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mathematics papers from different counties

                  Mathematics   CONTENTS 
  1. EMBU COUNTY COMMON EVALUATION
  2. KISII CENTRAL DISTRICT JOINT EVALUATION
  3. ALLIANCE GIRLS HIGH SCHOOL
  4. STAREHE BOYS CENTER
  5. ALLIANCE BOYS HIGH SCHOOL TRIAL EXAMINATION
  6. NAMBALE DISTRICT JOINT EXAMINATION
  7. MERU CENTRAL COMMON EVALUATION
  8. NYERI NORTH REGION JOINT EXAMINATION
  9. THIKA(FORM FOUR MID YEAR CONTINUOUS ASSESMENT TEST)
  10. KITUI WESTFORM FOUR EVALUATION EXAMINATION
  11. NYERI SOUTH FORM FOUR JOINT EVALUATION

121/1
MATHEMATICS
PAPER 1
2 ½ HOURS
July/August 2011
THIKA DISTRICT
FORM FOUR MID YEAR CONTINUOUS ASSESSMENT TEST.
Kenya Certificate of Secondary Education
MATHEMATICS
PAPER 1
2 ½ HOURS

INSTRUCTIONS TO CANDIDATES
(a) Write your name and admission number in the spaces provided above.
(b) Sign and write the date of examination in the spaces provided above.
(c ) This paper consists of TWO sections; Section I and section II.
(d) Answer all the questions in section I and ONLY FIVE questions from section II.
(e) All answers and working must be written on the question paper in the spaces provided below each question.
(f) Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
(g) Marks may be given for correct working even if the answer is wrong.
(h) Non-programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise
(i) Candidates should check the question paper to ascertain that all pages are printed as indicated and that no questions are missing.

FOR EXAMINER’S USE ONLY

SECTION I
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total

SECTION II
17 18 19 20 21 22 23 24 Total

SECTION 1 ( 50 MARKS )
Answer all questions in this section

  1. Without using a calculator, evaluate -5 (-23 + 41) – (-10)
    -3 + (-8) ÷ 2 x 4
    Leaving your answer as a mixed fraction. ( 3 marks )
  2. A year ago, Musango was four times as old as Wambua. In three years time,
    Musango will be three times as old as Wambua. Find their present ages ( 4 marks )
  3. Find the integral values of X satisfying the inequalities
    ½ (X + 3 ) > 2X + 4 ≥ X + 1 ( 3 marks )
  4. A translation maps a point P(1, 2) onto P1 (-2, 2). Point R is mapped onto R1 (-3, 3)
    under the same translation. Find the length of PR. ( 3 marks )
  5. Nairobi and Mtito Andei are 240km apart. A bus left Nairobi at 9.00 a.m and travelled
    towards Mtito–Andei at 80km/h. 45 minutes later, a car left Mtito-Andei for Nairobi
    at a speed of 100km/h. Calculate the distance covered by the bus before meeting
    the car. ( 3 marks )
  6. Solve for x in 9x + 32x – 1 = 53. ( 3 marks )
  7. The line 4ax = 5y + 12 and 7bx – 3y = 2 are parallel. Find the value of a
    b ( 3 marks )
  8. If the expression given is a perfect square, find the value of R.
    X2 – RX + 2R – 3. ( 3 marks )
  9. A Kenyan bank buys and sells foreign currencies as shown below.
    Buying ( in Ksh ) Selling (in Ksh)
    I S.A Rand 7.88 7.91
    I Saudi Riyal 19.75 20.00 Mr. Brown arrived in Kenya with 102000 S.A rand and changed the whole amount to Kenya
    shillings. While in Kenya, he spent Ksh. 203,760 and changed the balance to Saudi Riyal
    before leaving for Saudi Arabia. Calculate, without using Mathematical tables, the
    amount in Saudi Riyal that he received. ( 3 marks )
  10. Solve for P in the equation 3 log 8P + 5 = 2 logP8. ( 4 marks )
  11. In the figure below, triangle PQR is similar to triangle PST and QR//TS.
    Given that PQ : PT = 2 : 5, find the ratio of the area of triangle PQR to that of
    the trapezium QRST. ( 3 marks ) P T S Q R</code></pre></li>Using tables of square roots and reciprocals, evaluate 0.01960.246 Giving your answer correct to three significant figures. ( 3 marks )</code></pre></li>The sums of interior angles of two regular polygons of sides n-1 and n are in
    the ratio 2 : 3. Calculate; (i) The value of n. ( 2 marks ) (ii) The size of interior angle of each polygon. ( 2 marks ) Find the least number of steps in staircase if, when I go up 2 steps at a time, 3 steps
    at a time or 4 steps at a time, there is always 1 step remaining at the top. ( 3 marks )

. Solve for Ө in the range 00≤ Ө ≤3600 given that sin Ө = - ½ ( 2 marks ) Triangle ABC in the figure below is transformed to a triangle A1B1C1 by a rotation
of -600 about the point O. O. A B C</code></pre>Construct ∆A1B1C1 and measure the length of AA1. ( 4 marks ) SECTION II ( 50 MARKS )
Answer ONLTY five questions from this section The marks scored by a certain number of students in a maths contest are as shown below. Marks 45-49 50-54 55-59 60-64 65-69 70-74 75-79
Number of students 10 11 14 41 27 18 19 (a) Using an assumed mean of 62, calculate to 2 d.p. the mean of the marks. ( 4 marks ) (b) Calculate the median. ( 3 marks ) (c ) Calculate the standard deviation. ( 3 marks ) . A ship leaves port K for port M through port L. L is 200km on a bearing of 2300 from K.
M is 400km on a bearing of 1500 from L. (a) Using a scale of 1 : 4000000, draw a diagram to show the relative positions
of the three ports K, L and M. ( 3 marks ) (b) If the ship had sailed directly from K to M at an average speed of 40 knots, find how
long it would have taken to arrive at M. (Inm = 1.853km) ( 2 marks ) (c ) By further drawing on the same diagram, determine: (i) How far M is to the South of K. ( 2 marks ) (ii) The shortest distance from L to the direct path KM ( 2 marks ) (d) What is the bearing of K from M? ( 1 mark ) The figure below represents Musau’s piece of land PQR divided into two triangular plots. Q 320 125m 84m 70m P 65m S R </code></pre>(a) Calculate the area of the plot PQR in square metres correct to two decimal
places. ( 4 marks)
(b) In the year 2008, Musau used 65% of the plot QSR for grazing and 40%
of the plot QPS for horticultural farming. Calculate, to the nearest square
metre, the piece of land left unutilized. ( 2 marks ) (c ) In the year 2009, the land grazing changed in the ratio 5 :3 while that for
horticultural farming changed in the ratio 2 : 3. Calculate, to two decimal
places the total area of land used for both grazing and horticultural farming. ( 4 marks ) Vector OA = 3 and OB = 7
6 -6 Point C is on OB such that CB = 2 OC and D is on AB such that AD = 3DB. (a) Express CD as a column vector. ( 4 marks ) (b) Length CD ( 3 marks ) (c ) A point Q divides AB in the ratio 5 : -3. Find the position vector q of point Q. ( 3 marks) In the figure below, the circle with centre A, radius 18cm represents the number of
people in a certain town who are infected with HIV/AIDs. The circle with centre B,
radius 12cm represents the number of people in the same town infected with TB. Calculate to two decimal places
(a) The area of the sector subtending angle 650 at the centre ( π= 3.142) ( 2 marks ) (b) The area of the sector subtending angle 1000 at the centre. ( 2 marks ) (c ) The area representing people with both HIV/AIDs and TB. ( 4 marks ) (d) The minor arc length subtended by 1000. ( 2 marks ) The diagram below shows the speed –time graph of a bread delivery van traveling between
Emali and Sultan – Hamud. The van starts from rest and accelerates uniformly for 200 seconds.
It then travels at a constant speed for 350 seconds and then decelerates uniformly for 300 seconds. Speed (m/s) Time (sec)</code></pre>Given that the distance between the two towns is 15km, calculate the
(a) Maximum speed in km/h the van attained. ( 3 marks )
(b) Acceleration. ( 2 marks ) (c ) Distance the van travelled during the last 150 seconds. ( 2 marks ) (d) Time the van takes to travel the first half of the journey. ( 3 marks ) (a) Make a table of values for the function
y = 2x2 – 3x – 5 for integral values of x in the range -2 ≤ x ≤ 3. ( 2 marks ) (b) Plot the graph of y = 2x2 – 3x – 5 for -2 ≤ x ≤ 3
Use the scale 1cm rep. 2 units vertically
2cm rep. 1 unit horizontally ( 3 marks) (c ) Use your graph to solve the equation
2x2 – 3x – 5 = 0 ( 1 mark ) (d) By drawing a suitable straight line on the same axes, solve the equation
4x2 – 2x – 6 = 0 . ( 4 marks ) A rectangular tank whose internal dimensions are 2.4m by 2.5m by 3.7m is two thirds full of juice.
