INSTRUCTION TO STUDENTS:
- This paper consists of two Sections; Section I and Section II.
- Answer ALL the questions in Section I and only five questions from Section II.
- Show all the steps in your calculation, 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.
- KNEC Mathematical tables may be used, except where stated otherwise.
- Students should answer the questions in English.
SECTION I (50 MARKS)
Answer ALL the questions in this section in the spaces provided.
- Evaluate (3 Marks)
−12 ÷ 3 × 4 −( −15)
−5 ×6 ÷ 2+( −5) - Evaluate without using a calculator. (3 Marks)
- Use logarithms to evaluate. (4 Marks)
- Find the equation of the L1 in the form y=mx+c which is perpendicular to the line 3y + 2x = 6 and passes through the point (−3, 4). ( 3mks)
- The diagram below represents a prism of length 6cm whose cross-section is an equilateral triangle of sides 3cm. Draw a well labeled sketch of the net of the prism. (3 Marks)
- The figure below represents a kite ABCD, AB = AD = 15 cm. The diagonals BD and AC intersect at O. AC = 30cm and AO = 12 cm. Find the area of the kite (3mks)
- Mrs. Musyoka has Sh. 700 in Sh. 50 notes and Sh. 100 notes only. If she has a total of 11 notes find how many notes she has of each denomination. (3mks)
- In the figure below AD // BC. AC and BD intersect at E. Given that AE: EC = 1:5 and BD =12 cm, calculate the length of DE. (3 marks)
- Given that (3x−35) =cos (x+20) . Find (x+10) (3mks)
- Without using a mathematical tables or a calculator evaluate (3mks)
- Three bells ring at intervals of 9 minutes, 15 minutes and 21 minutes. The bells will next ring together at 11.00pm. Find the time the bells had last rung together. (3 Marks)
- The surface areas of two similar bottles are 12cm2 and 108cm2 respectively. If the bigger one has a volume of 810cm3. Find the volume of the smaller one. (3 Marks)
- In the figure below A′B′ is the image of AB under rotation. By construction, find and label the centre O of the rotation. Hence, determine the angle of the rotation.(4mks)
- Mr. Ombogo the principal of Chiga secondary would wish to cover the floor of the new administration block using the square tiles. The floor is a rectangle of sides 12.8m by 8.4m. Find the area of each of the largest tiles which can be used to fit exactly without breaking (3mks)
- The size of an interior angle of a regular polygon is (3x)° while the exterior angle is (x+20)°. Find the number sides of the polygon (3 Marks)
- In the figure below triangle ABO represents a part of a school badge. The badge has as symmetry of order 4 about O. Complete the figure to show the badge. (3mks)
SECTION II (50MKS)
Answer 5 questions only in this section
- The vertices of quadrilateral OPQR are O (0, 0), P (2, 0), Q (4, 2) and R (0, 3). The vertices of its image under a rotation are O’ (1, -1), P'(1, -3) Q'(3, -5) an R'(4, -1).
- On the grid provided, draw OPQR and its image O'P'Q'R' (2marks)
- By construction, determine the centre and angle of rotation. (3marks)
- On the same grid as (a) (i) above, draw O''P''Q''R'', the image of O'P'Q'R' under a reflection in the line y = x (3marks)
- From the quadrilaterals drawn, state the pairs that are:
- Directly congruent; (2marks)
- Oppositely congruent (2marks)
- A slaughter house bought a number of goats at Sh. 2000 each and a number of bulls at Sh. 15000 each. They paid a total of Sh. 190,000. If they bought twice as many goats and three bulls less, they would have saved Sh. 5000.
- If the number of goats and bulls bought were x and y respectively, write down two simplified equations involving the above information. (2mks)
- Solve the two equations above and hence find the number of each type of animals bought. (4mks)
- The slaughter house sold all the animals at a profit of 25% per goat and 30% per bull. Determine the total profit they made. (4 Marks)
-
- In a safari rally drivers are to follow route ABCDA. B is 250km from A on a bearing of 075° from A. C is on a bearing of 110° from A and 280km from B. The bearing of C from D is 040° and a distance of 300km. By scale drawing show the position of the point A, B, C and D. (4mks)
- Determine
- The distance of A from C. (2mks)
- The bearing of B from C. (1mk)
- The bearing of A from D. (1mks)
- The distance A from D (2mks)
- A saleswoman is paid a commission of 20% on goods sold worth over Ksh 100,000.She is also paid a monthly salary of Ksh 12,000.In a certain month, she sold 360 handbags at Ksh 500 each.
