Mathematics Paper 1 Questions and Answers - Form 3 Term 3 Opener Exams 2023

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Mathematics Paper 2 Questions and Answers - Form 3 Term 3 Opener Exams 2023

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Use reciprocals and square root tables to evaluate: (3mks)
0.3 + √0.498
0.0351
Peter bought a shirt and sold it to Kamau at a profit of 10%. Kamau sold the same skirt to Omondi at a price of Kshs. 2,700/= thus making a loss of 15%. Find the price at which Kamau bought the shirt from Peter. (3mks)
In the figure below, O is the centre of the circle and OB bisects angle ABC. Given that angle BAC=40o, find angle ABO. (3mks)
A,B and C are three points on a straight line (in that order) on horizontal ground. A and B are on the side of C. At point C stands a vertical tower 173.2M high. The distance from A to B is 100M and the angle of elevation of the top the tower from A is 30o. Find the angle of elevation of the top of tower from B. (3mks)
Solve for x: 2^x-3 x 8 ^x2+2= 128 (3mks)
Find the integral values of x for which : 5≤3x + 2, and 3x – 14< -2 (3mks)
Find the value of ‘a’ in the figure below, if its area is 128cm², (3mks)
The scale of a map is given as 1:20000. Find the actual area in hectares of a region represented by a triangle of sides 6cm, 7cm and 4cm. (4mks)
The figure below shows a prism ABCDEF with BC=BF=4cm and AF=3cm
1. Find AB (1mk)
2. Draw the net for the prism above. (3mks)
A triangular plot ABC is such that AB=72m, BC= 80m and AC=84M. calculate the acute angle between the edges AB and BC. (3mks)
Two towns A and B are 365Km apart. A bus left town A at 8:15a.m. and travelled towards town B at 60Km/hr. At 9.00 a.m. a car left town B towards town A at a speed of 100km/hr. the tow vehicles met at town C which lies between towns A and B. find the time of the day when the two vehicles met. (4mks)
A photograph is reduced in the ratio 3:5 for a newspaper. Further, the photograph is reduced in the ratio 4:5 for a textbook. Find the ratio of the newspaper size to the textbook size. (3mks)
The vertices of a square are P( -3,-3), Q( -1, -3) R(-1,1-) and S(-3,-1) if the square is first reflected in the line y=-x and followed by a translation of vector ^-1/₄ find the co-ordinates of the final image of the square. (3mks)
The line whose equation is 3y=2x-12 cuts the x-axis at point A and the y-axis at point B. if M is the mid point of AB, find the co-ordinates of A, B and M. (3mks)
Calculate the value of a and b in the figure below. (3mks)
The fourth number of four consecutive numbers is 2n+3. If their sum is 1766, find the first number. (2mks)