(a) Calculate the volume of the juice in litres. ( 3 marks ) (b) The juice is packed in small packets in a shape of a right pyramid with equilateral
triangle sides of 20cm. The height of each packet is 15cm. Full packets obtained are sold at Ksh. 50 per packet. Calculate; (i) The volume in cm3 of each packet to the nearest whole number. ( 3 marks ) (ii) The number of full packets of juice. ( 2 marks ) (iii) The amount of money realized from the sale of the juice. ( 2 marks ) 121/2
MATHEMATICS
PAPER 2
2 ½ HOURS
July/August 2011
THIKA DISTRICT
FORM FOUR MID- YEAR CONTINUOUS ASSESSMENT TEST
Kenya Certificate of Secondary Education
MATHEMATICS
PAPER 2
2 ½ HOURS INSTRUCTIONS TO CANDIDATES
(a) Write your name and admission number in the spaces provided above.
(b) Sign and write the date of examination in the spaces provided above.
(c ) This paper consists of TWO sections; Section I and section II.
(d) Answer all the questions in section I and ONLY FIVE questions from section II.
(e) All answers and working must be written on the question paper in the spaces provided below each question.
(f) Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
(g) Marks may be given for correct working even if the answer is wrong.
(h) Non-programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise
(i) Candidates should check the question paper to ascertain that all pages are printed as indicated and that no questions are missing. FOR EXAMINER’S USE ONLY SECTION I
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total SECTION II
17 18 19 20 21 22 23 24 Total This paper consists of 21 printed pages
Please Turn Over SECTION I ( 50 MARKS )
Answer all questions in this section Use logarithms correct to 4 decimal places to evaluate. 0.0249 x 5.825 3 0.0006748 ( 4 marks )</code></pre></li>Make X the subject of the formula ( 2 marks )
K = xt + 1
x The nth term of G.P is given by 5 x 2n-2
(i) Write down the first 4 terms of the G.P ( 1 mark ) 2. (ii) Calculate the sum of the first 6 terms. ( 2 marks ) A student converts the fraction 5/9 into decimal and truncates the result to
2 decimal places. Calculate the relative error in this approximation
(Give your answer in fraction form ) ( 3 marks ) The figure below shows a circle centre O, of radius 7cm. TP and TQ are tangents to
the circle at points P and Q respectively. OT = 16cm P 7cm O 16cm
T Q Calculate the length of chord PQ. ( 3 marks) 3. Triangle ABC is such that ABC = 900, AB = ( √3 + 1 )cm and AC = (√3 + 2 )cm.
Find cos A in simplified surd form. ( 4 marks ) If variables M, D, R are related by the equation M = KDR3 where K is constant,
find the percentage change in M if D is reduced by 10% and R is increased by 10%. ( 3 marks ) The table below is part of a taxation table for monthly income for the year 2010. Monthly taxable income in Ksh. Tax rate (%) per Ksh.
1 – 9680 10
9681 – 18800 15
18801 – 27920 20 Kyele paid sh. 2476 as tax during the month of February in this year. Calculate his monthly income. ( 3 marks ) Determine the equation of the tangent to the circle with equation. (x-5)2 +(y-5)2 = 5 at the point (6, 7) ( 4 marks ) The cost of three books and five pens is sh. 135. The cost of two similar books and
seven similar pens is sh. 145. Use the matrix method to find the cost of each item. ( 4 marks ) A point R (2,2) is mapped onto R1 (8,2) by a shear, x –axis invariant. Find the shear
factor hence or otherwise write down the shear matrix. ( 3 marks ) The graph in the figure below shows the variation of depth of water d(metres) in a
river with time t(hours) after a heavy down pour. .
Use the graph to find the rate of decrease in depth at t = 1.25 hours
( Give your answer correct to 2 significant figures) ( 3 marks ) The number of overtime hours worked per week by 130 men of the maintenance
department of a power and lighting company are as shown. (a) Find X ( 1 mark )
(b) Find the mean number of overtime hours giving your answer correct to 2dp. ( 3 marks ) No. of overtime hours 0 – 2.5 2.6 – 5.0 5.1 – 7.5 7.6 – 10.0 10.1 – 12.5 12.6 – 15.0
Frequency (f) 31 48 26 14 8 x P B Ө 90m 60m C A </code></pre>In the diagram A, B, C are all in a horizontal plane. Flag pole BP is 90m from A and 60m
from C. Angle ABC = 900. The angle of elevation of P from A is Ө where tan Ө = 1/3.
Calculate the height BP and the angle of elevation of P from C. ( 4 marks ) A square whose vertices are the points A(0, 4) B (1, 1), C (4, 2), D (3, 5) is mapped
onto the plane figure A1 B1 C1 D1 by matrix M = -1 0
4 3
Find the area of A1 B1 C1 D1. ( 3 marks ) Find the 4th term of the expansion of 1 + 1 x 6 ( 2 marks )
5 SECTION II ( 50 MARKS )
Answer only five questions in section II A bag contains 10 similar pens of which 6 are red and the rest are blue in colour.
Three pens are picked at random, one at a time from bag without replacement. (a) Draw a tree diagram to show the various outcomes. ( 2 marks )
(b) Find the probability that
(i) None of the pens picked is red. ( 2 marks) (ii) At least one of the pens picked is red. ( 2 marks ) (iii) Only one blue pen is picked. ( 2 marks ) (iv) The first two pens picked are of the same colour. ( 2 marks ) (a) On the grid provided, draw a graph of the function y = 3x2 + 8 for 0 ≤ x ≤ 5 ( 3 marks) (b) Calculate the mid-ordinates for 4 strips between x = 1 and x = 5 and hence use the mid-ordinate rule to approximate the area under the curve between x = 1, x = 5 and the x – axis . ( 3 marks ) (c ) Assuming the same area determined by integration to be the actual area, calculate the percentage error in using the mid-ordinate rule. ( 4 marks ) . (a) Complete the table below for the functions y = 2 cos x and y = sin 2x. ( 2 marks ) x0 -1800 -1500 -1200 -900 -600 -300 0 300
2x0 -300 -120 0
2 cos x 1.73 2
Sin 2x 0.87 0 x 600 900 1200 1500 1800
2x0 180 360
2 cos x 1 2
Sin 2x -0.87 0 (b) On the same axes, draw the graphs of y = 2 cos x and y = sin 2x for -1800 ≤ x ≤ 1800. ( 4 marks) ( c ) Use the graphs in (b) above to find; ( i) The values of x such that 2cos x – sin 2x = 0 . ( 1 mark ) (ii) State the amplitude and period of each graph. ( 2 marks ) (iii) Find the difference in the values of y when x = -45. ( 1 mark ) Use ruler and a pair of compasses only in this question.
(a) Construct triangle ABC with AB = 5cm, BC = 6cm and AC = 7cm. ( 2 marks ) (b) Construct the locus of a point O such that OA = OB = OC. ( 2 marks )
(c ) With O as centre and radius OB draw a circle. ( 1 mark ) (d) At B construct the tangent YBX such that X is on the same side of BO produced as C. ( 2 marks) (e) Construct the locus of points L through O such that the distance from L to
XY is constant. ( 2 marks ) (f) Locate point P such that P is on L and  ABY = BPA. ( 1 mark ) The derived function of a curve is 2x + 2
(a) Find the equation of the curve if it passes through ( -2, -3 ) ( 2 marks ) (b) Find the x intercepts of the curve. ( 3 marks ) (c ) The curve passes through point (O, y). Find the value of y. ( 1 mark ) (d) Find the value of y at the point where the gradient of the curve is O (zero). ( 2 marks ) E) In the space provided below, sketch the curve. ( 2 marks ) A drug company specializing in food supplements makes up a product by mixing two ingredients A and B each containing both zinc and iron. The mineral composition per gram of each ingredient is given in the table below. Ingredient Zinc (Units per gram) Iron (Units per gram)
A 20 50
B 40 75 The product is to be packaged in small bottles each containing x grams of A and y grams of B. The following conditions must be satisfied I: The mass of the product in each bottle must not exceed 10 grams II: Each bottle is to contain at least 200 units of zinc III: Each bottle is to contain at least 450 units of iron (a) Write down three inequalities in terms of x and y to express the above conditions. ( 3 marks ) (b) On the grid provided, draw the inequalities by shading the unwanted region. ( 3 marks ) (c ) Given that the costs per gram of ingredients A and B are sh. 60 and
sh. 100 respectively, draw a search line and use it to determine the minimum cost of one bottle of the product. ( 4 marks ) A plane flying at 200 knots leaves airport P(200N, 340W) and flies due South to an
airport Q(600S, 340W). (a) Calculate the distance covered by the plane in nautical miles. ( 3 marks ) (b) The plane stops at Q for 30 minutes before flying due East to an airport R(600S, 60E)
at the same speed. Calculate the total time taken to complete the journey from
airport P to airport R through airport Q. ( 5 marks ) (c ) If at the time of arrival, the local time at R is 6.00pm, what is the local time at P then ? (2 marks) The figure below shows a right pyramid VABCD with a square base of side 6cm.
