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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If A and B are two sets such that n (A ∪ B) = 50, n (A) = 28 and n (B) = 32, find n (A ∩ B). |
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Answer» Here, we have, n (A ∪ B) = 50 n (A) = 28 n (B) = 32 As we know that, n (A ∪ B) = n (A) + n (B) – n (A ∩ B) On substituting the values we get 50 = 28 + 32 – n (A ∩ B) 50 = 60 – n (A ∩ B) –10 = – n (A ∩ B) ∴ n (A ∩ B) = 10 |
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| 2. |
The solution of `8x = 6 (mod 14)` isA. [8],[6]B. [8],[14]C. [6],[13]D. [8],[14],[16] |
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Answer» Correct Answer - C `lambda=(8x-6)/(14), "where "lambda in I_(+)` `therefore 8x=14lambda+6impliesx=(14lambda+6)/(8)` `implies x=(7lambda+3)/(4)=lambda+(3)/(4)(lambda+1),"where "lambda in I` and here greatest common divisor of 8 and 14 is 2, so there are two required solutions. for `lambda = 3 and lambda = 7, x = 6, 13` or x = [6][13] |
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| 3. |
Find congruent solutions of `155-=7` (mod 4). |
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Answer» Since, `((155-7)/(4)=(148)/(4)=37)` and `a=155, b=7, m=4` `therefore lambda=(a-b)/(4)=(155-7)/(4)=(148)/(4)` [here, a=155, b=7] = 37 (integer) |
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| 4. |
Find all congruent solutions of `8x -= 6` (mod 14). |
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Answer» Given, `8x -= 6` (mod 14) `therefore lambda=(8x-6)/(14)`, where `lambdainI_(+)` `therefore 8x=14lambda+6` `implies x=(14lambda+6)/(8)` `implies x=(7lambda+3)/(4)` `=(4lambda+3(lambda+1))/(4)` `x=lambda+(3)/(4)(lambda+1)`,where `lambdainI_(+)` and here greatest common divisor of 8 and 14 is 2, so there are two required solutions. For `lambda = 3` and 7, x = 6 and 13. |
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| 5. |
Solve for X, if IX S 1. Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. Find the value of m and n. respectively. can be the (f A and B be two sets containing 3 and 6 elements What |
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Answer» Let A sn E are the sets So, the number of subsets of A =`2^m` number of subsets of B=`2^n` `2^m-2^n=56` `2^n(2^m/2^n-1)=56` `2^n(2^(m-n)-1)=8*7` `2^n(2^(m-n)-1)=2^3(2^3-1)` comparing on both sides n=3 m-n=3` `m=3+3=6`. |
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| 6. |
Two finite sets have m and n elements. The total number of subsets of first set is 56 more than the total number of subsets of the second set. Find the value of m and n. |
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Answer» Let A and B be two sets having m and n numbers of elements respectively. Number of subsets of A = 2m Number of subsets of B = 2n According to questions, 2m - 2n = 56 = × 7 2n (2m - n - 1) = 23 (23 − 1) Therefore, = 3 and m – n = 3 ∴ m = 6 |
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| 7. |
Two finite sets have m and n elements respectively. The total number of subsets of first set is 56 more than the total number of subsets of the second set. The values of m and n respectively are. (A) 7, 6 (B) 5, 1 (C) 6, 3 (D) 8, 7 |
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Answer» The correct answer is (C). Since, let A and B be such sets, i.e., n (A) = m, n (B) = n So n (P(A)) = 2m , n (P(B)) = 2n Thus n (P(A)) – n (P(B)) = 56, i.e., 2m – 2n = 56 ⇒ 2n (2m – n – 1) = 23 7 ⇒ n = 3 , 2m – n – 1 = 7 ⇒ m = 6 |
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| 8. |
Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The value of m and n isA. 7, 6B. 6, 3C. 5, 1D. 8, 7 |
