Instructions to candidates
- Write your name, admission number and class in the spaces provided above.
- Sign and write the date of examination in the spaces provided above.
- The paper contains two sections: Section I and Section II.
- Answer All the questions in section I and strictly any five questions from Section II.
- All answers and working must be written on the question paper in the spaces provided below each question.
- 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 if the answer is wrong.
- Non-programmable silent electronic calculators and KNEC mathematical tables may be used, unless stated otherwise.
SECTION A (50 MARKS)
Answer all the questions in this section
- Use logarithm table to evaluate. (4 mks)
∛0.52 × 0.312
2.122 - 200 cm3 of acid is mixed with 300 cm3 of alcohol. If the densities of acid and alcohol are 1.08g/cm3 and 0.8 g/cm3 respectively, calculate the density of the mixture.(3 mks)
- The coordinates of P and Q are P(5, 1) and Q(11, 4) point M divides line PQ in the ratio 2 : 1. Find the magnitude of vector OM. (3 marks)
- The table below shows income tax rates in a certain year.
Monthly income in Ksh Tax rate in each Ksh 1 – 9680 10% 9681 – 18800 15% 18801 – 27920 20% 27921 – 37040 25% Over 37040 30% - Make w the subject of the formulae. (3mks)
2x = √2w +8
3w-5 - A line passes through points (2, 5) and has a gradient of 2.
- Determine its equation in the form y=mx+c. (2mks)
- Find the angle it makes with x-axis. 1mk
- A quantity P is partly constant and partly varies as the cube of Q. When Q=1, P=23 and when Q =2, P= 44. Find the value of P when Q = 5. (3mks)
- The vertices of a triangle are A(1, 2) , B(3, 5) and C(4, 1). The co-ordinates of C’ the image of C under a translation vector T are (6, -2).
- Determine the translation vector T. (1mk)
- Find the co-ordinates of A’ and B’ under the translation vector T. (2mks)
- Expand (1 -x)4 using the binomial expansion. (1mk)
Use the first three terms of the expansion in (a) above to find the value of (0.98)4 correct to nearest hundredth. (2mks) - Find the centre and radius of a circle with equation:
χ² + y² - 6χ + 8y – 11 = 0 (3mks) - Two grades of coffee one costing sh.42 per kilogram and the other costing sh.47 per kilogram are to be mixed in order to produce a blend worth sh.46 per kilogram in what proportion should they be mixed. (3mks)
- Pipe A can fill an empty water tank in 3 hours while pipe B can fill the same tank in 5 hours. While the tank can be emptied by pipe C in 15 hours. Pipe A and B are opened at the same time when the tank is empty. If one hour later pipe C is also opened. Find the total time taken to fill the tank. (4 mks.)
- Simplify the expression: (3mks.)
(9t2- 25a2)
6t2+ 19at+15a2 - A business bought 300 kg of tomatoes at Ksh. 30 per kg. He lost 20% due to waste. If he has to make a profit 20%, at how much per kilogram should he sell the tomatoes. (3mks.)
- Evaluate without using a Mathematical table or a calculator. (2mks)
Log6216 +[Log42 - Log6] ÷ Log49 - Given that the ratio x: y = 2:3, find the ratio (5x-2y)∶ (x+y) (3 mk)
SECTION II (50mks)
Answer only five questions in this section in the spaces provide
- Draw the graph of y= x^3+2x^2-5x-8 for values of x in the range -4≤x≤3 (5mks)
x -4 -3 -2 -1 0 1 2 3 x3 -64 27 2x2 -5x -8 y -20
By drawing suitable straight line on the same axis, solve the equations.- x3+2x2-5x-8=0 (1mks)
- x3+2x2-5x-7=0 (2mks)
- 3+3x-2x2-x3=0 2mks
- A transformation represented by the matrix maps the points A(0, 0), B(2, 0), C(2, 3) and D(0, 3) of the quad ABCD onto A¹B¹C¹D¹ respectively.