VA = VB = VC = VD = 12cm. (a) Calculate
(i) The height of the pyramid correct to 2 d.p . ( 3 marks ) (ii) The angle between the planes VAD and VBC correct to 2 significant figures. ( 3 marks )
(b) B1 and C1 are points on VB and VC respectively such that
VB1 : VB = VC1 : VC = 1 : 3. Calculate the angle between planes ABCD
and AB1C1D. ( 4 marks ) 121/1
MATHEMATICS
PAPER 1
July/August 2011
THIKA DISTRICT
FORM FOUR MID YEAR CONTINUOUS ASSESSMENT TEST
MARKING SCHEME -5 (18) + +10
-3 + -4 x 4 M1 √ Removal of brackets -90 + 10
-3 + -16
-80
-19 M1 √ Order of operations 80 19 44/19 A1 3</code></pre></li>Musango now - x yrs
Wambua now - y yrs
x – 1 = 4 (y-1) M1
x + 3 = 3 ( y + 3 ) M1
x – 4y = -3
x – 3y = 6 M1 √ attempt to eliminate
-y = 9
x = 33 Musango = 33 yrs Wambua = 9 yrs A1 4 ½ ( x + 3 ) > 2x + 4 ………….. (i)
2x + 4 ≥ x + 1 ………….. (ii) x + 3 > 4x + 8
-3x > 5
x < 5/-3
x < -12/3 B1 2x + 4 ≥ x + 1
x ≥ -3 B1 -3 ≤ x < -12/3 Integral values = -2, -3 B1 3</code></pre></li> 4 Let T = x
 y 1 + x = -2 M1 2 y 2 x = -3 y 0 a + -3 = -3 b 0 3 M1 a = 0 b 3 PR = (1 – 0)2 + (2-3)2 = 12 + (-1)2 = 2 = 1.414 A1 3 Dist covered by bus before car starts
= ¾ x 80 M1
= 60km
Rem. Dist. = ( 240 – 60 )km
= 180km Time taken to meet = 180 = 1hr
180 Dist. Covered by bus = 60 + 80 M1
= 140km. A1 32x + 32x – 1 = 53 M1 √ Expressing in
terms of 3
2 x 32x = 54
32x = 27
32x = 33
2x = 3 M1 √ Equating powers
x = 3/2 A1 3 5y = 4ax – 12
y = 4ax – 12
5 5 3y = 7bx – 2
y = 7bx - 2 B1 equations in the form
3 3 y = mx + c Parallel lines have equal gradients
 4a = 7b
5 3 M1 equating gradients 12a = 35b
12a = 35
b a = 35 b 12 A1 03</code></pre></li>Last term = (Middle term )2
4 x 1st term 2R – 3 = R2 . X2
4 . X2 M1 4(2R – 3) = R2 R2 – 8R + 12 = 0 R2 – 6R – 2R + 12 = 0 R ( R – 6) -2 (R -6) = 0
(R – 6) (R -2) = 0 M1
R = 2 or R = 6 A1 102000 x 7.88 M1 Conversion
= Ksh. 803760 Rem. = 803760 – 203760
= Ksh. 600,000 In Riyal = 600000 M1
20
= 30000 Riyal A1 03 3log8P + 5 – 2logP8 = 0
3log8P + 5 – 2 1 = 0
log8P M1 Let log8P = t
3t + 5 – 2/t = 0 3t2 + 5t – 2 = 0
3t2 + 6t – t -2 = 0
3t (t+2) – 1(t+2) = 0
(3t – 1) (t + 2) = 0 M1
t = 1/3 or t = -2 But log8P = t
Log8P = 1/3 or log8P = -2 M1 1/3
8 = P or 8-2 = P
P = 2 or P = 1/64 A1 L.S.F of ∆s = 2 :5
A.S.F = 4 : 25 B1
Area of trap = 25 – 4 M1
= 21
Ratio 4 : 21 A1
3 1 = 0.4065 x 10
0.246 = 4.065 B1 0.14 x 4.065 M1
= 0.569 A1 C.A.O
03 [ 2(n-1) -4] x 900 2
(2n-4) x 900 = 3 M1 √ Expression for sum
2n – 6 = 2
2n – 4 3 6n – 18 = 4n – 8
2n = 10
n = 5 A1 n = 5 (2 x 5 – 4) x 900
5 = 5400 5 = 1080 B1 n -1 = 4 (2 x 4 – 4) x 900
= 3600
4
= 900 B1 4 L.C.M = 22 x 3
= 12 B1
No. of steps = 12 + 1
= 13 B1 02 Acute angle = 300 B1
Ө = 2100, 3300 16 B1 02 (a) B1
Marks (x) f t = x – 62 ft
47 10 -15 -150
52 11 -10 -110
57 14 -5 -70
62 41 0 0
67 27 5 135
72 18 10 180
77 19 15 285
f = 140 ft = 270 X = 62 + 270
140 M1 = 62 + 1.93 = 63.93 A1 C.F: 10, 21, 35, 76, 103, 121, 140 B1 (b) Median = 59.5 + 140 35 x 5
2 M1
41 = 59.5 + 35/41 x 5 = 59.5 + 4.27 = 63.77 A1 (c )
d = x – m d2 f Fd2
-16.93 286.62 10 2866.2
-11.93 142.32 11 1565.52
-6.93 48.02 14 672.28
-1.93 3.72 41 152.52
3.07 9.42 27 254.34
8.07 65.12 18 1172.16
13.07 170.82 19 3245.58
f = 140 fd = 9928.6 S = 9928.6
140 M1 = 70.9186 = 8.421 A1 10 1cm rep. 40km B1 √ Scale interpretation B1 K to L (√ bearing and √ distance ) B1 L to M (√ bearing and √ distance )</code></pre>(b) km = 12 x 40 = 259.04nm B1 (12.0  0.1)
1.853 Time taken = 259.04 = 6.476 hrs B1 40 . (c ) (i) 11.9 x 40 = 476km  4km B1 point X √ly located B1 distance 476km (ii) 4.2 x 40 = 168km  4km B1 point ly located B1 distance 168km. (d) 3540 10 B1 bearing 10 (a) ∆PQS , S = 130 B1 Total A = 130 (130 – 125) (130 – 65) ( 130 – 170) M1 + ½ x 70 x 84 sin 320 M1 = 2535000 + 1557.96 = 3150.13m2 A1 C.A.O </code></pre>(b) Unutilised = 35 x 1557.96 + 60 x 1592.17
100 100 = 1500.588 = 1501m2 A1 (c ) 5 x 65 x 1557.96 + 2 x 40 x 1592.17 M1M1
3 100 3 100 = 2112.368667 A1 = 2112.37m2 B1 10</code></pre></li>A (3, 6 ) a 3  B1 D O 1 b 1 C  2 B (7, -6) Q</code></pre>.
(a) CD = CO + OA + ¾ AB = - 1/3b + a + ¾ ( b – a ) M1     = - 1 7 + 3 + 3 4 M1 3 -6 6 4 -12 = -7/3 + 3 + 3 2 6 -9 = 32/3 -1 A1</code></pre>(b) CD = (32/3)2 + (-1)2 M1 = 14.44 M1 = 3.8 A1</code></pre>(c ) q = n a + m b
  
m + n m + n = -3 3 + 5 7 2 6 -2 -6 M1 = -9/2 + 35/2 M1 -9 -15 = 13 -24 A1 10</code></pre></li>(a) A = 65 x 3.142 x 18 x 18 M1
360
= 183.807
= 183.81cm2 A1 C.A.O (b) A = 100 x 3.142 x 12 x 12 M1
360
= 125.68cm2 A1 C.A.O (c ) For A, area of ∆ = ½ x 18 x 18 sin 650
= 146.82 Area of segment = 183.81 – 146.82 M1 = 36.99cm2 For B, area of ∆ = ½ x 12 x 12 sin 100 = 70.91cm2 Area of segment = 125.68 – 70.91 M1 = 54.77cm2 Total area = 36.99 + 54.77 M1 = 91.76cm2 A1 C.A.O (d) Length = 100 x 3.142 x 24 M1
360
= 20.95cm A1 10 (a) ½ h ( 850 + 350) = 15000 M1 √eqn.
