ x ?A and x ?B]
? A – (A – B) = f ? (A n B) = A n B
? (b) is also incorrect
The result (d) is correct as can be seen in the following Venn diagram
A ? B = (A – B) ? (A n B) ? (B – A)
9. (d) Let P = set of families buying A,
Q = set of families buying B
and R = set of families buying C.
similarly
, n(R) = 1,000
n(P Q) = 500, n (Q R) = 300
n (P R) = 400 and n (P Q R) = 200
(i) Number of families buying only A = n(P Q' R')
= n (P (Q R)') = n(P) –n(P (Q R))
Page 4
1. (a) From Venn-Euler’s Diagram.
2. (d) A = {(n, 2n) : n ? N} and B = {(2n, 3n)}: n ? N
Listing few members of each set
A = {(1, 2), (2, 4), (3, 6),....}
B = {(2, 3), (4, 6), (6, 9)......}
There is no member common to both these sets, hence.
A n B = f
3. (a) bN = {bx : x ?N}
cN = {cx : x ?N}
? bN n cN = {x : x is multiple of b and c both}
= { x: x is multiple of l.c.m. of b and c }
= { x : x is multiple of b c}
[given b and c are relatively prime ? l.c.m. of b andc = bc]
? bN n cN = {bc x : x ?N} = dN (Given)
? d = bc.
4. (d)
We have to find
Now
=
=
=
= 23 – 12 – 9 + 4 = 27 – 21 = 6
= 24 – 12 – 7 + 4 = 9
= 19 – 7 – 9 + 4 = 23 – 16 = 7
5. (c) We have
min n (A ? B) = max {n(A), n (B)} = max {3, 6} = 6
max n (A ? B) = n (A) + n (B) = 9
? 6 = n (A ? B) = 9
6. (a) See the following Venn diagram
n (I) = 29 + 23 = 52
n (F) = 100 – 52 = 48
?
?
7. (c) Suppose a ? X and a?A
? a? X ? A ? a? Y ? A
? a? Y and a?A ( Q X ?A = Y ?A)
? a?YnA ? YnA is non-empty
This contradicts that YnA = f
So, X = Y
8. (d) (a)
? A – B = A n B' ...(i)
x ?A and x ?B'
? x ?A' and x ?B' ? x?B' and x ?A'
? x?B'–A'
? A – B = B' – A' ...(ii)
Clearly (a) is not correct. Also from (i) (c) is not correct.
Next let x ?A – (A – B)
? x?A and x?A – B
? x?A and [x ?A or x ?B]
A – (A – B) = A n B
? [x ?A and x ?A] or [ x ?A and x ?B]
? A – (A – B) = f ? (A n B) = A n B
? (b) is also incorrect
The result (d) is correct as can be seen in the following Venn diagram
A ? B = (A – B) ? (A n B) ? (B – A)
9. (d) Let P = set of families buying A,
Q = set of families buying B
and R = set of families buying C.
similarly
, n(R) = 1,000
n(P Q) = 500, n (Q R) = 300
n (P R) = 400 and n (P Q R) = 200
(i) Number of families buying only A = n(P Q' R')
= n (P (Q R)') = n(P) –n(P (Q R))
= 4,000 – 500 – 400 + 200 = 3,300.
(ii) Number of families buying only B
[see (i)]
= 2,000 – 500 – 300 + 200 = 1,400.
(iii) Number of families buying none of A, B and
= 10,000 – [4,000 + 2,000 + 1,000 – 500 – 300 – 400 + 200]
= 10, 000 – 6,000 = 4,000.
Note : For sets A, B, we have
Replacing A by P and B by Q R, we have
n etc.
Hence all options are correct.
10. (c) n (A' n B') = n (A ? B)' = n (U) – n (A ? B)
= n (U) – [n (A) + n (B) – n (A n B)]
= 700 – [200 + 300 – 100] = 300
11. (b) C stands for set of students taking economics
a + b + c + d + e + f + g = 40; a + b + d + g = 16
b + c + e + g = 22; d + e + f + g = 26
Page 5
1. (a) From Venn-Euler’s Diagram.
2. (d) A = {(n, 2n) : n ? N} and B = {(2n, 3n)}: n ? N
Listing few members of each set
A = {(1, 2), (2, 4), (3, 6),....}
B = {(2, 3), (4, 6), (6, 9)......}
There is no member common to both these sets, hence.
