(a) Evaluate, without using mathematical tables or calculator, \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\), leaving your answer in standard form.
(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.
(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.
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Correct Answer: Option n
Explanation:
(a) \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\)
= \(\frac{3.69}{1.64} \times 10^{5 - (-3)}\)
= \(2.25 \times 10^{8}\)
(b) Interest from bank A : \(\frac{y}{100} \times N20,000 = N200y\)
Interest on bank B :\(\frac{1.5y}{100} \times N25,000 = N375y\)
\(\therefore N(200y + 375y) = N4,600\)
\(575y = 4,600 \implies y = 8\)
(a) \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\)
= \(\frac{3.69}{1.64} \times 10^{5 - (-3)}\)
= \(2.25 \times 10^{8}\)
(b) Interest from bank A : \(\frac{y}{100} \times N20,000 = N200y\)
Interest on bank B :\(\frac{1.5y}{100} \times N25,000 = N375y\)
\(\therefore N(200y + 375y) = N4,600\)
\(575y = 4,600 \implies y = 8\)