MATHEMATICS
PAPER 2
SECTION 1 (50 MARKS):
answer all questions in the section.
 Use Logarithms correct to four significant figures to evaluate. (4marks)
∛^{24.36×0.066547}/_{1.482} 
 Given the vectors a = 3i – j + 2k, b = 4i + 2j – k and p = 2a – b. Express p in terms of i, j and k. (2 marks)
 Hence calculate pcorrect to 3 significant figures. (1 mark)
 W varies directly as the cube of x and inversely as y. Find W in terms of x and y given that W = 80 when x = 2 and y = 5. (2 Marks)
 A cold water tap can fill a bath in 10 minutes while a hot water tap can fill it in 8 minutes. The drainage pipe can empty it in 5 minutes. The cold water and hot water taps are opened for 4 minutes. After four minutes all the three taps are opened. Find how long it takes to fill the bath. (3 Marks)
 Object A of area 10cm^{2} is mapped onto its image B of area 60cm2 by a transformation. Whose matrix is given by
p =
. Find the positive values of x (3 Marks)  Make P the subject of the formula in L = ^{2}/_{3} √^{x²  PT} /_{y} (3 Marks)

 Expand the expression (1+ ^{1}/_{2} x)^{5 }in ascending order powers of x, leaving the coefficients as fractions in their simplest form. (2 Marks)
 Use the first three terms of the expansion in (a) above to estimate the value of (1.05)^{5} (2 Marks)
 By rounding each number to the nearest tens, approximate the value of ^{2454 × 396}/_{66}
Hence, calculate the percentage error arising from this approximation to 4 significant figures. (3 Marks)  Without using a calculator or mathematical tables, express ^{√3}/_{1Cos 30º} in surd form and simplify (3 Marks)
 Solve for θ in the equation sin (3 θ + 120º = ^{√3}/_{2} for 0º ≤ θ ≤ 180º (3Marks)
 The equation of a circle is x² + y² + 6x 10y – 2 = 0. Determine the coordinates of the centre of the circle and its radius. (3 Marks)
 Kamau deposited ksh.50,000 in a financial institution in which interest is compounded quarterly. If at the end of second year he received a total amount of ksh79,692.40. Calculate the rate of interest p.a (3 Marks)
 Chords AB and CD in the figure below intersect externally at Q. if AB = 5cm BQ = 6cm and DQ = 4cm, calculate the length of chord CD. (3 Marks)
 Two containers have base area of 750cm2 and 120cm2 respectively. Calculate the volume of the larger container in litres given that the volume of the smaller container is 400cm3. (3 Marks)

 Find the inverse of the matrix (1 Marks)
 Hence or otherwise solve the simultaneous equations (3marks)
3x + 2y = 4
5x + 4y = 9  Solve for x in the equation. (3 Marks)
Log _{8} (x + 6) – Log_{8} (x – 3) = ^{2}/_{3}
 Find the inverse of the matrix (1 Marks)
SECTION II (50 MARKS)
Attempt ONLY FIVE questions from this section.
 In triangle ABC, shown below, AB = a AC = b point M lies on AB such that AM: MB = 2:3 and point N lies on AC such that AN: NC = 5:1 line BN intersects line MC at X.
 Express the following in terms of a and b
 BN (1 mark)
 CM (1 mark)
 Given that BX = kBN and CX = rCM where k and r are scalars
 Write two different expressions for AX in term of a, b, k and r (4 marks)
 Find the values of k and r (4 marks)
 Express the following in terms of a and b
 A particle moves such that t seconds after passing a given point O is given by
S = t ( t – 2 ) ( t – 1) Find its velocity when t = 2 second (3 marks)
 Find its minimum velocity. (3 marks)
 Find the time when the particles is momentarily at rest. (3 marks)
 Find its acceleration when t = 3 seconds. (1mark)
 The following table shows the rate at which income tax was charged during a certain year.
Monthly taxable income in Ksh. Tax rate % 0  9860
9861  19720
19721  29580
29581  39440
39441  49300
49301  59160
over 5916010
15
20
25
30
35
40
A civil servant earns a basic salary of Ksh.35750 and a monthly house allowance of sh.12500. The civil servant is entitled to a personal relief of sh.1062 per month. Calculate: Taxable income (2 marks)
 Calculate his net monthly tax (5 marks)
 Apart from the salary the following deduction are also made from his monthly income.
WCPS at 2% of the basic salary
Loan repayment Ksh.1325
NHIF sh.480
Calculate his net monthly earning. (3 marks)