- Draw the quadrilateral ABCD and its image A¹B¹C¹D¹. (3mks)
- Hence or otherwise determine the area of A¹B¹C¹D¹. (2mks)
- Another transformation maps A¹B¹C¹D¹ onto A¹¹B¹¹C¹¹D¹¹. Draw the image A¹¹B¹¹C¹¹D¹¹. (2mks)
- Determine the single matrix which maps A¹¹B¹¹C¹¹D¹¹ back to ABCD. (3mks)
- Draw the quadrilateral ABCD and its image A¹B¹C¹D¹. (3mks)
- In the figure below (not drawn to scale) AB = 8cm, AC = 6cm, AD = 7cm, CD = 2.82cm and angle CAB = 50°.
Calculate (to 2d.p.)- the length BC. (3 marks)
- the size of angle ABC. (3 marks)
- size of angle CAD. (3 marks)
- Calculate the area of triangle ACD. (2 marks)
- Three variables P, Q and R are such that P varies directly as Q and inversely as the square of R.
- When P = 18, Q = 24 and R = 4.
Find P when Q = 30 and R = 10. (3mks) - Express P in terms of Q and R. (1mk)
- If Q is increased by 20% and R is decreased by 10% find:
- A simplified expression for the change in P in terms of Q and R. (3mks)
- The percentage change in P. (3mks)
- When P = 18, Q = 24 and R = 4.
- A surveyor recorded the following information in his field book after taking measurement in metres of a plot.
To E
720 to F
240 G1000
880
640
480
400
200
320 to D
600 to C
40 to BFrom A - Sketch the layout of the plot. (4 mks.)
- Calculate the area of the plot in hectares. (6mks)
- A line L passes through points (-2, 3) and (-1,6) and is perpendicular to a line P at (-1,6).
- Find the equation of L. (2 mks)
- Find the equation of P in the form ax + by = c, where a, b and c are constant. (2 mks)
- Given that another line Q is parallel to L and passes through point (1,2) find the x and y intercepts of Q. (3 mks)
- Find the point of intersection of lines P and Q. (3 mks)
- The figure below shows a square ABCD point V is vertically above middle of the base ABCD. AB = 10cm and VC = 13cm.
Find;- the length of diagonal AC (2mks)
- the height of the pyramid (2mks)
- the acute angle between VB and base ABCD. (2mks)
- the acute angle between BVA and ABCD. (2mks)
- the angle between AVB and DVC. (2mks)
- The diagram below represents a conical vessel which stands vertically. The which stands vertically,. The vessels contains water to a depth of 30cm. The radius of the surface in the vessel is 21cm. (Take π=22/7).
- Calculate the volume of the water in the vessels in cm3 (3mks)
- When a metal sphere is completely submerged in the water, the level of the water in the vessels rises by 6cm.
Calculate:- The radius of the new water surface in the vessel; (2mks)
- The volume of the metal sphere in cm3 (3mks)
- The radius of the sphere. (3mks)
MARKING SCHEME
No | Working | ||||
1 |
→ 0.3305 |
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2 | Mass of acid = 200cm3 x 1.08g/cm3 =216g Mass of alcohol = 300cm3 x 0.8g/cm3 =240g Total volume of the mixture = 200 + 300 = 500cm3 Density of the mixture = Total mass Total volume =216g + 240g 500cm3 = 456g 500cm3 = 0.912g/cm3 |
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3 | |||||
4 | Taxable income = Ksh32500 Income Tax(sh) 9680 x 10/100 = 968 + 9120 x 15/100 = 1368 + 9120 x 20/100 = 1824 + 4580 x 25/100 = 1145 Total tax payable = 5305 Tax due = Tax payable - Relief = sh 5305 - 1056 = sh4249 |
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5 | 2x = √2w +8 3w-5 (2x)2 = 2w +8 3w - 5 4x2(3w - 5) = 2w + 8 12x2w - 20x2 = 2w + 8 12x2w - 2w = 8 + 20x2 w = 8 + 20 2 12x2 - 2 |