SECTION B

At the beginning of a certain year Gaitho deposited sh. 100,000 in an investment account which earned compound, interest at 15% per annum. At the beginning of each subsequent year, he deposited a further sh. 5000 in the same account. Determine:
1. How much money he had in the account after 5 years. (7mks)
2. The percentage interest earned over the five years. (3mks)
The unshaded region in the figure below is reflected in the x-axis.
1. Write the coordinates of A’ and B’ the images of A and B after the reflection. (2mks)
2. Show the new region R’ after the reflection, in the same axes, hence calculate its area. (2mks)
3. Write down the inequalities which satisfy the new region. (6mks)
1. Draw a circles, centre O, radius 4cm and mark a point A’ which is 9cm from the centre. (1mk)
2. Construct two tangents from A to touch the circle at points B and C. measure the lengths of the tangents AB and AC. (3mks)
3. Calculate the area of triangle ABO. (2mks)
4. On the same drawing, construct a triangle APO which has the same area as triangle ABO and in which has the same area as triangle ABO and in which AP=PO. Measure and write down the length of AP. (4mks)
A box contains 15 white, 9 pink and 3 brown cloth pegs. The pegs are identical except for the colour, two pegs are picked at random, one at a time, without replacement.
1. Draw a probability space to show all the possible outcomes. (2mks)
2. Find the probability that:
  1. A white peg and a brown peg are picked. (4mks)
  2. Both pegs are of the same colour. (4mks)
The figure below shows the motion of a car for various parts of a journey named A,B,C and D.
1. Calculate the acceleration for the parts.
  1. AB (2mks)
  2. BC (2mks)
  3. CD (2mks)
2. Calculate the total distance covered by the car in kilometers. (4mks)
P and Q are points whose coordinates are (-2, 4) and (x,y) respectively. B is another point (2,0) such that PQ = 3QB. Find.
1. 1. X and Y (3mks)
  2. The magnitude of BQ (2mks)
2. OABC is a parallelogram, O is the origin, A is (6, 4) and B is (4, 8)
  1. Express OC and AB as column vectors. (3mks)
  2. Given that M is the mid. Point of BC, write the coordinates of M. (2mks)
The area A cm² of a cylinder depends partly on r and partly on r², where r is the radius of the base, when r=1cm, A=7cm² and when r =2cm, A=16cm²,
1. Find an expression for A in terms of r. (4mks)
2. Calculate the radius when the area is 115cm², give your answer to 1 d.p. (4mks)
3. Find the radius for which the two parts are equal. (2mks)
Water flows through a cylindrical pipe of diameter 4.2cm at a speed of 50m/min.
1. Calculate the volume of water delivered by the pipe per minute in litres. (3mks)
2. A cylindrical storage tank of depth 3m is filled by water form this pipe at the same rate of flow. Water begins flowing into the empty storage tank at 8.30a.m. And is full at 3.10p.m. calculate the area of cross section of this tank in m². (4mks)
3. A family consumes the capacity of this tank in one month. The cost of water is sh. 40 per thousand litres plus a fixed basic charge of sh. 1650. Calculate this family’s water bill for a month. (3mks)

MARKING SCHEME

Use recipricals and square root tables to evaluate! (3mks)
0.3 + √0.498
0.0351
0.3 x 1 √49.8
3.51 x 10² + 10²
0.3 x 0.2849 x 10² + 7.057
10
8.547+ 0.7057
= 9.253
Peter bought a skirt and sold it to Ramay at A Profit of 10%, Kamay sold the same shirt to Omondi at a price of Kshs. 2,700, thus making a Loss of 15%. Find the price at which Ramau bought the shirt from Peter, (3mks)
Let Peters buying price be x
Let Kamar's buying price be 110x
100
Let Omondis buying price be 85 (110x)
100 100
0.85 x 1.1x = 2700
x = 2700
= Sh 2887.70
In the figure below, O is the centre of the arcle and OB bisects angle ABC. Given that angle BAC = 40° find angle ABO (3 marks) ∴ ∠ OBC = 50°
∴ ∠ ABO = 50° SINCE OB is angle ABC bisector
∠BOC = 80 angel at centre twice angle at circumference
∠OBC = 180 − 80 i.e base angle of 2 isoscles ∠BOC
A, B and C are three points on a staright line(In that Order) on horizontal ground. A and B are on the Side of C. At point C stands a vertical tower 173.2 m high. The distance from A to B is 100m and the angle of elevation of the top of the tower from A is 30. Find the angle of elevation of the top of tower from B, (3mks)

