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WAEC 2008 · Paper 2 · Q7

  1. (a)

    Solve, correct to two decimal places, the equation 4x2=11x+214x^2 = 11x + 21.

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

  2. (b)

    A man invests £1500 for two years at compound interest. After one year, his money amounts to £1560. Find the: (i) rate of interest; (ii) interest for the second year.

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

  3. (c)

    A car costs ₦300,000.00. It depreciates by 25%25\% in the first year and 20%20\% in the second year. Find its value after 2 years.

Worked solution (try it first)

(a)

  1. Rearrange to the standard form: 4x2−11x−21=04x^2 - 11x - 21 = 0, so a=4a = 4, b=−11b = -11, c=−21c = -21.
  2. Use the formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
  3. The discriminant is 121+336=457121 + 336 = 457, and 457=21.378\sqrt{457} = 21.378.
  4. So x=11±21.3788x = \frac{11 \pm 21.378}{8}.
  5. x=32.3788=4.05x = \frac{32.378}{8} = 4.05 or x=−10.3788=−1.30x = \frac{-10.378}{8} = -1.30, to two decimal places.

(b)(i)

  1. The interest in the first year is £1560 − £1500 = £60.
  2. Rate =601500×100%=4%= \frac{60}{1500} \times 100\% = 4\%.

(ii)

  1. Compound interest: the second year's interest is on the new amount, £1560.
  2. Interest =4100×1560=62.40= \frac{4}{100} \times 1560 = 62.40, so £62.40.

(c)

  1. After the first year the car is worth 75%75\% of ₦300,000: 0.75×300 000=225 0000.75 \times 300\,000 = 225\,000.
  2. The second year's 20%20\% comes off the new value: 0.80×225 000=180 0000.80 \times 225\,000 = 180\,000.
  3. The car is worth ₦180,000.00 after 2 years.

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