(a) Solve, correct to two decimal places, the equation \(4x^{2} = 11x + 21\). (b) A man invests £1500 for two years at compound interest. After one year, hi...
Question 1 Report
(a) Solve, correct to two decimal places, the equation \(4x^{2} = 11x + 21\).
(b) A man invests £1500 for two years at compound interest. After one year, his money amounts to £1560. Find the :
(i) rate of interest ; (ii) interest for the second year.
(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.
Answer Details
(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).
Using the formula with \(a=4,\ b=-11,\ c=-21\):
\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).
\(x\approx 4.05\) or \(x\approx -1.30\)
(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).
(i) First-year interest \(=1560-1500=\pounds60\).
\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\](ii) Interest for the second year is charged on \(\pounds1560\):
\[\frac{4}{100}\times1560=\pounds62.40\](c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).
After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).
After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).
Value after 2 years \(=N180{,}000.00\)
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