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Answer» Correct Answer - B Since, `2^(m)-2^(n)=56=8xx7=2^(3)xx7` `implies 2^(n)(2^(m-n)-1)=2^(3)xx7` `implies n=3and2^(m-n)=8=2^(3)impliesn=3andm-n=3` implies n = 3 and m - 3 = 3 implies n = 3 and m = 6 |
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| 9. |
If U = {2, 3, 5, 7, 9} is the universal set and A = {3, 7}, B = {2, 5, 7, 9}, then prove that:(i) (A ∪ B)’ = A’ ∩ B’(ii) (A ∩ B)’ = A’ ∪ B’ |
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Answer» (i) (A ∪ B)’ = A’ ∩ B’ Now, firstly let us consider the LHS A ∪ B = {x: x ∈ A or x ∈ B} = {2, 3, 5, 7, 9} (A∪B)’ means Complement of (A∪B) with respect to universal set U. Therefore, (A∪B)’ = U – (A∪B)’ U – (A∪B)’ is defined as {x ∈ U: x ∉ (A∪B)’} U = {2, 3, 5, 7, 9} (A∪B)’ = {2, 3, 5, 7, 9} U – (A∪B)’ = ϕ Then, RHS A’ means Complement of A with respect to universal set U. Therefore, A’ = U – A (U – A) is defined as {x ∈ U: x ∉ A} U = {2, 3, 5, 7, 9} A = {3, 7} A’ = U – A = {2, 5, 9} B’ means Complement of B with respect to universal set U. Therefore, B’ = U – B (U – B) is defined as {x ∈ U: x ∉ B} U = {2, 3, 5, 7, 9} B = {2, 5, 7, 9} B’ = U – B = {3} A’ ∩ B’ = {x: x ∈ A’ and x ∈ C’}. = ϕ ∴ LHS = RHS Thus verified. (ii) (A ∩ B)’ = A’ ∪ B’ Now, firstly let us consider the LHS (A ∩ B)’ (A ∩ B) = {x: x ∈ A and x ∈ B}. = {7} (A∩B)’ means Complement of (A ∩ B) with respect to universal set U. Therefore, (A∩B)’ = U – (A ∩ B) U – (A ∩ B) is defined as {x ∈ U: x ∉ (A ∩ B)’} U = {2, 3, 5, 7, 9} (A ∩ B) = {7} U – (A ∩ B) = {2, 3, 5, 9} (A ∩ B)’ = {2, 3, 5, 9} Then, RHS A’ means Complement of A with respect to universal set U. Therefore, A’ = U – A (U – A) is defined as {x ∈ U: x ∉ A} U = {2, 3, 5, 7, 9} A = {3, 7} A’ = U – A = {2, 5, 9} B’ means Complement of B with respect to universal set U. Therefore, B’ = U – B (U – B) is defined as {x ∈ U: x ∉ B} U = {2, 3, 5, 7, 9} B = {2, 5, 7, 9} B’ = U – B = {3} A’ ∪ B’ = {x: x ∈ A or x ∈ B} = {2, 3, 5, 9} ∴ LHS = RHS Thus verified. |
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| 10. |
Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X,show that A = B.(Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law) |
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Answer» Let A and B be two sets such that A ∩ X = B ∩ X = f and A ∪ X = B ∪ X for |
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| 11. |
Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n are (a) 7, 6 (b) 6, 3 (c) 5, 1 (d) 8, 7 |
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Answer» (b) 6, 3 ∵ The number of subsets of a set containing p elements = 2p, Here, 2m – 2n = 56 The values of m and n satisfying the given equations from the given options are 6, 3. 26 – 23 = 64 – 8 = 56. |
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| 12. |
Match List-I with List-II and select the correct answer using the codes given below for the lists:List - IList - II(A) (E – A) ∪ (Ε – A′)1. ϕ(Β) (Ε – [(Α ∪ A′) – (Α ∩ A′)]2. A(C) (Ε ∩ (Α – A′) ∪ Α3. A′(D) [(E – ϕ) ∪ (ϕ – E)] – A4. ΕHere A′ is the complement set of A, E is the universal set and ϕ is an empty set. Codes: ABCD(a)4123(b)4321(c)2341(d)2143 |
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Answer» (a) A- 4, B- 1, C- 2, D- 3 (A) (E – A) ∪ (E – A′) = A′ ∪ A = E (B) E – {(A ∪ A′) – (A ∩ A′)} = E – { E – ϕ } = E – E + ϕ = ϕ (C) {E ∩ (A – A′) } ∪ A = {E ∩ A} ∪ A = A ∪ A = A (D) {(E – ϕ) ∪ (ϕ – E)} – A = {E ∪ ϕ} – A = E – A = A′ |
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| 13. |
