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WAEC 2013 · Paper 2 · Q9

  1. (a)

    Solve: x2+1x=112\dfrac{x}{2} + \dfrac{1}{x} = 1\frac12.

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

  2. (b)

    The range of the numbers pp, 6, 8, 11 and qq, arranged in ascending order, is 16. If the mean is 9, find the value of (2p+q)(2p + q).

Worked solution (try it first)

(a)

  1. Write 1121\frac12 as 32\frac32 and multiply every term by 2x2x, the LCM of the denominators: x2+2=3xx^2 + 2 = 3x.
  2. Rearrange: x2−3x+2=0x^2 - 3x + 2 = 0.
  3. Factorise: (x−1)(x−2)=0(x - 1)(x - 2) = 0, so x=1x = 1 or x=2x = 2.

(b)

  1. The numbers are in ascending order, so pp is the smallest and qq the largest.
  2. Range: q−p=16q - p = 16.
  3. Mean: p+6+8+11+q=9×5=45p + 6 + 8 + 11 + q = 9 \times 5 = 45, so p+q=20p + q = 20.
  4. Add the two equations: 2q=362q = 36, so q=18q = 18.
  5. Then p=20−18=2p = 20 - 18 = 2.
  6. 2p+q=2(2)+18=222p + q = 2(2) + 18 = 22.

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