INSTRUCTIONS TO CANDIDATES
- The paper contains TWO sections: Section A and B
- Answer ALL questions in section I and STRICTLY ANY FIVE questions from section II.
- All working and answers must be written on the question paper in the spaces provided below each question.
- Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
- Marks may be awarded 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.
SECTION A (50MKS)
ANSWER ALL THE QUESTIONS IN THIS SECTION
- Simplify (1³/₇ − ⁵/₈) + ²/₃ of 1¹/₅ (4mks)
¾ + 1⁵/₇ ÷ ⁴/₇ of 2¹/₃ - A straight line ax + by = 16 passes through A (2, 5) and B (3, 7). Find the values of a and b (3mks)
- Simplify 2−x−x2 (3mks)
3x2−2x−1 - Solve for X where 0 ≤ x ≤ = 90° (2mks)
sin 2x − cos (x−30) = 0 - Solve for X in (3mks)
2x − 4 ≤ 3x + 2 < 10 − x
Hence represent your solution on a number line - Two similar cylindrical solids have heights of 18 cm and 24 cm. The volume of the larger cylinder is 320cm3, find the volume of the smaller cylinder (4mks)
- Solve for X (3mks)
83x − 2 x 16½x = ¼ - A quantity P varies jointly as Q and inversely as on the square root of R. If Q is increased by 10% and R is reduced by 19%, find the percentage change in P (3mks)
- Okedi sold goods whose marked price is sh. 340,000 at a discount of 2%. He was paid sh. 16660 as commission for the total sales. Calculate the percentage rate of commission (3mks)
- The interior angle of a regular polygon is three and a half times the exterior angle. Determine the sides of the polygon (3mks)
- Give that A = , B = ;find matrix C where AC = B (3mks)
- Amoit bought 2 pens and 5 exercise books at a cost of sh. 275. Allan bought 4 such pens and exercise books from the same shop at a cost of sh. 415 by letting sh. X and y to be the costs of a pen and a book respectively, find the cost of each item (4mks )
- Okech left some money in his will to be shared amongst his wife, son and daughter in the ratio 4:3:2 respectively. If the daughter received sh. 120,000 less than the mother’s share, find the total amount of money Okech left in his will. (2mks)
- Use tables of reciprocals to find the reciprocal of 0.3758. Hence find the value 3√0.125 correct to 4.S.f (4mks)
0.3758 - A major sector of a circle subtends an angle of 150 at the centre. The radius of the circle is 7cm and the centre is at O as shown
If the sector is folded into a conical shape, calculate the radius of the cone correct to 1 d.p (3mks) - A Kenyan bank buys and sells currencies at the exchange rates below
Currency Buying (ksh) Selling (ksh) 1 euro
1 us dollar147.87
74.22148.00
74.50
SECTION II (50MKS)
ANSWER ANY FIVE QUESTIONS IN THIS SECTION
- The diagonals of a rectangle P, Q, R, S intersect at (5, 3). Given that the equation of line PQ is 4y − 9x =13 and that of line PS is y − 4x =5
- The co-ordinators of P (3mks)
- The co-ordinates of R (2mks)
- The equation of line RQ (2mks)
- The equation of a perpendicular line drawn to meet PR at (5,3) (3mks)
- A bus left Malaba town at 6.00am and travelled at an average speed of 80km/h towards Nairobi which is 510km away. At 6.30am a salon car left Nairobi the same day following the same route and travelled at average speed of 100km/h towards Malaba. After 1 hour, the car had a puncture which took 15minutes to repair before proceeding with the journey;
Determine- The distance covered by the bus in 30minutes (1mks)
- The time of the day when car met the bus. (6mks)
- The distance from Nairobi to the point where the car met with the bus (2mks)
- The time of the day to the nearest minute when the bus got to Nairobi (1mk)
- Points P, Q and R are a straight line on a level ground. An electricity pole is erected at P with a point X and Y on it. From point X, the angle of depression of point Q is 48° while the angle of depression of R from Y which is 3m above X is 60°
- Illustrate the position of X, Y, P and R by sketching. (1mk)
- Hence calculate to 1 d.p.
- The length XP (3mks)
- The distance YQ (2mks)
- The distance PQ (2mks)
- The angle of elevation of Y from R given that PR = 8cm (2mks)
-
- The figure shows a velocity- time graph of a car
- Find the total distance covered by the car in metres (3mks)
- Calculate the deceleration of the car (3mks)
- A lorry left kisumu at 8.00am and travelled towards the Nakuru at an average speed of 72km/h. At 8.30am a matatu left kisumu and followed the lorry at an average speed of 96km/h.