1200h = 30000
h= 25 m/s A1 In km/h = 3.6 x 25 = 90 km/h B1 (b) a = 25 m/s M1
200s = 1/8 m/s2 A1 or 0.12m/s2 (c ) S = ½ x 150 x 12.5 M1
= 937.5m A1 (d) ½ x 25 ( 200 + t + t ) = ½ x 15000 M1 eqn. 2t + 200 = 7500 12.52t = 400 t = 200 A1 Total time = 200 + 200 = 400 sec. B1 x -2 -1 0 1 2 3
y 9 0 -5 -6 -3 4 B2 all values √
allow B1 for 4 value √ S1 √ Scale P1 All pts √ly plotted √ if B1 earned C1 Smooth curve</code></pre></li> . (c ) x = -1 or 2.5 B1 (both ) (d) O = 4x2 – 2x – 6 O = 2x2 – x – 3 M1 Subtraction y = 2x2 – 3x – 5 y = -2x – 2 A1 eqn of line L1 √ line drawn x = -1 or x = 1.5 B1 (both) 10 (a) V = 2/3 x 2.4 x 2.5 x 3.7m3 M1 = 14.8m3 M1 = 14,800 litres A1 (b) (i) 20 20 h 600 10 10 h = 20 sin 60 = 17.32 M1Volume = 1/3 x ½ x 20 x 17.32 x 15 M1 = 866cm3 A1 (ii) No. of pockets = 14.8 x 1000000 866 M1 = 17,090 A1(iii) Money = sh. 50 x 17,090 M1 = Sh. 854,500 A1 10</code></pre></li> 121/2
July/August 2011
THIKA DISTRICT
FORM FOUR MID YEAR CONTINUOUS ASSESSMENT TEST
MATHEMATICS
PAPER 2
Kenya Certificate of Secondary Education
MARKING SCHEME No. Log
0.0249 2.3962
5.825 0.7653 + M1 All logs √
1.1615
3√ 0.0006748 1/3 x 4.8292 M1 √attempt to add/sub
= -6 + 2.8292
3 2.9431 M1 √ attempt to divide by 3 1.6535 0.2184 A1 C.A.O 04 kx = xt + 1
kx – xt = 1 M1 Xmult and collect
like terms
x ( k – t) = 1 x = 1 k – t A1 02</code></pre></li>(i) nth term = 5 x 22-n
n = 1 = 5 x 2 = 10
n = 2 = 5 x 20 = 5
n = 3 = 5 x ½ = 2.5
n = 4 = 5 x ¼ = 1.25  10, 5, 2.5, 1.25 …. B1 √ all terms (ii) Sn = a (1 – rn )
1 – r S6 = 10 (1 – 24) 1 – 2 M1 √ substitution = 10 ( 1 - 16 ) -1 = 10 x -15 -1 = 150 A1 √ 3</code></pre></li>5 becomes 0.55 = 55 B1 for 0.55
9 100 Rel. error = 5 - 55
9 100 M1
5/9 = 1 ÷ 5 180 9 = 1 100 A1 03</code></pre></li>PT = (16)2 – (7)2 = 256 - 49 = 207 = 14.39cm Cos Ө = 14.39 = 0.5756 M1 25 Ө = cos -1 ( 0.5756) = 54.860Sin 35.14 = h 7 PQ = 2h = 2 ( 7 sin 35.14 ) M1 = 2 ( 4.029) = 8.058cm A1 03 A √3 + 1 √3 + 2 B C</code></pre>Cos A = √3 + 1 √3 + 2 M1 √ Fraction = √3 + 1 x (√3 - 2√3 + 2 (√3 – 2) M1= 3 – 2√3 + √3 – 2 M1 removing surd from 3 – 4 denominator = √3 - 1 A1 04</code></pre></li>M = KDR3
M1 = 90/100 x D x ( 110/100 x R )3 x K M1 √ new M M1 = K x 0.9D x ( 1.1r)3
= 1.1979 KDR3 Change in M = 1.1979KDR3 – KDR3 M1 √change in M = 0.1979KDR3 % change in M = 0.1979KDR3 x 100 KDR3 = 19.8% A1 1st bracket : 10/100 x 9680 = 968 Remaining tax = 2476 – 968 = 1508 M1 1st and 2nd brackets 2nd bracket: 15/100 x 9120 = 1368 3.
Remaining tax = 1508 – 1368 = 140 20/100x X = 140 M1 x = sh. 700 Income = 700 + 9120 + 9680 = sh. 19500 A1 03 Centre O is (5, 5), D is (6, 7)  Gradient of OD = 7 – 5
6 - 5 M1 √ gradient of OD
= 2  The gradient of the tangent at D = - ½ M1 √ Gradient of tangent
at D Choosing a point on the tangent P(x, y)
D (6, 7) and P (x, y)
Gradient = y – 7
x – 6 y – 7 = - ½ x – 6 M1 √ expression -1(x – 6) = 2 ( y – 7 ) -x + 6 = 2y – 14 2y + x = 20 A1 √ tangent equation 3x + 5y = 135
2x + 7y = 145 M1 √ equations 3 5 x = 135
2 7 y 145 Inverse = 1 7 -5
11 -2 3 B1 inverse x = 1 7 -5 135 y 11 -2 3 145 M1 x = 20 y 15 x = 20, y = 15 A1 04 </code></pre></li>Alternatively
Shear factor = 8 – 2 M1 1 K 2 = 8 M1
2 0 1 2 2
= 3 A1 2 +2k = 8
k = 3 A1 Matrix = 1 3  1 3
0 1 B1 0 1 B1 03 Rate of decrease
Tangent drawn at t = 1.25 B1 see graph Rate = 2.2 – 0.8 M1 check for √ reading
0.5 – 1.55 = 1.4 -1.05 = -1.333 Rate of decrease = 1.3 m/h A1 Accept 1.2, 1.4
Accept answer with
negative sign
03 (a) x = 130 – 127
= 3 B1 Hours Mid pt f fx
0 – 2.5 1.25 31 38.75 B1 √ fx column
2.6 – 5.0 3.75 48 180.0
5.1 – 7.5 6.25 26 162.5
7.6 – 10.0 8.75 14 122.5
10.1 – 12.5 11.25 8 90.0
12.6 – 15.0 13.75 3 41.25
f =130 fx = 635 Mean (x) = fx = 635 M1 √ substitution f 130 = 4.88 (2dp) A1 √ A.O 14. Tan Ө = 1/3 </code></pre> 1/3 = BP
AB M1 √ relation 1 = BP
3 90 3BP = 90 BP = 30m Tan C = BP
60 Tan C = 30 60 M1 √expression for PCB Tan C = 0.5 C = 26.60 A1 03</code></pre></li>Length of side of the square = √ 10 M1
 The area of ABCD = 10 sq units
Area scale factor = det. = -3 – 0
= 3
 Area of A1B1C1D1
= 10 x 3 M1
= 30 sq units A1 20.13 . ( 1/5x)3 M1
20 x 1/125 x3
20/125x3 4/25x3 A1 02</code></pre></li>R …..RRR 4/8 R 4/8 B ……RRB B1 first two sets 5/9 R …. RBR 5/8 R 4/9 B</code></pre>6/10 3/8 B ….RBB B1 Last set
R R ….BRR
4/10 B 6/9 5/8
3/8
B …..BRB 3/9 R …..BBR B 6/8 2/8 B …..BBB (b) (i) P (no red) = P (BBB)
= 4/10 x 3/9 x 2/8 M1 = 1/30 A1 (ii) P (at least red) = 1 – P ( no red) = 1 = -1/30 M1 = 29/30 A1</code></pre>(iii) P(one blue) = P (RRB or RBR or BRR)
= ( 6/10 x 5/9 x 4/8 ) + ( 6/10 x 4/9 x 5/8) + (4/10 x 6/9 x 5/8) M1
= ½ A1 (iv) P (first two same) = P(RRR or RRB or BBR or BBB)
= (6/10 x 5/9 x 4/8) + (6/10 x 5/9 x 4/8) + (4/10 x 3/9 x 6/8) + (4/10x3/9x2/8) M1
= 7/15 A1
10 (a)
x 0 1 2 3 4 5
y 8 11 20 35 56 83 (b)
x 1.5 2.5 3.5 4.5
Y 14.75 26.75 44.75 68.75 A = 1 ( 14.75 + 26.75 + 44.75 + 68.75) M1
= 155 sq. units A1 (c ) ∫ (3x2 + 8) dx
1
= 5
= x3 + 8x M1
1 = {125 + 40 } - { 9 } = 156 A1 % error = 156 – 155 x 100 M1 156 = 1/156 x 100 = 0.006410 x 100 = 0.6410% A1 10</code></pre></li> . (a)
x0 -1800 -1500 -1200 -900 -600 -300 00 300 600 900 1200 1500 1800
2x -3600 -3000 -2400 -1800 -1200 -600 0 600 1200 1800 2400 3000 3600
2 cos x -2 -1.73 -1 0 1 1.73 2 1.73 1 0 -1 -1.73 -2
Sin 2x 0 0.87 0.87 0 -0.87 -0.87 0 0.87 0.87 0 -0.87 -0.87 0 B2 all values B1 10 values (i) x = -900 and x = 900 B1 (ii) y = sin 2x; amplitude = 1
period = 1800 B1 y = 2 cos x ; amplitude = 2 period = 3600 B1 (iii) Difference 2.4  0.1 B1 10 (b) (a) dy = 2x + 2
dx y = x2 + 2x + C M1 √ Integration at ( -2, -3 ) -3 = 4 + ( -4) + c c = -3  y = x2 + 2x – 3 A1 √ equation (b) x2 + 2x – 3 = 0 M1
( x + 3) ( x – 1) = 0 M1 Factorisation
or equivalent
x = -3 and x = 1 A1 (both)
(-3, 0) and (1, 0) (c ) y = 0 + 0 – 3
y = -3 B1 (d) 2x + 2 = 0
x = -1 B1 for x = -1
y = (-1)2 – 2 – 3
y = -4 B1 for y = -4 (e) y x -3 -1 0 1 -3 -4 (-1, -4) B2 Curve thro’ all Intercepts and stationary point (a) x + y ≤ 10 …………. (i) B1
20x + 40y ≥ 200 ………. ..(ii) B1
Or x + 2y ≥ 10 50x + 75y ≥ 450 ………. (iii) B1 Or 2x + 3y ≥ 18 (b) B1 x + y ≤ 10 B1 x + 2y ≥ 10 B1 2x + 3y ≥ 18</code></pre>(c ) 60x + 100y = c B1 objective function
x = 6, y = 2 or ( 6, 2) L1 Search line drawn
C = 60 x 6 + 100 x 2
= sh. 560 B1 (a) Ө = 60 + 20 = 800 B1
PQ = 60 x 80 M1
= 4800nm A1 (b) Ө = 6 + 34 = 400 B1 QR = 60 x 40 cos 600 M1 = 1200nm A1 Total time = 1200 + 4800 + ½ M1 200 = 30 ½ A1 (c ) Time difference = 40 x 4 = 160 min B1 Time at P = 6.00 2.40 - = 3.20pm B1 10 (a) (i)
C
AC = √ 62 + 62 M1
O
= √ 72 A B AO = ½ √ 72
VO = 122 – ( ½ √ 72)2 M1 = 144 – 72/4 VO = 11.22cm A1 C.A.O (ii) V Ө 11.22cmM 3cm O Tan Ө = 3 11.22 M1 follow through if Ө = 14.970 A1 Cosine rule is used. Reqd angle = 2 x 14.97 ≃ 300 B1 C.A.O </code></pre>(b) l.s.f = 1/3 M1 2cm N1 7.48cm M1N1 = 1/3 x 6 = 2cm B1 angle Ө identified Ө</code></pre>M 4cm x 2cm N XN1 = 2/3 x 11.22 = 7.48 B1 MX and XN1 obtained
MX = 4cm ( any two sides of
∆MXN1) Tan Ө = 7.48 4 M1 Ө = 61.860 A1 10</code></pre></li> NAME……………………………………………………. INDEX NO………………. SCHOOL……………………………………………. CANDIDATE SIGN……………… 121/1
MATHEMATICS
PAPER I
TIME: 2 ½ hours
JULY/AUGUST 2011
KITUI WEST
FORM FOUR EVALUATION EXAMINATION 2011
Kenya Certificate Of Secondary Education INSTRUCTIONS TO CANDIDATES a) Write your name and index number in the spaces provided.
