Question Library

a. From the figure: |PZ| = |PQ| + |QY| + |YZ| |PQ| = |RS| = 3 cm (opposite sides of a parallelogram) |YZ| = |RS| = 3 cm (opposite sides of a parallelogram) |QY| = 2 cm (given) |PZ| = |PQ| + |QY| + |YZ| |PZ| = 3 + 2 + 3 = 8 cm b. radius of latitude r = R cos \alpha, where \alpha is the angle 56º r = 6400 cos 56^\circ = 3578.8 km < PLQ = 95º - 25^\circ = 70^\circ Length of arc PQ = \(\frac{\theta}{360}\) \times 2\pir = \(\frac{70}{360}\) \times 2 \times (\(\frac{22}{7}\)) \times 3578.8 = \(\frac{39367.2}{9}\) = 4374 km

a(i) Σfx =2 + 6 +12 + 16 + 20 + 24 + 35 + 24 +18 +10 = 167 Mean = \(\frac{Σfx}{Σf}\) =\(\frac{167}{32}\) = 5.22 ii. Median = Middle Position Value = \(\frac{\text{N}}{2}\)th position = \(\frac{32}{2}\) position = 16 th position .: Median = 5 iii. Mode = most people => highest frequency .: Mode = 7 b. People scoring at least 8 marks => frequency of x > = 8 = 3 + 1 + 2 = 6 Percentage = \(\frac{6}{32}\) x 100% = 18.75%

ai From the four-figure/ mathematical table 2sin63.35º = 2(0.8938) = 1.7876 ii. From the four-figure/ mathematical table, log cos 44.74º = log 0.7103 = \(\bar{1}\).8515 iii. log k - log(k - 2) = log 5 = log \(\frac{\text{k}}{k - 2}\) = log 5 (law of logarithms) log \(\frac{\text{k}}{k - 2}\) = log 5 \(\frac{\text{k}}{k - 2}\) = 5 cross multiply 5(k - 2) = k 5k - 10 = k 5k - k = 10 4k = 10 k = \(\frac{10}{4}\) = \(\frac{5}{2}\). b. \(\frac{3.68^2 \times 6.705}{\sqrt{0.3581}}\) = 151.8. See the steps in the diagram above.

a. p = x + ym\(^3\) ym\(^3\) = p - x m\(^3\) = \(\frac{p - x}{\text{y}}\) \(m = \sqrt[3]{\frac{p - x}{y}}\). b. x\(^2\) - 6x + 7 = 0 Using the completing the square method x\(^2\) - 6x = - 7 Add the square of half the coefficient of x to both sides x\(^2\) - 6x + (-3)\(^2\) = -7 + 9 x\(^2\) - 6x + (-3)\(^2\) = 2 (x - 3)\(^2\) = 2 x - 3 = \pm \(\sqrt{2}\) x = 3 + \(\sqrt{2}\) or 3 - \(\sqrt{2}\) x = 3 + 1.4142 or 3 - 1.4142 x = 4.1 or 1.6 to 1 decimal place. c. Let the two consecutive positive odd numbers be \(n\) and \(n + 2\). The equation for their product is: \(n(n + 2) = 195\) Expanding gives: n\(^2\) + 2n - 195 = 0 Using the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 2\), and \(c = -195\): \(b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-195) = 4 + 780 = 784\) Now substituting: \(n = \frac{-2 \pm \sqrt{784}}{2} = \frac{-2 \pm 28}{2}\) This results in: \(n = \frac{26}{2} = 13 \quad (\text{positive solution})\) Thus, the two numbers are: \(13 \quad \text{and} \quad 15\)

ai. ∠Y = 270º - (195^\circ + 30^\circ) = 270^\circ - 225^\circ = 45º ∠Z = 180º - (30^\circ + 45º) = 105º Let the distance of Y from Z be k \(\frac{\text{k}}{sin30º} = \frac{20}{sin 105º}\) k sin 105º = 20 sin 30º k = \(\frac{20 sin 30^\circ}{ sin 105^\circ}\) = \(\frac{20 \times 0.5 }{sin 105^\circ}\) = \(\frac{10}{0.9659}\) = 10.35 ⇒ k = 10.4 km (3. s.f.) ii. Let the distance of Z from X be t Using the sine rule \(\frac{\text{t}}{sin 30º} = \frac{20}{sin 105º}\) t sin 105^\circ = 20 sin 45^\circ t = \(\frac{20 sin 45º}{sin 105º}\) t = 14.6km to 3. s.f bi. Latitudinal difference = 36^\circ + 36^\circ = 72^\circ Distance PQ = Length of arc PQ = \(\frac{\theta}{360}\) \times 2\piR = \(\frac{72}{360}\) \times 2 \times \(\frac{22}{7}\) \times 6400 km = \(\frac{88 \times 640}{ 7 }\) = \(\frac{56320}{7}\) = 8045.7 km = 8050 km (3 s.f.) ii. Speed of aircraft = 800 km/hr Time taken = \(\frac{\text{Total distance}}{\text{Speed}}\) = \(\frac{8050}{800}\) = 10.05 hrs

