(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 |