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) 

𝐿2 ⟹ 𝑥 ≥ −1

𝐿3 ⟹ 5𝑦 < −3𝑥 + 15 

11. A dealer sells a certain spare part for Kshs 650, making a profit of 30%. The manufacturer reduces the price to the dealer by Kshs 50 and the dealer reduces his selling price by the same amount. Find the dealer’s new percentage profit.(3mks)

𝑜𝑙𝑑 𝑏𝑢𝑦𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒  =\(\frac{100}{130}\times 650 =500\)

𝑛𝑒𝑤 𝑏𝑢𝑦𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 = 500 − 50 = 450

𝑛𝑒𝑤 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 = 650 − 50 = 600

%𝑝𝑟𝑜𝑓𝑖𝑡 = \(\frac{600-450}{450}\times100\)

\(=33.33 \% \)

12. A taxi travelling at 20m/s accelerates uniformly and in 4 seconds, its velocity is 30m/s. it maintains this velocity for another 5 seconds before decelerating uniformly to rest after 3 seconds. Calculate the total distance travelled by the taxi during the journey.(3mks)

Distance = area under the curve 

= ( ½ x 4 ( 20+30) + 5 x 30 + (½ x 3 x 30) m 

= 100 + 150 + 45 

= 295m 

13. The length of a rectangle is (𝑥 + 3) cm. If the width of the rectangle is two-thirds its length and the perimeter is 40 cm, find its width. (3 mks) 

Perimeter = \(2(x+3+\frac{2}{3}(x+3))=40\)
 \((2x+6+\frac{4}{3}(x+3))=40\)

Multiplying through by 3:

\(6x+6+4(x+3)=120\) 

\(6x+18+4x+12=120\)  

\(10x+30=120\) 

\(10x=120-30\)

\(10x=90\)   

\(x=9cm\)  

length = 9+3=12

width = \(\frac{2}{3}\times12 \) 

\(x=8cm\) 

14. Ali traveled a distance of 5km from village A to village B in the direction of N600E. He then changed direction and traveled a distance of 4km in the direction of 1350 to village C.

a) Using a scale of 1cm to represent 1.0 km represent the information on an accurate diagram.(2marks)

• Check for the correct diagram 

b) Using the scale drawing in (a) above determine

(i) The distance between A and C        (1mk)

\(7km \)

(ii)The bearing of A from C  (1mk)

\(226^\circ\)  

15. The figure below is a rhombus ABCD of sides 4cm. BD is an arc of circle centre C. Given that ABC = \(138^\circ\). Find the area of the shaded region correct to 3 significant figures.  \( Take \: \pi=\frac{22}{7}\)

Area of the rhombus = \(\frac{1}{2} \times  4^2 \times \sin 42^\circ\)

\(=10.71 cm^2\)

Area of the sector \(=\frac{42}{360}\times \frac{22}{7}\times4^2\)

\(=5.867\:𝑐𝑚^2 \)

Shaded Area \(= 10.71 − 5.867\)

\(= 4.56\:𝑐𝑚^2\) 

16. The figure below is a part of the sketch of a triangular prism ABCDEF. 

• Transfer the angle F to E 

• Transfer length FB to EC

• Using the length of EC construct the triangle  CDE using a pair of compasses

Join AD, DE, and DC using broken  lines