b) This paper consists of two sections I and II.
c) Answer all the questions in section I and only five from section II.
d) All answers and working must be written on the question paper in the space provided below each question.
e) Show all the steps in your calculations giving your answer at each stage in the space blow each question.
f) Marks may be given for correct working even if the answer is wrong.
g) Use calculators and KNEC mathematical tables except where stated otherwise. FOR EXAMINERS USE ONLY
SECTION 1
TOTAL 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 SECTION 11
17 18 19 20 21 22 23 24 TOTAL SECTION A
Answer all questions in this section in space provided. Use logarithm table to evaluate. (4 marks) 3 (0.0246)2 x 142 0.002 x 1.14 A two digit number is such that the sum of the ones and the tens digit is ten. If the
digits are reversed, the new number formed exceeds the original number by 54.
Find the number.
(3 marks) A straight line L1 passes through P(2,1) and is perpendicular to straight line L2,
whose equation is 2y – x + 4 = 0. Find the equation of L1. (3 marks) Alice chepchumba on her cycling practice cycled on a bearing of 120o for 5.5km, then
on a bearing of 200o for 8km finally he turned northwards for 13.5km, by scale
drawing determine her final position from starting point. (4 marks) Make t the subject of the formula. (3 marks) x = 1 + kt
kt – 1 The volume of a rectangular tank is 256cm3. The dimensions are as in the figure. ¼ x x-8
16cm Find the value of x (3 marks) Solve for x 125-x x 52(x-2) = 25 (x+2) (3 marks) Find the equation of the normal to the curve X2 = 4y at the point (6,9) (3 marks) The acceleration of a particle in M5-2 is given by the expressions 3t –4
Find:-
(i) an expression for velocity Vms1 (1 mark) (ii) an expression for distance 5 metres from a fixed point O. Given that S=0 when V=3 and t=0 (2 marks) 10 On Monday this currency exchange rate was
1 Euro (E) = Kshs.95.65
1 US dollar($) = Ksh.76.50
A gentle man Tourist decided to exchange half of his 2400E into Dollars.
Calculate to 2 decimal places the number of dollars he received. (3 marks) Two coils which are made by winding copper wire of different gauges and length have the same mass. The first coil is made by winding 270 metres of wire with cross sectional diameter 2.8mm while the second coil is made by winding a certain length of wire with cross-sectional diameter 2.1mm. Find the length of wire in the second coil . (4 marks) a) Find the greatest common divisor of the term. (1 mark) 144x3y2 and 81xy4 b) Hence factorise completely this expression 144x3y2-81xy4 (2 marks) a) Find the range of values x which satisfied the following inequalities simultaneously. (2 marks) 4x – 9 < 6 + x
8 – 3x < x + 4 b) Represent the range of values of x on a number line. (1 mark) Find the value of x in the equation. Cos(3x - 180o) = √3 in the range Oo < x < 180o (3 marks) 2 The length and width of a rectangle are stated as 18.5cm and 12.4cm respectively. Both measurements are
given to the nearest 0.1cm. a) Determine the lower and upper limit of each measurement. (1 mark) b) Calculate the percentage error in the area of the rectangle. (3 marks) A surveyor recorded the measurement of field in a field book using lines AB = 260m as shown below. B
130 R40
70 Q10
50 P20
S50 10
A a) Use a suitable scale to draw the map of the field. (2 marks) b) Find the area of the field. (2 marks) SECTION B (50 MARKS)
Answer only five questions in this section in the spaces provided. A trader sold an article at sh.4800 after allowing his customer a 12% discount on the marked price of the
article. In so doing he made a profit of 45% . a) Calculate
(i) the marked price of the article. (3 marks) (ii) the price at which the trader had bought the article (2 marks) b) If the trader had sold the same article without giving a discount. Calculate the percentage profit he would have made. (3 marks) c) To clear his stock, the trader decided to sell the remaining articles at a loss of 12.5% (Calculate the price at which he sold each article. (2 marks) Three quantities P, Q and R are such that P varies directly as the cube of Q and inversely as the square of R. a) Given that P = 16 when Q = 2 and R = 3. Determine the value of R when P= 288 and Q = 4
(5 marks) b) Q decreases by 30% while R increases by 40%. Find the percentage decrease or increase in P.
(5 marks) All employees of silver springs enterprises pay income tax at the rate shown in the table below. Taxable income (p.a) Rate sh. Per K₤
1 – 3780
3781-7560
7561-11,340
11,341-OVER 2
3
4
5 Mr. Mooka earns a basic salary of sh.12,150 and a house allowance of sh.2800 per month. He is
entitled to a family relief of sh.450 per month. A part from income tax the following deductions are
also made from his monthly pay. a) Servicing loan payment sh.450
b) Hospital fund sh.260
c) Sacco contribution sh.120 Determine Mr. Mooka’s net monthly income. (10 marks) A passenger train travelling at 25km/h is moving in the same direction as a truck travelling at 30km/h. The
railway line runs parallel to the road and the truck takes 1 ½ minutes to over take the train completely.
a) Given that the truck is 5 metres long determine the length of the train in metres. (6 marks) b) The truck and the train continue moving parallel to each other at their original speeds. Calculate the distance between them 4 minutes and 48 seconds after the truck overtakes this train. (4 marks) 22.5cm The diagram represent a solid frustum with base radius 21cm and top radius 14cm. The frustum is 22.5cm high and is made of a metal whose density is 3g/cm3 π = 22/7. a) Calculate
(i) The volume of the metal in the frustrum. (5 marks) (ii) The mass of the frustrum in kg. (2 marks) b) The frustrum is melted down and recast into a solid cube. In the process 20% of the metal is lost. Calculate to 2 decimal places the length of each side of the cube. (3 marks) The table below shows the masses to the nearest kg of all the students of marigu-ini secondary. School. Masses (kg) No. of students
30-34 5
35-39 7
40-44 10
45-49 10
50-54 19
55-59 20
60-64 20
65-69 6
70-74 2
75-79 1 a) Taking the assumed mean A=52kg
Calculate: (i) The actual mean mass of the students. (3 marks) (ii) The standard deviation of the distribution. (3 marks) b) Draw a cumulative frequency curve and use it to estimate the number of students whose masses lie
between 44.5kg and 59.5kg. (4 marks) 1 a) Given the transformation matrices T1 = 2 1 and T2 = 3 1 -1 -2 1 3 and that transformation T1 followed by T2 can be replaced by a single transformation T, write down the matrix for T. (3 marks) b)Find the inverse of matrix T (2 marks)</code></pre></li> c) The points A11(7,-11), B11(-7,-13), C11(-8,16) and D11(8,8) are the images of points A, B, C and
D respectively under transformation T1 followed by T2
Write down the co-ordinates of A, B, C, and D. (5 marks) 24. Draw the graph of y = 2x2 + x – 2 and use it to solve the equations (10 marks) a) 2x2 + x – 2 = 5 b)2x2 + x – 5 = 0 c)2x2 +2x – 3 = 0 Name ………………………………………..……………………………. Index no. …………………………….. 121/2
MATHEMATICS
PAPER 2
JULY / AUGUST2011
2 ½ HOURS KITUI WEST FORM FOUR EVALUATION EXAMINATION 2011
Kenya Certificate of Secondary Education
MATHEMATICS
PAPER 2
2 ½ HOURS INSTRUCTIONS TO CANDIDATES
(a) Write your name and Index number in the spaces provided above
(b) This paper consists of two sections. Section I and section II.
(c ) Answer ALL the questions in section I and only FIVE questions in section II.
(d) All answers and working must be written on the question paper in the spaces provided below each question.
(e) Non-programmable silent calculators and KNEC maths tables may be used except
here stated otherwise. FOR EXAMINER’S USE ONLY SECTION I 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total SECTION II 17 18 19 20 21 22 23 24 Total Use mathematical tables to evaluate. 3 4.68 x 0.13242 5 log 7 ( 3 marks ) Make y the subject of the formular
m = d + ky
d – ky ( 2 marks ) Expand ( 1 – 2x)6 upto the term in x3. Use your expansion to estimate 0.986. ( 3 marks ) In a geometric sequence, the third term is 20 and the fourth term is 5. Find
(a) The two possible values of the common ratio. ( 2 marks ) (b) The first term. ( 1 mark ) Let x = 8
3 - √5 (a) Write x in the form a + b √5 where a and b are integers. ( 2 marks ) (b) Use √5 = 2.236068 to write x correct to 6 decimal places. ( 2 marks ) Solve the equation ( log x)2 – log x – 2 = 0 ( 3 marks ) Three representatives are to be selected randomly from a group of 7 girls and 8 boys.