Tan 30 = 173.2 AC = 300m
AC BC = 300 – 100
∴ AC = 173.2 = 200m
Tan 30
Solve for x; 2^x-3 x 8 ^x2+2= 128 (3mks)
2^x–3 x 2^{3(x2 + 2)} = 2⁷
2(^{(x − 3)}) + ^{(3x2 + 6)}) = 2⁷
∴ x – 3 + 3x²+6 = 7
3x² + x − 4 = 0
3x − 4 = −12
(+4,−3)
3x² + 4x − 3x −4 = 0
x(3x + 4) −1(3x − 4) = 0
(x −1)(3x + 4) = 0
∴ x = 1
3x =−4
x = − 1¹/₃
Find the integral values of x for which (3 marks)
5 ≤ 3x + 2, and 3x −14<−2
1. 3 ≤ 3x
  1 ≤ x
2. 3x < 12
  x < 4
  1 ≤ x < 4
  values are 1,2, 3
Find the value of a' in the figure below, if it's area is 128 cm²,
½h(15 + 17) = 128
32  h = 128
2
∴h = 8cm
Sin 60 = 8
a
∴ a = 8
sin 60
= 9.238cm
The Scale of a map is given as 1:20000, Find the actual area in hectares of a region represented by a triangle of Sides 6cm, 7cm, and 4cm
S = ½(6+7+4)
= 8.5
A = √8.5 x 2.5 x 1.5 x 4.5
= 11.98cm²
1:20000
∴ 1cm² → (2000)²
A = 11.98 x (20000)²
Also
1m → 100cm 1 ha → 10000m²
∴ 1m² → 10000cm² ∴ 11.98 x 40000
? → (20000)²cm^{2 →→→}10000
? = 40000m² = 47.9062
→ 11.98 x (20000)² = 47.91 ha
= 11.98 x 20000² ÷ 10000
The figure below shows a prism ABCDEF with BC=BF= 4cm and AF=3cm.
1. Find AB
  √4² + 3²
  √25
  = 5cm
2. Draw the net for the prism above.
A triangular plot ABC is such that AB=72m, BC=80m and AC=84m calculate the acute angle between the edges AB and BC,

84² = 72²+ 80− 2 x 80 X 72 Cos B
7056 = 5184 + 6400−11520 COSB
7056 =11584 −11520 Cos B
− 4528 = −11520 CosB
∴ CoSB = 0·3931
B = 66·83°
Two towns A and B are 365 km apart. A bus left town A at 8·15 a.m. and travelled towards town B at 60 km/hr. At 9.00a.m., a car left town B towards town A at a speed of 100 km/hr. The two vehicles met at town C which lies between towns A and B.
Find the time of the day when the two vehicles Met, (4mrks)
Distance by bus in 45 min
45 x 60 = 45km Alternative
60 Let distance coverd by bus to meet be x
Relative distance = 365 − 45 ∴ Distance by car = (320 - x)
= 320km Time : x = 320 − x
Relative speed = 60+100 60 100
= 160 km/hr. 100x = 19200 − 60x
Time taken to meet = RD   160x = 19200
RS ∴ x = 120km
= 320   Time taken = 120
160 60
= 2 hrs = 2hrs

∴ Time they meet = 9.00+2 = 11.00am   ∴ Time to meet = 9.00 + 2 = 11.00 a.m.
A photograph is reduced in the ratio 3:5 for a newspaper. Further, the Photograph is reduced in the ratio, 4:5 for a textbook. Find the ratio of the newspaper size to the textbook size, (3 Marks)
Photograph: Newspaper = (3:5) x 4
Photograph: Textbook = (4:5)x 3
P : N = 12:20
P : T = 12:15
→ P : N : T= 12: 20:15
∴ Newspaper : Textbook = 20:15
=4:3
The Vertices of a square are P(-3,-3), G(-1, -B), R(-1,-1) and S(3,-3). If the square is first reflected in the Line Y=-X and followed by a translation of vector (-1), find the co-ordinates of the furial image of the square

Reflection Y = −x
(a,b) → (−b,−a)
P(-3, -3) → p¹(3, 3)
Q(-1,-3) → Q¹(3, 1)
R(−1,−1) → R¹(1,1)
S(−3,−1) → S¹(1,3)

Translation (⁻¹ ₄)
p¹ (3,3) + T(⁻¹₄) = p" (2, 7)
Q¹(3, 1) + T(^-1₄) = Q'' (2,5)
R (−1, +1) + T(⁻¹₄) = R" (0,5)
S (1₁,3) +T(⁻¹₄) = 5" (0, 7)
The Line whose equation is 34=2x-12 cuts the x-axis at point A and the y-axis at point B. If M is the und-point of AB, find the Co-ordinates of A, B and M. (3mks)
3y-2x=-12 Let mid-point
−3 y + 2 x = 1 (6 + 0 0 + −4)
12 12 2 , 2
− y + x = 1
4 6 → (3, −2)
Double intercepts ∴ m = (3, −2)
(6,0) and (0,−4)
∴ A = (6,0)
B = (0-4)
Calculate the value of a and b in the figure
∠SRT=180−b is. angles on a straight line
⇒ 180− b + a + 80 = 180
∴ a − b= −80
Also:
3a+b = 180 (opp. ∠s of cyclic quad.)
→ (i) a − b = − 80
(ii) 3a + b = 180
Solving Simultaneously
4а = 100
∴ a =25°
→ 3 x 25 + b = 180
∴ b=105°
The fourth number of four consecutive numbers is 2n+3. If their sum is 1766, find the first number. (2 Marks)
4th = 2n+3
3rd = 2n+3-1
= 2n+2
2nd = 2n+2-1
= 2n+1
1st = 2n+1−1
= 2n
→ 2n+ 2n+1+2n+2+2n+3 =1766
8n + 6 =1766
∴ 8n =1760
n = 220
∴ 1st no. = 2 x 220
= 440