U is a universal set and n (U) = 160. A, B and C are subset of U. If n(A) = 50, n(B) = 70, n(B ∪ C) = ϕ , n(B ∩ C) = 15 and A ∪ B ∪ C = U, then n (C) equals. (a) 40 (b) 50 (c) 55 (d) 60 |
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Answer» (c) 55 A ∪ B ∪ C = U and A ∩ (B ∪ C) = ϕ ⇒ (B ∪ C) = A′= U – A ⇒ n (B ∪ C) = n (U) – n (A) = 160 – 50 = 110 Now, n (B ∪ C) = n (B ) + n (C) – n (B ∩ C) ⇒ 110 = 70 + n (C) – 15 ⇒ n (C) = 110 + 15 – 70 = 55. |
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| 14. |
If ξ is the universal set and P is a subset of ξ, then what is P ∩ { (P– ξ) ∪ (ξ– P)} equal to(a) ϕ (b) P′ (c) ξ (d) P |
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Answer» (a) ϕ Given set = P ∩ {( P – ξ) ∪ ( ξ – P)} = P ∩ {ϕ ∪ P′} = P ∩ P′ = ϕ |
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| 15. |
If `X={8^n-7n-1 | n in N}` and `Y={49n-49| n in N}`. ThenA. `X sub Y`B. `Y sub X`C. `X = Y`D. `X nn Y = phi` |
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Answer» Correct Answer - A `X = (8^(n) - 7 n - 1| n in N } = { 0 , 49, 490, "….."}` `Y = {49n - 49 | n in N} = {0, 49, 98, 147, "….."}` Clearly, every elements of X is in Y but every element of Y is not in X. `:. X sub Y` |
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| 16. |
If X = {8n – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N}. ThenA. X ⊂ YB. Y ⊂ X C. X = YD. X ∩ Y = ϕ |
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Answer» A. X ⊂ Y Given: X = {8n – 7n – 1 | n ∈ N} and Y = {49n – 49 | n ∈ N} X = {8n – 7n – 1 | n ∈ N} On putting n = 1, 2, 3…. X = {0, 49, 490………} Y = {49n – 49 | n ∈ N} On putting n = 1, 2, 3, 4,….,11,… Y = {0, 49, 98, 147,………, 490………} Clearly, X ⊂ Y |
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| 17. |
Match each of the set on the left in the roster form with the same set on the right described in set-builder form:(I) {1, 2, 3, 6}(A) {X: X IS A PRIME NUMBER AND A DIVISOR OF 6}(II) {2, 3}(B) {X: X IS AN ODD NATURAL NUMBER LESS THAN 10}(III) {M, A,T, H, E, I,C, S}(C) {X: X IS NATURAL NUMBER AND DIVISOR OF 6}(IV) {1, 3, 5, 7, 9}(D) {X: X IS A LETTER OF THE WORD MATHEMATICS} |
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Answer» (i) All the elements of this set are natural numbers as well as the divisors of 6. Therefore, (i) matches with (c). (ii)It can be seen that 2 and 3 are prime numbers. They are also the divisors of 6. Therefore, (ii) matches with (a). (iii) All the elements of this set are letters of the word MATHEMATICS. Therefore, (iii) matches with (d). (iv) All the elements of this set are odd natural numbers less than 10. Therefore, (iv) matches with (b). |
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| 18. |
Match each of the set on the left in the roster form with the same set on the right described in set-builder form: |
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Answer» (i) All the elements of this set are natural numbers as well as the divisors of 6. |
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| 19. |
Match each of the sets on the left in the roster form with the same set on the right described in the set builder form: i.{A,P,L,E} i. {x:x + 5 = 5, x z} ii.{5,- 5} ii. {x:x is a prime natural number and a divisor of 10} iii. {0} iii. {x:x is a letter of the word “RAJASTHAN”} iv. {1,2,5,10} iv. {x:x is a natural and divisor of 10} v. {A,H,j,R,S,T,N} v. {x:x2 – 25 = 0} vi.{2,5} vi. {x:x is a letter of word “APPLE”} |