- Calculate the saleswoman’s earnings that month. (3 mks)
- The following month, the saleswoman’s monthly salary was increased by 10%.Her to total earnings that month were Ksh17, 600.
Calculate the total amount of money received from the sales of handbags that month.(5mks) - The number of handbags sold that month. (2 mks)
- Using a ruler and a pair of compasses only, draw a parallelogram ABCD, such that angle DAB = 75°. Length AB = 6.0cm and BC = 4.0cm.
From point D, drop a perpendicular to meet line AB at N. (7 Marks)- Measure length DN. (1 Mark)
- Find the area of the parallelogram. (2 Marks)
-
- Two cubes of length 5cm and 7cm are melted and cast into a single cube.
Determine the:- Volume of the new cube (3mks)
- Length of the new cube correct to 1 decimal place(2mks)
- Surface area of the new cube (2mks)
- Suppose that it was instead cast into a cylinder of radius 3.5 cm. what would the height be to the nearest cm? Take = 22/7 ,(3 mks)
- Two cubes of length 5cm and 7cm are melted and cast into a single cube.
- The diagram below shows vertical telephone pole RS supported by wires SP and SQ pegged at points P and Q respectively on a level ground. Points P and Q are on the same straight line from the base R of the pole. The angles of elevation of S from P and Q are 33.9° and 48.2° respectively. Given that PR = 5 m, calculate:
- The distance QR (4 marks)
- The length of the wires SP and SQ (4 marks)
- If the cost of the pole and labour is sh. 1600 and the cost of 1 meter of the wire is sh. 233. Find the total cost of the installation. (2 marks)
- A glass in the form of a frustum of a cone, is represented by the diagram below. The glass contains water to a height of 9 cm. The bottom of the glass is a circle of radius 2 cm while the surface of the water is a circle of radius 6 cm.
- Calculate the volume of the water in the glass. (3Mks)
- When a special marble is submerged into the water in the glass, the water level rises by 1 cm.
Calculate :- the volume of the marble (4 marks)
- the radius of the marble (3 marks)
MARKING SCHEME
- −4x4+15 √M1
−15 − 5
= −1 √M1
−20
= 1/20 √A1 - (17/7 − 11/6 ) x 6/5
2/3 × 9/4 − 8/7
Numerator = (102 − 77) x 6
42 5
= 25/42 x 6/5 = 5/7 √M1
Denominator = 3/2 − 8/7
= 21 − 16 = 5 √M1
14 14
5/7 ÷ 5/14 = 5/7 x 14/5
= 2 √A1 - No. log all logs √ M1
43.52 1.6387
0.08792 2.9441 + √operations M1
0.5828
785.3 2.8950 –
3.6878÷4 √division by 4 M1
2.642 x 10-1 1.4220 Answer √ A1 - 3y + 2x = 6
y = −2x/3 + 2
m1 = −2/3
m2 = 3/2 √B1
y − 4 = 3 = 3 √M1
x + 3 2
2( y − 4) = 3(x + 3)
2y − 8 = 3x + 9
2y = 3x + 17
Y = 3x/2 + 17/2 √A1 -
- BO = √(152 − 122) = 9 √B1
Area of kite = ½ x 30 x 9 x 2 √M1
= 270cm2√A1 - Let sh.50 notes be x and sh.100 notes be y
X + y = 11 ....... (i)
50x + 100y = 700 Both equations √M1
⇔
5x + 10y = 70 ....... (ii)
X + y = 11 x 5
5x + 10y = 70 √Attempt to solve M1
