1. The sum of four consecutive odd integers is less than 64. Determine the first four such integers.  (3mks) 

 \(𝑥 + 𝑥 + 2 + 𝑥 + 4 + 𝑥 + 6 < 64 \)

\(4𝑥 < 48\)

\( 𝑥 < 12 \)  

 \( 11,13,15,17 \)

2. Solve the equation  \(\frac{2}{t-1}  - \frac{1}{t+2}= \frac{1}{t} \)  (3mks)

\(\frac{2t+4-t+1} {(t-1)(t+2)} =  \frac{1}{t}\)

\( \frac{t+5}{t^2+t-2} = \frac{1}{t}\)

\(t^2+5t=t^2+t-2\)

\(4t=-2\)

\(t=\frac{-1}{2}\)

3. Moses has twenty shillings more than Jane. After he spends a quarter of his money and Jane \(\frac{1}{5}\) of hers, they find that Jane has 10 shillings more than Moses. How much money did both have? (4 mks)

Before spending:

Moses \(=20+x\)

Jane  \(=x\)

After spending:

Moses \(= \frac{60}{4} + \frac{3}{4}x\)

Jane =\( \frac{4}{5}x\)

\( \frac{4}{5}x   - ( \frac{60}{4} + \frac{3}{4}x ) = 10\) 

\(16x-15x-300 =200\)

\(x-300=200\)

\(x=500\)

\(Jane: 500\) and \(Moses: 520\)

4. The sum of interior angles of two regular polygons of side \(n-1\) and \(n\) are in the ratio 4:5. Calculate;

(i) The size of interior angle of the polygon with side \(n-1\) (2 marks)

\( \frac{(2(n-1)-4)90}{(2n-4)90}=\frac{4}{5}\)

\( \frac{2n-6}{2n-4} =\frac{4}{5}\)

\( \frac{2(n-3)}{2(n-2)} =\frac{4}{5}\)  

\(5n-15 = 4n-8\)

\(n=7\)

For a polygon with n-1 sides i.e 6 sides, the interior angle is given by:

\(\frac{(2(6)-4)90 ^{\circ}}{6}= 120  ^{\circ}  \)

(ii) The size of the exterior angle of the polygon with side \(n-1\)(1 mark)

5. Find the greatest common factor of \(𝑥^3𝑦^2\)  and \(4𝑥𝑦^4 \). Hence factorise completely the expression   \(x^3y^2-4xy^4\) (3mks)

\(𝐺𝐶𝐷 = 𝑥𝑦^2  \)

\( =𝑥𝑦^2(𝑥^2 − 4𝑦^2) \) 

\(= 𝑥𝑦^2(𝑥^2 − (2𝑦)^2) \)

\(= 𝑥𝑦^2(𝑥−2𝑦)(𝑥+2𝑦) \)

6. Without using a calculator, solve for n in the equation   

\(1-(\frac{1}{3})^n=\frac{242}{243}\) 

\( (\frac{1}{3})^n=1 - \frac{242}{243}\)

\((\frac{1}{3})^n=\frac{1}{243}\) 

\((\frac{1}{3})^n=(\frac{1}{3})^5\)

\(\therefore n=5\)     

7. Given that \( OA = (\begin{smallmatrix}-2 \\ 10\end{smallmatrix})\)  and  \( 𝑂𝐵 =  (\begin{smallmatrix}x \\ -2\end{smallmatrix})\)   and that the magnitude of AB is 13 units, find the possible values of x.       (3mks) 

Solution

\(AB = (\begin{smallmatrix}x \\ -2\end{smallmatrix})-  (\begin{smallmatrix}-2 \\ 10\end{smallmatrix})\) 

\(= (\begin{smallmatrix} x+2 \\ -12\end{smallmatrix})\) 

\(\sqrt{(x+2)^2 + (-12)^2}=13\)

When we square both sides, we obtain:  

\((x+2)^2 + (-12)^2=169\) 

\(x^2 + 4x+4+144=169\) 

\(x^2 + 4x-21=0\)    

\(x^2 + 7x-3x-21=0\) 

\(x(x + 7)-3(x+7)=0\)   

\((x-3)(x+7)=0\)  

\((x-3)=0 \:  or \: (x+7)=0\)  

\(x=3 \: or \: x=-7\)       

8. Three numbers p,q and r are such that \(p^3 × q^2 × r = 2250\). Find p, q and r.(3 marks)

\(p^3 × 𝑞^2 × r = 5^3 × 3^2 × 2\) 

\(p = 5 \: q = 3 \:and \: r = 2 \)

9. A bus starts off from Kitale at 9.00 a.m. and travels towards Kakamega at a speed of 60km/hr. At 9.50 a.m., a matatu leaves Kakamega and travels towards Kitale at a speed of 60 km/h. If the distance between the two towns is 150km, how far from Kitale will the two vehicles meet? ( 3mks)

At 9.50 am, the bus has travelled:

\((\frac{50}{60} \times 60)=50km)\)

The distance between the two vehicles at 9.50 a.m.:

150 - 50 = 100km

Rel . speed = 120 km/ h.

Time taken to meet = \(\frac{100}{120}\)

\(= 50 min\)  

Distance covered by the bus:

\((\frac{50}{60} \times 60)=50km\)

Distance from Kitale to the meeting point

\(=50+50=100km\) 

10. Find the inequalities that satisfy the region R shown in the figure below. (3mks)