WAEC 2016 · Paper 2 · Q1

(a)
Without using mathematical tables or calculators, evaluate $\dfrac{0.09 \times 1.21}{3.3 \times 0.00025}$ , leaving the answer in standard form (scientific notation).
(b)
A principal of GH¢ 5,600.00 was deposited for 3 years at compound interest. If the interest earned was GH¢ 1,200.00, find, correct to 3 significant figures, the interest rate per annum.

Worked solution (try it first)

(a)

Write each decimal as a whole number times a power of 10: $\frac{9 \times 10^{-2} \times 121 \times 10^{-2}}{33 \times 10^{-1} \times 25 \times 10^{-5}} = \frac{9 \times 121}{33 \times 25} \times 10^{-4 - (-6)}$ .
$\frac{9 \times 121}{33 \times 25} = \frac{1089}{825}$
$= 1.32$ , and $10^{2} = 100$ , so the value is $132 = 1.32 \times 10^2$ .

(b)

The amount after 3 years is $5600 + 1200 = 6800$ .
Compound interest: $5600\left(1 + \frac{r}{100}\right)^3 = 6800$ , so $\left(1 + \frac{r}{100}\right)^3 = \frac{6800}{5600}$
$\approx 1.2143$ .
Take the cube root: $1 + \frac{r}{100} \approx 1.06686$ , so $r \approx 6.69\%$ per annum (3 significant figures).

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