 Taking the radius of the earth, R= 6370km and π = 22/7, calculate the shortest distance between the two cities P(60oN, 29oW) and Q(60oN, 31oE) along the parallel of latitude. (3marks)
 If it is 1200hrs at P, what is the local time at Q. (3marks)
 An aeroplane flew due south from a point A (60oN, 45oE) to a point B. the distance covered by the aeroplane was 8000km. determine the position of B. (4marks)

 A married couple intends to have 3 children. They consult an expert who tells them that the probability of a male birth is 0.55.
 Draw a tree diagram to represent this occurrence. (2 marks)
 Find the probability that
 All the three children will be female. (2 marks)
 At least a male is born. ( 2 marks)
 At least 2 will be females, giving your answer to 3 s.f. ( 4 marks )
 An arithmetic progression (AP) has the first term a and the common difference d.
 Write down the third, ninth and twenty fifth terms of the AP in terms of a and d. (1mark)
 The AP above is increasing and the third, ninth and twenty fifth terms form the first three consecutive terms of a Geometric Progression (G.P) The sum of the seventh and twice the sixth terms of the AP is 78. Calculate:
 the first term and common difference of the AP. (5marks)
 the sum of the first nine terms of the AP. (2marks)
 The difference between the fourth and the seventh terms of an increasing AP. (2marks)

 Draw ∆PQR whose vertices are P(1,1)Q(3,2) and R(0,3) on the grid provided (2marks)
 Find and draw the image of ∆PQR under the transformation whose matrix is and label the image P’Q’R’ (2mks)
 P’Q’R’ is then transformed into P11 Q11 R11 by the transformation with the matrix Find the coordinates of P^{11}Q^{11}R^{11} and draw P^{11}Q^{11}R^{11} (3marks)
 Describe fully the single transformation which maps PQR onto P^{11}Q^{11}R^{11} and find the matrix of this transformation (3marks)
 For a sample of 100 bulbs, the time taken for each bulb to burn was recorded. The table below shows the result of the measurements.
Time
(in hours)1519 2024 2529 3034 3539 4044 4549 5054 5559 6064 6569 7074 Number
of bulbs6 10 9 5 7 11 15 13 8 7 5 4  Using an assumed mean of 42, calculate
 the actual mean of distribution (4 Marks)
 the standard deviation of the distribution (3 Marks)
 Calculate the quartile deviation (3 Marks)
 Using an assumed mean of 42, calculate
MARKING SCHEME