Calculate the probability of selecting two girls and one boy. ( 3 marks ) Two variables x and y are such that x varies partly as y and partly as the square root of y.
Given that x = 30, when y = 9 and x = 14 when y = 16, find x when y = 36. ( 3 marks ) . Given that cos 2x0 = 0.8070. Find x where O0  x  3600. ( 3 marks ) Given that A = 3i + 2j + 6k and B = 2i – 2j – 5k. Find
     
(a) AB ( 2 marks ) (b) AB ( Leave your answer in root form ) ( 2 marks ) . A minor sector of a circle of radius 28cm includes an angle of 1350 at the centre.
Convert 1350 into radians. Hence or otherwise find the area of the sector. ( 3 marks ) The rate of depreciation of a car is 12% per annum. Find its value 3 years ago if its
present value is 320,000 shillings. Give your answer to the nearest shillings. ( 3 marks ) In the figure below, find the length of UT. (3 marks ) X 3cm Y 4cm T 5cm V U The equation of a certain circle drawn in a Cartesian plane is given by x2 + 2x + y2 – 16y = 42.
Find the radius and the coordinates of the centre of the circle. ( 4 marks ) A rectangular card measures 3.5cm by 1.4cm. Find
Absolute error in the area of the card. ( 3 marks ) Solve for the values of x and y given that
x – y = 5
x2 – y2 = 5 ( 3 marks ) SECTION II An engineer wants to design a roof for a CDF sponsored classroom in Kathithyamaa
secondary school. The roof takes the shape shown below.
F
E 10cm D 8cm C A 12cm
B
Calculate
(a) The angle between faces CDEF and ABCD. ( 3 marks )
(b) The space occupied by the roof. ( 3 marks )
(c ) The angle between plane BEC and ABCD. ( 4 marks ) is projected at time t = 0 along the x-axis so that at time t seconds its velocity is
Vm/s and its displacement from the origin O is x metres, where x = 24t + 9t2 – 2t3 + 77, t 0 Find (a) An expression for V in terms of t. ( 1 mark ) (b) The value of t when the particle is at rest and the displacement of P from O at this time. ( 4 mks ) (c ) An expression for acceleration a m/s2 of P in terms of t. ( 1 mark ) (d) The distance traveled by P during the fifth second. ( 2 marks ) (e) Show that the particle P passes through O when t = 7 and find the acceleration of P at this time. ( 2 marks ) In the figure below TA is a tangent to a circle. AB and CD are parallel. AC and BD meet
at X BAS = 770 and DCB = 1230.
C 1230 B x D 770T A S (a) Find ABD. ( 2 marks )
(b) Find DAT. ( 2 marks )
(c ) Show that ∆XCD is isosceles. ( 3 marks )
(d) Let TB = xcm and TD = ycm. Find TA in terms of x and y. ( 3 marks ) (a) Complete the table for y = ½ sin ( x – 200) and y = cos 2x
x0 -90 -60 -30 0 30 60 90 120 150
½ sin (x-200) -0.49 -0.17 0.49
Cos 2x -1 1 -0.50 (b) On the same set of axis on the grid provided draw the graphs of y = ½ sin ( x – 200)
and y = cos 2x. Use a scale of 2cm to represent 1 unit on y –axis and 1cm to
represent 300 on the x-axis. ( 4 marks ) (c ) Use your graph to solve the equation ½ sin (x – 20)0 – cos 2x = 0 ( 2 marks ) (d) For what range of values of x is the function cos 2x  ½ sin ( x – 200) ( 1 mark ) R is a transformation defined by +900 turn about (0, 0), M is a transformation represented
by the matrix 1 0 . Triangle PQR where P (3, -1) Q (5, -1) and R ( 5, -4) is mapped onto
0 -1
P1 Q1 R1 under R. (a) Plot PQR and P1Q1R1 on the same axes. ( 3 marks )
(b) Triangle PQR is mapped onto P2Q2R2 under the transformation M.
Plot P2Q2R2 under the same grid. ( 2 marks )
(c ) P3Q3R3 is the triangle whose vertices are P3(-3, -1) Q3(-7, -1) and R3 (-7, 5).
Draw P3Q3R3 on the same grid. ( 1 mark )
(d) Obtain a 2 x 2 matrix N, that maps PQR onto P3Q3R3, hence describe fully the
transformation represented by matrix N. ( 4 marks ) The figure below shows a rectangle and a smaller rectangle inside it. 6cm 3cm 4cm 8cm</code></pre>(a) If a point is selected at random inside the bigger rectangle. What is the probability
that it lies in the shaded region? ( 2 marks )
(b) A bag contains 4 yellow balls and 3 black balls. Two balls are picked at random
one after the other from the bag without replacement. By using a tree diagram,
find the probability that the second ball picked from the bag is black ball. ( 4 marks )
(c ) A certain form 3 in a mixed class has 18 girls. If the probability that a student
selected at random from this class is a girl is 2/5.
(i) How many students are there in the class? ( 2 marks )
(ii) What is the probability of a student selected at random for the class being a boy. ( 2 mks ) . The table show the income tax rates
Total income k£p.a Rate in sh. per k£
1 – 2100 2
2101 – 4201 3
4201 – 6300 5
6301 – 8400 7
8401 and over 9 Norah is salaried employee. She earns a basic salary of Kshs. 11,550 per month and a house
allowance of Kshs. 2,045 monthly. She claims Kshs. 275 relief per month. Calculate (a) Norah’s taxable income per annum in k£. ( 3 marks )
(b) Her P.A.Y.E per month to the nearest 10ct. ( 5 marks )
(c ) Her net monthly income ( 2 marks ) The height to the nearest cm, of 40 boys is as shown in the distribution table Height (cm) 130 – 139 140 – 149 150 – 159 160 – 169 170 – 179 180 - 189
Frequency 1 3 7 13 10 6 Using assumed mean of 164.5 or otherwise, find (a) The mean ( 5 marks ) (b) The standard deviation of the distribution. ( 5 marks ) JULY/AUGUST 2011 FORM FOUR MID-TERM EXAMINATION 2011
121/1
MATHEMATICS
PAPER 1
MARKING SCHEME Q WORKING MRKS COMMENTS
1 No.
(0.00246)2
142 0.002
1.14 3.3527 Std. form
(2.46 x 10-2)2
1.42 x 102 2.0 x 10-3
1.14 x 10o 3.3527 x 10o
Log
2.3909
x 2
4.7818
+2.1523
2.9341 3.3010
+0.0569
3.3579 2.9341
-3.3579
1.5762
1.5762
3 0.5254 1M 1M 1M A1 Correct logs addition Correct logs addition Correct logs substractions Correct answer
4
2 Let the digits be x and y
The number becomes xy
= 10x +y
and x + y = 10
Reserved yx = 10y + x
(10y + x) – (10x + y) = 54
10y + x – 10x – y = 54
9y – 9x = 54
y – x = 6
y – x = 6
y + x = 10
2y = 16
y = 8 x = 8-6 = 2
.
. . The number is 28 M1 1M A1 Splitting of ones & tens and the reverse Solving of the simultaneous eqn. Answer
3
3 Eqn of L2 2y = x – 4
Y = ½ x – 2
Hence M1 = ½
And ½ x M2 = -1
M2 = -1 x 2 = -2
Y = mx + c
Y = -2x + c
But L1 passes through P(2,1)
Hence 1 = -2 x 2 + c
C = 5 Y = -2x + 5 1M 1M Gradient of L2 Use of gradient of L2 to get gradient of L1
3
4 1cm represent 1km N D
208o
4KM 120o 5.5km B 200o 13.5km 8KM C Bearing 030o 4km from starting point 1M 1M 1M A1 Bearing of starting point A Use of scale correctly and plotting of points Use of bearing correctly
5 X2 = 1+kt
Kt –1 x2(kt-1) = 1 + kt
ktx2 – x2 = 1 + kt2
ktx2 – kt = 1 + x2
t(kx2 – k) = 1 + x2 t= 1 + x2
kx2 - k M1 M1 A1 Squaring both sides factorisation
3 6 Volume ¼ x (x-8) 16=256
¼ x (x-8) = 16
x2 – 8x – 64 = 0
use of formula = -(-8) + (-8)2 – 4x1x-64
2 x 1 = 8 + 64 + 256
2
= 8 + √320 = 12.9
2
M1 M1 A1
Volume in terms of x(quadratic eqn.) Use of inte 7 125-x x 52(x-1) = 25x+2 5-3x x 52x –2 = 52x + 4
-3x + (2x-2) = 2x + 4
-x –2 = 2x + 2 -4 = 3x -4/3 = x M1 M1 A1 Simplification Addition of power
8 Curve equation x2 = 4y
Y = x2
4
dy = 2x = ½ x
dx 4
But at the point x = 6
dy = ½ x 6 = 3
dx
m = 3, p (6,9)
Assume Q (xy)
y-9 = 3
x – 6 = y = 3x -9 M1 M1 A1 Defferention Substitution of x Equation
9 dv = 3t – 4
dt V= ∫3t – 4dt
= 3t2 - 4t + c
2
Since V= 3 when t = 0 Then 3(0)2 - 4(0) + C = 3
2
C = 3
S = ∫vdt = 3t2 – 4t + 3) dt
2
= t3 – 2t2 +3t + k
2
= S = 0, when t = 0
k = 0
s = t3 – 2t2 + 3t
2 M1 1M 1M A1 Use of gradient in intergration Substitution of t=0 Integration
4
10 ½ of 2400E = 1200E
In ksh. = 1200E x 95.65
= Ksh.114,780
Number of dollar = Kshs.114,780
76.50
= sh1500.39
M1 M1 A1
3
11 Mass = Density x volume
But Density is constant.
x y Vol (270000 x 2.8): x2.1 = 270000 x 2.8 = 360m
2.1 M1 M1 M1 A1
4 a) G.C.D of 81xy4 and
144x3y2 is 9xy2
144x3y2 – 81xy4
9xy2(16x2 – 9y2)
9xy2(4x – 3y) (4x + 3y)
M1 M1 A1
3 a) 4x – 9 < 6 + x
x < 5
8 – 3x < x + 4
1 < x
b) 1 < x < 5
M1 M1
A1 a) The least unit of measurement is 0.1cm
.
