INSTRUCTIONS TO THE CANDIDATES:
- This paper consists of two sections: Section l and Section II.
- Answer all questions in section l and any Five in Section II
- Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
- Marks may be given for correct working even jf the answer is wrong.
- Non- programmable silent electronic calculators and KNEC Mathematical tables may be used.
SECTION I (50 Marks)
Answer ALL Questions in this Section
- Evaluate without using tables or calculator. (3 Marks)
- Simplify . (3 Marks)
x + 4 − 5x + 20
x − 4 x² − 16 - Evaluate without using log tables. (3 Marks)
log₁₀343 − log₁₀125
log₁₀7/5 - Solve the following inequalities and represent the solution on a single number line. (3 Marks)
−2x + 3 < 5
4 − 3/2x ≥ −8 - The scale of a map is 1: 200. Calculate the actual area of a triangular field whose sides are 6cm, 8cm and 10cm on the map. (4 Marks)
- 24 mean each working 10 hours a day take 4 days to complete a piece of work. How many more days will it take 15 men each working 8 hours a day to complete the same piece of work. (3 Marks)
- Find the value of in (½)x × (1/8)1 −x = 32 (3 Marks)
- The figure below shows a circle ABCD, centre O. BO is perpendicular to the diameter AOD and ∠COD = 42°.
Calculate:- ∠CAD (2 Marks)
- ∠BAC (2 Marks)
- Given that sin x = 2/3 find without using tables tan x (2 Marks)
- A three digit number is such that twice the hundreds digit is more than the tens digit by 2. The units digit is thrice the hundreds digit. When the digits are reversed, the number is increased by 594. Find the number. (4 Marks)
- If 2m + 3n : 5m − n = 3 2 calculate the ratio m : n. (3 Marks)
- A dealer sells a car battery to a customer at a profit of 22%. The customer sells it to a friend for Sh.4800, at a profit of 8%. Find: -
- How much it cost the dealer to buy the battery. (2 Marks)
- How much it cost the dealer to buy the battery. (2 Marks)
- Given vectors and are parallel, find the value of a, hence calculate . (3 Marks)
- Find the greatest common factor of x3 y2 and 4xy4 . Hence factorise completely the expression.
x3y2 − 4xy4 (3 Marks) - Express 2.4545………. in the form p/q where p and q and are integers and q ≠ 0. (3 Marks)
- Given that a = 2.5 and b = 6.15, find the percentage error in b − a. (3 Marks)
SECTION II (50 Marks)
Answer any FIVE Questions in this Section.
- Towns A and B are 580km apart. A matatu started from town A and travelled towards town B at an average speed of 90km/h at 7.30 a.m. One and a half hours later a car travelled from town B towards town A at an average speed of 120km/h.
-
- At what time did the two vehicles meet? (2 Marks)
- How far from B did the two vehicles meet? (2 Marks)
- If the matatu took a total of 30 minutes to drop and pick travellers along the way, calculate the time it arrived at town B. (2 Marks)
- A rally driver starts from B towards A at 10.00 a.m. at an average speed of 180km/h.
- At what time did the rally driver overtake the car? (2 Marks)
- How far from A did the rally driver overtake the car? (2 Marks)
-
- The figure below shows marks obtained by 60 students in a test.
Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 No. of pupils 2 2 3 8 23 13 6 2 1 - Calculate the mean mark. (3 Marks)
- Calculate the median. (3 Marks)
- State the modal class. (1 Mark)
- Represent the information on a histogram. (3 Marks)
- The figure shows the cross-section of a cylindrical tank containing some oil and lying horizontally. The tank is 4m long. O is the centre of the circle, radius 14m. ∠AOB=120°.
Calculate:- The length of chord AB. (2 Marks)
- The area of segment ABC. (3 Marks)
- The volume of oil in the tank. (2 Marks)
- The area of the tank in contact with the oil. (3 Marks)
- Given that y = x2 + 3x − 5, complete the following table and use it to answer the questions that follow. (2 marks)
X −6 −5 −4 −3 −2 −1 0 1 2 3 Y - Using a scale of 1cm to represent 1 unit on both axes plot the graph of y = x2 + 3x − 5 . (3 Marks)
- On the same grid draw the line y = 2 − x . (3 Marks)
- Use your graph to estimate the roots of the equation. (2 Marks)
x2 + 4x − 7 - Give the least value of y for the curve y = x2 + 3x − 5. (1 Mark)
- Give the range of values of x for which y = x2 + 3x − 5 is less than or equal to 2 − x. (1 Mark)
- The vertices of a triangle are at A(3,1), B(5,3) and C(5,0). Under a rotation R, the images of A and B are A'(0,0) and B'(−2,2) respectively.