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Answer» When we are dealing with the set builder - roster form match the following it is always better to go in the direction in which we need to convert set builder to roster rather than roster to set builder. For i. {x:x + 5 = 5, x z} X is such that x+5 = 5 ⇒ x = 5 - 5 ⇒ x = 0 Roster form = {0} i.e third option For ii. {x:x is a prime natural number and a divisor of 10} All natural numbers that are divisor of 10 are = 1,2,5,10 Numbers that are prime are = 2,5 Roster form = {2,5} i.e. sixth option. For iii. {x:x is a letter of the word “RAJASTHAN”} All letters means no letter should be repeated RAJASTHAN = R,A,J,S,T,H,N ∴ Roster form will be {A,H,J,R,S,T,N} i.e. fifth option For iv. {x:x is a natural and divisor of 10} All natural numbers that are divisor of 10 are = 1,2,5,10. Roster form= {1,2,5,10} i.e. Fourth option. For v. {x:x2 – 25 = 0} X2-25 = 0 X =√25 X = ± 5. ∴ Roster form will be {5,- 5} i.e. second option for vi. {x:x is a letter of word “APPLE”} All letters means no letter should be repeated APPLE = A,P,L,E ∴ Roster form will be {A,E,L,P} i.e. first option. |
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| 20. |
`R_(1)` on Z defined by `(a,b)inR_(1) " iff "|a-b|le7, R_(2)` on Q defined by `(a,b)inR_(2) " iff "ab=4 and R_(3)` on R defined by `(a, b)inR_(3)" iff "a^(2)-4ab+3ab^(2)=0` Relation `R_(2)` isA. reflexive and symmetricB. symmetric and transitiveC. reflexive and transitiveD. equivalence |
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Answer» Correct Answer - A We have, (a, b) `in R_(1)` iff |a-b| `le 7`, where `a, b in z` Reflexivity Let `a in z` implies a - a = 0 `implies |a - a| le 7` `implies 0 le 7` `implies (a, a) in R_(1)` `therefore` The relation `R_(1)` is reflexive. Symmetry `(a, b) in R_(1)` `implies |a-b|le7implies|-(b-a)|le7` `implies |b-a|le7implies(b, a)in R_(1)` `therefore` The relation `R_(1)` is symmetric. Transitivity We have (2, 6), (6, 10) `in R_(1)` because `|2-6|=4le7 and |6-10|=4le7` Also, |2-10| = `8cancelle7` `therefore (2, 10) cancelinR_(1)` Hence, the relation `R_(1)` is not transitive. |
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| 21. |
`R_(1)` on Z defined by `(a,b)inR_(1) " iff "|a-b|le7, R_(2)` on Q defined by `(a,b)inR_(2) " iff "ab=4 and R_(3)` on R defined by `(a, b)inR_(3)" iff "a^(2)-4ab+3b^(2)=0` Relation `R_(3)` isA. reflexiveB. symmetricC. transitiveD. equivalence |
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Answer» Correct Answer - A We have, (a, b) `in R_(3)` iff `a^(2) - 4ab + 3b^(2) = 0` where a, b `in R` Reflexivity `therefore a^(2)-4a.a+3d^(2)=4a^(2)-4a^(2)=0` `therefore (a, a) in R_(3)` `therefore` The relation `R_(3)` is reflexive. Symmetry `(a, b) in R_(3)` `implies a^(2)-4ab+3b^(2)=0`, we get a = b and a = 3b and `(b, a) in R_(3)` implies `b^(2) - 4ab + 3a^(2) = 0` we get b = a and b = 3a `therefore (a, b) in R_(3) cancelimplies(b, a)in R_(3)` `therefore` The relation `R_(3)` is not symmetric. Transitivity We have `(3, 1), (1, (1)/(3))inR_(3)` because `(3)^(2)-4(3)(1)+3(1)^(2)=9-12+3=0` and `(1)^(2)-4(1)((1)/(3))+3((1)/(3))^(2)=1-(4)/(3)+(1)/(3)=0` Also, `(3, (1)/(3))cancelinR_(3)`, because `(3)^(2)-4.(3)((1)/(3))+3((1)/(3))^(2)=9-4+(1)/(3)=(16)/(3)ne0` `therefore` The relation `R_(3)` is not transitive. |
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| 22. |
Let `R` be the relation on the set R of all real numbers defined by a `R b` Iff `|a-b| le1.` Then `R` isA. reflexive and symmetricB. symmetric onlyC. transitive onlyD. anti-symmetric only |
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Answer» Correct Answer - A `because|a-a|=0lt1impliesaRa,AAainR` `therefore` R is reflexive Again, aRb implies `|a-b|le1` `implies |b-a|le1impliesbRa` `therefore` R is symmetric. Again, 1 R 2 and 2R1 but `2 ne 1` `therefore` R is not anti-symmetric Further, 1R2 and 2R3 but `1 cancelR 3" "[because|1-3|=2gt1]` `therefore` R is not transitive. |
