(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 (Scienti...
Question 1 Report
(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.
Answer Details
(a) Evaluate \(\dfrac{0.09\times1.21}{3.3\times0.00025}\).
Numerator: \(0.09\times1.21=0.1089\).
Denominator: \(3.3\times0.00025=0.000825\).
\[\frac{0.1089}{0.000825}=132\]In standard form: \(132=1.32\times10^{2}\).
\(=1.32\times10^{2}\)
(b) Principal \(P=\text{GH}\!\cent5600\); interest earned in 3 years \(=1200\), so the amount is \(A=5600+1200=6800\).
Compound interest: \(A=P(1+r)^{3}\).
\[6800=5600(1+r)^{3}\Rightarrow (1+r)^{3}=\frac{6800}{5600}=1.2143\]\[1+r=\sqrt[3]{1.2143}=1.0669\Rightarrow r=0.0669\]Rate \(\approx 6.69\%\) per annum (3 s.f.).
Download The App On Google Playstore
Everything you need to excel in your exams
