WAEC 2023 · Paper 2 · Q12

  1. (a)

    In the diagram, PQ‾∥MN‾\overline{PQ} \parallel \overline{MN}, ∣PR∣=∣QR∣|PR| = |QR|, ∠QMN=53∘\angle QMN = 53^\circ and ∠MNP=(32y−13)∘\angle MNP = \left(\frac32 y - 13\right)^\circ. Find: (i) ∠PRQ\angle PRQ; (ii) the value of yy.

    53°(3y/2 − 13)°RQPMN

    Separate values with commas, e.g. 3, −2

  2. (b)

    A bowl contains 15 yellow and green balls of the same size. If the probability of selecting a yellow ball is 25\frac25, find the number of: (i) yellow balls; (ii) green balls in the bowl.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. PQ∥MNPQ \parallel MN, and QMQM crosses both, so the alternate angles are equal: ∠PQR=∠QMN=53∘\angle PQR = \angle QMN = 53^\circ.
  2. ∣PR∣=∣QR∣|PR| = |QR|, so triangle PQRPQR is isosceles and ∠QPR=∠PQR=53∘\angle QPR = \angle PQR = 53^\circ.
  3. Then ∠PRQ=180∘−53∘−53∘\angle PRQ = 180^\circ - 53^\circ - 53^\circ
    =74∘= 74^\circ.

(ii)

  1. PNPN also crosses the parallel lines, so ∠MNP=∠QPN=53∘\angle MNP = \angle QPN = 53^\circ (alternate angles).
  2. So 32y−13=53\frac32 y - 13 = 53, 32y=66\frac32 y = 66 and y=44y = 44.

(b)(i)

  1. Yellow balls =25×15=6= \frac25 \times 15 = 6.

(ii)

  1. Green balls =15−6=9= 15 - 6 = 9.

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