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.