INSTRUCTIONS
 This paper consists of two sections. Answer all the questions in section 1 and any 5 questions in section 2.
 KNEC mathematical tables and nonprogrammable calculators may be used when necessary
SECTION 1
Answer all the questions in this section
 Evaluate 36 –8×–4 –15÷ –3 (3mk)
–3×–3–8(–6+ –2)  Simplify a+b – 2a–b (3mk)
2 3  Find the greatest number which divides 181 and 236 and leaves a remainder of 5 in each case (2mk)
 A rectangle measures 20cm by 15cm.If each dimension is increased by 2.5cm, by what percentage is
 the perimeter of the rectangle increased (2mk)
 the area of the rectangle is increased (2mk)
 The angle of elevation of the top of a tree from a point P on horizontal ground is 30º. From another point Q 8 metres from the base of the tree, the angle of elevation of the top of the tree is 48º.
 Calculate to one decimal place the height of the tree. (1mk)
 Calculate the distance between P and Q (2mk)
 Given that cos θ= –0.8070 , find θ for 0≤θ≤720 (3mk)
 A piece of wire 40cm is bent to form a rightangled triangle whose hypotenuse is 17cm long. Find the lengths of the other two sides of the triangle (4mk)
 Solve for x in log 5–2+log(2x+10)=log(x–4) (3mk)
 Solve the quadratic equation by completing of squares giving your answer to 3sf (4mk)
11x^{2} –13x+3=0  Rationalize the denominator and simplify (4mk)
4√5+3√2
2√2–√5  Use a calculator to work out
(1mk)
(1mk)
 A tourist from Kenya left for Ethiopia. He exchanged sh 9898 into Ethiopian Birr at the rate of 1 Eth.Birr=ksh7.95. He spent ¾ of the money he got and converted the balance back to Kenyan money at the rate of 1Eth.Birr=ksh7.98 Calculate what he finally got to 2dp (3mk)
 Simplify the expression (3mks)
 Three types of tea costing sh203,sh146 and sh197 per kg are blended in the ratio of 2:5:k. Find the value of k,if the blend when sold at sh221 per kg gives 30% profit (3mk)
 A two digit number is such that 4 times the units digit exceeds the tens digit by1. If the digits are reversed, the number formed is decreased by 45. Find the number. (3mk)
 A triangular field has dimensions 21m by 52m by 47m.
 calculate the area of the field to the nearest m^{2}. (2mk)
 calculate the length of a straight ditch dug from the largest angle meeting the opposite side at right angles. (1mk)
SECTION 2
Answer any 5 questions in this section
 A bookseller bought a number of cartons of books at a cost of ksh 57600 from Kagumo bookstore. Had he bought the same books from Kerugoya bookstore, it would have cost him ksh 480 less per carton. This would have enabled him to buy 4 extra cartons of books for the same amount of money. By taking x to be the number of cartons of books he actually bought;
 write an expression in x
 for the cost of each carton he bought at Kagumo bookstore (1mk)
 for the cost of each carton had he bought from Kerugoya bookstore (1mk)
 find the value of x (6mk)
 The bookseller later sold all the books he had bought each carton at ksh 720 more than he had paid for it. Determine the percentage profit he made (2mk)
 write an expression in x
 A cylindrical metal bar of diameter 14cm and length 2m is melted and moulded into spherical balls. In the process, 5% by volume of metal is lost and what remains makes balls of radius 3.5cm.
 calculate the volume of metal used to make the balls. (3mk)
 find to the nearest whole number the number of balls made (3mk)
 find the total surface area of the metal bar (2mk)
 find the total surface area of the balls made (2mk)

 Draw the graph of y=(2x–3)(x–1) for the interval –2 ≤ x ≤ 4 (6mk)
 use your graph to solve
 2x^{2}–5x+3=0 (1mk)
 2x^{2}=3x+2 (3mk)

 Plot triangle ABC with coordinates A(1, 1), B(3, 1) and C(1,3 ) (1mk)
 Plot A′B′C′ the image of ABC under an enlargement scale factor 2 centre A and write down it’s coordinates (2mk)
 Plot A′′B′′C′′ the image of A′B′C′ under a reflection in the line x+y=0 and write down it’s coordinates (3mk)
 A′′B′′C′′ is then reflected in the line y=0 to give A′′′B′′′C′′′.Give the coordinates of A′′′B′′′C′′′ (2mk)
 Describe fully a rotation that maps A′′′B′′′C′′′ onto A′B′C′ (2mk)
 Three businessladies Wanjiku, Muthoni and Njoki decided to buy a lorry. The .marked price of the lorry was 2.8million shillings. The dealer agreed that the ladies could pay a deposit of 60% of the marked price and the rest to be paid within a year.The ladies raised the deposit in the ratio of 3:2:5 respectively. At the end of the year the lorry had realized 2.08million shillings which the three shared in the ratio of their contribution. However, they were required to contribute for the balance of the lorry from these earnings again in the ratio of their original contributions.