. . error 0.1 = + 0.05
2
Limits of length = 18.5 + 0.05
Upper limit = 18.5 + 0.05 = 18.55cm
Lower limit = 18.5 – 0.05 = 18.45cm
Limits of width = 12.4 + 0.05
Upper limits = 12.4 + 0.05 = 12.4cm
Lower limits = 12.4 + 0.05 = 12.35CM a) Working area = 18.5 x 12.4 = 229.4cm2
Max.Area = 18.55 x 12.45
= 230.9475cm2 Min. Area = 18.45 x 12.35
= 227.8575cm2
Error area = 330 .9475 – 229.4
= 1.5475cm2 Average error = 1.5415 + 1.5425
2
= 1.545cm2 . .
. % error = 1.545 x 100
22.94 = 0.67% (2 d.pls) B1 B1 M1 A1 ALL Sides Correct average error Expression of error in % Let 3x – 180 =
0 < x < 180
cos = √3
2
= 30o
0 – 180< < 180 x 3 – 18
-180 < < 360
= 30o, 330,
30 = 3x – 180
210 = 3x
70, = x
330 = 3x – 180
510 = x
3
170 = x
x = 70, 170 M1 M1 A1 3
16 a) 1cm rep. 10m R P d 4cm B Q C A 1CM A 1cm 4cm 2cm 6cm 3cm B
F e
5cm b) Area
A = ½ x 5 x 2 = 5cm2
B = ½ X 2(2+ 1) = 3cm2
C = ½ x 6 (1+4) = 15cm2
D = ½ x 3 x 4 = 6cm
E= ½ x 5 x 15 = 75/2
= 37.5 f = ½ x 1 x 5 = 2.5
Total = 69cm2 Area = 69 x 1000000
10000 = 690m2 B1 B1 M1 A1 Correct scales Correct drawing
4
17 SECTION B (50 MARKS) Selling price = 88/100 of marked price
(a)
(i) 4800 = 88/100 of m.p
4800/88 x 100 = m.p
= sh.5454.54 (ii) 145/100 of buying price = 4800
buying price = 48000 x 100
145
= 3310.34 (b) 5454.54 – 3310.34 x 100 3310.34 = 0.6477 x 100 = 64.77% C) 87.5 of 3310.34
100 = 2,896.55 18 (a) P Q3
R2 P = KQ3
R2 K = PR2 = 16 x 32 = 18
Q 3 23 R = KQ3 = 18x(4)3 = 2
P 288 B) P = KQ3
R2 Q decreases to 0.7
R increases to 1.4
P = (0.7)3 x 2 = 0.35
(1.4)2 P decreases by 65% M1 M1 M1 M1
A1 M1 M1
M1
M1 M1
10 Taxable income
Salary sh12,150
House allowance sh 2800
Total sh 14950 Sh. 14950 x 12 = 179400
8970 p.a 1-3780 = 3780 x 2 = 7580 x 2 = 7560
3781 – 7560 = 3780 x 3 = 11340
7561 – 11,340 = 1410 x 4 = 5640 +
Total sh. 24540 Tax per month
Sh. 24540 :- 12
= SH2045
450 –
sh
Deductions
Loan sh 450
Hospital sh 260+
Sacco sh 120
830
Tax 1595
2,425 net salary
sh. 14,950
2,425
sh 12,525 a) Distance moved by front of truck
= length of train + length of truck
=relative of truck
= (30 – 25) = 5km/hr.
Distance = speed x time
5km/hr x 1.5
60 = 0.125km
= 125m
length of the train
= 125 – 5 = 120m b) Relative speed remain constant
c) Time = 4min & 48 sec.
d) 5km/hr = 50 1.39m/sec
36 distance = speed x time
= 1.39 x 288 = 400.32m
M1 M1 M1
A1 M1 A1 M1
A1 M1 A1 x x 14 22.25cm 2.25cm a) I) Volume
Ratio x/14 = 22.5 + x
21 21x = 14(22.5 + x)
21x = 3.15 + 14x
21x – 14x = 315
7x = 315
x = 45 volume of whole cone
1/3 x 22/7 x 21 x 21 x 67.5 = 31,185cm
volume of small cone
1/3 x 22/7 x 14 x 14 x 45 = 9,240cm3 volume 31,185 – 9240
= 21945cm3 ii) mass of frustrum
mass = 21945cm3 x 3g
= 65835g cm3
1000 a(ii) mass in kg 65,835g 1000 = 65.835kg b) 20% of 65.835kg
= 14.167kg 65,835 – 13.167kg
= 52.668kg. 52.668 c 1000g = V x 3g
52,668 X cm3 = V cm3 V= 17,556cm3
b3 = 3 √ 17,556 = 25.99cm Length 25.99cm 10 Mass
(kg) y x x-52 d=x-52
10 fd fd2 cf
30-34
35-39
40-44
45-49
50-54
55-59
60-64
65-69
70-74
75-79 5
7
10
10
19
20
20
6
2
1
Eƒ100 32
37
42
47
52
57
62
67
72
77 20
15
10
5
0
5
10
15
20
25 -4
-3
-2
-1
0
-1
-2
-3
-4
-5
-20
-21
-20
-10
0
20
40
18
8
5
Eƒd=20 80
63
40
10
0
20
80
54
32
25
Eƒd2=
404
5
12
22
32
51
71
91
97
99
100
6 marks a(i) mean = A + cEfd
Ef = 52 + 5 x 20
10 = 52kg (ii) S.D = C Efd2 - Efd
Ef Ef = 5 x 404 – 20 100 100 = 5 4.04 – 0.2 = 5 3.84 = 5 x 1.959 = 5 x 1.96 = 9.8 M1 M1 M1 A1 23 T=T2T1 = 3 1 2 1
1 3 -1 -2 = 6 + -1 3 + -2
2 + -3 1 + -6 = 5 1 = 1 -5 -1
1 -5 5 x –5 –1 -1 5 = 1 -5 -1
24 -1 5 = 5/24 1/24 1/24 -5/24 c) 5/24 1/24 7 -7 -88 1/24 -5/24 -11 -13 168 = 24/24 -48/24 -24/24 48/24 62/24 58/24 58/24 -32/24 = 1 -2 -1 2 31/24 29/24 -11/3 -4/3 A(1, 31/24) B(-2, 29/24) C(1, -11/3) D(2, -4/3) M1 A1 M1 M1 A1 M1 M1 A1 B2 Bo if Bracket not included.
