Edufocus.co.ke

Quality Exams

Form 4 Mathematics Paper 1 MECS Joint Exam

Published on November 6th 2023 | 5 mins , 959 words

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

 \(π‘₯ + π‘₯ + 2 + π‘₯ + 4 + π‘₯ + 6 < 64 \)

\(4π‘₯ < 48\)

\( π‘₯ < 12 \)  

 \( 11,13,15,17 \)


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 \(π‘₯^3𝑦^2\)  and \(4π‘₯𝑦^4 \). Hence factorise completely the expression   \(x^3y^2-4xy^4\) (3mks)

\(𝐺𝐢𝐷 = π‘₯𝑦^2  \)

\( =π‘₯𝑦^2(π‘₯^2 βˆ’ 4𝑦^2) \) 

\(= π‘₯𝑦^2(π‘₯^2 βˆ’ (2𝑦)^2) \)

\(= π‘₯𝑦^2(π‘₯βˆ’2𝑦)(π‘₯+2𝑦) \)

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  \( 𝑂𝐡 =  (\begin{smallmatrix}x \\ -2\end{smallmatrix})\)   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 \(p^3 Γ— q^2 Γ— r = 2250\). Find p, q and r.(3 marks)


\(p^3 Γ— π‘ž^2 Γ— r = 5^3 Γ— 3^2 Γ— 2\) 

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


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) 


Inequality Graph
𝐿1 ⟹ 3𝑦 β‰₯ π‘₯ βˆ’ 5

𝐿2 ⟹ π‘₯ β‰₯ βˆ’1

𝐿3 ⟹ 5𝑦 < βˆ’3π‘₯ + 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)


π‘œπ‘™π‘‘ 𝑏𝑒𝑦𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’  =\(\frac{100}{130}\times 650 =500\)

𝑛𝑒𝑀 𝑏𝑒𝑦𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 500 βˆ’ 50 = 450

𝑛𝑒𝑀 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 650 βˆ’ 50 = 600

%π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ = \(\frac{600-450}{450}\times100\)

\(=33.33 \% \)

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 

= ( Β½ x 4 ( 20+30) + 5 x 30 + (Β½ x 3 x 30) m 

= 100 + 150 + 45 

= 295m 


13. The length of a rectangle is (π‘₯ + 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)


scale drawing


  • 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.  \( Take \: \pi=\frac{22}{7}\)

Rhombus

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

\(=10.71 cm^2\)

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

\(=5.867\:π‘π‘š^2 \)

Shaded Area \(= 10.71 βˆ’ 5.867\)

\(= 4.56\:π‘π‘š^2\) 


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


net of a triangular prisms
Complete the sketch by drawing the hidden edges using broken lines. (3 marks)


  • 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 


sketch

Download File