 calculate amount to be paid as deposit (1mk)
 how much did each contribute to pay for the deposit? (3mk)
 how much did Njoki receive at the end of the year? (1mk)
 calculate the total amount Muthoni and Njoki contributed to pay for the balance. (3mk)
 how much money did Wanjiku remain with after paying her share of the balance? (2mk)

 Make a table and draw the graph of y= sinx–cosx for x in the range 0°≤x≤360°, with y values to 2dp and x intervals of 30° (6mk)
 use the graph to find the value of y when
 x=75° (1mk)
 x=255° (1mk)
 find the values of x for which y= –0.9 (2mk)
 The figure below is a segment of a circle centre O radius r units.CM is the perpendicular bisector of AB.
Given that CM=1cm and AB=2√3cm, calculate the radius of the circle centre O from which the segment was cut (3mk)
 calculate the angle that chord AB subtends at the centre of the circle (2mk)
 hence calculate
 the length of arc ACB (2mk)
 the area of the segment AMBC (3mk)
 A rectangular sheet of metal which measures 120cm by 0.8m is 1.5mm thick and is made of material whose density is 2.2 g/cm^{3}. From each of the four corners of the rectangle, a square of side 10cm is cut off and the remaining part folded to form an open cuboid.
 calculate
 the capacity of the cuboid in cm^{3} to the nearest whole number (3mk)
 the mass of the empty cuboid in kg to the nearest whole number (3mk)
 the cuboid is filled with a liquid whose density is 0.75g/cm^{3}. Calculate the mass in kg of the cuboid when full of the liquid (2mk)
 calculate the mass of metal lost in kg (2mk)
 calculate
Marking Scheme
SECTION 1
Answer all the questions in this section
 Evaluate 36 –8×–4 –15÷ –3 (3mk)
–3×–3–8(–6+ –2)
numerator 36 + 32 + 5 = 73 M1
denominator 9 − 8(−) = 9+64=73 M1
Ans ^{73}/_{73} =1 A1  Simplify a+b – 2a–b (3mk)
2 3
3(a+b) −2(2a−b) M1
6
3a + 3b − 4a +2b M1
6
=5b − a or −a+5b A1
6 6  Find the greatest number which divides 181 and 236 and leaves a remainder of 5 in each case (2mk)
GCD of 181−5 and 236−5
= M1
The number is 11 A1  A rectangle measures 20cm by 15cm.If each dimension is increased by 2.5cm, by what percentage is
 the perimeter of the rectangle increased (2mk)
Original perimeter = 2(20+15) =70
Increase = 4 × 2.5 =10 M1
%Inc = ^{10}/_{70} × 100 =14.29% A1  the area of the rectangle is increased (2mk)
Original area =20×15=300
Area with increase =(2.5 +20) (2.5 + 15) = 393.75
Increase= 93.75 M1
%inc= 93.75 ×100 =31.25% A1
300
 the perimeter of the rectangle increased (2mk)
 The angle of elevation of the top of a tree from a point P on horizontal ground is 30º. From another point Q 8 metres from the base of the tree, the angle of elevation of the top of the tree is 48º.
