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(a) \(3 \times 9^{1 + x} = 27^{-x}\) \(3 \times (3^{2})^{1 + x} = (3^{3})^{-x}\) \(3 \times 3^{2 + 2x} = 3^{-3x} \) \(3^{2x + 2 + 1} = 3^{-3x} \implies 3^{2x + 3} = 3^{-3x}\) \(\implies 2x + 3 = - 3x\) \(-5x = 3 \implies x = \frac{-3}{5}\) (b) \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\) = \(\log_{10} (\frac{\sqrt{35} \times \sqrt{2}}{\sqrt{7}}\) = \(\log_{10} (\frac{\sqrt{35 \times 2}{\sqrt{7}}\) = \(\log_{10} \sqrt{10}\) = \(\log_{10} 10^{\frac{1}{2}}\) = \(\frac{1}{2} \log_{10} 10 = \frac{1}{2}\)

(a) \(17.57^{2} - 12.43^{2}\) Using the difference of two squares method, = \((17.57 + 12.43)(17.57 - 12.43)\) = \((30.00)(5.14)\) = \(154.2\) (b) Given: D and C are points on the major arc of circle ADCB. To prove that < ADB = < ACB. Construction : Join A and B to O, the centre of the circle . Proof: With the lettering \(< AOB = 2x_{1}\) (angle at the centre is twice that subtended at the circumference) But \(< AOB = 2x_{2}\) (the same theorem applies here) \(\therefore 2x_{1} = 2x_{2} \implies x_{1} = x_{2}\) \(\therefore < ADB = < ACB\)

(a) \(< QPB = 90^\circ - 65^\circ = 25^\circ\) Also, \(< TQP = 90^\circ + 25^\circ = 115^\circ\) \(\therefore < QTP = 180^\circ - (25^\circ + 115^\circ) = 40^\circ\) Using sine rule, \(\frac{PQ}{\sin 40^\circ} = \frac{PT}{\sin 115^\circ}\) \(PT = \frac{20 \sin 115^\circ}{\sin 40^\circ} = \frac{20 \sin 65^\circ}{\sin 40^\circ}\) = \(28.199m \approxeq 28.2m\) (to 3 sig. fig) (b) Height of the tower = ST \(\sin 65 = \frac{ST}{PT} = \frac{ST}{28.199}\) \(ST = 28.199 \times \sin 65^\circ = 25.557m\) \(\approxeq 25.6m\) (to 3 sig. figs)

(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two. I'll be using the latter method. \(11111111_{two} = 1 \times 2^{7} + 1 \times 2^{6} + 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0}\) = \(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\) \(101_{two} = 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0} \) = \(4 + 0 + 1 = 5_{ten}\) \(\frac{255}{5} = 51_{ten}\) 2 51 2 25 r 1 2 12 r 1 2 6 r 0 2 3 r 0 2 1 r 1 0 r 1 = \(110011_{two}\) (b) (i) Length of arc AXB = \(\frac{105}{360} \times 2 \times \frac{22}{7} \times 6cm\) = \(11cm\) \(\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\) (ii) Area of sector = \(\frac{105}{360} \times \frac{22}{7} \times 6cm \times 6cm\) = \(33 cm^{2}\)

(a) x f fx 15 1 15 25 4 100 35 12 420 45 24 1080 55 18 990 65 8 520 75 3 225 70 3350 \(Mean = \frac{\sum fx}{\sum f} = \frac{3350}{70} = 47.8571\) \(\approxeq 47.9\) (to 3 sig. figs) (b)(i) Mode = 45 (score with the highest frequency) (ii) Range = Highest mark - Lowest mark = 75 - 15 = 60.

