Mathematics Paper 1 Questions and Answers - Cekana Mock Exams 2023

INSTRUCTIONS

• The paper consists of two sections. Section I and Section II.
• Answer ALL the questions in Section I and any FIVE questions in Section II.
• Show all the steps in your calculations, giving your answer at each stage in the spaces provided
  below each question.
• Marks may be given for correct working even if the answer is wrong.
•  Non-programmable silent electronic calculators and KNEC Mathematical tables may be used
  except where stated otherwise.
• Candidates should answer the questions in English.

QUESTIONS

SECTION I (50 MARKS)
Attempt All Questions in this section

1. Find the value of y given that (3 marks)
   −2(5+3)−9÷3+5 =1
    −3x+5y+2y×4
2. The length of a minute hand of a clock is 3.5cm. What will be the time if from 10.15am it sweeps through an area of 19.25cm²? (4marks)
3. Use reciprocal, square and square root table to evaluate to 4 significant figures, the expression (3marks)
   
4. Given that sin (90-x)⁰ =0.8, where x is an acute angle, find without using mathematical table the value of 2 tan x+cos(90-x) (3marks)
5. Find the equation of the tangent and the normal to the curve y=2x³-3x²+6  at the point (2,10) (4marks)
6. Simplify 2y²+3xy−2x² (3marks)
                    x²−4y²
7. Find greatest integral value of x which satisfies (3marks)
   2x+3 < 8−3x <5x+6
      2          5          3
8. One of the roots of the equation x²+(k+1)x+28=0 is 4. Find the values of k and hence the second root (4marks)
9. Solve for x in the following without using a calculator or mathematical table. (3marks)
   9^x(27^(x-1))=tan30°
10. A shear parallel to the x-axis maps point (1, 2) onto a point (5, 2). Determine the shear factors and hence state the shear matrix (invariant line is y=0) (3marks)
11. A solid is in the shape of a right pyramid on a square base on side 8cm and height 15cm. A frustum whose volume is a third of the pyramid is cut off. Determine the height of the frustum. (3 marks)
12. The interior angle of a regular polygon is 20° more than three times the exterior angle. Determine the number of sides of the polygon (2marks)
13. Three fifths of work is done on the first day. On the second day 3⁄4 of the remainder is completed. If the third day 7⁄8 of what remained id done, what fraction of work still remain to be done. (3marks)
14. A two-digit number is such that the sum of the digits is ten. If the digits are reversed, the new number formed is less than the original number by 18. Find the number (3marks)
15. In the figure bellow, ∠CBD= 2 ∠CAD. Find the value of x (3marks)
    
16. Two boys, Ababu and Chungwa, on the same side of a tall building are 100m apart. The building and the two boys are in a straight line and the angles of elevation from the boys to the top of the building are 30° and 20° respectively calculate the height of the building. (3marks)

SECTION II (50 MARKS)
Attempt only five Questions from this section

17. A business lady bought 100 quails and 80 rabbits for sh25600. If she had bought twice as many rabbits as half as many quails she would have paid sh7400 less. She sold each quail at a profit of 10% and each rabbit at a profit of 20%.
    1. Form two equations to show how much she bought the quails and the rabbits. (2marks)
    2. Using matrix method, find the cost of each animal. (5mrks)
    3. Calculate the total percentage profit she made from sale of the 100 quails and 80 rabbits. (3marks)
18. A(3,7) B(5,5), C(3,1), D(1,5) are vertices of a quadrilateral
    1. On the grid provided below, plot ABCD on a Cartesian plan (2marks)
    2. A¹B¹C¹1D¹ is the image of ABCD under a translation T  plot A¹B¹C¹D¹and state its coordinates (2marks)
    3. Plot A¹¹B¹¹C¹¹D¹¹ the image of A¹B¹C¹D¹ after a rotation about (-1,0) through a positive quarter turn. State its coordinates. (3marks)
    4. A¹¹¹B¹¹¹C¹¹¹D¹¹¹ is the image of A¹¹B¹¹C¹¹D¹¹ after a reflection in the line y=x+2. Plot A¹¹¹B¹¹¹C¹¹¹D¹¹¹ and state its coordinates (3marks)
       
19. The figure below shows a velocity time graph of a journey of a car. The car start from rest and accelerates at 2 3⁄4 for t seconds until it is 22m/s
    
    Brakes are applied bringing it uniformly to rest. The total journey is 847m long. Find.
    1. The value of t, the acceleration time (2marks)
    2. The distance travelled during the first t seconds. (2marks)
    3. The value of x, the deceleration time (4marks)
    4. The rate of deceleration (2marks)
20. The diagram below shows a histogram representing the mass of some pupil in a school.
    
