WAEC 2017 · Paper 2 · Q2

  1. (a)

    Solve the equation 23(3x−5)−35(2x−3)=3\frac23(3x - 5) - \frac35(2x - 3) = 3.

  2. (b)

    In the diagram, PP, UU, TT, SS lie on a straight line and PP, QQ, RR on another. ∠STQ=m\angle STQ = m, ∠TUQ=80∘\angle TUQ = 80^\circ, ∠UPQ=r\angle UPQ = r, ∠PQU=n\angle PQU = n and ∠RQT=88∘\angle RQT = 88^\circ. Find the value of (m+n)(m + n).

    rn88°80°mPQRUTS
    The values of r, m and n are not given; the drawing uses r = 20°.
Worked solution (try it first)

(a)

  1. Multiply every term by 15, the LCM of 3 and 5: 10(3x−5)−9(2x−3)=4510(3x - 5) - 9(2x - 3) = 45.
  2. Expand: 30x−50−18x+27=4530x - 50 - 18x + 27 = 45.
  3. So 12x−23=4512x - 23 = 45, 12x=6812x = 68 and x=6812=523x = \frac{68}{12} = 5\frac23.

(b)

  1. ∠TUQ=80∘\angle TUQ = 80^\circ is an exterior angle of △PUQ\triangle PUQ, so it equals the two opposite interior angles: r+n=80∘r + n = 80^\circ (1).
  2. At TT, the angle inside △PQT\triangle PQT is 180∘−m180^\circ - m (angles on a straight line).
  3. ∠RQT=88∘\angle RQT = 88^\circ is an exterior angle of △PQT\triangle PQT: r+(180∘−m)=88∘r + (180^\circ - m) = 88^\circ, so m−r=92∘m - r = 92^\circ (2).
  4. Add (1) and (2): the rr terms cancel, giving m+n=172∘m + n = 172^\circ.

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