2004 - JAMB Mathematics Past Questions and Answers - page 4

ac - 2bc - a² + 4b²

ac - 2bc - (a² - 4b²)

ac - 2bc - (a² -(2b)²)

c(a - 2b) - {(a - 2b)(a + 2b)}

(a - 2b)(c - (a + 2b))

(a - 2b)(c - a - 2b)

(a=\frac{1}{2}\r=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{6}\times\frac{2}{1}=\frac{1}{3}\S_n = \frac{a}{1-r}\S_n = \frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{2}\times\frac{3}{2}=\frac{3}{4})

L ∝ T²

L = KT²

K = L/T² (when T = 4s, L = 64 cm)

K = 64/4²

K = 64/16 = 4

∴ ∠ = 4T²

∠ = 4 * 9²

= 324 cm

First teacher received (\frac{2}{5}) ∴Remainder (= 1 - \frac{2}{5} = \frac{3}{5})

2nd teacher received; (\frac{2}{15} of \frac{3}{5} = \frac{2}{15} \times \frac{3}{5} = \frac{2}{25})

3rd teacher will receive; (1-\frac{2}{5}-\frac{2}{25}=\frac{25-10-2}{15}=\frac{13}{25})

6log_x2 - 3log_x3 = 3log₅0.2

= logx²⁶ - 3logx³³ = log5^(0.2)3

= logx(64/27) = log5(1/5)³

logx(64/27) = log5(1/125)

let logx(64/27) = y

∴x^y = 64/27

and log5(1/125) = y

∴ 5^y = 1/125

5^y = 125^-1

5^y = 5^-3

∴ y = -3

substitute y = -3 in x^y = 64/27

implies x^-3 = 64/27

1/x³ = 64/27

64x³ = 27

x³ = 27/64

x³ = ³√27/64

x = 3/4

451₆ - P₇ = 305₆

P₇ = 451₆ - 305₆

P₇ = 142₆

convert 142₆ = 1 * 6² + 4 * 6¹ + 2 * 6⁰

= 36 + 24 + 2

= 62

Convert 62₁₀ to base 7

62/7 = 8 R 6

8/7 = 1 R 1

1/7 = 0 R 1

∴P₇ = 116₇

³√4^2x = 16

this implies that (³√4^2x)³ = (16)³

4^2x = 4^2*3

4^2x = 4⁶

∴ 2x = 6

x = 3

(\frac{\frac{1}{10}\times\frac{2}{3}+\frac{1}{4}}{\frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}}\Numerator \hspace{1mm}\frac{1}{10}\times\frac{2}{3}+\frac{1}{4} = \frac{1}{5}+\frac{1}{4}=\frac{4+15}{60}=\frac{19}{60}\denominator\hspace{1mm}= \frac{\frac{1}{2}}{\frac{3}{5}}-\frac{1}{4}=\frac{1}{2}\times\frac{5}{3}-\frac{1}{4}=\frac{5}{6}-\frac{1}{4}=\frac{10-3}{12}=\frac{7}{12}\frac{Numerator}{denominator}=\frac{\frac{19}{60}}{\frac{7}{12}}=\frac{19}{60}\times\frac{12}{7}=\frac{19}{35})

(\frac{1}{\sqrt{3}+2}=\frac{1}{\sqrt{3}+2}\times \frac{\sqrt{3}-2}{\sqrt{3}-2}=\frac{\sqrt{3}-2}{(\sqrt{3})^{2} -2^{2}}=\frac{\sqrt{3}-2}{3-4}=\frac{\sqrt{3}-2}{1}=-\sqrt{3}+2=2-\sqrt{3})