    
    1. Prepare a frequency distribution table of the data (3marks)
    2. From the table above, estimate
       1. The mean mass of the pupils to 3 s.f (3marks)
       2. The median mass (3marks)
       3. How many pupils were 40kg and above (1marks)
21. In the figure below OY:YA=1:3 AX:XB=1:2 OA=a and OB=b . N is the point of intersection of BY and OX
    
    
    1. Determine
       1. OX (2marks)
       2. BY (1mark)
    2. Give that  and ON=hOX, express in two ways in terms of a b,h and k (3marks)
    3. Find the value of h and k and hence show that O,N and X are collinear (4marks)
22. Using a ruler and a pair of compass only, construct triangle XYZ where XY is 6cm and XYZ is 135° and YZ=7cm (2marks) 
    1. Measure XZ (1mark)
    2. Drop a perpendicular from Z to meet line XY at K , measure ZK. (2marks)
    3. Bisect line XY and let the bisector meet line XZ at Q (1mark)
    4. Join Q to Y and measure angle XQY (2marks)
    5. Find the area of triangle XYZ (2marks)
23. Four towns are situated in such that a way that town Q is 500km on a bearing of 120° from P. Town R is 240 km on a bearing of 210° from town P, while town S is due north of town Q and due east of town P
    1. Draw a sketch diagram showing the relative position of P,Q,R and S (scale:1cm :100km) (3marks)
    2. Find by calculation
       1. The distance QR (1mark)
       2. The distance QS (2marks)
       3. The angle PRSQ (2marks)
       4. The area of triangle PQS (2marks)
24. The figure below shows a sketch of the graph of y=x²+6 .
    
    
    1. Estimate the area bounded by the curve, the x-axis and the line x =−4 and x=4 using
       1. The trapezium rule with 8 sides (3marks)
       2. The mid- ordinates rule with and strips (3marks)
    2. What percentage error is caused by estimating the area of the curve using the mid ordinates rule as in a (ii) above (4marks)