No. Log 24.36
0.066547
1.48
10^{1 }x 9.045
0.90451.3867
2.8231
0.2098
0.1703
X 2
0.3406
0.2098
0.3406
1.8692 x ^{1}/_{3}
3 /3 + 2.8692
3
=1.9564 
 P = 2 (3i – j + 2k) – 4i + 2j – k)
= 6i – 2j + 4k – 4i – 2j + k
= 2i – 4j + 5k  /p/ = √2^{2} + 4^{2} + 5^{2}
= 6.708
=6.71
 P = 2 (3i – j + 2k) – 4i + 2j – k)
 W = Kx^{3}
y
80 = ^{8}/_{5} k
K = 50
W = 50 x^{3}
y  (^{1}/1_{0}+ ^{1}/_{8}) x 4 = ^{9}/_{40} x 4 = ^{9}/_{10}^{1}/1_{0} + ^{1}/_{8}  ^{1}/_{5} = ^{(4+58)}/_{40} = ^{1}/_{40}∴Time taken = ^{1/}_{10} ÷ ^{1}/_{40}
= ^{40}/_{10} = 4 Min  ASF = determinant of the matrix
^{60}/_{10} = x(x + 3) – 12
6 = x^{2} + 3x – 12
(x^{2} – 3x) + (6x – 18) = 0
x(x3) + 6 (x – 3) = 0
(x + 6) (x3) = 0
x = 6 or 3  ^{4}/_{9} ^{(X²PT)}/_{y} = L^{2}
x^{2}PT = ^{9}/_{4} L^{2}y
PT = ^{9}/_{4} L^{2}  x^{2}
P = x²  ⁹/₄L²y
T 
 1.15 [^{1}/_{2} x]^{0 }+ 5.14 [1/2 x]^{1}+ 10.1^{3} [^{1}/_{2} x]^{2} + 10.1^{2} [^{1}/_{2} x]^{3} + 5.1^{1} + [^{1}/_{2} x]^{4}+ 1.1^{0}+[^{1}/_{2} x]^{5}1 + ^{5}/_{2} x + ^{5}/_{2} x^{2} + ^{5}/_{4} x^{3} + ^{5}/_{16} x4 + ^{1}/_{32} x^{5}
 [1.05]^{5} = 1 + ^{5}/_{2} × 0.1 +^{ 5}/_{2}× 0.1^{2}= 1.275
 2450 × 400 = 14,000
70
2454×396= 14724
66
% error = (1472414000)/14724 x 100
≈ 4.917  √3 = √3
1Cos30º 1^{√3}/_{2}= (2√3  (2+√3)
(2√3)  (2+√3)
= 2√3 (2+ √3)
43
= 4√3+6  Sin (3θ+120)=^{√3}/_{2}3θ+120 = 60,120,420,480
3θ+120 = 120,420,480
θ=0º,100º,120º  x^{2} + 6x + y^{2} – 10y = 2
complete the square
x^{2} + 6x + (^{6}/_{2})^{2} + y^{2} – 10y +(^{10}/_{2})^{2} = 2 + (^{6}/_{2})^{2} +(^{10}/_{2})^{2}x^{2} + 6x + 9 + y^{2} – 10y + 25 = 36
(x + 3)^{2} + (y – 5)^{2} = 36
(x + 3)^{2} + (y – 5)^{2} = 6^{2}
(x – h)^{2} + (y – k) = r^{2}Center (3,5)
Radius = 6  50,000 (1 + ^{r}/_{100})^{8}. =79, 692.4
50,000 50 000
= ³√(1 + ^{r}/_{100})^{8 }= ⁸√1.593848
(1 + ^{r}/_{100} )= 1.06
^{r}/_{100}= 0.06
r = 6% per interest period
per annum = 6 x 4 = 24%  AQ x BQ = CQ x DQ
11 x 6 = (x + 4) 4
66 = 4x + 16
^{4x}/_{4} = ^{50}/_{4}
x = 12.5  A.S.F = ^{750}/_{120} = ^{25}/_{4}L.S.F = √^{25}/_{4}= ^{5}/_{2} ✓
V.S.F = (^{5}/_{2})^{3}= ^{125}/_{8}
V.S.F = ^{125}/_{8} = ^{V}/_{400cm³}
V = ^{125}/_{8} x 400cm^{3}
Volume = ^{6250}/_{1000}
= 6.25 litres 
 Log_{8} (^{x+6}/_{x3}) = Log_{8}8^{²/₃(x+6)}/_{(x3)} = 4
x + 6 = 4x – 12
3x = 18
x = 6 

 BN = BA + AN
= a +^{5}/_{6} b
= ^{5}/_{6} b – a  CM = b + r CM
= b + r (^{2}/_{5} a  b)
 BN = BA + AN

 AX = b + rCM
= b – rb +^{2}/_{3} ra
= ( 1 – r)b + ^{2}/_{5} ra
AX = a + k (^{5}/_{6} b – a)
= a – ka +^{5}/_{6} kb
= (1 – k)a +^{5}/_{6} kb  ( 1 – r)b + ^{2}/_{5} ra= ( 1 k)a + ^{5}/_{6} kb
^{5}/_{6} k = 1 – r – ( i)
^{2}/_{5} r = 1 – k  (ii)
K =^{6}/_{5} – ^{6}/_{5} r
^{2}/_{5} = 1 (^{6}/_{5} – ^{6}/_{5} r)= 1 – ^{6}/_{5} + ^{6}/_{5} r
4r = 1 → r = ¼
^{5}/_{6} k = ¾
k = ^{9}/_{10}
 AX = b + rCM


