Correct Answer: Option B

Explanation

Using the venn diagram above

μ = 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,
\(n(W ∩ M^1)\) +\( n(W^1 ∩ M)\) + \(n(W ∩ M)\) + \(n(W ∪ M)^1\) = \(μ\)
∴ (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.
 

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