(a) \(\log_{10} (3x - 1) - \log_{10} 2 = 3\) \(\log_{10} (\frac{3x - 1}{2}) = 3\) \(\implies \frac{3x - 1}{2} = 10^{3} = 1000\) \(\implies 3x - 1 = 2000\) \(3x = 2000 + 1 = 2001\) \(x = \frac{2001}{3} = 667\) (b) No Log 0.8972 \(\bar{1}.9528\) + 3.536 \(0.5485\) = 0.5013 - 0.00249 \(\bar{3}.3962\) Antilog - 35.70 \(3.1051 \div 2 = 1.5526\) \(\therefore \sqrt{\frac{0.8972 \times 3.536}{0.00249}} = 35.70\)

(a) 1st bag contains 12 white (W) balls and 8 black (B) balls; 2nd bag contains 10 white (W) balls and 15 black (B) balls. P(two black balls from the 1st bag and two black balls from 2nd bag) = \((\frac{8}{20} \times \frac{7}{19}) \times (\frac{15}{25} \times \frac{14}{24})\) = \(\frac{14}{95} \times \frac{7}{20}\) = \(\frac{49}{950}\) (b) P(picking exactly one white ball) = P(WB from 1st bag and BB from 2nd bag) + P(BW from 1st bag and BB from 2nd bag) + P(BB from 1st bag and WB from 2nd bag) + P(BB from 1st bag and BW from 2nd bag) = \(\frac{12}{20} \times \frac{8}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{12}{19} \times \frac{15}{25} \times \frac{14}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{10}{25} \times \frac{15}{24} + \frac{8}{20} \times \frac{7}{19} \times \frac{15}{25} \times \frac{10}{24}\) = \(\frac{42}{475} + \frac{42}{475} + \frac{7}{190} + \frac{7}{190}\) = \(\frac{238}{950}\)

(a) (b) /QY/ = 4.8 cm

(a) Age in years No of people Cum Freq 10 8 8 15 19 27 20 10 37 25 7 44 30 6 50 (b) (c)(i) Median age = 14.5 years (ii) Number of people who are at most 15 years of age = 27 people (iii) Number of people who are between 20 and 25 years = people who are 25 years - people who are 20 years = 44 - 37 = 7 people

(a) x -1 0 1 2 3 4 5 6 y 10 -1 -8 -11 -10 -5 4 17 (b) (c)(i) \(2x^{2} - 9x = 4\) \(2x^{2} - 9x - 1 = 4 - 1 = 3\) At y = 3, x = 0.2 and 4.3. (ii) Gradient of \(y = 2x^{2} - 9x - 1\) at x = 3 = \(\frac{BC}{AB} = \frac{9.5}{4.3}\) = \(2.2\)

(a) \(T_{n} = a + (n - 1)d\) (nth term of an A.P) \(T_{4} = a + 3d = 37 ... (1)\) \(T_{6} = a + 5d = 12 + 37 = 49 ... (2)\) (2) - (1) : \(2d = 12 \implies d = 6\) \(a + 3d = 37 \implies a + 3(6) = 37\) \(a + 18 = 37 \implies a = 37 - 18 = 19\) \(T_{7} = a + 6d = 19 + 6(6) = 19 + 36 = 55\) (b) \(P = {1, 2, 3, 4} ; Q = {3, 5, 6}\) (i) \(P \cap Q = {3}\) (ii) \(P \cup Q = {1, 2, 3, 4, 5, 6}\) (iii) \((P \cap Q) \cup Q = {3, 5, 6}\) (iv) \((P \cap Q) \cup P = {1, 2, 3, 4}\)

(a) Difference in longitude = 47^\circ + 25^\circ = 72^\circ Distance /XY/ = \(\frac{\theta}{360^\circ} \times 2\pi r\) Where \(r = R \cos 32^\circ\) \(\therefore /XY/ = \frac{72}{360} \times 2 \times \frac{22}{7} \times 6400 \times \cos 32\) = \(2 \times \frac{22}{7} \times 1280 \times 0.8480\) = \(6822.765km\) \(Speed = \frac{Distance}{Time}\) = \(\frac{6822.766}{11} = 620.25 km/h\) (b) \ \(

Source: Myschool Past Questions Archive (Record #54560)