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Two or more sets are said to be equivalent when they contain equal elements, denoted by ≡ Original: {1,2,3,4,5}. ≡ {4,3,1,5,2}

The Venn diagram has three overlapping circles labeled P, Q, and R. The shaded region has cross-hatching and covers only the part that belongs to Q but not to P or R (the exclusive part of Q outside the overlaps with P and R). In set notation, the shaded region is \( Q \cap (P \cup R)^c \)

65 = 30 -x + x + 45 - x 65 - 75 = 10 = x ( number of students offering both subjects) Pr( \(\text{students offering both subjects}\)) = \(\frac{10}{65}\) = \(\frac{2}{13}\)

P = {Primes numbers b/w 0 and 10} = { 2, 3, 5, 7} Q = { Odd numbers b/w 0 and 10} = {1, 3, 5, 7, 9} Let W = { Union of elements in P not in Q and elements in Q not in P} = {1, 2, 9}

For a set with n elements, the total number of subsets is given by: 2\(^n\) where n = the number of members of the set = 5 so, 2\(^5\) = 32 (every set is an improper subset of itself)

144 = 42 - x + 30 - x + 34 + 4x + 3x + 6 + x + 2x 144 = 112 + 8x 8x = 144 - 112 = 32 x = 4. The number of students that registered for physics and mathematics = 4x + 2x = 4 x 4 + 2 x 4 = 24.

Let F be the set of people who can speak French and E be the set of people who can speak English. Then, n(F) = 400 n(E) = 350 n(F ∪ E) = 500 We have to find n(F ∩ E). Now, n(F ∪ E) = n(F) + n(E) – n(F ∩ E) ⇒ 500 = 400 + 350 – n(F ∩ E) ⇒ n(F ∩ E) = 750 – 500 = 250. ∴ 250 people can speak both languages.

A = { 1, 2, 3, 4, 5, 6} B = { 2, 4, 6, 8 } A - B = { 1, 3, 5 } B - A = { 8 } ∴ (A - B) ⋃ (B - A) = {1, 3, 5, 8}

Using the venn diagram above \mu = 30 n(W) = 15 n(M) = 13 \(n(W ∪ M)^1 = 6\) Let x = number of students that study both woodwork and metalwork i.e. n(W ∩ M) = x Number of students that study only woodwork,\(n(W ∩ M^1)\) = \(15 - x\) Number of students that study only metalwork, \(n(W^1 ∩ M)\) = \(13 - x \) Bringing all together, \(n(W ∩ M^1)\) +\( n(W^1 ∩ M)\) + \(n(W ∩ M)\) + \(n(W ∪ M)^1\) = \(\mu\) ∴ (15 - x) + (13 - x) + x + 6 = 30 ⇒ 34 - x = 30 ⇒ 34 - 30 = x ∴ x = 4 \(n(W ∩ M^1)\) = \(15 - 4 = 11\) ∴ The number of students that study woodwork but not metalwork is 11.

ai. To find all the elements in the set M, we have to solve the inequality. 2n - 3 \le 37 2n = 37 + 3 = 40. 2n \le 40 n \le \(\frac{40}{2}\) n \le 20 so, the set M contains all positive numbers n such that n is less than or equal to 20 Therefore, set M elements are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} (\alpha) Pr(multiple of 3) = \(\frac{multiples of 3}{ Total number of element in M}\) But the multiples of 3 in set M are = {3, 6, 9, 12, 15, 18} = 6 ஃ Pr(multiples of 3) = \(\frac{6}{20} = \frac{3}{10}\) (\beta) Pr( factors of 10) = \(\frac{№ of factors of 10}{ Total № of set M elements}\) factors of 10 in set M = { 1, 2, 5, and 10} = 4 ஃ Pr( factors of 10) = \(\frac{4}{20} = \frac{1}{5}\) bi. Let the total amount shared = y Gapson's age = \(\frac{3}{2}\times Kontor's age\). Since kontor is 20yrs old then Gapson's age = \(\frac{3}{2}\times20\) = 30years. sharing ratio = 20 : 30 = 2 : 3 total ratio = 2 + 3 = 5 Since Kontor received Le200,000 \(\frac{2}{5} \times y\) = 200,000 \(\frac{2y}{5} = 200,000\) cross multiplying, we have 2y = 5 x 200,000 = 1000,000 y = \(\frac{1000,000}{2}\) = 500,000. ஃ Total amount shared = Le500,000. ii. Gapson's share = Total - kontor's share = 500,000 - 200,000 = Le300,000.

-The shaded/stripped portion in the diagram is set F only. -This implies that sets E and G though present in the Universal set, are not included in the expression. -Mathematically; we have E' n F n G'

n(PuQ) = n(p) + n(q) - n(PnQ) 40 = 20 + 30 - n(PnQ) n(PnQ) = 50 - 40 = 10

n(AuB) = n(A) + n(B) - n(AnB) n(AuB) = 8 + 12 - 3 = 17