 Calculate to one decimal place the height of the tree. (1mk)
Tan 48 = ^{h}/_{8}
h = 8 tan 48 =8.885 B1  Calculate the distance between P and Q (2mk)
Tan 30 = 8.885 M1
PF
PF = 8.885 = 15.39
tan30
PQ = 15.39 − 8 =7.389 A1
 Calculate to one decimal place the height of the tree. (1mk)
 Given that cos θ= –0.8070 , find θ for 0≤θ≤720 (3mk)
θ = cos^{−1} (−0.8070)
Acute θ =36.20°
2^{nd} & 3^{rd} quad
θ = 180 − 36.20 = 143.8°
θ = 180 + 36.20 =216.2°
Beyond 360° θ = 143.8 + 360 = 503.8°
θ = 216.2 + 360 = 576.2°  A piece of wire 40cm is bent to form a rightangled triangle whose hypotenuse is 17cm long. Find the lengths of the other two sides of the triangle (4mk)
a+b = 23.....(i)
a^{2} + b^{2} =289....(ii) M1
Substitute (i) in (ii)
a^{2} + (23 − a)^{2} =289
a^{2} + 529 − 46a + a^{2} =289
2a^{2} − 46a + 240 =0
a^{2} − 23a +120 =0 M1
a^{2} − 8a − 15a + 120 = 0 M1 Attempt to solve correctly
a(a − 8) −15(a−8) = 0
a=15 or 8
b=8 or 15 A1  Solve for x in log 5–2+log(2x+10)=log(x–4) (3mk)
log 5 − 2 log10 + log(2x+10) = log(x−4)
Log (^{5}/_{102 }× (2x +10) ) =log(x−4)
^{x}/_{10} + ½ = x − 4
^{−9}/_{10}x = ^{−9}/_{2}
x = −9 × 10
2 −9
=5_{}  Solve the quadratic equation by completing of squares giving your answer to 3sf (4mk)
11x^{2} –13x+3=0
x^{2} − ^{13}/_{11} = ^{−3}/_{11 }x^{2} −^{13}/_{11} + (^{−13}/_{11})^{2} = ^{−3}/_{11} + (^{−13}/_{11})^{2}
(x − ^{13}/_{11})^{2} = ^{−3}/_{11 }+ ^{169}/_{121}
=^{136}/_{121}
x − ^{13}/_{11} =±√^{136}/_{121}
either x = √^{136}/_{121} + ^{13}/_{11}
=11.66 +13 =2.242
11 11
Or x = −√^{136}/_{121 }+ ^{13}/_{11}
=−11.66 +13 = 0.1218
11  Rationalize the denominator and simplify (4mk)
4√5+3√2
2√2–√5
(4√5+3√2)(2√2+√5)
(2√2–√5) (2√2+√5)
Num
8√10+20+12+3√10
=32+11√10
den
8+2√10 − 2√10 − 5
=3
Ans = 32+11√10
3  Use a calculator to work out
(1mk)
=0.6816 B1
(1mk)
=0.2030 B1
 A tourist from Kenya left for Ethiopia. He exchanged sh 9898 into Ethiopian Birr at the rate of 1 Eth.Birr=ksh7.95. He spent ¾ of the money he got and converted the balance back to Kenyan money at the rate of 1Eth.Birr=ksh7.98 Calculate what he finally got to 2dp (3mk)
1 Eth. Birr =sh 7.95
? =sh 9898
=Birr 1245.03 M1
Bal after spending
=¼ × 1245.03 = 311.26 M1
1 Eth. Birr =sh7.98
311.26 = ?
=311.26 × 7.98
=sh 2483.80 A1  Simplify the expression (3mks)
num (3t−5a)(3t+5a) M1
den 6t^{2} +9at+10at+15a^{2}
=3t(2t+3a) +5a(2t+3a)
=(3t+5a)(2t+3a) M1
(3t−5a)(3t+5a) = 3t−5a
(3t+5a)(2t+3a) 2t+3a A1  Three types of tea costing sh203,sh146 and sh197 per kg are blended in the ratio of 2:5:k. Find the value of k,if the blend when sold at sh221 per kg gives 30% profit (3mk)
130% =221
100% = ?
=22100 =sh 170
130
2 (203) + 5 (146) + K (197) =170
7+k 7+k 7+k
406+730+197k =170(7+k)
1136+197k =1190+170k
27k = 54
k=2  A two digit number is such that 4 times the units digit exceeds the tens digit by 1. If the digits are reversed, the number formed is decreased by 45. Find the number. (3mk)
let the number be xy
4y − x =1......(i)
10x+y−(10y+x) =45
10x+y−10y−x =45
9x−9y =45......(ii) M1
From (i) x=4y−1, subst in (ii)
9(4y−1)−9y =45
36y−9−9y =45
27y = 54
y=2 M1
x = 4 × 2−1
=8−1 =7
The no. is 72 A1  A triangular field has dimensions 21m by 52m by 47m.
