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If Na = {an : N € N }, Then N5 ∩ N7 =
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Given N(U) = 20, N(A) = 12, N(B) = 9, N(A∩B) = 4, Where U Is The Universal Set A And B Are Subsets Of U, Then N((A∪B)^c) =
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If Two Sets A And B Are Having 99 Elements In Common, Then The Number Of Elements Common To Each Sets A*B And B*A Are
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In A City 20 Percent Of The Population Travels By Car, 50 Percent Travels By Bus And 10 Percent Travels By Both Car And Bus.
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For Any Two Sets A And B, A - (A - B) Equals
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A Class Has 175 Students. The Following Data Shows The Number Of Students Obtaining One Or More Subjects.
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The Number Of Non-empty Subsets Of The Set { 1, 2, 3, 4} Is
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Given The Sets A = {1,2,3}, B = {3,4}, C = {4, 5, 6}, Then
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The Set A = { X : X € R,x^2 = 16 And 2x = 6 } Equals
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If A, B And C Are Non-empty Sets, Then (A-B) ∪ (B-A) Equals
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Out Of 800 Boys In A School, 224 Played Cricket, 240 Played Hockey And 336 Played Basketball. Of The Total,
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In A Class Of 100 Students, 55 Students Have Passed In Mathematics And 67 Students Have Passed In Physics. Then
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If The Sets A And B Are Defined As A={(x,y) : Y = E^x , X € R } ; B = { (x,y) : Y = X, X € R}, Then
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In A Class Of 55 Students, The Number Of Students Studying Different Subjects Are 23 Ij Mathematics, 24 In Physics, 19 In Chemistry
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If X = {4^n - 3 - 1 : N € N} And Y = {9 (n -1) : N € N}, Then X ∪ Y Is Equal To
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If | X^2 - 3x -1 |<| 2x^2 + 3x +4 | + | X^2 + 6x +5 |, Then Complete Solution Set Of X Is
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Number Of Integers Satisfying | X - 1 |/x +2 > 1 Is
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Solution Set Of Inequality X - √1-| X | < 0 Is
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