Consider LHS, A U B = { x / { x / x A or x B } ε ε = { x / x ¢ A or x ¢ B } = { x / x ¢ (A B) } ∩ = A B ∩ 6. In a survey of 260 college students, the following data were obtained: 64 had taken a mathematics course, 94 had taken a computer science course, 58 had taken a business course, 28 had taken both a mathematics and a business course, 26 had taken both a mathematics and a computer science course, 22 had taken both a computer science and a business course, and 14 had taken all three types of courses. (8m) How many of these students had taken none of the three courses? How many had taken only a computer science courses jun 2013 Solution: Given: U=260 |A|=64 |B|=94 |C|=58 |A C| = 28 |A B| = 26 |B C| = 22 |A B C| = ∩ ∩ ∩ ∩ ∩ 14 |A U B U C| =? |B1| =? |A U B U C| = A + B + C |A B| |B C| |A C| + |A B C| – ∩ – ∩ – ∩ ∩ ∩ = 64 + 94 + 58 26 22 28 + 14 – – – = 154 iii. |A U B U C| = U – AUBU C| = 260 154 = 106 iv. |B1| = |B| - |B C| - B – ∩ A| + |A B C| ∩ ∩ ∩ = 94 22 26 + 14 – – = 60 7. For any two sets A and B, prove the following (2m) jun 2013 A (A B) = A B – – ∩ Solution: A (A B) = A (A B1) [i.e A B = A B1] – – – ∩ – ∩ = A (A B1)1 [i.e A B = A B1] ∩ ∩ – ∩ = A (A1 U B) [Demorgan’s Law] ∩
Consider LHS, A U B = { x / { x / x ε A or x ε B } = { x / x ¢ A or x ¢ B } = { x / x ¢ (A ∩ B) } = A ∩ B 6. In a survey of 260 college students, the following data were obtained: 64 had taken a mathematics course, 94 had taken a computer science course, 58 had taken a business course, 28 had taken both a mathematics and a business course, 26 had taken both a mathematics and a computer science course, 22 had taken both a computer science and a business course, and 14 had taken all three types of courses. (8m) How many of these students had taken none of the three courses? How many had taken only a computer science courses jun 2013 Solution: Given: U=260 |A|=64 |B|=94 |C|=58 |A ∩ C| = 28 |A ∩ B| = 26 |B ∩ C| = 22 |A ∩ B ∩ C| = 14 |A U B U C| =? |B1| =? |A U B U C| = A + B + C – |A ∩ B| – |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C| = 64 + 94 + 58 – 26 – 22 – 28 + 14 = 154 iii. |A U B U C| = U – AUBU C| = 260 – 154 = 106 iv. |B1| = |B| - |B ∩ C| - B ∩ A| + |A ∩ B ∩ C| = 94 – 22 – 26 + 14 = 60 7. For any two sets A and B, prove the following (2m) jun 2013 A – (A – B) = A ∩ B Solution: A – (A – B) = A – (A ∩ B1) [i.e A – B = A ∩ B1] = A ∩ (A ∩ B1)1 [i.e A – B = A ∩ B1] = A ∩ (A1 U B) [Demorgan’s Law]
DMS-QUESTIONS-WITH-SOLUTION.docx