Questions
INSTRUCTIONS TO CANDIDATES
- This paper consists of two sections: Section I and Section II.
- Answer ALL questions from section I and ANY FIVE from section II
- Show all the steps in your calculation, giving your answer at each stage in the spaces below each question.
- Non – Programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.
SECTION I: Answer all questions from this section
- Use logarithm tables to evaluate (4 Marks)
3√(45.3 × 0.00697 / 0.534) - Solve for x in the equation 2Sin2x - 1 = Cos2x + Sinx for 00≤x≤3600 (3 Marks)
-
- Expand (1+3/x)5 upto the fifth term (2 Marks)
- Hence use your expansion to evaluate the value of (2.5)5 to 3 d.p. (2 Marks)
- Make p the subject of the formula (3 Marks)
E = √(p-3u / y-3xp) - The figure below shows a rectangular based right pyramid. Find the angle between the planes ABCD and ABV. (2marks)
- A object A of area 10cm2 is mapped onto its image B of area 60cm2 by a transformation whose matrix is given by P= x 4 3 x+3 . Find the possible values of x (3 Marks)
- The position vectors of A and B are a = 4i + 4j – 6k and b = 10i + 4j + 12k. D is a point on AB such that AD:DB is 2:1. Find the co-ordinates of D (3 Marks)
- A dealer has two types of grades of tea, A and B. Grade A costs Sh. 140 per kg. Grade B costs Sh. 160 per kg. If the dealer mixes A and B in the ratio 3:5 to make a brand of tea which he sells at Sh. 180 per kg, calculate the percentage profit that he makes (3 marks)
- A variable Z varies directly as the square of X and inversely as the square root of Y. Find the percentage change in Z if X increased by 20% and Y decreased by 19% (3 Marks)
- By rounding each number to the nearest tens, approximate the value of 2454 x 396/66 Hence calculate the percentage error arising from this approximation to 4 significant figures (3 Marks)
- Find the centre and radius of the circle whose equation is 2x2 + 2y2 – 8x + 12y – 2 = 0 (3 Marks)
- In the figure below AB = 8cm and O is the centre of the circle. Determine the area of the circle if angle OAB= 150 (3 Marks)
- Pipe A can fill a tank in 2 hours; pipes B and C can empty the tank in 5 hours and 6 hours respectively. How long would it take:
- To fill the tank if A and B are left open and C closed (2 Marks)
- To fill the tank with all the pipes open (2 Marks)
-
- Find the inverse of the matrix 4 3 3 5 (1 Mark)
- Hence solve the simultaneous equation below using matrix method (3 Marks)
4x + 3y = 6
5y + 3x – 5 = 0
- Evaluate by rationalizing the denominator and leaving your answer in surd form.(2 Marks)
√8
1+Cos 45° - Form the three inequalities that satisfy the given region R (3 Marks)
SECTION II – 50 MARKS
Answer any FIVE questions from this section
-
- P, Q and R are three quantities such that P varies directly as the square of Q and inversely as the square root of R.
- Given that P = 12 when Q = 24 and R = 36, find P when Q =27 and R = 121. (3 Marks)
- If Q increases by 10% and R decreases by 25%, find the percentage increase in P. (4 marks)
- If Q is inversely proportional to the square root of P and P = 4 when Q = 3. Calculate the value of P when Q = 8. (3 marks)
- P, Q and R are three quantities such that P varies directly as the square of Q and inversely as the square root of R.
-
- complete the table for the curves y = 3sin (2x + 300) and y = Cos2x ,use the range 0 ≤ x ≤ 1800
x 0 15 30 45 60 75 90 105 120 135 150 165 180 y= 3sin(2x+30) 1.5 3 1.5 -1.5 -2.60 -1.00 1.5 y= Cos2x 1 0 -0.866 -0.866 -0.5 0.866 1 - Using the scale Horizontal axis 1cm represent 300, vertical axis 1cm represent 1 unit, draw the graphs of y = 3 Sin (2x + 30) and y = Cos 2x (4 Marks)
- Use your graph to solve the equation 3Sin (2x + 30) = Cos 2x (1 Mark)
- Determine the following from your graph
- Amplitude of y = 3Sin (2x + 30) (1 Mark)
- Period of y = 3 Sin (2x + 30) (1 Mark)
- Period of y = Cos 2x (1 Mark)
- complete the table for the curves y = 3sin (2x + 300) and y = Cos2x ,use the range 0 ≤ x ≤ 1800
- The 2nd and 5th terms of an arithmetic progression are 8 and 17 respectively. The 2nd, 10th and 42nd terms of the A.P. form the first three terms of a geometric progression. Find:
- The 1st term and the common difference. (3 Marks)
- The first three terms of the G.P and the 10th term of the G.P. (4 Marks)
- The sum of the first 10 terms of the G.P. (3Marks)
- The probability of passing KCSE depends on the performance in the KCPE. If the candidate passes the KCPE, the probability of passing KCSE is4/5. If the candidate fails in the KCPE, the probability of passing KCSE is3/5. If a candidate passes KCSE the probability that he/she will get employed is5/8. If he/she fails KCSE the probability of getting employed is1/3. The probability of passing KCPE is2/3.