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6a) | y - 5 = 2 x - 2 y - 5 = 2x - 4 y = 2x + 1 |
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6b) | θ = Tan-1(gradient) = Tan-1 2 = 63.43 |
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7 | p = a + bQ3 23 = a + b -44 = a + 8b -21 = -76 b = 3 23 = a + 3 → a = 20 P= 20 + 3Q3 When Q = 5 P = 20 + 3(5)3 = 395 |
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8 | |||||
9 | (1 - x)4 = 1(1)4(-x)1 + 4(1)3(-x)1+6(1)2(-x)2 + 4(1)3(-x) + 1(1)4(-x)4 = 1 - 4x + 6x2 - 4x3 + x4 1 - x = 0.98 → x = 0.02 (0.98)4 = 1 - 4(0.02) + 6(0.02)2-4(0.02)2 + (0.02)4 = 1 - 0.08 + 0.0024 - 0.000032 =0.922 ≈0.902 |
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10 | x2 + y2 - 6x + 8y - 11 = 0 x2 - 6x + (-6/2)2 + y2 + 8y + (8/2)2 = 11 + (-3)2 + (4)2 (x -3)2 + (y +4)2 = 36 (x -3)2 + (y + 4)2 = 62 Center → (3, -4) Radius = 6 units |
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11 | Let the ratio be x:y in kg respectively Cost of the mixture = sh(42x + 47g) Total mass of mixture = (x + y)kg Cost per kg of the mixture = Total cost of the mixture Total mass sh46= 42x + 47y x + y 46x + 46y = 42x + 47y 4x = y x = 1 → x:y = 1:4 y 4 |
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12 | Rate of work of each pipe A = 1/3 B = 1/5 C = 1/15 Rate of work of A and B = 1/3 + 1/5 = 8/15 per hour Work done in 1 hr= 8/15 x 1 = 8/15 of the volume Volume still empty = 1 - 8/15 =7/15 Rate of work of A, B & C = 1/3 + 1/5 - 1/15 =7/15 Time taken to fill =7/15 ÷ 7/15 = 1hr Total time = 1 + 1 = 2hrs |
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13 | Numerator 9t2 - 25a2 = (3t + 5a)(3t - 5a) Denominator 6t2 + 19at + 15a2 6t2 + 10at + 9at + 15a2 2t(3t + 5a)+3a(3t + 5a) (2t + 3a)(3t + 5a) Both combined (3t + 5a)(2t - 5a) = (3t - 5a) (2t + 3a)(3t + 5a) (2t + 3a) |
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14 | Buying price = 300kg x sh30 per kg = sh9,000 After the loss = 80/100 x 300kg = 240kg 9000 = 100% sp = 120% sp =9000 x 120 = sh10,800 100 sp per kg =10800 240 =ksh45 |
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15 |
Log6216 +[Log42 - Log6] ÷ Log49 |
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16 | x:y = 2:3 x = y = k 2 3 x = k x =2k 2 y = k y =3k 3 5x - 2y = 5(2k)-2(3k) x + y 2k + 3k =10k - 6k 5k = 4k 5k 5x - 2y = 4 = 5x - 2y:x + y = 4:5 x + y 5 |
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17 | |||||
18 | |||||
19 |
a) a2 = b2 + c2 - 2bcCosA b) b = a c) a2 = d2+c2 -2dcCosA d)Area = ½dcSinA |
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20 |
a) P x Q b)P = 12 Q c) d) P = 1.48(k Q ) |
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21 |
a) 1cm rep. 100m b)Area 1 = ½ x 400 x 240 = 48 000m2 |
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22 |
a) Gradient of L = b) Gradient of P x 3 = -1 c) Gradient of Q = 3 d) Solve P & Q simultaneously x + 3(5) = 17 |
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23 |
a) AC =√102 + 102 b) √132 - 7.072 c) G.θ = 7.07 d) Tan x = ( 10.91 ) e) Tan xº = 5 |
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24 |
a) Volume of water =1/3 x 22/7 x 212 x 30 b i) h = r ii) New volume = 1/3 x 22/7 x 212 x 36 iii) 4/3 x 22/7 x r3 = 10 090.08 |
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