WAEC Mathematics 2023 Paper

Using the venn diagram above

μ = 30

n(W) = 15n(M) = 13$n(W \cup M)^1 = 6$Let x = number of students that study both woodwork and metalworki.e. n(W ∩ M) = xNumber of students that study only woodwork,$n(W \cap M^1)$ = $15 - x$Number of students that study only metalwork, $n(W^1 \cap M)$ = $13 - x$Bringing all together,$n(W \cap M^1)$ +$n(W^1 \cap M)$ + $n(W \cap M)$ + $n(W \cup M)^1$ = $\mu$∴ (15 - x) + (13 - x) + x + 6 = 30⇒ 34 - x = 30⇒ 34 - 30 = x∴ x = 4$n(W \cap M^1)$ = $15 - 4 = 11$∴ The number of students that study woodwork but not metalwork is 11.