Question Library

\(cos\theta = \frac{adj}{hyp}=\frac{300}{600}=\frac{1}{2}\\ \theta = cos^{-1}(0.5000)=60^{\circ}\) The bearing of P from \(Q = \theta + 180 = 60 + 180 = 240^{\circ}\)

\(A=\pi r^2 \Rightarrow 77 = \frac{22}{7} \times \frac{r^2}{1}\Rightarrow r^2 = \frac{77\times 7}{22} \Rightarrow r^2 = \frac{49}{2}\\ \Rightarrow r = \sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}. Since D = 2r ∴ D = 2 \times \frac{7}{\sqrt{2}} = \frac{14}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 7\sqrt{2}cm\)

2x - 3y = 22 ---- eqn I 3x + 2y = 7 ---- eqn II multiply eqn I by 2 4x - 6y = 44 ---- eqn III multiply eqn II by 3 9x + 6y = 21 ---- eqn IV Adding eqn III and IV => 13x = 65 => x = 5 Substituting 5 for x in eqn II 3x5 + 2y = 7 => 2y = -8 y = -4

\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 8y - 4 - 9y + 3 = 12 \\ \Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)

\(log_9x= 1.5\Rightarrow x = 9^{1.5} = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=3^3=27\)

\(a = 2\frac{1}{2}, nth = a + (n-1)d \Rightarrow 25 = 2\frac{1}{2} + (n-1)2\frac{1}{2}\\ 25 = \frac{5}{2}+(n-1)\frac{5}{2} \Rightarrow 22\frac{1}{2} = \frac{5n-5}{2}\Rightarrow \frac{45}{2} = \frac{5n-5}{2}\\ 45 = 5n - 5 \Rightarrow 5n = 50 \Rightarrow n = 10\)

Substituting for x in the equation \(2(2)^2 + (k+2)2+k = 0 \Rightarrow 8 +2k + 4 + k =0 \Rightarrow 3k =-12; k=-4\)

mean \(=\frac{1+3+4+8+8+4+7}{7}=\frac{35}{7}=5\)

T.S.A of s closed cylinder = \(2\pi r(r+h)\) =\(\frac{2}{1}\times \frac{22}{7} \times \frac{7}{2}\left(\frac{7}{2}+\frac{10}{1}\right) = \frac{22}{1}\left(\frac{27}{2}\right) = 27 \times 11 =297cm^2\)

The perimeter of the sector \(=2r+\frac{\theta}{360}\times 2\pi r \\ \Rightarrow 28 + \frac{300}{360} \times \frac{2}{1} \times \frac{22}{7}\times \frac{14}{1} = \frac{220}{3}+28\\ 73.133+28=101.33cm\)

No expanded working has been attached for this question yet. The verified answer is Option A.

PI = {1, 7, 15} ; QI = {1, 13, 11} PI ∩ QI = {1}

Since 36km in 21 minutes 1km in \(\frac{21}{36}\)minutes 60km in \(\left(\frac{21}{36}\times 60\right)\)minutes = \(\frac{70}{2}\)min = 35 mins

\(\%Increase = \frac{Actual \hspace{1mm}increase}{initial \hspace{1mm}ratio}\times 100\%\) Actual increase = 7; Initial ratio = 40 \(=\left(\frac{7}{40} \times 100\%\right)=\frac{35}{2}=17.5\%\)

Exterior angle \(=\frac{360}{n}\) where n = no of sides \(=\frac{360}{9}=40^{\circ}\)