(a) Without using Mathematical tables or calculators, evaluate (frac{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 was deposited for 3 years at compound interest. If the interest earned was GH¢1,200, find, correct to 3 significant figures, the interest rate per annum.
Explanation
(a) (frac{0.09 times 1.21}{3.3 times 0.00025} = frac{9 times 10^{-2} times 121 times 10^{-2}}{33 times 10^{-1} times 25 times 10^{-5}})
= (frac{33 times 10^{-4}}{25 times 10^{-6}})
= (1.32 times 10^{2})
(b) Amount, A = Principal + Interest
= GH¢(5,600 + 1,200)
= GH¢6,800
Compound Interest formula : (A = P(1 + frac{r}{100})^{3})
(6,800 = 5,600(1 + frac{r}{100})^{3})
((1 + frac{r}{100})^{3} = frac{6800}{5600} = 1.214)
(1 + frac{r}{100} = sqrt[3]{1.214} = 1.06686)
(frac{r}{100} = 1.06686 – 1 = 0.06686)
(r = 0.06686 times 100 = 6.686 approxeq 6.69) (3 significant figures)