(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.

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(a) Solve, correct to two decimal places, the equation $4 x^{2} = 11 x + 21$ . (b) A man ...

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(a) Solve, correct to two decimal places, the equation

4 x^{2} = 11 x + 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)

4 x^{2} = 11 x + 21 ⟹ 4 x^{2} - 11 x - 21 = 0

a = 4, b = - 11, c = - 21

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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}

x = \frac{11 - 21.378}{8}

x = \frac{32.378}{8}

x = \frac{- 10.378}{8}

x = 4.047 or -1.297

x ≊ 4.05 or -1.30

.(2 decimal place).
(b) (i)

I = \frac{P R T}{100}

I = £ (1560 - 1500) = £ 60

60 = \frac{1500 \times R \times 1}{100}

60 = 15 R ⟹ R = 4

(ii) Interest for the second year
=

\frac{1560 \times 4 \times 1}{100}

£ 62.40

(c)

Initial price of the car	$N 300, 000.00$
Depreciation after first year (25%)	$\frac{25}{100} \times 300, 000 = N 75, 000$
Value after 1st year	N225,000.00
Depreciation after 2nd year(20%)	$\frac{20}{100} \times 225, 000 = N 45, 000$
Value after 2nd year	N180,000.00

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