Financial Arithmetic - SS3 Mathematics Past Questions and Answers - page 1
An account having \(N12,000\) earns a compound interest of \(20\%\) per annum. What is its value after \(5\) years? \((a)N30,500\ (b)N29,600\ 9\ (c)N29,860\ (d)N25,200\)
N30,500
\(N29,600\ 9\)
N29,860
N25,200
\[A = P(1 + R)^{T}\]
\[P = 12000\ \ \ R = 20\% = 0.2\ \ \ \ \ T = 5\]
\[A = 12000{(1 + 0.2)}^{5}\]
\[A = 12000{(1.2)}^{5}\]
\(A = N29,859.84\ \cong N29,860\ \ \)
William bought \(1,000\) \(50k\) shares at \(75k\) per share with a dividend of \(15\%\). What is the total dividend?
N750
N75
N7,500
N7.50k
\[15\%\ dividend\ on\ the\ face\ value\ of\ shares\ held\ = \ 10\%\ of\ (1,000 \times 50k)\]
\[= 15\%\ of\ 50,000k\]
= 7,500k
= N75
Mr Wilkerson bought some Nigerian government bonds worth N2,500,000. If the bonds mature after 10 years at an interest rate of 8%, what is the maturity value of his investment?
N4,000,000
N2,000,000
N4,500,000
N2,500,000
\[Bond\ interest\ (coupon) = \ 2500000 \times 0.08 \times 10 = N2,000,000\]
\(Maturity\ value = N2,500,000 + N2,000,000 = N4,500,000\)
Matthew deposits N15,000 at the end of each quarter for 15 years with Ivory Bank which pays an annual interest rate of 10% compounded quarterly. Find the worth of the account.
\[A = P\left( 1 + \frac{R}{n} \right)^{nT}\]
\[P = 15000\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 10\% = 0.1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 4\ \ \ \ \ \ \ \ \ \ \ \ \ T = 15\]
\[A = 15000\left( 1 + \frac{0.1}{4} \right)^{4(15)}\]
\[A = 15000(1 + 0.025)^{60}\]
\[A = 15000(1.025)^{60}\]
\[A = N65,996.85\]
A company wants to accumulate N5,000,000 in 5 years’ time to make a purchase. A sinking fund is established by making fixed monthly payments into an account paying 15% per annum compounded monthly. How much should each payment be?
\[S = 5000000,\ i = \frac{0.15}{12} = 0.0125,\ n = 5 \times 12 = 60\]
\[R = \frac{5000000 \times 0.0125}{(1 + 0.0125)^{60} - 1}\]
\[R = \frac{5000000(0.0125)}{1.10718} = 56,449.72\]
\[R = N56,449.72\]
Kyle got a \(\$ 250,000\) housing loan at an annual interest rate of \(10\%\). If the loan is amortized for a \(30\ year\ term\), what is the monthly payments Kyle is expected to pay?
\[P = 250000\ ,\ i = \frac{0.1}{12} = 0.0083333,\ n = 30 \times 12 = 360\]
\[R = \frac{250000(0.0083333)}{1 - \left( \frac{1}{1 + 0.0083333} \right)^{360}} = \$ 2,193.92\]
The value of a factory machine depreciates by 15% of its value in the beginning of each year. If the machine cost N7.5 million, find the book value after 3 years.
|
\[\mathbf{1}\mathbf{st}\mathbf{\ year:}\] |
|
|
|---|---|---|
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\[Value\ of\ car:\] |
\[N7,500,000\] |
|
|
\[15\%\ depreciation:\] |
\[(\frac{15}{100} \times 7500000)\] |
\[N1,125,000\] |
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\[Value\ after\ one\ year:\] |
\[N6,375,000\] |
|
|
\[\mathbf{2}\mathbf{nd}\mathbf{\ year:}\] |
||
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\[Value\ of\ car:\] |
\[N6,375,000\] |
|
|
\[15\%\ depreciation:\] |
\[(\frac{15}{100} \times 1200000)\] |
\[N956,250\] |
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\[Value\ after\ 2\ years:\] |
\[N5,418,750\] |
|
|
\[\mathbf{3}\mathbf{rd}\mathbf{\ year:}\] |
||
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\[Value\ of\ car:\] |
\[N5,418,750\] |
|
|
\[15\%\ depreciation:\] |
\[(\frac{15}{100} \times 960000)\] |
\[N812,812.50\] |
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\[Value\ after\ 3\ years:\] |
\[N4,605,937.50\] |
\[Therefore,\ book\ value\ after\ 3\ years\ = \ N4,605,937.50\]