SECTION B.

At the begining of a certain year Gaitho deposited sh, 100,000 in an investment account which earned compound interest at 15% Per annum. At the beginning of each subsequent Year, he deposited a further Sh.5000 in the Same account. Determine.
1. How much money he had in the account after 5 years,
  Year 1 amount = 100,000 × 115
  = Sh. 115,000
  Year 2 amount = (115000+-5000) x 115
  100
  = Sh. 138,000
  Year 3 Amount = (138000 + 5000) x 115
  100
  = Sh. 164,450
  Year 4 amount = (164,450 +5000) 45
  100
  = Sh 194,867.50
  Year 5 amount: S (194867.50 + 5000) x 115
  100
  = Sh.229847.68
  ∴ Amount in the a/c. Sh.229,847.63
2. The Percentage profit interest earned over the five years. (3 marks)
  Total amount mrested
  100,000 + (4 x 5000)
  100,000 + 20,000
  = Sh 120,000
  Total after 5 years = Shs. 229,847.63
  ∴ interest amount = 229,847.63 −120,000
  = 109847.63
  ∴ % age interest = 109847.63 x 100
  120000
  = 91.54%
The Unshaded region in the figure below is reflected in the x-axis.
1. Write the coordinates of A' and B', the images of A and B after the Reflection (2 Marks)
  A' (−7, −7)
  B' (7,−7)
2. Show the new region R' after the reflection, in the same axes Hence calculate its area (2mark)
  A = ½ x 7 x 14 = 49 sp. units.
3. Write down the inequalities which satisfy the new region, (6 Marks)
  Let L₁ be A' B'
  y = −7
  ∴ y ≥ − 7
  L₂ be OA'
  y = − x
  ∴ y ≤ − x
  L₃be OB'
  y = x
  y ≤ x
1. Draw a circle, centre 0, radius HCM and mark a Point A which is Centre 9cm from the centre (1Mark)
2. Construct two tangents from A to touch the circle at points B and C. Measure the Lengths of the tangents AB and AC (3 Marks)
3. Calculate the area of triangle ABO (2 marks)
  A = ½ x AB x BO
  = ½ x 8 x 4
  = 16 cm²
4. On the same drawing, construct a triangle APO which has the same area as triangle ABO and in which AP = PO. Measure and write down the Length of AP. (4mrks)
  Let h be height of ∠ APO
  ∴ ½ x h x AO = 16
  ½h x 9 = 16
  ∴ h = 16 x 2 =3.6cm
  9
  ∴ P is 3.6cm from M, on the I bisector of AO
A box contains 15 white, 9 pink and 3 brow, cloth Pegs. The pegs are identical except for the Colour, Two pegs are picked at random, one at a time, without replacement.
1. Draw a probability space to show all the possible outcomes. (2 Marks)
2. Find the Probability that:
  1. A white peg and a brown peg are picked. (4 marks)
    P(NB) + P(BW)
    (15 x 3) + (3 x 15)
    27 26 27 26
    5 + 5
    78 78
    10 = 5
    78 39
  2. Both pegs are of the same colour
    P(WW) + P(PP) + P(BB)
    15 x 14 + 9 x 8 + 3 x 2
    27 26 27 26 27 26
    210 + 72 + 6
    702 702 702
    288 = 16
    702 39
The figure below shows the motion of a car for various parts named A, B, C and D. of a journey named A, B, C and D,
1. Calculate the acceleration for the parts (2 Marks)
