WAEC 2014 · Paper 2 · Q10

  1. (a)

    (i) Solve the inequality 12x−56(x+2)≤1+x\frac12x - \frac56(x + 2) \le 1 + x. (ii) Illustrate the solution on a number line.

    Model answer
    −3−2−10123

    (i) x≥−2x \ge -2. (ii) A solid dot at −2-2 (it is included) with an arrow to the right.

  2. (b)

    From a point PP on level ground and directly west of a pole, the angle of elevation of the top of the pole is 45∘45^\circ, and from a point QQ east of the pole, the angle of elevation of the top of the pole is 58∘58^\circ. If ∣PQ∣=10 m|PQ| = 10\text{ m}, calculate, correct to 2 significant figures, the: (i) distance from PP to the pole; (ii) height of the pole.

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

Worked solution (try it first)

(a)(i)

  1. Multiply every term by 6: 3x−5(x+2)≤6+6x3x - 5(x + 2) \le 6 + 6x.
  2. Expand: 3x−5x−10≤6+6x3x - 5x - 10 \le 6 + 6x, so −2x−10≤6+6x-2x - 10 \le 6 + 6x.
  3. Collect terms: −8x≤16-8x \le 16.
  4. Divide by −8-8, which reverses the inequality sign: x≥−2x \ge -2.

(ii)

  1. On a number line, put a solid dot at −2-2 (because −2-2 is included) and an arrow pointing to the right.

(b)

  1. The pole stands between PP (to the west) and QQ (to the east).
  2. Let the distance from PP to the foot of the pole be xx m.
  3. Then QQ is (10−x)(10 - x) m from it.
  4. The height hh from each side: h=xtan⁡45∘=xh = x\tan 45^\circ = x, and h=(10−x)tan⁡58∘h = (10 - x)\tan 58^\circ.
  5. Set them equal: x=(10−x)(1.6003)x = (10 - x)(1.6003).
  6. So x+1.6003x=16.003x + 1.6003x = 16.003, 2.6003x=16.0032.6003x = 16.003 and x≈6.154x \approx 6.154.

(i)

  1. PP is about 6.2 m from the pole.

(ii)

  1. The height is h=x≈6.2h = x \approx 6.2 m (2 significant figures).

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