Question Library

First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9) If 2(mod 9) y(mod 6), then y = 2 by comparism

\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\) = \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\ = \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\) = \(\frac{p^2 - 2pr}{2p(p - 2r)}\) = \(\frac{p(p - 2r)}{2p(p - 2r)}\) = \(\frac{1}{2}\)

In the diagram, o(angles in same segment) o - o o (same reason as above) o

n(red balls) = 5 n(blue balls) = 4 n(\(\iff\)) = 9 Hence, prob (R1, B2) = \(\frac{5}{9} \times \frac{4}{9}\) = \(\frac{20}{81}\)

Y = x2 + 2x + k (given) y = o when x = 2 thus 0 = 22 + 2 x 2 + k 0 = 4 + 4 + k given k = -8

0, 1, 1, 2, 3, 5, 8. 0 + 1 = 1 1+ 1 = 2 1+ 2 = 3 2 + 3 = 5 3 + 5 = 8 5 + 8 = 13 8 + 13 = 21 13 + 21 = 34

Lower quartile, Q1 = \(\frac{1}{4}\)Nth term = \(\frac{1}{4}\) x 600th term = 150th term = 15.5 + 3 + 0.5 = 19

Let a represent an interior angle; e represent an exterior angle. A section of the polygon is down in the diagram. \(\frac{e}{a}\) = \(\frac{l}{11}\) given a = 11e a + e = 180o(angles on a straight line) 11e + e = 180o 12e = 180o e = \(\frac{180^o}{12}\) = 15o Hence, number of sides = \(\frac{360^o}{\tect{size of one exterior angle}\) = \(\frac{360^o}{14^o}\) = 24

Halima's age = n years old Her brother's age = \(\frac{n}{2}\) + 5

From the diagram sin 70o = \(\frac{x}{10}\) x = 10sin 70o = 9.3969 Hence, length of chord MN = 2x = 2 x 9.3969 = 18.7938 = 19cm (nearest cm)

In the diagram give, \(\alpha\) 50o(alternative angles) tan\(\alpha\) = \(\frac{h}{6}\) h = 6tan\(\alpha\) = 6tan 50o = 6 x 1.1918 = 7.1508 = 7.15m (2d.p)

In the diagram, \(\beta\) = p(vertically opposite angles) m + \(\beta\) = n(sum of interior opp. angles) m + p = n

(a) \(\frac{0.09 \times 1.21}{3.3 \times 0.00025} = \frac{9 \times 10^{-2} \times 121 \times 10^{-2}}{33 \times 10^{-1} \times 25 \times 10^{-5}}\) = \(\frac{33 \times 10^{-4}}{25 \times 10^{-6}}\) = \(1.32 \times 10^{2}\) (b) Amount, A = Principal + Interest = GH¢(5,600 + 1,200) = GH¢6,800 Compound Interest formula : \(A = P(1 + \frac{r}{100})^{3}\) \(6,800 = 5,600(1 + \frac{r}{100})^{3}\) \((1 + \frac{r}{100})^{3} = \frac{6800}{5600} = 1.214\) \(1 + \frac{r}{100} = \sqrt[3]{1.214} = 1.06686\) \(\frac{r}{100} = 1.06686 - 1 = 0.06686\) \(r = 0.06686 \times 100 = 6.686 \approxeq 6.69\) (3 significant figures)

(a) \(7(x + 4) - \frac{2}{3}(x - 6) \leq 2[x - 3(x + 5)]\) Multiply through by 3 to clear the fraction : \(21(x + 4) - 2(x - 6) \leq 6[x - 3(x + 5)]\) \(21x + 84 - 2x + 12 \leq 6x - 18x - 90\) \(19x + 96 \leq -12x - 90\) \(19x + 12x \leq -90 - 96\) \(31x \leq - 186\) Hence, \(x \leq -6\). (b) Let : t represent number of tricycles; and b represent number of taxicabs Then \(t + b = 20 ... (1)\) \(2t + 4b = 66 .... (2)\) From (1), b = 20 - t. Put that into eqtn (2), we have: \(2t + 4(20 - t) = 2t + 80 - 4t = 66\) \(80 - 66 = 14 = 2t \) \(t = 7\) The transport company has 7 tricycles.

(a) In the diagram above, \(\alpha = 28^\circ\) (angles in alternate segment) \(46^\circ + \beta + \alpha = 180^\circ\) (angles on a straight line) \(46^\circ + \beta + 28^\circ = 74^\circ + \beta = 180^\circ\) \(\beta = 180^\circ - 74^\circ = 106^\circ\) \(x + \beta = 180^\circ\) (opposite angles of a cyclic quadrilateral) \(x + 106^\circ = 180^\circ\) \(x = 180^\circ - 106^\circ = 74^\circ\) Hence, < VUS = 74^\circ. (b) Total surface area = curved surface area + area of closed end = \(2\pi rh + \pi r^{2}\) = \(\pi r(2h + r)\) \(462 = \frac{22r}{7}((2 \times 7) + r)\) \(\frac{462 \times 7}{22} = r(14 + r)\) \(147 = 14r + r^{2}\) \(r^{2} + 14r - 147 = 0\) \(r^{2} - 7r + 21r - 147 = 0\) \(r(r - 7) + 21(r - 7) = 0\) \((r + 21)(r - 7) = 0\) \(r = \text{-21 or 7}\) The radius cannot be negative, therefore, r = 7 cm.