5x + 5y = 55
5y = 15
y = 3
x + 3 = 11
x = 8
sh50 = 8 notes
sh100 = 3 notes Both √A1 - AE = DE = 1 √M1
EC BE 5
DE = 1 √M1
12-DE 5
5DE = 12 – DE
6DE = 12
DE = 2cm √A1 - 3x – 35+x + 20 = 90 √M1
4x = 105
X = 26.25 √A1
Tan (26.25+10) = Tan36.25 = 0.7332 √B1 -
= M1
= M1
= 3 x 15
4
= 45/4 = 11¼ or 11.25 √A1 -
LCM = 3 x 3 x 5 x 7 = 315 min √ determination of LCM
= 5h 15min M1
Last rung at 11.00p.m – 5h 15min √MI
= 5.45p.m √A1 - ASF = 12/108 = 1/9
⇒ LSF = 1/3 √B1
V.S.F =(1/3)3 = 1/27
1/27 = V/810 √M1
⇒ V = 810 = 30cm3 √A1
27 -
-
GCD = 2x 2 x 2 x 5 = 40cm √A1
Area = 40 x 40 = 1600cm2√B1 - 3x + x + 20 = 180°√M1
4x = 160°
x = 40°√A1
n = 360° = 9 √ B1
40° -
-
-
- Centre (0, −1) ∠ of rotation = −90°
-
- OPQR & O'P'Q'R' ✓Both B2
- O'P'Q'R' & O''P''Q''R'' ✓any 2 B2
OR
OPQR & O''P''Q''R''
-
-
- 2000x + 15000y = 190000
2x + 15y = 190 ..... (i) √B1
2 x 2000x+ (y−3) 15000 = 185000
4x + (y−3)15 = 185
4x + 5y−45 = 185
4x + 5y = 230 .........(ii) √B1 - 2x + 15y = 190
4x + 5y = 230
4x + 30y = 380 √attempt to solve M1
4x + 5y = 230
25y = 150
y = 6 √AI
4x + 5x6 = 230 √M1
4x = 200
X = 50 √A1
Goats = 50
Bulls = 6 - Profit from goats = 25/100 x 2000x50 = sh25000 √M1
Profit from Bulls = 30/100 x 15000x6 = sh27000 √M1
Total profit = 25000 + 27000 √M1
= sh52,000√AI
- 2000x + 15000y = 190000
-
-
- Handbag sales = 360 x 500 = 180000 √BI
Commission = 20/100 x 80,000 = sh16000 √BI
Total earnings = 12000 + 16000 = 28000 √BI - Sales woman salary = 110/100 x 12000 = sh13200 √BI
Commission = 17600 – 13200 = 4400 √MI
Money received from Handbag sales = 4400 x 100 + 100,000 √√ M2
20
= sh120,000√A1 - No. of handbags = 120,000 √M1
500
= 240 √A1
- Handbag sales = 360 x 500 = 180000 √BI
-
- DN = 3.9cm
- Area = ½ × 6 × 3.9
= 11.7cm2
-
-
- Vol = 53 + 73√ M1
= 125 + 343 √M1
= 468 √A1 - L3 = 468
L = 3√468 √M1
= 7.764
≅ 7.8cm √A1 - SA= 6a2
= 6(7.8)2
= 365.04cm2
- Vol = 53 + 73√ M1
- 22/7 x 3.52 x h = 468 √M1
h = 468 x 7
22x3.52
= 468 x 7
22x12.25
= 12.16 √A1
≅12cm √B1
-
-
- Tan 33.9 = SR/5 ⇒ SR = 5 tan 33.9 √M1
Tan 48.2 = SR ⇒ SR = (5−x) tan 48.2 √M1
5−x
5 x 0.672 = 1.118(5−x) √M1
3.36 = 5.59 – 1.118x
1.118x = 2.23
x= 2.23 = 1.995 √A1
1.118 - Cos 33.9 = 5/SP
SP = 5 = 5 √M1
Cos33.9 0.8300
= 6.024m
Cos 48.2 = 1.995 √M1
SQ
SQ = 1.995 = 1.995 √M1
Cos48.2 0.6665
= 2.993 √A1 - Cost of wire = (6.024+2.993) x 233
= 9.017 X 233 = 2100.96
≅ 2101 √B1
Total cost = 2101 + 1600
= sh3701 √B1
- Tan 33.9 = SR/5 ⇒ SR = 5 tan 33.9 √M1
-
- Volume of water = 6 = 2/x M1
9+x
x = 4.5 M1
; volume
= 1/3x 3.142(62 x 13.5 − 23 x 4.5)
= 508.94 − 18.25 = 490.09 A1 -
- Volume of sphere M1
Top radium r/14.5 = 2/4.5 r=6.440 A1
Volume = 1/3 x 3.142 (6.4442 x 14.5 − 62 x 13.5) M1
= 121.6 = 121 11/15 A1 - 4/3Лr3 = 121.6 M1
r3 = 121.6 x ¾x M1
r = 3.073 A1
10Mrks
- Volume of sphere M1
- Volume of water = 6 = 2/x M1
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