 S = t (t^{2} – t – 2t + 2)
= t^{3} – 3t^{2} + 2t
^{ds}/_{dt} = 3t^{2} – 6 t + 2
v = 3 ( 4)  6 ( 2) + 2
v = 2m/s  dv/dt=0
dv/dt=6t6
6t – 6 = 0
t = 1
v= 3(1)^{2} – 6(1) + 2
= 1m/s  3t2 – 6t + 2 = 0
t =( 6±√(364×3×2))/(2×3)
t = (6± 3.464)/6
t= 1.577 or 0.4227  a = 6 ( 3) – 6 = 12m/s^{2}
 S = t (t^{2} – t – 2t + 2)

 taxable income
35750 + 12500 = 48250= sh.48250  9860 × ^{10}/_{100} = sh.986
9860× ^{15}/_{100} = sh.1479
9860×^{ 20}/_{100} = sh.1972
9860× ^{25}/_{100} = sh.2465
8810×^{ 30}/_{100} = sh.2643
Total= 9545
95451062
=sh.8483pm  WCPS =^{2}/_{100} × 35750 = 715
Total deduction=(8483 + 715 + 1325 + 480)
= 11003
Net salary = 48250 11003
= sh.37247 p.m M1
 taxable income


 P(60º N,29º W) and Q (60º ,31º E)
longitudinal diff = 60º
Shorter distance = ^{60}/_{360 }× 2 × ^{22}/_{7 }× 6370 cos60
= 3335.3 km
 P(60º N,29º W) and Q (60º ,31º E)
 Time difference = 4 × 60 = 240 min
= 4hrs
Time at Q = 1200 + 0400 = 1600hrs  ^{(θ+60)}/_{360} × 2 × ^{22}/_{7} × 6370=8000
θ+60=71.92
θ=11.92
B(11.92ºS,45ºE)




 P(FFF)= ^{45}/_{100} × ^{45}/_{100} × ^{45}/_{100}
=91125/1000000
= 0.091125  1P(FFF)
1 – 0.091125
= 0.908875  P(MFF)or P(FMF)or P(FFM)
=(^{55}/_{100} × ^{45}/_{100} × ^{45}/_{100} ) +(^{45}/_{100} × ^{55}/_{100} × ^{45}/_{100)} + ^{45}/_{100} × ^{45}/_{100} × ^{55}/_{100} )
= 0.111375 + 0.111375 + 0.111375 = 0.334125
=0.334
 P(FFF)= ^{45}/_{100} × ^{45}/_{100} × ^{45}/_{100}


 a+2d, a+8d, a+24d

 a + 8d = a + 24d
a + 2d a + 8d I
a+6d+2(a+5d)=78 II
16d=10aI
3a+16d=78 II
3a+10a=78
3a=78
a=6
16d=10×6
d=3.75  S__{q}=^{9}/_{2} (2×6+(91)3.75) =189
 ⇒[6+6(3.75)][6+3(3.75)] =11.25
 a + 8d = a + 24d



 It is an enlargement centre origin (0,0) scale factor 2

Class Mid point X f t = ^{XA}/_{C} ft t^{2} ft^{2} cf 1519 17 6 5 30 25 150 6 2024 22 10 4 40 16 160 16 2529 27 9 3 27 9 81 25 3034 32 5 2 10 4 20 30 3539 37 7 1 7 1 7 37 4044 42 11 0 0 0 0 48 4549 47 15 1 15 1 15 63 5054 52 13 2 26 4 52 76 5559 57 8 3 24 9 72 84 6064 62 7 4 28 16 112 91 6569 67 5 5 25 25 125 96 7074 74 4 6 24 36 144 100 Σf=100 Σft=28 Σft^{2}=873 
 x = 42 + (^{28}/_{100} ×5) = 43.4
 873/100  (^{28}/_{100})^{2} = 0.7856
 x = 42 + (^{28}/_{100} ×5) = 43.4
 [49.5+ ^{(7563)}/_{13})5.29.5 ]× ^{1}/_{2}
= 12.31

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