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| 23. |
Statement-1: The relation R on the set `N xx N` defined by (a, b) R (c, d) `iff` a+d = b+c for all a, b, c, d `in` N is an equivalence relation. Statement-2: The intersection of two equivalence relations on a set A is an equivalence relation. |
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Answer» (i) (a,b) R (a,b) implies a + b = b + a `therefore` R is reflexive. (ii) (a, b) R (c, d) implies a + d = b + c `implies c+b=d+aimplies(c,d)R(a,b)` `therefore R` is symmetric. (iii) (a, b) R (c, d) and (c, d) R (e, f) implies a + d = b + c and c + f = d + e `impliesa+d+c+f=b+c+d+e` `impliesa+f=b+eimplies(a,b)R(e,f)` `therefore R` is transitive. Thus, R is an equivalence relation on `NxxN`. |
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| 24. |
Prove that the relation R defined on the set N of natural numbers by xRy `iff 2x^(2) - 3xy + y^(2) = 0` is not symmetric but it is reflexive. |
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Answer» (i) `2x^(2)-3x.x+x^(2)=0,AAx inN`. `therefore xRx,AAx inN`, i.e. R is reflexive. (ii) For `x=1,y=2,2x^(2)-3xy+y^(2)=0` `therefore 1R2` But `2.2^(2)-3.2.1+1^(2)=3ne0` So, 2 is not related to 1 i.e., `2cancelR1` `therefore R` is not symmetric. |
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| 25. |
Let L be the set of all straight lines in the Euclidean plane. Two lines `l_(1)` and `l_(2)` are said to be related by the relation R iff `l_(1)` is parallel to `l_(2)`. Then, the relation R is notA. reflexiveB. symmetricC. transitiveD. equivalence |
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Answer» Correct Answer - A::B::C::D Relation R on the set of all straight lines in the plane is of parallel line. A line is parallel to itself. So, R is reflexive. If `l_(1)` is parallel to `l_(2)`, then `l_(2)` is parallel to `l_(1)`. `therefore` R is symmetric relation. `[l_(1), l_(2) in L]` Let `l_(1), l_(2), l_(3) in L` `l_(1)` is parallel to `l_(2)` and `l_(2)` is parallel to `l_(3)`. Then, `l_(1)` is parallel to `l_(3)` `therefore R` is transitive relation. So, R is equivalence relation. |
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| 26. |
Let w denote the words in the english dictionary. Define the relation R by: R = `{(x,y) in W xx W` | words x and y have at least one letter in common}. Then R is: (1) reflexive, symmetric and not transitive (2) reflexive, symmetric and transitive (3) reflexive, not symmetric and transitive (4) not reflexive, symmetric and transitive |
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Answer» Given relation `R` such that `R = {(x, y) in W xx W |` the word x and y have at least one letter in common`}`, where W denotes set of words in English dictionary. Clearly `(x ,x ) in R` for all `x in W`. `:.` `(x, x)` has every letter common, therefore `R` is reflexive. Let `(x, y) in R` then `(y, x) in R` as `y` and `x` have at least one letter in common, this implies, `R` is symmetric. Let `(x,y) in R` and `(y,z) in R`, then it is not necessary that both `x` and `z` have at least one letter in common. Let `x = ABC, y = CDE, z = EFG` As we can see from this example `x` and `z` have no common letters. `:. (x,z) !in R`. So, `R` is not transitive. `:.` option `(1)` is correct. |
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| 27. |
If AB = CD, BC ∪ AD = Φ, then ABCD is A) Rhombus B) Square C) Parallelogram D) Trapezium |
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Answer» Correct option is C) Parallelogram |
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| 28. |
