JAMB Mathematics Past Questions & Answers - Page 65

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)

If x = 3 and y = 100,

then, \(\frac{100}{1} = \frac{k}{3^2}\)

\(\frac{100}{1} = \frac{k}{9}\)

k = 100 x 9 = 900

Substitute 900 for k in

y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\)

= \(yx^2 = 900\)

\(\sqrt{108} + \sqrt{125} - \sqrt{75}\)

= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\)

= \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\)

= \(\sqrt{3} + 5\sqrt{5}\)

54 \(_{eight}\) and 67\(_{eight}\) = 1438

  Starting with normal addition, 4 + 7 gives 11

  (it is more than the base, 8) 8 goes in 11 just 1 time, remaining 3, the remainder will be written, and the 1 will be added to the sum of 5 and 6 which gives 12 altogether, 8 goes in 12 one time remaining 4, the remainder 4 was written and then the 1 that was the quotient was then written since nothing to add the 1 to.

  So answer is 143 in base eight

Cos θ = \(\frac{adj}{hyp}\)

  = \(\frac{300}{600}\)

  = 0.5

  θ = Cos - 10.5

  = 60

  ∠ RPQ = ∠ PQs

  So the bearing of P from Q is 180 + 60 = 240\(^o\)

  Answer is D

Using law of indices

25\(\frac{1}{2}\) × 8\(\frac{-2}{3}\)

  = √25 x (\(\sqrt[3]{8}\)) -2

  = 5 x 2-2

  = 5 x \(\frac{1}{2^2}\) =

 \(\frac{5}{4}\) = \(\frac{11}{4}\)

  Answer is A