- Draw the two triangles on the Cartesian plane. (2 Marks)
- Find the centre and angle of rotation (2 marks)
- State the coordinates of C' the image of C. (1 mark)
- Triangle A'B'C' is reflected in the line y = −2 to get A''B''C''. Draw triangle A''B''C''. State the coordinates of A''B''C''. (2 marks
- ΔA'''B'''C''' is the image of ΔA''B''C'' under an enlargement centre (0, −5) and scale factor = 2. Draw ΔA'''B'''C'' and state its coordinates. (3 marks)
- Three straight lines L1, L2 and L3 are such that;
L1 cuts the y-axis at y = 5 and has a gradient of 2.
L2 is perpendicular to L1 at the point where L1 cuts the x-axis.
L3 is parallel to L2 and passes through the point (1, 2).- Find the equations in the form of y = mx + c of
- L1 (2 Marks)
- L2 (2 Marks)
- L3 (2 Marks)
- Determine the coordinates of the point at which L3 is perpendicular to L1. (3 Marks)
- Find the equations in the form of y = mx + c of
- Use a ruler and a pair of compasses only to construct a triangle ABC with AB = 3.8 cm, BC = 4.2 cm and ∠ABC=75°. (2 marks)
- Measure AC. (1 Mark)
- Drop a perpendicular from B to meet AC at N. Measure BN and AN. (2 Marks)
- Calculate the area of ΔABN. (2 Marks)
- Construct a perpendicular bisector of AB and BC to meet at O. Use OA as the radius to draw a circle. Find the area of the circle. (3 Marks)
- In the figure below AN = ¼ AB and OM = 2/3 OB. OA = a and OB = b. The lines AM and ON intersect at point X.
- Express in terms of a and b .
- ON
- AM
- OX
- Express in terms of a and b .
MARKING SCHEME
= 4- (x + 4)(x + 4) − 5(x + 4)
(x − 4) (x + 4)
(x + 4)(x + 4 − 5)
(x − 4) (x + 4)
(x + 4) (x − 1)
(x − 4) (x + 4)
(x − 1)
(x − 4)
Alternative
(x + 4)(x + 4) − 5(x + 4)
x2 − 16
= x² + 8x + 16 − 5x − 20
(x − 4) (x + 4)
= x² + 3x − 4
(x − 4) (x + 4)
= x² + 4x − x − 4
(x − 4) (x + 4)
= (x + 4) (x − 1) = (x − 1)
(x − 4) (x + 4) (x − 4) - log₁₀7³ − log₁₀5³
log₁₀(7/5)
3log₁₀7 − 3log₁₀5 = 3 log₁₀ (⁷/₅)
log₁₀7/5 log₁₀ (⁷/₅)
= 3 - −2x < 2
x > −x > −1
8 − 3x ≥ −16
−3x ≥ −24
x ≤ 8
−1 < x ≤ 8 - L.S.F = 1/200
A.S.F = 1/400000
Area of triangular field, S = (6 + 8 + 10)½ = 12
= √(12(12−6)(12−8)(12−10))
= √576 = 24
Actual area of the triangular field = 24cm2 × 40000
= 960,000cm2 - Men Hours Days
24 10 4
15 8 ?
1 man working 1 hr takes 24 × 10 × 4 = 960 days
∴ 15 men 8 hrs will take 960 days
15 × 8
= 8 days
∴ 8 − 4 = 4 more days - (2−1)x × (2−3)1−x = 25
2−x−3+3x = 25
22x − 3 = 25
2x − 3 = 5
x = 4 -
- ∠OCD = 180 − 42 = 69°
2
∠OCD = 69°
∠CAD = 180° − (90 + 69) = 21° - ∠AOC = 180° − 42° = 138°
∠BOC = 138° − 90° = 48°
∠OCB = 180° − 48° = 66°
2
∠BCA = 66° − 21 = 45°
- ∠OCD = 180 − 42 = 69°
-
m = √(32 − 22) = √5
tan x = 2/√5 = 2/√5 × √5/√5
= 2√5
5 - Let the number be x y z
2x = y + 2 .......(i)
Z = 3x .........(ii)
100z + 10y + x − 594 = 100 + 10y + z .....(iii)
