WAEC 2019 · Paper 2 · Q5✱✱

  1. (a)

    In the diagram, MNPRMNPR is a circle with centre OO. The reflex angle at OO is 196∘196^\circ and ∠NMO=52∘\angle NMO = 52^\circ. Find the value of mm (=∠OPN= \angle OPN).

    196°52°mOMPNR
  2. (b)

    A farmer uses 25\frac25 of his land to grow cassava, 13\frac13 of the remainder for plantain and the rest for yam. Find the part of the land used for yam.

Worked solution (try it first)

(a)

  1. The reflex angle at OO is 196∘196^\circ, so the angle MOPMOP on the other side is 360∘−196∘=164∘360^\circ - 196^\circ = 164^\circ.
  2. The angle at the circumference is half the angle at the centre: ∠MNP=82∘\angle MNP = 82^\circ.
  3. OM=OPOM = OP (radii), so triangle OMPOMP is isosceles: ∠OMP=∠OPM\angle OMP = \angle OPM
    =180∘−164∘2= \frac{180^\circ - 164^\circ}{2}
    =8∘= 8^\circ.
  4. In triangle MNPMNP: ∠NMP=52∘+8∘\angle NMP = 52^\circ + 8^\circ
    =60∘= 60^\circ and ∠NPM=m+8∘\angle NPM = m + 8^\circ.
  5. So 60∘+82∘+m+8∘=180∘60^\circ + 82^\circ + m + 8^\circ = 180^\circ, and m=30∘m = 30^\circ.

(b)

  1. Cassava uses 25\frac25, leaving 35\frac35.
  2. Plantain uses 13\frac13 of the remainder: 13×35=15\frac13 \times \frac35 = \frac15.
  3. Yam gets the rest: 1−25−15=251 - \frac25 - \frac15 = \frac25 of the land.

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