 calculate the area of the field to the nearest m^{2}. (2mk)
Area
S=½(47+21+52) =60
A= √(60(60−52)(60−47)(60−21))
= √(60×8×13×39) =493.3m^{2}
=493m^{2}  calculate the length of a straight ditch dug from the largest angle meeting the opposite side at right angles. (1mk)
½ × 52 ×h =493
h= 493 × 2 = 18.96m
52
 calculate the area of the field to the nearest m^{2}. (2mk)
SECTION 2
Answer any 5 questions in this section
 A bookseller bought a number of cartons of books at a cost of ksh 57600 from Kagumo bookstore. Had he bought the same books from Kerugoya bookstore, it would have cost him ksh 480 less per carton. This would have enabled him to buy 4 extra cartons of books for the same amount of money. By taking x to be the number of cartons of books he actually bought;
 write an expression in x
 for the cost of each carton he bought at Kagumo bookstore (1mk)
57600 B1
x  for the cost of each carton had he bought from Kerugoya bookstore (1mk)
57600 − 480 B1
x
 for the cost of each carton he bought at Kagumo bookstore (1mk)
 find the value of x (6mk)
No. of cartons bought from Kerugoya bookstore = x + 4 M1
(x + 4) ( 57600 − 480) = 57600 M1
( x )
57600 − 480x + 230400 − 1920 = 57600 M1
x
55680x − 480x^{2} + 230400 =57600x M1
−1920x −480x^{2} +230400 =0
x^{2} + 4x − 480 =0 M1
x^{2} − 20x+ 24x − 480 =0
x( x−20) + 24(x−20) =0
x=20
x=−24 A1
No of cartons = 20 B1  the bookseller later sold all the books he had bought each carton at ksh 720 more than he had paid for it. Determine the percentage profit he made (2mk)
720 × 20 × 100 =25%
57600
 write an expression in x
 A cylindrical metal bar of diameter 14cm and length 2m is melted and moulded into spherical balls. In the process, 5% by volume of metal is lost and what remains makes balls of radius 3.5cm.
 calculate the volume of metal used to make the balls. (3mk)
^{22}/_{7} × 7 ×7 × 200 =30,800cm^{3}^{ } M1
Remaining = ^{95}/_{100} × 30800 M1
= 29,260cm^{3} A1  find to the nearest whole number the number of balls made (3mk)
^{4}/_{3} × ^{22}/_{7} × 3.5^{3} = volume of 1 ball
No. of balls = 29260 × 21 M1
4×22×3.5^{3}
=162.86 A1
=162 balls B1  find the total surface area of the metal bar (2mk)
2 × ^{22}/_{7} × 7^{2} =308
C.S = 2× ^{22}/_{7} × 7 ×200 =8800 M1
T.S.A =8800 + 308
=9108cm^{2} A1  find the total surface area of the balls made (2mk)
162 × 4 × ^{22}/_{7} × 3.5^{2} M1
=24948cm^{2} A1
 calculate the volume of metal used to make the balls. (3mk)

 Draw the graph of y=(2x–3)(x–1) for the interval –2 ≤ x ≤ 4 (6mk)
y=2x^{2} − 2x − 3x +3
y=2x^{2} − 5x +3x −2 −1 0 1 2 3 4 2x^{2} 8 2 0 2 8 18 32 −5x 10 5 0 −5 −10 −15 −20 3 3 3 3 3 3 3 3 y 21 10 3 0 1 6 15  use your graph to solve
 2x^{2}–5x+3=0 (1mk)
y=0
x=0  2x^{2}=3x+2 (3mk)
y=2x^{2} − 2x − 3x +3 B1
−y=2x^{2} − 5x +3
0=2x^{2} − 3x −2
y =−2x +5x 0 −1 1 y 5 7 3
y=−0.45
 2x^{2}–5x+3=0 (1mk)
 Draw the graph of y=(2x–3)(x–1) for the interval –2 ≤ x ≤ 4 (6mk)

 Plot triangle ABC with coordinates A(1, 1), B(3, 1) and C(1,3 ) (1mk)
 Plot A′B′C′ the image of ABC under an enlargement scale factor 2 centre A and write down it’s coordinates (2mk)
A′(1,1) B′(5,1) C′(1,5)
B1 B1 B1  Plot A′′B′′C′′ the image of A′B′C′ under a reflection in the line x+y=0 and write down it’s coordinates (3mk)
A′′(−1,−1) B′′(−1,−5) C′′(−5,−1)
L1 B1 B1  A′′B′′C′′ is then reflected in the line y=0 to give A′′′B′′′C′′′.Give the coordinates of A′′′B′′′C′′′ (2mk)
A′′′(−1,−1) B′′′(−1,5) C′′′(−5,−1)
B1 B1  Describe fully a rotation that maps A′′′B′′′C′′′ onto A′B′C′ (2mk)
Rotation −90° about (0,0)
B1 B1
 Plot triangle ABC with coordinates A(1, 1), B(3, 1) and C(1,3 ) (1mk)
 Three businessladies Wanjiku, Muthoni and Njoki decided to buy a lorry. The .marked price of the lorry was 2.8million shillings. The dealer agreed that the ladies could pay a deposit of 60% of the marked price and the rest to be paid within a year.The ladies raised the deposit in the ratio of 3:2:5 respectively. At the end of the year the lorry had realized 2.08million shillings which the three shared in the ratio of their contribution. However, they were required to contribute for the balance of the lorry from these earnings again in the ratio of their original contributions.