24 a) y = (2 x 2 + x-2) 2x2 + x – 2 = 5
y = 5 At y=5 x=-2.1 or x = 1.6 c) 2x2 + x – 5 = 0
Add three both sides
Y =3. At these points
X = 1.9 or x = 1.3 (1d.p) d) Add (t-2) to both sides
Y = -x + 1
At the points X = -1.8 or x = 0.8 (1d.p)
M1 A1 M1 A1 M1 A1 121/2 FORM FOUR MID TERM EXAMINATION 2011
Kenya Certificate of Secondary Education
MATHEMATICS
PAPER 2 MARKING SCHEME No Log
4.68 0.6702
0.13224 1.1209 x 2 2.2418 + M1 All logs correct
2.9120 5 0.6990 0.8451 + 1.5441 3.3679 x 1/3 M1 1.1226 </code></pre>.1326
0.1326 A1 3 m ( d – ky) = d + ky
md – mky = d + ky M1 -mky – ky = d – md
Mky + ky = md – d
y ( mk + k) = d(m-1) y = d (m-1) A1
mk + k 02 ( 1 – 2x) 6 (1)6 (2x)0 – (1)5 (2x) + 14(2x)2 – (1)3 (2x)3 …. 1 – 2x + 4x2 – 8x3 + ………….. B1 (0.98)6 = ( 1 – 0.02)6
1 – 2 (0.02) + 4(0.02)2 – 8(0.02)3 M1 1 – 0.04 + 0.0016 – 0.000064 0.961536 A1 3</code></pre></li>(A) 3rd term = ar2
5th term = ar4 ar2= 20 ……. (i) ar4 = 5 ……. (ii)a = 20 r2 Subsistituting 20 (r4) = 5 M1 r2 20r2 = 5 r2 = 5/20 r2 = ¼ r = ½ or – ½ A1 </code></pre>(b) a = 20
r2 a = 20 = 20 ( ½ )2 ¼ 20 x 4/1 = 80 B1 3</code></pre></li>(a) 8 3 + √5 3 - √5 3 + √5 M1 24 + 8 √5 9 – 5 24 + 8 √5 4 6 + 2√5 A1 (b) 6 + 2 ( 2.236068 M1
6 + 4.472136
10.472136 A1 4 (Log x)2 – log x -2 = 0
Let log x = t t2 – t – 2 = 0 M1 t2 + t – 2t – 2 = 0
t ( t + 1) -2 (t + 1) = 0 (t – 2 ) ( t + 1 ) = 0 M1
t = 2 or t = -1 A1 3 Sample space = 8 + 7 = 15
P(2 girls ) = 2/15
P ( 1 boy) = 1/15 B1 P ( 2 girls and 1 boy )
2/15 x 1/15 = 2/225 M1A1 3 x X y x = ky
x X √y x = m√y x = ky + m√y
30 = 9k + 3m ……. (i) x 4
14 = 16k + 4m ……(ii) x 3 B1 120 = 36k + 12m
42 = 48k + 12m M1 for any method used
78 = -12k K = -78 12 K = -13 2 30 = 9(-13/2) + 3m 60 = -117 + 6m
177 = 6m
m = 177
6 x = -13/2y + 177/6 √y
x = -13 (36) + 177 6
2 6 -13 (18) + 177 - 234 + 177 - 57 A1 3</code></pre></li>Cos 2x0 = 0.8070
2x0 = cos-1 (0.8070)
= 36.1960 B1
2x = 36.1960
≂ 36.200, 323.800
396.2, 683.8 B1 x = 18.10, 161.90. 198.10 341.90 A1 3</code></pre></li>(a) AB = B – A
(I – 2j - 5k) – ( 3i + 2j + 6k) B1
I – 2j – 5k – 3i – 2j – 6k
-2i - 4j – 11k B1 (b) AB = √i2 + j2 + k2 = √ (-2)2 + (-4)2 + (-11)2 M1√ 4 + 16 + 121 √ 20 + 121 √141 A1 4</code></pre></li>1C = 57.290
? = 1350 135 57.29 = 2.35C B1 Area = Ө r2
360
4 1
135 x 22 x 28 x 28 M1
360 7
90 1 27 3
135 x 22 x 28
90 18 2
11
3 x 22 x 28
2 1 33 x 28 = 924cm2 A1 3 A = P ( 1 + r/100)n A = 320,000 [ 1 + 12/100)3 M1M1 = 320,000 [1.12)3= 449,577 A1 3</code></pre></li>Let UT = a TY x TX = TU x TV
4 x 7 = a x ( 5 + a) M1
28 = 5a + a2
a2 + 5a – 28 = 0 a = b  √b2 – 4ac
2a a = -5  √52 -4 (1) (-28)
2(1) a = -5t √25 + 112 a = -5  √137 a = -5  11.7 a = 6.7cm M1 2 a = 3.35cm A1 3</code></pre></li>x2 + 2x + (2/2)2 + y2 – 16y + (-16/2)2 = 16 + 1 + 64 M1 (x + 1)2 + (y-8)2 = 16 + 1 + 64 = 81 A1 for the equation Radius √81 units =9 units B1 Cordinates of centre = ( -1, 8 ) B1 4</code></pre></li>3.45  3.5  3.55
1.35  1.4  1.45 Max area = 3.55 x 1.45
= 5.1475
M1
Min area = 3.45 x 1.35
= 4.6575 A.E = 5.1475 – 4.6575 M1 2 = 0.245 A1 3</code></pre></li>x = 5 + y M1
( 5 + y)2 – y2 = 5 M1
25 + 10y + y2 – y2 = 5
25 + 10y = 5
+ 10y = -20
y = -2
x = 5 – 2
x = 3, y = -2 A1 for both x and y SECTION II ( 50 MARKS ) (a)
B F
10m D 8m Ө C 12mA B</code></pre>The required angle is BCF = Ө
C2= f2 + b2 – 2 fb cos C Cos C = f2 + b2 – C2
2fb = 102 + 122 – 82 2 x 10 x 12 M1 180 = 0.75 240 M1 Ө = cos -1 ( 0.75) = 41.410 A1 (b) Cross sectional area of the roof
½ x 10 x 12 sin 41.410
½ x 10 x 12 x 0.6614 M1 Space occupied by the roof ( volume) ½ x 10 x 12 x 0.6614 x 24 M1 = 952.4m3 A1 (c )
E F D C x H G A B</code></pre></li> The required angle is HGE
EH = FG = 10 sin Ө
Tan x = EH
AB Tan x = 10 sin 41.41 24 = 0.2756 x = tan -1 ( 0.2756) M1 15.410 A1 10 x = 24t + 9t2 – 2t3 + 77 (a) √ = dx = 24 + 18t – 6t2 B1
dt (b) When √ = 0
24 + 18t – 6t2 = 0 B1
6t2 – 18t – 24 = 0
( t + 1) ( t – 4) = 0 M1 t = 4 secs A1 When t = 4
x = ( 24 x 4) + ( 9 x 16) – (2 x 43) + 77
= 96 + 144 – 128 + 77
= 189m B1 (c ) a = d√ = 18 – 12t B1
dt (d) When t = 5
x = ( 24 x 5) + ( 9 x 25) – (2 x 125) + 77
= 120 + 225 – 250 + 77
= 172m B1  Distance traveled = ( 189 – 172) m = 17m B1 (e) When t = 7, x = 0
 x = ( 24 x 7) + ( 9 x 49) – (2 x 73) + 77
= 168 + 441 – 686 + 77
= 0m B1 a = 18 – 12t = 18 – ( 12 x 7) = -66m/s2 B1 10</code></pre></li>(a) BAD = ( 180 – 123) = 570 ( opposite B1B1
angles of a cyclic quadrateral DAT = 180 – ( SAB + BAD) = 1800 – ( 77 + 57)0 B1 1800 – 1340 = 460 B1 Angles on a straight line (c ) ∆ X CD
XDC = DCX which are base angles of the ∆ B1
therefore line DX = XC hence B1
triangle XCD is isosceles B1 (d) If TB = xcm and TD = ycm
Then TA2 = TD x TB M1 TA2 = y X x M1 TA = √ yx A1 10</code></pre></li>x -90 -60 -30 0 30 60 90 120 150 ½ sin (x – 20) -0.47 -0.49 -0.38 -0.17 0.09 032 0.47 0.49 0.38
Cos 2x -1 -0.5 0 1.0 .05 -0.50 -1 -0.5 0.5 . (a) R 1 0
0 -1 P Q R P1 Q1 R1 M1 1 0 3 5 5 = 3 5 5 0 -1 -1 -1 -4 1 1 4 (d) a b 3 5 5 = -3 -7 -7
C d -1 -1 -4 -1 -1 5 M1 3a – b = -3 - 5a – b = -7 -2a = 4 M1 a = 4 -2 a = -2 3a – b = -3 ( 3x – 2) –b = -3 M1
-6 = b = -3 -6 + 3 = b b = = -3 3c – d = -1 5c – d = -1 -2c = 0 c = 0 3c – d = -1 -d = -1 d = 1 A1 (a) 48 – 12
48 M1
36 or ¾
48 A1
(b)
Y
3/6
B
Y 3/6 B1 4/7 4/6 Y3/7 B 2/6 B B14/7 x 3/6 + 3/7 x 2/6 12/42 + 6/42 18/42 or 3/7 A1 (c) (i) Let the no. of students be x M1
18 = 2
x 5 = 45 A1
x = 18 x 5 = 45
2 (ii) 45 – 18 45 M1 27/45 = 3/5 A1 10</code></pre></li>(a) Taxable income per month M1
= employee’s salary + allowance
11,550 + 2045
= Ksh. 13,595 Annual taxable income in kt B1 13,595 x 12 20 = kt 8157 A1 (b) 1st 2100 = 2100 x 2 = Ksh. 4200 M1
Next 2100 = 2100 x 3 = Kshs 6,300 M1
Next 2100 = 2100 x 5 = Kshs 10,500
Rem 1857 = 1857 x 7 = Kshs. 12999
Income tax = Kshs. 33,999
Per month = Kshs. 33999
12 M1 = Kshs. 2833.25 B1 Less relief = Kshs. 275.00 P.A.Y.E = Kshs. 2,558.25 A1 (c ) Net monthly income = Taxable Income – tax = ( Salary + house allowance – tax Kshs. 13595 – 2833.25 Kshs. 10,761.75 A1 10 Class Mid-points t –x – 164.5 f ft ft2 B2 130 – 139 135.5 -3 1 -3 9
140 – 149 144.5 -2 3 -6 12
150 – 159 154.5 -1 7 -7 7
160 – 169 164.5 0 13 0 0
170 – 179 174.5 1 10 10 10
180 - 189 184.5 2 6 12 24
40 f 6 ft2 = 62 t = ft = 6 = 0.15
f 40 M1 x = t + C + A = 0.15 x 10 + 164.5 M1 = 166.0cm A1 (b) Variance of t = ft2 – (f)2
f = 62 - ( 0.15)2 40 = 1.55 – 0.0225 M1 = 1.5275Varies of x = variance of t x C A4 1.5275 x 100
= 15.28
Standard deviation = √variance
= √ 15.28 M1
= 3.909 A1
10