99z − 99x = 594
z − x = 6 .....(iv)
z − x = 6
z − 3x = 0
2x = 6
x = 3
z − 3(3) = 0
z = 9
2(3) = y + 2, y = 4
∴ The number is 349 - 2m + 3n = 3
5m − n 2
2(2m + 3n) = 3(5m − n)
4m + 6n = 15m − 3n
−11m = −9n
m = −9 = 9
n −11 11
m : n = 9 : 11 - 108/100x = 4800
x = 4800 × 100
108
x = Sh. 4444.40
Customers buying price = Sh.4444.40 - Dealers buying price
122/100x = Sh 4444.40
x = Sh 4444.40 × 100
122
x = Sh. 3643.0
- 108/100x = 4800
- PQ = kRS
6k = 2, k = 3
3(a − 1) = 3
a − 1 = 1
a = 2
= √(62 + 12 + 152)
= √262
=16.19 (2dp) - Greatest common factor is xy2
x3y2 − 4xy4 = xy2(x2 − 4y4)
= xy2(x − 2y)(x + 2y) - Let r = 2.4545 ....(i)
100r = 245.4545 ......(ii)
(ii) − (i). 99r = 243
r = 243 = 27
99 11 - 2.45 ≤ a ≤ 2.55
6.145 ≤ b ≤ 6.155
Limits b − a, maximum difference
6.155 − 2.45 = 3.705
6.145 − 2.55 = 3.595
Working difference = 6.15 − 2.5 = 3.65
Absolute error = 3.65 − 2.595 = 0.055
% error = 0.055 × 100% = 1.51%
3.65 -
-
- 3/2 × 90 = 135km
580 − 135 = 445km
210
= 2.119048
9.00 + 2.07 = 11.07a.m - 2.119048 × 120 = 254.29km from B
- 3/2 × 90 = 135km
- 580/90 = 6.444hrs
6hrs 27 min
6 hrs 57mins
7.30 + 6hrs 57 mins
= 2.27 p.m -
- 1 × 120 = 120/60km
180 − 120
= 2 hrs
10.00 + 2hrs = 12.00 noon - 180 × 2 = 360km
580 − 360 = 220km from A
- 1 × 120 = 120/60km
-
-
class ƒ cƒ x ƒx 10-19
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-992
2
3
8
23
13
6
2
12
4
7
15
38
51
57
59
6014.5
24.5
34.5
44.5
54.5
64.5
74.5
84.5
94.529
49
103.5
356
1253.5
383.5
447
169
94.560 3340 - 3340 = 55.67
60 - 30 + 31 = 30.5
2
49.5 + (30.5 − 15) × 10
23
49.5 + 6.739 = 56.24 - 50 - 59
-
- 3340 = 55.67
-
-
sin 60° = MB
1.4
MB = 1.4 sin 60°
= 1.212m
∴ Chord AB = 2.424m - Area of segment ABC = 120/360 × 22/7 × 1.42 − ½ × 1.42 sin 120°
= 2.053 − 0.849
=1.204m2 - Volume of oil = 1.204 × 4
= 4.816m3 - Area in contact with oil
= 120/360 × 2 × 22/7 × 1.4 + 2(1.204)
= 2.933 + 2.408
= 5.341m2
-
-
y = x2 + 3x − 5
X −6 −5 −4 −3 −2 −1 0 1 2 3 Y 13 5 −1 −5 −7 −7 −5 −1 5 13 - y = 2 − x
x −6 3 y 8 −1 - x = 5.2 ± 0.1 or 1.3 ± 0.1
- − 7.2 ± 0.1
- −4.2 ≤ x ≤ 1
-
-
-
- (0, 5) g = 2
y − 5 = 2
x − 0
y − 5 = 2x
y = 2x + 5 - y = 2x + 5 = 0
x = − 2.5
( −2.5, 0)
g1 = 2
g2 = −1/2
y − 0 = ½
x − −2.5
2y = −1(x + 2.5)
2y = −x − 2.5
y = −1 x − 5
2 4 - g2 = −1/2
g3 = −1/2
y − 2 = −1/2
x − 1
2(y − 2) = −(x − 1)
2y − 4 = −x + 1
2y = − x − 5
y = −1/2x − 5/2
- (0, 5) g = 2
- y = 2x + 5
y = −1/2x − 5/2
2x + 5 = −1/2x − 5/2
4x + 10 = − x − 5
5x = −15
x = − 3
y = 2( −3) + 5
= − 1
( −3,−1)
-
-
-
AC = 5cm - BN = 3.1cm
AN = 2.2 cm - A = ½ × 5 × 3.1
= 7.75cm2 - r = 2.6cm
A = 22/7 × 2.6 × 2.6
= 21.25cm2
-
-
-
- ON = a + ¼(b − a)
= ¾a + ¼b - AM = a + 2/3b
2/3b − a - OX = hON
= h(¾a + ¼b)
= ¾ha + h/4b
OX = a + k(2/3b − a)
= (1 − k)a + 2/3k 2/3b
- ON = a + ¼(b − a)
-
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