- Draw a well labelled tree diagram to represent the above information. (2 Marks)
- Using the tree diagram, find the probability that a candidate:-
- Passes the KCSE (2 Marks)
- Gets employed (2 Marks)
- Passes KCSE and get employed (2 Marks)
- Passes KCPE and does not get employed (2 Marks)
- The heights of 100 maize plants were measured to the nearest centimeter and the results recorded in the table shown below.
Height x (cm) Frequency d d2 fd fd2 cf 25 - 29 5 -15 5 30 - 34 12 -24 17 35 - 39 18 -1 1 -18 35 40 - 44 30 0 0 0 65 45 - 49 17 1 1 50 - 54 11 2 55 - 59 7 3 - Complete the table (2 Marks)
- Calculate to 2 d.p.
- The mean (2 Marks)
- The standard deviation (2 Marks)
- Using the data above plot an ogive and use it to find the quartile deviation (4 Marks)
- The figure below shows rectangular plot ABCD with AB =60m and BC=45m. PN is a vertical pole of length 30m to which four taut wire PB1, PC1,PD and PA are attached
Calculate- length of the projection of PCon the plane ABCD (2mrks)
- the angle PC made with the base ABCD (3mks)
- The angle between the planes PBC and ABCD (3Mrks)
- If point A is to be the North of point C. calculate the bearing of B from A (2mks)
-
- Construct a parallelogram ABCD in which AB = 9cm, AD = 5cm and angle BAD = 60°. Measure the length AC (3 Marks)
- Show the locus of point P which moves so that it is equidistant from A and C.(1 Mark)
- Show the locus of point Q which moves such that angle BQD = 900.(2 Marks)
- The position of point X such that AX ≥ XC and angle BXD = 900(2 Marks)
- Shade the region inside the parallelogram such that AX≥XC and angle BXD ≥900 (2 Marks)
-
- Draw ∆PQR whose vertices are P(1,1), Q(-3,2) and R(0,3) on the grid provided (1 Mark)
- Find and draw the image P’Q’R’, image of ∆PQR under the transformation whose matrix is (3 Marks)
- P’Q’R’ is then transformed into P’’Q’’ R’’ by the transformation of matrix
Find the co-ordinates of P’’Q’’ R’’ and draw the image (3 Marks) - Describe fully the single transformation which maps PQR onto P’’Q’’ R’’. Find the matrix of this transformation (3 Marks)
Marking Scheme
-
No. Log 45.3 0.00697
8.450 x 10-1
0.84501.6651 3.8432
1.5083
1.7275
1.7808
3
-1-2 + 2.7808
3 3T.9269
- 2Sin2 x – 1 = 1 – Sin2x + Sinx
3Sin2x – Sinx – 2 = 0
(3Sinx + 2) (Sin x – 1) = 0
Either
Sinx = 0.6667 or Sin x = 1
x = 90°, 221.81, 318.19 -
- 1 + 5 x 3/x + 10 x 3²/x +10x 3³/x
+ 5 x 34/x …
1 + 15/x + 90/x2 + 270/x3 + 402/x4 + …. - (2.5)2 = (1+ 3/x)5
1.5 = 3/x
x = 2
1 + 15/2 + 90/4 + 270/8 + 405/16
1 + 7.5 + 22.5 + 33.75 + 25.3125 + ….
