1992 - JAMB Mathematics Past Questions and Answers - page 1

To find n if 34n = 10011₂, convert both sides to base 10

= 3n + 4 = (1 x 2⁴) + (0 x2³) + (0 x 2²) + (1 x 2¹) + 1 x 2^o

= 3n + 4 = 16 + 0 + 0 + 2 + 1

3n + 4 = 19

3n = 15

n = 5

% error in Area = (\frac{\pi(5.01) - \pi(5)^2 \times 100%}{\pi(5)^2})

= (\frac{\pi 26.01 - 25 \times 100%}{\pi(25)})

= (\frac{0.01}{25}) x 1004

= 4.04

log_baⁿ = log_an^a

x(a^{(\frac{1}{n})})x = a

a^{(\frac{x}{n})} = a¹

(\frac{x}{n}) = 1

x = n

(\frac{4^{2x}}{4^{3x}}) = 2

4^{2x - 3x} = 2

4^-x = 2

(2²)^-x

= 2¹

Equating coefficients: -2x = 1

x = -(\frac{1}{2})

(\frac{(1.25 \times 10^{-4}) \times (2.0 \times 10^{-1})}{(6.25 \times 10^5)}) = (\frac{1.25 \times 2}{6.25}) x 10^{4 - 1 - 5}

(\frac{2.50}{6.25}) x 10^-2 = (\frac{250}{625}) x 10^-2

0.4 x 10^-2 = 4.0 x 10^-3

5(\sqrt{18}) - 3(\sqrt{72}) + 4(\sqrt{50}) = 5(3(\sqrt{2})) - 3(6(\sqrt{2})) + 4(5(\sqrt{2}))

15(\sqrt{2}) - 18(\sqrt{2}) + 20(\sqrt{2}) = 35(\sqrt{2}) - 18(\sqrt{2})

= 17(\sqrt{2})

x = 3 - (\sqrt{3})

x² = (3 - (\sqrt{3}))²

= 9 + 3 - 6(\sqrt{34})

= 12 - 6(\sqrt{3})

= 6(2 - (\sqrt{3}))

∴ x² + (\frac{36}{x^2}) = 6(2 - (\sqrt{3})) + (\frac{36}{6(2 - \sqrt{3})})

6(2 - (\sqrt{3})) + (\frac{6}{2 - \sqrt{3}}) = 6(- (\sqrt{3})) + (\frac{6(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})})

= 6(2 - (\sqrt{3})) + (\frac{6(2 + \sqrt{3})}{4 - 3})

6(2 - (\sqrt{3})) + 6(2 + (\sqrt{3})) = 12 + 12

= 24

x = (all prime factors of 44) and y = (all prime factors of 60)

∴ x = (2, 11), y = (2, 3, 5)

X ∪ Y = (2, 3, 5, 11),

X ∩ Y = (2)

U = (1, 2, 3, 6, 7, 8, 9, 10)

E = (10, 4, 6, 8, 10)

F = (x : x² = 2⁶, x is odd)

∴ F = (\phi) Since x² = 2⁶ = 64

x = + which is even

∴ E ∩ F = (\phi) Since there are no common elements