A n B = f
3. (a) bN = {bx : x ?N}
cN = {cx : x ?N}
? bN n cN = {x : x is multiple of b and c both}
= { x: x is multiple of l.c.m. of b and c }
= { x : x is multiple of b c}
[given b and c are relatively prime ? l.c.m. of b andc = bc]
? bN n cN = {bc x : x ?N} = dN (Given)
? d = bc.
4. (d)
We have to find
Now
=
=
=
= 23 – 12 – 9 + 4 = 27 – 21 = 6
= 24 – 12 – 7 + 4 = 9
= 19 – 7 – 9 + 4 = 23 – 16 = 7
5. (c) We have
min n (A ? B) = max {n(A), n (B)} = max {3, 6} = 6
max n (A ? B) = n (A) + n (B) = 9
? 6 = n (A ? B) = 9
6. (a) See the following Venn diagram
n (I) = 29 + 23 = 52
n (F) = 100 – 52 = 48
?
?
7. (c) Suppose a ? X and a?A
? a? X ? A ? a? Y ? A
? a? Y and a?A ( Q X ?A = Y ?A)
? a?YnA ? YnA is non-empty
This contradicts that YnA = f
So, X = Y
8. (d) (a)
? A – B = A n B' ...(i)
x ?A and x ?B'
? x ?A' and x ?B' ? x?B' and x ?A'
? x?B'–A'
? A – B = B' – A' ...(ii)
Clearly (a) is not correct. Also from (i) (c) is not correct.
Next let x ?A – (A – B)
? x?A and x?A – B
? x?A and [x ?A or x ?B]
A – (A – B) = A n B
? [x ?A and x ?A] or [ x ?A and x ?B]
? A – (A – B) = f ? (A n B) = A n B
? (b) is also incorrect
The result (d) is correct as can be seen in the following Venn diagram
A ? B = (A – B) ? (A n B) ? (B – A)
9. (d) Let P = set of families buying A,
Q = set of families buying B
and R = set of families buying C.
similarly
, n(R) = 1,000
n(P Q) = 500, n (Q R) = 300
n (P R) = 400 and n (P Q R) = 200
(i) Number of families buying only A = n(P Q' R')
= n (P (Q R)') = n(P) –n(P (Q R))
= 4,000 – 500 – 400 + 200 = 3,300.
(ii) Number of families buying only B
[see (i)]
= 2,000 – 500 – 300 + 200 = 1,400.
(iii) Number of families buying none of A, B and
= 10,000 – [4,000 + 2,000 + 1,000 – 500 – 300 – 400 + 200]
= 10, 000 – 6,000 = 4,000.
Note : For sets A, B, we have
Replacing A by P and B by Q R, we have
n etc.
Hence all options are correct.
10. (c) n (A' n B') = n (A ? B)' = n (U) – n (A ? B)
= n (U) – [n (A) + n (B) – n (A n B)]
= 700 – [200 + 300 – 100] = 300
11. (b) C stands for set of students taking economics
a + b + c + d + e + f + g = 40; a + b + d + g = 16
b + c + e + g = 22; d + e + f + g = 26
b + g = 5; e + g = 14; g = 2
Go by backward substitution
e = 12, b = 3, d + f = 12, c + e = 17 ? c = 5;
a + d = 11
a +d + f = 18 ? f = 7 ? d = 12 – 7 = 5
12. (c) Numbers which are divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40,
45, 50, 55, 60, 65, 70, 75, 80 they are 16 in numbers. Now,
Numbers which are divisible by 7 are 7, 14, 21, 28, 35, 42, 49, 56,
63, 70, 77 they are 11 in numbers.
Also, total odd numbers = 40
Let C represents the students who opt. for cricket, F for football and H
for hockey.
? we have n(C) = 40, n(F) = 16, n(H) = 11
Now, C n F = Odd numbers which are divisible by 5.
C n H = Odd numbers which are divisible by 7.
F n H = Numbers which are divisible by both 5 and 7.
n(C n F), 8, n(C n H) = 6,
n(Fn H) = 2, n (C n F n H) = 1
We Know
n(C ? F ? H) = n(C) + n(F) + n(H)
– n(C n F) – n(C n H)
– n(F n H) + n(C n H n F)
n(C ? F ? H) = 67 – 16 + 1 = 52
? n(C' n F' n H')
= Total students – n(C ? F ? H)
n(C' n F' n H')= 80 – 52 = 28
13. (c) a + e + f + g = 23
b + d + f + g = 15
c + d + e + g = 20
f + g = 7; d + g = 5
e + g = 4
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