(a) Solve, correct to two decimal places, the equation \(4x^{2} = 11x + 21\). (b)
(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.
Explanation
(a) \(4x^{2} = 11x + 21 \implies 4x^{2} - 11x - 21 = 0\)
\(a = 4, b = - 11 , c = -21\)
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
\(x = \frac{-(-11) \pm \sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\)
\(x = \frac{11 \pm \sqrt{121 + 336}}{8}\)
\(x = \frac{11 \pm \sqrt{457}}{8}\)
\(x = \frac{11 \pm 21.378}{8}\)
\(x = \frac{11 + 21.378}{8}\) or \(x = \frac{11 - 21.378}{8}\)
\(x = \frac{32.378}{8}\) or \(x = \frac{-10.378}{8}\)\
\(x = \text{4.047 or -1.297}\)
\(x \approxeq \text{4.05 or -1.30}\).(2 decimal place).
(b) (i) \(I = \frac{PRT}{100}\)
\(I = £(1560 - 1500) = £60\)
\(60 = \frac{1500 \times R \times 1}{100}\)
\(60 = 15R \implies R = 4%\)
(ii) Interest for the second year
= \(\frac{1560 \times 4 \times 1}{100}\)
= \(£62.40\)
(c)
| Initial price of the car | \(N300,000.00\) |
| Depreciation after first year (25%) | \(\frac{25}{100} \times 300,000 = N75,000\) |
| Value after 1st year | N225,000.00 |
| Depreciation after 2nd year(20%) | \(\frac{20}{100} \times 225,000 = N45,000\) |
| Value after 2nd year | N180,000.00 |