Determine the time of the day when the matatu caught up with the lorry (4mks)
- The figure shows a velocity- time graph of a car
- The table below shows marks scored by 48 students in a geography exam.
Marks % 30-39 40-49 50-59 60-69 70-79 80-89 Students 6 10 x 9 12 2 - Determine the value of x (2mks)
- State the modal class (1mk)
- Calculate the (3mks)
- Mean mark
- Median mark (4mks)
-
- Complete the table below for the equation Y = x2 + 3x − 6 where −7 ≤ × ≤ 4
x −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 y 4 −6 −2 - Using the scale 1 cm to represent 1 unit on the X- axis and 1cm to represent 2 units on the Y - axis, draw the graph of y = x2 + 3x − 6 for −7≤ ×≤4 (4mks)
- Use your graph to solve for x in
x2 + 3x − 6 =0 (2mks) - State the;
- Turning point of the curved (1mk)
- Equation of the line symmetry (1mk)
- Complete the table below for the equation Y = x2 + 3x − 6 where −7 ≤ × ≤ 4
- the figure shows triangle ABC inscribed in a circle where AC = 10cm, BC = 7cm and AB =11cm
Calculate correct 1 d p ( use π = 22/7 )- The size of the angle CAB (4mks)
- The radius of the circle (2mks)
- Hence, find the area of the shaded region (4mks)
- ABCDEFGA is a belt tied around two wheels whose centres are O and Q forming a pulley system. Given that Q =36cm, AO = 5cm BQ = 7cm. calculate correct 1 d.p (Take π = 22/7 )
- Angle AOQ (3mks)
- The length of the belt in contact with
- The wheel whose centre is O (2mks)
- The wheel whose centre is Q (2mks)
- The length of AB, hence the total length of the belt (3mks)
MARKING SCHEME
- Num (¹⁰/₇ − ⁵/₈) + ²/₃ of ⁶/₅
= 80 − 35 + 4
56 5
= 45/56 + 4/5
= 225 + 224
280
= 449
280
Den ¾ + ¹²/₇ ÷ ⁴/₇ of ⁷/₃
= ¾ + ¹²/₇ × ⁷/₄ of ¾
= ¾ + ¹²/₇ × ¾
= 21 + 36
28
= 57/28
∴ 449/280 × 28/57 = 449/570 - 2a + 5b = 16
3a + 7b = 16
6a + 15b = 48
6a + 14b = 32
b = 16
a = −32 - 2 − x − x2 = 2 − 2x + x − x2
= 2(1 − x) + x ( 1 − x)
= (2 + x)(1 − x)
3x2 − 3x + x − 1
= 3x(x − 1) +1(x − 1)
=(3x+1)(x −1)
∴ =(2 + x) (1 − x2)
(3x + 1)(x − 1)
= − 2 + x
3x + 1 - 2x + x − 30 = 90
3x = 120
x = 40 - 2x − 4 ≤ 3x + 2
−6 ≤ x
3x + 2 < 10 − x
4x < 8
x < 2 - LSF = 18/24 = ¾
VSF = (¾)3 = 27/64
27/64 = x/320
x = 135cm3 - 23(3x − 2) x 24(½x) = 4−1
23(3x − 2) x 24(½x) = 2−2
9x − 6 + 2x = −2
11x = 4
x = 4/11 - P ∝ Q
√R
P = KQ
√R
New P = 1.1 P
√0.81
= 1.1 P = 1.2222P
0.9
%Increase = 0.2222 × 100%
= 22.22% - Sales = 100 − 2 × sh 340,000
100
= 98/100 × sh 340,000
= Sh 333,200
%Commission = 16660 × 100%
333200
=5% - Let ext angle = x
3½x + x = 180°
4.5x = 180
x = 40° - AC = B
C = A−1 = B - 2x + 5y = 275
4x + 7y = 415
8x + 20y = 1100
8x + 14y = 830
6y = 270
y = 45
x = 25 - W : S : D
4 : 3 : 2
Let total amount = x
(4/9 − 2/9)x = sh 120,000
2/9x = 120,000
x = Sh 540,000 - 0.3758 = 1
0.3758
= 1
1 × 3.758 × 10−1
