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