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Arrange in ascending order 89, 100, 108, 119, 120,130, 131, 131, 141, 161 Median = \(\frac{120 + 130}{2}\) = 125

\(\frac{1}{3}\) (x + 1) - 1 > \(\frac{1}{5}\)(x + 4) = \(\frac{x + 1}{3} - 1\) > \(\frac{x + 4}{5}\) \(\frac{x + 1}{3} - \frac{x + 4}{5} -1\) > 0 = \(\frac{5x + 5 - 3x - 12}{15}\) 2x - 7 > 15 2x > 22 = x > 11

\(\frac{0.021741 \times 1.2047}{0.023789}\) = \(\frac{0.022 \times 1.2}{0.024}\) (to 2 s.f) = \(\frac{0.0264}{0.024}\) = 1.1

Let x rep. numbers of boys that can work at \(\frac{5}{4}\) the rate at which the 10 girls work For 1 hrs, x boys will work for \(\frac{\frac{1}{5} \times 10}{4}\) x = \(\frac{5}{4}\) x 10 = 8 boys 8 boys will do the work of ten girls at the same rate 4 + 8 = 12 bous cut the field in 5 hrs for 3 hrs, \(\frac{12 \times 5}{3}\) boys will be needed = 20 boys.

Circumstances of latitude 0\(^o\)s where R is the radius of the earth 2\(\pi\)r cos \(\theta\)

ABCD is the floor. By pathagoras AC\(^2\) = 144 + 81 = \(\sqrt{225}\) AC = 15cm Height of room 8m, diagonal of floor is 15m Therefore, the cosine of the angle which a diagonal of the room makes with the floor is EC\(^2\) = 15\(^2\) + 8\(^2\) cosine \(\frac{adj}{Hyp} = \frac{15}{17}\) EC\(^2\) = \(\sqrt{225 + 64}\) EC = \(\sqrt{289}\) = 17

Let x represent number of goals \(\begin{array}{c|c|} x & F & Frequency(fx)\\\hline 0 & 4 & 0 \\\hline 1 & 3 & 3\\\hline 2 & 15 & 30 \\\hline 3 & 16 & 48\\ \hline 4 & 1 & 4 \\\hline 5 & 0 & 0 \\\hline 6 & 1 & 6 \\\hline & \sum Fx = 91& \end{array}\) \(\sum fx\) = 91

No of students offering physics are \(\frac{12}{360}\) x 150 = 5

log\(_{10}\)2 = 0.3010 and log\(_{10}\)3 = 04771 log\(_{10} 4.5 = log_{10}\) (\(\frac{3 \times 3}{2}\)) log\(_{10}\) 3 + log\(_{10}\) 3 - log\(_{10}\)2 = 0.4471 + 0.771 - 0.3010 = 0.6532

234 - 4x\(^2\) = 18\(^2\) - (2x)\(^2\) = (18 - 2x)(18 + 2x) 2x + 18 = 2x + 18 = (2x + 18) 18 - 2x = 2(9 - x) or -2(x - 9)

Rice = 75g, Margarine = 40g, Meat = 105g Bread = 20g Total = 240 Angle of sector represented by meat = \(\frac{105}{240} \times \frac{360^o}{1}\) = 157.5

The lenght of Arc AB = \(\frac{Q}{360}\) 2\(\pi\)r = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x 10.5 = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times 3}{30} \times \frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm

\(\frac{27}{5} \times 3^{-3} \times \frac{(1)^{-1}}{5}\) = \(\frac{27}{5} \times \frac{1}{3^3} \times \frac{1}{\frac{1}{5}}\) \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{5}{1}\) = 1

Coca-Cola = 10 bottles, Fanta = 8 bottles, Spirite = 6 bottles Total = 24 P(Coca-Cola) = \(\frac{10}{24}\); P(not Coca-Cola) 1 - \(\frac{10}{24}\) \(\frac{24 - 10}{24} = \frac{14}{24} = \frac{7}{12}\)

Since PQRS is quadrilateral 2y + 2x + QPS = 360\(^o\) i.e. (y + x) + QPS = 360\(^o\) QPS = 360\(^o\) - 2 (y + x) But x + y + 68\(^o\) = 180\(^o\) There; x + y = 180\(^o\) - 68\(^o\) = 112\(^o\) QPS = 360 - 2(112\(^o\)) = 360\(^o\) - 224 = 136\(^o\)