The length of side AB of quadrilateral ABCD having CD = 4, BC + AD = 10 units and inradius as a finite nonzero number will be: |
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Answer» Let the incircle touches the quadrilateral `ABCD` at points `P,Q,R and S`. Please refer to video for the diagram. We know tangents from a common point outside a circle is alays equal. `:. DP = DS->(1)` `CS = CR = CD-DS = 4-DS->(2)` (As CD = 4) `BQ = BR->(3)` `AQ = AP->(4)` Also, we are given, `BC+AD = 10` From `(1),(2),(3),(4)` `=>BR+CR+AP+DP = 10` `=>BQ+4-DS+AQ+DS = 10` `=>AQ+BQ = 6` `=>AB = 6` |
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| 29. |
Two finite sets have m and n elements. The number of elements in the power set of first set is 48 more than the total number of elements in power in power set of the second set. Then, the values of m and n are: A. 7, 6 B. 6, 8 C. 6, 4 D. 7, 4 |
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Answer» Let A and B be the set which contain m and n elements respectively. Then n(P(A) ) = 2m and n(P(B) ) = 2n Also given that,n(P(A) ) = n(P(B) ) + 48 ⇒ 2m = 2n + 48 Above equation is only true when m = 6 and n = 4 |
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| 30. |
Let A and B be two sets such that : n (A) = 20, n (A∪ B) = 42 and n (A ∩ B) = 4. Find n (B – A) |
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Answer» n (B – A) = n (B) – n (A ∩ B) = 26 – 4 = 22 |
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| 31. |
Show that A∩B = A∩C need not imply B = C |
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Answer» Let A = {1, 2}, B = {1, 3}, C = {1, 4}. A ∩ B = {1} and A ∩ C = {1} A ∩ B = A ∩ C But B ≠ C. |
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| 32. |
Using properties of sets, show that(i) A∪(A∩B) = A(ii) A∩(A∪B) = A |
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Answer» (i) A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) ∵ distributive law = A ∩ (A ∪ B) = A (ii) A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) ∵ distributive law = A ∪ (A ∩ B) = A. |
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| 33. |
Suppose that A denotes the collection of all complex numbers whose square is a negative real number. Which one of the following statements is correct ?A. `A sube R`B. `A supe R`C. `A={x+iy|x^(2)inR|x, yinR}`D. `A={iy|yinR}` |
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Answer» Correct Answer - D If A denotes the collection of all complex number whose square is a negative number, then Square of a complex number is a negative real number only if has no real part and has only imaginary part. Hence, A={iy : `y in R`) |
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| 34. |
If `0 lt a lt 1,` the value of `log_10 a` is negative. This is justified byA. Negative power of 10 is less than 1B. Negative power of 10 is between 0 and 1C. Negative power of 10 is positiveD. Negative power of 10 is negative |
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Answer» Correct Answer - B `0ltalt1` `"Let "log_(10)a=-x` `impliesa=10^(-x)` `10^(-x)` can have values only between 0 and 1 |
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| 35. |
The sum of all the solutions to the equation `2log_10x-log_10(2x-75)=2`a. 30b. 350c. 75d. 200 |
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Answer» Given equation is , `2log_10x-log_10(2x-75) = 2` Here, solution should satisfy two conditions. (i) `x gt 0` and (ii) `2x-75 gt 0 => x gt 75/2` Now, we will solve the given equation, `2log_10x-log_10(2x-75) = 2` `=>log_10x^2-log_10(2x-75) = 2` `=>log_10(x^2/(2x-75)) = 2` `=>(x^2/(2x-75)) = 10^2` `=>(x^2/(2x-75)) = 100` `=> x^2-200x+7500 = 0` `=>x^2-150x-50x+7500=0` `=>(x-150)(x-50) = 0` So, `x = 150` and `x = 50` As both these values satisfying the above conditions (i) and (ii), both are the solutions. `:. ` Sum of all solutions `= 150+50 = 200` |