  1. AB
    Gradient AB
    = 45
    3
    = 15 ms⁻²
  2. BC
    Gradient BC
    = 0
    5
    = 0ms⁻²
  3. CD
    Gradient CD
    Gradient CD
    = − 45
    6 = − 7.5ms⁻²
2. Calculate the total distance covered by the car in Kilometres. (4 Marks)
  Area (a) = ½ x 3 x 45
  = 67.57m
  Area (b) = 45×5
  = 225 m
  Area (c) = ¹/₂x 45 x 63
  = 135 m
  Total area = 427.5 m
1. P and Q are points whose coordinates (−2, 4) and (x, y) respectively. B is another point (2,0) Such that PQ = 3QB,
  1. Find x and y (2) − (x)   ∴ x + y = 6 − 3x    (3 Marks)
    Let PQ = Q − P 0 y 4x + y = 6   ∴ x = 5
    = (x) − (−2) (2 − x) and 4
    y 4   → −y → y − 4 = –3y   → = 1¼
    = ( x+ 2) ( x + y) = 3(2 − x ) ∴ 4y = 4
    y − 4 y − 4 −y y = 1
    ∴ QB = B − Q 4x + 1 = 6
    4x = 5
  2. The magnitude of BQ
2. OABC is a parallelogram. O is the origin, A is (6, 4) and B is (4,8).
  1. Express OC and AB as column vectors (3 marks),
  2. (ii) Given that M is the mid. point of BC, write the coordinates of M. (2 Mark)
The area A CM2 of a cylinder depends partly on r and Partly on r², where r is the radius of the base. When r=1cm, A=7cm² and when r= 2cm, A = 16 cm²,
1. Find an expression for A in terms of r
  A = Cr + Kr²
  (i) (7 = c + r) x 2 ∴ k = 1
  (ii) 16 = 2c + 4k substitution in (i)
  14 = 2c + 2k → 7 = c + 1
  16 = 2c + 4k c = 6
  (ii) − (i) ∴ A = 6r + r²
  2 = 2k
2. Calculate the radius when the area is 115 cm². Give our answer to 1 d.p.
  115 = 6r + r²
  r²+ 6r − 115 = 0
  r = −b ± √b² − 4ac
  2a
  = − 6 ± √6² − 4 x 1 x −115 → 16.27
  2 x 1 2
  − 6 ± √36 + 460 8.135
  2 8.1cm
  − 6 ± 22.27
  2
3. Find the radius for which the two parts are equal,
  A = 6r + r²
  For part 6r equals parf r²
  6r = r²
  ∴ r = 6cm
Water flows through a cylindrical Pipe of diameter 4.2 cm at a speed of 50m/min.
1. Calculate the Volume of water delivered by the pipe per minute in Litres (3 marks)
  Vol = πr²h
  = (0.021² x 22 x 50)³m
  7
  = 0·0693m³
  1000L → 1m³
  ? → 0·0693m³
  = 69.27L
2. A Cylindrical storage tank of depth 3m is filled water from this pipe at the same rate of flow, Water begins flowing into the empty storage tank at 8.30 am and is full at 3.10 p.m. Calculate the area of cross section of this tank in m² (4 Marks)
  Time taken = 3·10 − 8:30
  = 6 ²/₃ hrs,
  Vol in 1 min = 0·0693m³∴ 6²/₃ hrs = ?
  6²/₃x 60 x 0.0693m³ = 27.72m³Vol· of tank − Cross section area x depth.
  = A x 3
  27·72 = 3A
  ∴ A = 27.72
  3
  = 9.24m³
3. A family consumes the capacity of this tank in one month. The cost of water is sh. 40 per thousand litres plus a fixed basic charge of sh.1650. Calculate this family's water bill for a month. (3 Marks)
  1m³ = 1000L
  27.72m³ →?
  27.72 x 1000
  = 27.720L
  ∴ capacity of tank = 27.720L
  1000L → sh 40.
  27720L → ?
  27.720 x 40
  ∴ Bill = 1108.8+1650
  = 2,758.80