= 1 × 0.2661 × 101
= 2.661
3√0.125 = 0.05
∴ = 2.661 × 0.5
= 1.3305
= 1.331 - Area = 150/360 × 22/7 × 7 × 7
= 64..17cm2
∴ 64.17 = 22/7 × R × 7
2.917 = R
2.9 = R - Amount on arrival = 24000 × 147.87
= Sh. 3,548,880
Balance = Sh. 3,548,880 − Sh 200,000
= Sh 3,348,880
Dollars received = 3348880
74.50
= 44951dollars -
- y − 4x = 5 ......(i)
4y − 9x = 13 .....(ii)
4y − 16x = 20
− 4y − 9x = 13
−7x = 7
x = −1
y = 1
Hence P (−1,1) - x = 2(5) − (−1) = 11
y = 2(3) −1 = 5
Hence R(11,5) - SP // RQ
Hence M1 = M2 = 4
∴ 4 = y − 11
x − 5
y − 11 = 4(x−5)
y = 4x − 20 + 11
y = 4x − 9 - PR ∴ P(−1,1) and (5,3)
Grad = M1 = 3−1 = 2 = 1
5+1 5 3
M2 = −3
Hence −3 = y − 3
x −5
y−3 = −3(x − 5)
y = −3x + 15 + 3
y = − 3x + 18
- y − 4x = 5 ......(i)
-
- Distance covered = 80 × ½h ½h + 1h + ¼h = 1¾h
= 40km - Relative distance = 510 − (80 × 1¾ + 100 × 1)
= 510 − (80 × 7/4 + 100)
= 510 − (140 + 100)
= 510 − 240
= 270km
Time taken = 270km = 270
(100 + 80)km/h 180
= 1½hrs
Time of day = 6.00am + ½hr + 1hr + ¼hr + 1½hr
= 6.00am + 3hr 15mins
= 9.15a.m - Distance from NRB = 100 × 1 + 100 × 3/2
= 100 + 150
= 250km - Time taken = 510km = 6.375hr
80km/h
= 6hr 23mins
Time of day = 6.00am + 6hrs 23 mins
= 12.23pm
- Distance covered = 80 × ½h ½h + 1h + ¼h = 1¾h
-
-
- Tan 42 = h/t
h1 = t tan 42°
Tan 30 = h
t + 3
h2 = tan 30(t+3)
Hence h1 = h2
t tan 42 = tan 30(t+3)
0.9004t = 0.5774(t+3)
0.9004t = 0.5774t + 1.7322
0.9004t − 0.5774t = 1.7322
0.323t = 1.7322
t = 1.7322
0.323
= 5.363
≅ 5.4m - Cos 30° = 8.362
YR
YR = 8.362
cos 30
= 9.656 = 9.7m - PQ = (3 + 5.362) tan 30°
= 8.362 tan 30
= 4.828 = 4.8m - Tan x = 8.362
8
x = tan−1 (8.362)
8
= 46.27°
= 46.3°
- Tan 42 = h/t
-
-
x −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 y 22 12 4 −2 −6 −8 −8 −6 −2 4 12 22
-
-
- Area of √14(14−11)(14−10)(14−7) Alternative method
Δ ABC = √(14 × 3 × 4 ×7 Cos A = 11² + 10² − 7²
=√1156 2 × 11 × 10
= 34.292 = 172/220 = 0.7818
A = ½ab sin θ A = Cos−1 (0.7818)
34.292 = ½ × 11 × 10 sin A = 38.57
Sin A = 34.292 × 2 = 38.6°
110
= 0.6235
A = sin−1 0.6235
= 38.57
= 38.6° - Radius = a
2 sin A
= 7
2 sin 38.57
= 5.614
= 5.6 - Shaded Area = πr2 − ½ab sinθ Area of circle and evidence of subtraction
= 22/7 × 5.614² − 34.292
= 99.05 − 34.292
= 64.758
= 64.8
- Area of √14(14−11)(14−10)(14−7) Alternative method
-
Sin x = 2/36 = 0.05556
x = sin−1 (2/36)
= 3.185°- < AOQ = 90 + 3.185
= 93.185 = 93.2
<AOE = 360 − 2 × 93.185
= 360 − 186.37
= 173.63 ≅ 173.6 -
- Arc length = θ/360 × 2πr
= 173.63 × 2 × 22 × 5
360 7
= 15.16 ≅ 15.2 - Arc length = 186.37 × 2 × 22 × 5
360 7
= 22.78 ≅ 22.8
- Arc length = θ/360 × 2πr
- AB = √(362 − 22)
= √1292
= 35.94 ≅ 35.9
Total length = 15.16 + 22.78 + 2 × 35.94
= 109.82
- < AOQ = 90 + 3.185
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