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| 36. |
In a group of 25 students, 13 can speak English, 12 can speak Hindi and 6 speak neither. How many can speak (i) Both English and Hindi ? (ii) Only Hindi ? (iii) Exactly one of the two languages ? |
| Answer» Correct Answer - (i) 6 , (ii) 6 , (iii) 13 | |
| 37. |
If `n(A) = 15` and `n(B)= 17`, then find `n(A uu B uu C)` from the following Venn diagram. (The number in each region represents the number of elements of that region) |
| Answer» Correct Answer - 37 | |
| 38. |
If a set has 512 non-empty proper subsets, then find the cardinal number of the set. |
| Answer» Correct Answer - 9 | |
| 39. |
If `A {2,{3,4},5{6,7,8}}`. Then write all the non-empty proper subsets of A. |
| Answer» Correct Answer - `{2},{{3,4}},{5},{{6,7,8}},{2,{3,4}},{2,5},{2,{6,7,8}},{{3,4}5},{{3,4}},{6,7,8}},{2,{3,4},5},{2,{3,4},{6,7,8}},{{3,4},5{6,7,8}}` i.e, 14 | |
| 40. |
Let F1 be the set of parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane. Then F1 may be equal to(A) F2 ∩ F3 (B) F3 ∩ F4(C) F2 ∪ F5 (D) F2 ∪ F3 ∪ F4 ∪ F1 |
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Answer» Answer is D Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram. F1 = F2 ∪ F3 ∪ F4 ∪ F1 |
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| 41. |
Let `F_(1)` be the set of parallelograms, `F_(2)` the set of rectangle , `F_(3)` the set of rhombuses, `F_(4)` the set of squares and `F_(5)` the set of trapeziums in a plane. Then, `F_(1)` may be equal toA. `F_(2) nn F_(3)`B. `F_(3) nn F_(4)`C. `F_(2) uu F_(5)`D. `F_(2) uu F_(3) uu F_(4) uu F_(1)` |
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Answer» Correct Answer - D Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram So, `F_(1)` is either of `F_(1), F_(2), F_(3)` and `F_(4)`. `:. F_(1) = F_(2) uu F_(3) uu F_(4) uu F_(1)` |
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| 42. |
If a set contains n elements, then write the number of elements in its power set. |
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Answer» Let A be the set with n elements. Then power set of A contains all the subsets of A. Each member of A has two possibilities either present or absent. ⇒ Total possible subsets of A are 2×2×2×…n times= 2n ∴ number of elements in power set of A are 2n. 2n elements are present in power set. |
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| 43. |
Write the number of elements in the power set of null set. |
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Answer» Power set contains all the subsets of a set. Null set has no member in it and each set is a subset of itself (i.e. null set is the only subset of null set) ∴ number of elements in power set of null set is1. if null set, then we consider it a subset of itself as it contains no elements, so no. elements in its power set would be.If you think by formula, take n=0; 2^n=2^(0)=1 element will be present |
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| 44. |
A set of points is such that each point is three times as far away from the y-axis as it is from the point (4,0).Then locus of the points is: |
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Answer» Let `(h,k)` is the point that is 3 times as far away from `y`-axis as it is from the point (4,0). Then, `h = 3sqrt((h-4)^2+k^2)` `=>h^2 = 9(h^2+16-8h)+k^2` `=>8h^2-72h+k^2+144 - 0` `=>8(h^2-9h+81/4-81/4)+k^2 = -144` `=>8(h-9/2)^2+k^2 = 162-144` `=>8(h-9/2)^2+k^2 = 18` `=>(h-9/2)^2/(9/4)+k^2/18 = 1` If, we replace `(h,k)` by `(x,y)`, we can see that this is an equation for ellipse. |
