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WAEC 2013 · Paper 2 · Q1

  1. (a)

    Simplify, without using tables or a calculator: 34(338+158)218−112\dfrac{\frac34\left(3\frac38 + 1\frac58\right)}{2\frac18 - 1\frac12}.

  2. (b)

    Given that log⁡102=0.3010\log_{10}2 = 0.3010 and log⁡103=0.4771\log_{10}3 = 0.4771, evaluate, correct to 2 significant figures and without using tables or a calculator, log⁡101.125\log_{10}1.125.

Worked solution (try it first)

(a)

  1. Work out the top and bottom separately.
  2. Top: 338+158=53\frac38 + 1\frac58 = 5, so 34×5=154\frac34 \times 5 = \frac{15}{4}.
  3. Bottom: 218−112=178−1282\frac18 - 1\frac12 = \frac{17}{8} - \frac{12}{8}
    =58= \frac58.
  4. Divide: 154÷58=154×85\frac{15}{4} \div \frac58 = \frac{15}{4} \times \frac85
    =6= 6.

(b)

  1. 1.125=98=32231.125 = \frac98 = \frac{3^2}{2^3}.
  2. So log⁡1.125=2log⁡3−3log⁡2\log 1.125 = 2\log 3 - 3\log 2
    =2(0.4771)−3(0.3010)= 2(0.4771) - 3(0.3010)
    =0.9542−0.9030= 0.9542 - 0.9030
    =0.0512= 0.0512, which is 0.0510.051 to 2 significant figures.

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