90.063
- 1 + 5 x 3/x + 10 x 3²/x +10x 3³/x
- E = √((P-3u)/ (y-3xp))
E2 = P-3u
y-3xp
E2Y – 3E-2XP = P – 3u
P + 3E2XP = E2Y+3u
P = E²y+3u
1+3E²x -
Cos θ = 1.5/√60
= 0.1936
θ = Cos- 0.1936
θ = 78.84° - A.S.F = Determinant of matrix
60/10 = x (x + 3) – 12
x2 + 3x – 12 = 6
x2 + 3x – 18 = 0
x2 – 3x + 6x – 18 = 0
x(x – 3x) + 6(x – 3) = 0
(x + 6) (x – 3) = 0
X = - 6 or x = 3 -
(4 6 -6) + 2/3 (10 -4 4 -4 12 --6)
(4 4 -6 )+ 2/3 (6 0 18)
(4 4 -6) + 4 0 12
8i + 4j + 6k
Co-ordinates of D are(8, 4, 6) - Cost price = 3 x 140+5 x 160 = 152.50
8
Profit = 180 – 152.50 = 27.50
% Profit = 27.50152.50 x 100 = 18.03% - Zα(X2/√Y)⇒ Z = kx2/√y
Z1 = 1.2²kx²
√(81/100) √y
1.44 kx²
0.9√y
1.6z
% Change = 1.6Z - Z x 100%
Z
z 1.6-1 x 100
z
= 60% - 2450 x 400 = 14000
70
2454 x 396 = 14724
66
% error 14724 - 14000 x 100
14724
= 4.917 - x2 – 4x + y2 + 6y – 1 = 0
x2 – 4x + 4 + y2 + 6y + 9 = 14
(x – 2)2 + (y + 3)2 = 14
Centre (2, -3)
Radius = 3.742 - ∠OBA = 150
∠AOB = 180 – (15 + 15) = 1500
∠ACB = 75°
2R = 8
Sin 75
R = 8
2 X 09695
r = 4.146
Area of circle
3.142 x 4.1462
54.01 -
- 1/2 - 1/5 = 3/10
Required time 10/3
31/3 or 3 hrs 20 min - 1/2 - 1/5 - 1/6 = 4/30
Required time 30/4
71/2 hrs
- 1/2 - 1/5 = 3/10
-
- Det. 20 – 9 = 11
Inverse ⇒ 1/11 (5 -3 -3 4) - (4 3 3 5)(x y) = (6 5)
(5/11 -3/11 -3/11 4/11)( 4 3 3 5)(x y) = (5/11 -3/11 -3/11 4/11)( 6 5)
(1 0 0 1)( x y)= (15/11 2/11)
x = 15/11, y = 2/11
- Det. 20 – 9 = 11
- √8
1 + Cos 45°
2√2
1+ 1/√2
2√2 (1-1/√2)
(1+ 1/√2)(1- 1/√2)
2√2-2
12
2(√2-1) = 4(√2- 1)
1/2 - L1 y = x
y > x
L2 x/4 + y/4 = 1
x + y = 4
x + y ≤ 4
L3 x/-1 + y/3 = 1
-3x + y = 3
y – 3x = 3
y – 3x ≤ 3 - P= KQ²/√R² - constant
- K = P√R/Q² = 12 x √ 36 = 1/8
24²
Hence P = 1/8 Q²/√R
When Q = 27, R = 121
P = 1/8 x 27²/√121 = 729/88 - Q1 = 1.21Q²
R1 = 0.866025403 √R
P1 = Kx 1.21Q² = 1.397187651
(1.397187651 - 1)KQ²/√R x 100%
New value of New change = 39.7187651%
Hence increase of 39.72% - Q= K1/√P
K =Q/ √P
=3/√4
= 6 Eqn = Q = 61/√P
P = (K1/Q)²
=(6 x 1/8)² - (6/8)²
=9/16
- K = P√R/Q² = 12 x √ 36 = 1/8
-
-
- a + d = 8
a + 4d= 17
3d=9
d = 3
∴ a = 5 - 2nd = 8
10th = 5 + 9 × 3 = 32
42nd = 5 + 41 × 3 = 128
∴GP is 8, 32, 128, - - - -
a = 8
r = 4
nth term of G.P = arn – 1
∴10th term = 8(4)9 = 2097152 - Sn - a(rⁿ-1)
r-1
S10 = 8(4¹⁰ - 1)
4 -1
= 2796200
- a + d = 8
-
-
-
- (2/3 x 4/5) + (1/3 x 3/5)
8/15 + 3/15
= 11/15 - (2/3 x 4/5 x 5/8)+ (2/3 x 1/5 x 1/3) +
(1/3 x 3/5 x 5/8) +( 1/3 x 2/5 x 1/3)= 197/360 - (2/3 x 4/5 x 5/8) + (1/3 x 3/5 x 5/8) = 11/24
- (2/3 x 4/5 x 3/8) + (2/3 x 1/5 x 2/3) = 13/45
- (2/3 x 4/5) + (1/3 x 3/5)
-
-
-
Height X (cm) f d d2 fd fd2 cf 25 – 29
30 – 34
35 – 39
40 – 44
45 – 49
50 – 54
55 - 595
12
18
30
17
11
7-3
-2
-1
0
1
2
39
4
1
0
1
4
9-15
-24
-18
0
17
22
2145
48
18
0
17
44
635
17
35
65
82
93
100Σf=100 Σfd=3 Σfd2= 235
-
- AC = √602 + 452 = 75
NC = 1/2AC
= 37.5m
tan r = 30/37.5
r = tan 1(30/37.5) = 38.66°
tan r= 30/30 = 45°
n² = a² + b² - 2ab Cos n
bo² - 37.5² CosN
Cos-1 = -0.28
<N = 106.3° - Therefore the bearing of B from A = 360° - 106.3° = 253.7°
-
-
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