(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.
(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.
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Correct Answer: Option n
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)
(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)