a. x f cum. frequency End point 11-20 68 68 20.5 21-30 184 252 30.5 31-40 294 546 40.5 41-50 402 948 50.5 50-60 480 1428 60.5 61-70 310 1738 70.5 71-80 164 1902 80.5 81-90 94 2000 90.5 bi. Cut-off mark = 64% ii. Pr(student scoring at least 45%) = \(\frac{1075}{2000}\) = 0.5375

a. Total ball = 5 blues + 3 blacks + 2 red balls = 10 balls Selection with replacement Pr(1st red ball) = \(\frac{2}{10}\) Pr(2nd red ball) = \(\frac{2}{10}\) Therefore, Pr(two red balls) = \(\frac{2}{10}\) x \(\frac{2}{10}\) = \(\frac{4}{100}\) = \(\frac{1}{25}\) = 0.04 b. Pr( of selecting 2 blue) = \(\frac{5}{10}\) x \(\frac{5}{10}\) = \(\frac{25}{100}\) = \(\frac{1}{4}\) Pr(of selecting 2 black balls) = \(\frac{3}{10}\) x \(\frac{3}{10}\) = \(\frac{9}{100}\) Pr( 2 blue or 2 black balls) = \(\frac{1}{4}\) + \(\frac{9}{100}\) = \(\frac{25}{100}\) + \(\frac{9}{100}\) = \(\frac{34}{100}\) = 0.34. c) 1 black, and 1 red in any other means it can be 1 black, and 1 red, and 1 black = \(\frac{3}{10}\) x \(\frac{2}{10}\) + \(\frac{2}{10}\) + \(\frac{3}{10}\) = \(\frac{6}{100}\) + \(\frac{6}{100}\) = \(\frac{12}{100}\) = 0.12.

\(\frac{16^{2 + x}}{4} = 64^x\) 16\(^{2 + x} = 4 \times 64^x\) 4\(^{2(2 + x)} = 4^1 \times 4^{3(x)}\) (Equate coefficients since the bases are the same) 2(2 + x) = 1 + 3x 4 + 2x = 1 + 3x 2x - 3x = 1 - 4 -x = -3 x = 3.

Length of Arc = \(\frac{\theta}{360}\) x 2\(\pi\)r Length of arc = 33cm, diameter = 28cm → r = 14cm 33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7}\times14\) \(\theta\) = \(\frac{33 \times 180}{22 \times 2}\) \(\theta\) = 135º

Length of a chord = 2rSin\(\frac{\theta}{2}\) \(\theta\) = 60º not 30º and radius = 9cm Therefore, Length of the chord = 2 x 9 Sin\(\frac{60º}{2}\) = 2 x 9 x Sin 30º = 2 x 9 x \(\frac{1}{2}\) = 9cm.

The area of a triangle given two sides and the included angle = \(\frac{1}{2}\) x a x b x Sin\(\theta\) a = b = 8cm \(\theta\) = 60º Area = \(\frac{1}{2}\) x 8 x 8 x Sin 60º = \(\frac{1}{2}\) x 64 x \(\frac{\sqrt{3}}{2}\) ( Sin 60º = \(\frac{\sqrt{3}}{2}\) ) = \(\frac{64\sqrt{3}}{4}\) = 16\(\sqrt{3}cm^2\)

The construction of angle 30º (If you take a 60º angle and bisect it, you would divide it into two angles of 30º each.)

\(\overline{MN} || \overline{KL}\), \(\triangle\) LXK is similar to \(\triangle\)MXN So, the ratio of the areas of the two similar triangles equals the square of the ratio of their corresponding sides Therefore, \(\frac{ Area of \triangle MXN}{Area of \triangle LXK} = \frac{12^2}{9^2}\) \(\frac{16}{Area of \triangle LXK} = \frac{144}{81}\) Area of \(\triangle\)LXK = \(\frac{ 16 \times 81}{144} = 9cm^2\)

10111.11\(_{two}\) convert the number to base 10. 10111.11\(_{two}\) = 1 x 2\(^4\) + 0 x 2\(^3\) + 1 x \(^2\) + 1 x 2\(^1\) + 1 x 2\(^0\) + 1 x 2\(^{-1}\) + 1 x 2\(^{-2}\) = 16 + 0 + 4 + 2 + 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) = 23 + \(\frac{3}{4}\) = 23\(\frac{3}{4}\)

Subtract 14256\(_{seven}\) from 20045\(_{seven}\) = 2456\(_{seven}\) 2 0 0 4 5 1 4 2 5 6 0 2 4 5 6