Published on October 17th 2023 | 2 mins , 347 words
Mathematics Paper 1
An aircraft company bought eight aircrafts for eighteen billion, nine hundred and seventy-five million, twenty-eight thousand, two hundred and fourty
(a) Write the total cost of the eight aircrafts in figures
18,975, 028,240
(b) Calculate the cost of each aircraft
\(\frac{18,975, 028,24} {8}\)
\(2,371,878,530\)
2. Solve for x in the equation
\(\frac{3}{x+1}+ \frac{2}{x+5}= \frac{1}{x-2} \)
\(\frac{3x+15+2x+2}{(x+1)(x+5)}= \frac{1}{x-2} \)
\((x-2)(5x+17)\)
\(=(x+1)(x+5)\)
\(5x^{2}+17x-10x+34\)
\(=x^{2}+6x+5\)
\(4x^{2}+x-39=0\)
\(x=\frac{-1\pm\sqrt{1-16\times39}}{8}\)
\(x=3 \: or \: x=-3.25\)
3a) The number 16200 is given as \(2^{x}\times 3^{y}\times5^{z}\)
Find the value of \(x+y+z\)
\(16200= 2^{3}\times 3^{4}\times5^{2}\)
x=3,y=4 and z=2
x+y+z=3+4+2=9
b) When a number N is multiplied by 16200, a perfect cube is obtained. Find the least value of N
\(2^{3}\times 3^{4}\times5^{2}\)
$$2^{0}\times 3^{2}\times5^{1}$$
\(=45\)
4. Given that \(\sin\alpha^{\circ}=\frac{1}{\sqrt{5}}\) where \(\alpha \) is an acute angle, find without using mathematical tables
(a) \(cos \alpha^{\circ} \) in the form of \(a\sqrt{b}\) where \(a\) and \(b\) are rational numbers
\(=\frac{2\sqrt{5}}{\sqrt{5}\sqrt{5}}\)
\(=\frac{2\sqrt{5}}{5}\)
(b) \(\tan (90-\alpha)^{\circ}\)
\(=\frac{2}{1}=2\)
5. The area of a rhombus is \(60cm^{2}\). If the shorter diagonal is 8 cm. Find the perimeter of the rhombus (4mks)
Area = \(\frac{1}{2}\times D \times d\)
60 = \(\frac{1}{2}\times D \times 8\)
\(\Rightarrow 4D=60\)
D = 15 cm
\(s^{2}= 7.5^{2} + 4^{2}\)
\(s^{2}=72.25 \)
\(s = 8.5 cm\)
P = 4s
P = \(4 \times 8.5 \)
P = 34 cm
6. A 63 kg metal of density, 7000 \(kg/m^{3}\) is moulded into a rectangular pipe with external dimensions of 12 cm by 15 cm and internal dimensions of 10 cm by 12 cm. Calculate the length of the pipe in metres.
Volume of material used to make the pipe \(V_{1}\) = \(\frac{mass}{densiry}\)
\(V_{1}\) = \(\frac{63}{7000}\)
= 0.009 \(m^{3}\)
= \(0.009 \times 1000000 \: cm^{3}\)
= \(9000 \: cm^{3} \)
Cross-sectional area of the rectangular pipe
= \((12 \times 15)\) - \((10 \times 12)\)
= \( 60 \: cm^{2}\)
But Volume equals cross-sectional area \(\times\) length
This implies that length of the pipe therefore = \(\frac{volume}{cross-sectinal \: area}\)
= \(\frac{9000}{60}\)
= 150 cm
= 1.5 m