Marking Scheme

1.	num: -2(8)-3+5 -16-3+5 =-14 den: 15y+-8y =7y -14⁄y =1 -14=7y -14⁄7=y y= -2	m1 m1 a1
		3 marks
2.	19.25= θ⁄360 ×π × 3.52     θ= 19.25×360/ π×3.52      =180   time= 180⁄360 × 60=30min   ∴time=10.15+30 10.45+30 =10.45am	m1 m1 m1 a1
		4marks
3.		b1 b1 b1	üreciprocal üsquare üsquare root
		3marks
4	sin (90-x)=8⁄10  tanx =6⁄8 cos(90-x) = 6⁄10 2tanx + cos(90-x) =2(6⁄8)+ 6⁄10 = 6⁄4 + 6⁄10 =2 1⁄10	b1 m1 a1	Diagram
		3marks
5.	dy⁄dx =6x2 - 6x atx =2  dy⁄dx =12  equation of tangent  y-10/x-2 =12 y=12x-14 grade of normal =-1⁄12 equation⇒y-10/x-2 = -1/12 y=-1/12x+10 1⁄6	m1 a1 b1 b1
6.	num: 2y2 +3xy − 2x2 (2y-x)(y+2x) den: x2-4y2=(x-2y)(x+2y) (2y-x)(y+2x)/ (x-2y)(x+2y) =-y-2x/ x+2y	m1 m1 a1
7.i)            ii	2x+3 < 8-3x      2         5 10x +15 <16-6x 16x<1 x< 1 1⁄6 8.3x < 5x+6     5          3 24-9x<25x+30 -6<34x -3/ 17 < x 3/ 19 < x < 1⁄16 integral value ∴x=0	b1 b1 b1
		3marks
8		M1 m1 a1 b1
		4marks
9	32x+3(x-1) =3 - 1⁄2 5x - 3 = - 1⁄2 5x = 2 1⁄2 x=0.5	b1 m1 a1	ü power
10		m1A1 b1
		3marks
11.	vs.ƒ = 1/ 2⁄3 = 3⁄2 ls.ƒ = 3√ 3⁄2 =1.145 1.145 =15/ 15-h 17.175-1.145h =15 h=1.90cm	b1 m1 a1
		3marks
12	exterior=x∴interior =3x+20 x+3x+20=180 x=40° no of sides = 360⁄40 =9	m1 a1
		2marks
13.	3⁄5 + 3⁄4 of 2⁄5 =9⁄10 3rd day=7⁄8 × 1⁄10 =7⁄80 9⁄10 + 7⁄80 =79⁄80 work remaining⇒1- 79⁄80 = 1⁄80	m1 m1 a1
		3marks
14.	x+y =10 (10x+y)-(10y+x)=18 9x-9y=18 x-y=2 x+y=10/ 2x  =12 x =6 ∴y=4 number is 64	b1 m1 a1	Both equations Attempt to solve
		3marks
15.	x-2x+x+20∠CBD=x+30+x+2x+ ∠CBD/ 2 4x+20+∠CBD=180or4x+30+∠CBD/ 2 =180 ∠CBD=180-20-4xor∠CBD=360-60-8x ⇒160-4x=300-8x 4x=140 x=35°	m1 m1 a1
		3marks
16.	tan 20 =     h    ⇒=(100+x)tan 20               100+x tan 30 =h/x ⇒ h=xtan 30 xtan30=(100+x)tan20 0.5774x=36.40+0.3640x x=  36.40        0.2134 =170.57 h=170.5757 tan30 98.48m	M1 A1 b1
		3marks
17.a)	100q+80r=25600 5q+4r=1280 50q+160r=18200 5q+160r=18200 5q+16r=1820	b1 b1	Equation ü
b)		b1 m1 m1 m1 a1	Invers Pre multiplication Evidence multiplication Simplification both
c)	100q+80r=100(400)+80(45) =43600 110/ 100  100 × 400 +120/ 100 × 80 × 45  %=(48320-43600/ 48320) × 100 =9.768%	m1 m1 a1	Buying price Selling price
		10marks
18.	A'(-3,-2)B'(-1.-4)C'(-3,-8)D'(-5,-4) A''(-1,2)B''(3,0)C''(-7,-2)D''(-3,-4) A'''(-4,3)B'''(-2,5)C'''(-4,9)D'''(-6)	b2 b1 B1 B1 B1 B1 B1 B1 B1	üABCD üA1B1C1D1 üline y=x+2
19.a)	a= v-u/ t 2 3⁄4 =22-0/t t=22/ 2.75= 8s	m1 a1
b)	d=1⁄2 × 8 × 22 =88m	m1 a1
c)	847 = 88 +22 × 32 + 1⁄2 × 22 × x 847-792+11x  x= 55⁄11 =5s	m1m1 m1 a1	addition
d)	d= v-u/ t =0-22/5 =-4.4ms-2	m1 a1
		10marks
20.a)	Mass Kg x f fx cf 0-5 6-15 16-35 36-50 2.5 10.5 26.5 43 40 100 80 90 100 1050 2120 3870 40 140 220 310       7140	b1 b2 b1 b1 b1	üclasses üfrequencies for any 2üfrequencies üfx üc.f
b) (i)	x=∑ ƒx/ ∑ƒ  = 7140                             310 =23.0kg	m1 a1
(ii)		M1 a1
(iii)	10.5 × 6 =63pupils	b1
21.a)i)      (ii)	OX=OA +AX =a+1⁄3(b-a) =2⁄3 a+ 1⁄3 b BY=BO+OY =-b+14 a	m1 a1 b1
b)      i            ii	BN=-kb+1⁄4 ka ON=b-kb+1⁄4 ka =b(1-k)+1⁄4 ka ON=h(2⁄3 a+ 1⁄3 b) =2⁄3 ha+ 1⁄3 hb	m1 a1 b1
c)	b(1-k)+1⁄4 ka =2⁄3 ha+1⁄3 b 1-k=1⁄3 h 1⁄4 k=2⁄3 h 1- 8⁄3 h=1⁄3 h 9⁄3 h=1 h=1⁄3 k=8⁄3 × 1⁄3 =8⁄9 ON=hOX ON=1⁄3 OX ∴ON // OX O is common hence O,N,X are collinear	m1 m1 a1 b1	both
22		b1 b1	ü construction of 135o at y ü xyz
a)	xz=12cm±0.1	b1
b)	ZK=4.9cm±0.1	b1 b1	ü┴ from Z
c)		b1	ü┴ bisector
d)	∠XQY=133o±1	b1 b1	-QY ü angle
e)	area = 1⁄2 × 6 × 4.9 =14.7cm2	M1 A1
		10marks
23.a)		b1 b1 b1	ü location of Q ü location of R ü location of S
b)i		b1
ii)	sin 30 = QS/ 500 QS=500sin30 =250km	m1 a1
iii)	tan R =500⁄240 R= tan-1 (500⁄240) =64.39°	M1 a1
iv)	area= 1⁄2 × 500 × 250sin60 =54126.6km2	m1 a1	Follow through
24.a) i)	x -4 -3 -2 -1 0 1 2 3 4 y=x2+6 22 15 10 7 6 7 10 15 22  A= 1×2×1[(22+22)+2(15+10+7+6+7+10+15)] = 12 [44+140] =92sq unit	b1 m1 a1	-all üordinates
ii)	x -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 y=x2+6 18.25 12.25 8.25 6.25 6.25 8.25 12.25 18.25  A= 1[18.25+12.25+8.25+6.25+6.25+8.25+12.25+18.25] =90sq units	b1 m1 a1	ordinates
b)		m1 m1 m1 a1	Corr integration