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| 45. |
Prove that:`sin^2(pi/18)+sin^2(pi/9)+sin^2([7pi]/18)+sin^2([4pi]/9)=2` |
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Answer» `L.H.S. = sin^2(pi/18)+sin^2(pi/9)+sin^2((7pi)/18)+sin^2((4pi)/9)` Now, As `sin x = cos((pi/2)-x)` `:. sin^2((7pi)/18) = cos^2(pi/2-(7pi)/18) = cos^2(pi/9)` `:.sin^2((4pi)/9) = cos^2(pi/2-(4pi)/9) = cos^2(pi/18)` So, putting these values, `L.H.S. = sin^2(pi/18)+sin^2(pi/9)+cos^2(pi/9)+cos^2(pi/18)` `=(sin^2(pi/18)+cos^2(pi/18))+(sin^2(pi/9)+cos^2(pi/9))` `1+1 = 2` |
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| 46. |
Prove that `4cos12^@cos48^@cos72^@=cos36^@` |
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Answer» Here, we will use the following rules, `2cosAcosB = cos(A+B)+cos(A-B)` `cosA+cos B = 2cos((A+B)/2)cos((A-B)/2)` `L.H.S. = 4cos12^@cos48^@cos72^@` `=2cos72^@(cos(48+12)^@+cos(48-12)^@)` `=2cos72^@(cos60^@+cos36^@)` `=2cos72^@(1/2+cos36^@)` `=cos72^@ +2cos72^@cos36^@` `=cos72^@ +cos(72+36)^@+cos(72-36)^@` `=cos72^@+cos108^@+cos36^@` `=2cos((108+72)/2)^@cos((108-72)/2)^@+cos36^@` `=2cos90^@cos36^@+cos36^@` `=0+cos36^@ = cos36^@=R.H.S.` |
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| 47. |
Find the value of x and y if (i) `3x+(2x-y)i=6-3i` (ii) ` 3x + 5yi = 5i` |
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Answer» (i) `3x+(2x-y)i = 6-3i` Comparing real and imaginary part both sides, `=>3x = 6 and 2x-y = -3` `=> x = 2 and 2(2)-y = -3` `=>x = 2 and y = 7` (ii) `3x+5yi = 5i` Comparing real and imaginary part both sides, `=>3x = 0 and 5y = 5` `=> x= 0 and y = 1` |
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| 48. |
If A and B be two sets containing 6 and 3 elements respectively, what can be the minimum number of elements in `AuuB` ? Also, find the maximum number of elements in `AuuB`. |
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Answer» We have, `n(AuuB)=n(A)+n(B)-n(AnnB),n(AuuB)` is minimum or maximum according as `n(AnnB)` is maximum or minimum, respectively. Case I If `n(AnnB)` is minimum i.e., `n(AnnB)` = 0 such that A = {a, b, c, d, e, f} and B = {g, h, i} `therefore n(AuuB)=n(A)+n(B)=6+3=9` Case II If `n(AnnB)` is maximum i.e., `n(AnnB) = 3`, such that A = {a, b, c, d, e, f} and B = {d, a, c} `therefore n(AuuB)=n(A)+n(B)-n(AnnB)=6+3-3=6` |
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| 49. |
The number of elements of the power set of a set containing n elements isA. `2^(n-1)`B. `2^(n)`C. `2^(n)-1`D. `2^(n+1)` |
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Answer» Correct Answer - B Let set A contains n elements. Power set of a is the set of all subsets. `therefore` Number of subsets of `A=.^(n)C_(o)+.^(n)C_(1)+.^(n)C_(2)+...+.^(n)C_(n)=2^(n)` `therefore` Power set of A contains `2^(n)` elements. |
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| 50. |
If A and B two sets containing 2 elements and 4 elements, respectively. Then, the number of subsets of `A xx B` having 3 or more elements, isA. 220B. 219C. 211D. 256 |
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Answer» Correct Answer - B `because A xx B` has 8 elements. `therefore` Number of subsets = `2^(8) = 256` Number of subsets with zero element = `.^(8)C_(0) = 1` Number of subsets with one element = `.^(8)C_(1) = 8` Number of subsets with one elements = `.^(8)C_(2) = 28` Hence, Number of subsets of `AxxB` having 3 or more elements `=256-(1+8+28)=256-37=219` |
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