(a) Out of 30 candidates applying for a post, 17... | WAEC SSCE - General Mathematics - 2009 (Essay)

Question 1 Report

(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both?

(b) In triangle PQR, M and N are points on the side PQ and PR respectively such that MN is parallel to QR. If < PRQ = 75°, PN = QN and < PNQ = 125°, determine :

(i) < NQR ; (ii) < NPM.

Answer

(a) \(n=30\); degrees \(=17\), diplomas \(=15\), neither \(=4\), so \(30-4=26\) have at least one.

\(|D\cup P|=|D|+|P|-|D\cap P| \Rightarrow 26=17+15-|D\cap P| \Rightarrow |D\cap P|=32-26=\mathbf{6}\).

Six candidates have both a degree and a diploma.

(b) In \(\triangle PQR\), \(M\) on \(PQ\), \(N\) on \(PR\), \(MN\parallel QR\); \(\angle PRQ=75^{\circ}\), \(PN=QN\), \(\angle PNQ=125^{\circ}\).

Triangle \(PNQ\) is isosceles (\(PN=QN\)), so \(\angle NPQ=\angle NQP\).

\(\angle NPQ+\angle NQP+\angle PNQ=180^{\circ}\Rightarrow 2\angle NPQ=180^{\circ}-125^{\circ}=55^{\circ}\Rightarrow \angle NPQ=\angle NQP=27.5^{\circ}\).

In \(\triangle PQR\): \(\angle P=27.5^{\circ}\), \(\angle R=75^{\circ}\), so \(\angle PQR=180^{\circ}-27.5^{\circ}-75^{\circ}=77.5^{\circ}\).

(i) \(\angle NQR=\angle PQR-\angle NQP=77.5^{\circ}-27.5^{\circ}=\mathbf{50^{\circ}}\).

(ii) \(\angle NPM\) is the angle at \(P\) between \(PM\) (on \(PQ\)) and \(PN\) (on \(PR\)), i.e. \(\angle QPR=\mathbf{27.5^{\circ}}\).

Read lesson note on Loci (WAEC)

Read lesson note on Sets (WAEC)

Answer Details

(a) \(n=30\); degrees \(=17\), diplomas \(=15\), neither \(=4\), so \(30-4=26\) have at least one.

\(|D\cup P|=|D|+|P|-|D\cap P| \Rightarrow 26=17+15-|D\cap P| \Rightarrow |D\cap P|=32-26=\mathbf{6}\).

Six candidates have both a degree and a diploma.

(b) In \(\triangle PQR\), \(M\) on \(PQ\), \(N\) on \(PR\), \(MN\parallel QR\); \(\angle PRQ=75^{\circ}\), \(PN=QN\), \(\angle PNQ=125^{\circ}\).

Triangle \(PNQ\) is isosceles (\(PN=QN\)), so \(\angle NPQ=\angle NQP\).

\(\angle NPQ+\angle NQP+\angle PNQ=180^{\circ}\Rightarrow 2\angle NPQ=180^{\circ}-125^{\circ}=55^{\circ}\Rightarrow \angle NPQ=\angle NQP=27.5^{\circ}\).

In \(\triangle PQR\): \(\angle P=27.5^{\circ}\), \(\angle R=75^{\circ}\), so \(\angle PQR=180^{\circ}-27.5^{\circ}-75^{\circ}=77.5^{\circ}\).

(i) \(\angle NQR=\angle PQR-\angle NQP=77.5^{\circ}-27.5^{\circ}=\mathbf{50^{\circ}}\).

(ii) \(\angle NPM\) is the angle at \(P\) between \(PM\) (on \(PQ\)) and \(PN\) (on \(PR\)), i.e. \(\angle QPR=\mathbf{27.5^{\circ}}\).

Read lesson note on Loci (WAEC)

Read lesson note on Sets (WAEC)

Loci

Sets

View Answer

(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both? (b) In triangl...

Download The App On Google Playstore

(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both? (b) In triangl...

Related Questions

Download The App On Google Playstore