 calculate amount to be paid as deposit (1mk)
^{60}/_{100} × 2800000 = sh. 1,680,000 B1  how much did each contribute to pay for the deposit? (3mk)
Wanjiku = ^{3}/_{10} × 1680000 =504,000 B1
Muthoni = ^{2}/_{10} × 1680000 =336,000 B1
Njoki = ^{5}/_{10} × 1680000 =840,000 B1  how much did Njoki receive at the end of the year? (1mk)
^{5}/_{10} × 2080000 = 1,040,000 B1  calculate the total amount Muthoni and Njoki contributed to pay for the balance. (3mk)
bal = ^{40}/_{100} × 2800000 = 1,120,000 M1
^{7}/_{10} × 1,120,000 = 784,000 M1 A1  how much money did Wanjiku remain with after paying her share of the balance? (2mk)
Received = ^{3}/_{10} × 2080000 = 624,000
Paid ^{3}/_{10} × 1120000 = 336,000 M1
remained = 288,000 A1
 calculate amount to be paid as deposit (1mk)

 Make a table and draw the graph of y= sinx−cosx for x in the range 0°≤x≤360°, with y values to 2dp and x intervals of 30° (6mk)
x 0 30 60 90 120 150 180 210 240 270 300 330 360 Sin x 0 0.5 0.87 1 0.87 0.5 0 −0.5 −0.87 −1 −0.87 −0.5 0 Cos x 1 0.87 0.5 0 −0.5 −0.87 −1 −0.87 −0.5 0 0.5 0.87 1 y −1 −0.37 0.37 1 1.37 1.37 1 0.37 −0.37 −1 −1.37 −1.37 −1
B1 B1 B1 S1 P1 C1  use the graph to find the value of y when
 x=75° (1mk)
0.75 ± 0.05 B1  x=255° (1mk)
0.675 ± 0.05 B1
 x=75° (1mk)
 find the values of x for which y= –0.9 (2mk)
6°, 264° ±6° B1 B1
 Make a table and draw the graph of y= sinx−cosx for x in the range 0°≤x≤360°, with y values to 2dp and x intervals of 30° (6mk)
 The figure below is a segment of a circle centre O radius r units.CM is the perpendicular bisector of AB.
Given that CM=1cm and AB=2√3cm, calculate the radius of the circle centre O from which the segment was cut (3mk)
r^{2}=(r − 1)^{2} + (√3)^{2} M1
r^{2} =r^{2} − 2r + 1 + 3 M1
2r = 4
r=2cm A1^{}  calculate the angle that chord AB subtends at the centre of the circle (2mk)
Cos θ = ½ = 0.5
θ =60º M1
Angle at centre =120° A1  hence calculate
 the length of arc ACB (2mk)
^{120}/_{360} × 2 × ^{22}/_{7} ×2 =4.190 M1 A1  the area of the segment AMBC (3mk)
^{120}/_{360} × ^{22}/_{7} × 2^{2} − ½ ×2^{2} sin 120 M1 M1
=4.190 − 1.7321 = 2.4579 A1
 the length of arc ACB (2mk)
 calculate the radius of the circle centre O from which the segment was cut (3mk)
 A rectangular sheet of metal which measures 120cm by 0.8m is 1.5mm thick and is made of material whose density is 2.2 g/cm^{3}. From each of the four corners of the rectangle, a square of side 10cm is cut off and the remaining part folded to form an open cuboid.
 calculate
B1 the capacity of the cuboid in cm^{3} to the nearest whole number (3mk)
99.7 × 59.7 × 9.85 M1
=58628cm^{3 } A1  the mass of the empty cuboid in kg to the nearest whole number (3mk)
Outer volume = 100 × 60 × 10
= 60,000
Inner volume =−58,628 M1
Volume of material 1372cm^{3} M1
Mass = 1372 × 2.2 =3kg A1
1000
 the capacity of the cuboid in cm^{3} to the nearest whole number (3mk)
 the cuboid is filled with a liquid whose density is 0.75g/cm^{3}. Calculate the mass in kg of the cuboid when full of the liquid (2mk)
Mass of liquid
=0.75 × 58628 = 43.971kg M1
1000
Total mass = 46.971kg A1  calculate the mass of metal lost in kg (2mk)
Volume = 4 × 10 × 10 × 0.15 M1
=60cm^{3}
Mass = 60 × 2.2 =132g A1
 calculate
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