Mathematics Questions and Answers - Form 4 Mid Term 2 Exams 2023

Share via Whatsapp

Next Topic »
English Questions and Answers - Form 4 Mid Term 2 Exams 2023

Download PDF for future reference Get on Whatsapp for 50/-

QUESTIONS

SECTION A

The equation of a circle is given by X² + Y² – 10Y + 16 = 0.
Find the radius of a circle and its Centre. (4mks)
The point (5,2) undergoes transformation 3 2 followed by translation -6. Determine the Co-ordinates of the image. (3mks)
Using a ruler and a pair of compasses only construct a parallelogram.
1. PQRS in which PQ=6 QR=4 and angle SPQ=75º (3mks)
2. Determine the perpendicular distance between PQ and SR. (1mk)
1. Find the values between 0º and 360º which satisfy the equation 2 Ө = -0.5. (4mks)
2. Determine the amplitude, period and phase angle of the following equation. (3mks)
  Y = 6 Sin (^x/₂ – 90)º
The first term of AP is 2 and the sum of 10 terms of the AP is 650. Find the value of d. (2mks)\
The figure below shows a rhombus PQRS with PQ= 9cm and ‹SPQ=60⁰. SXQ is a circular arc, centre P.

Calculate the area of the shaded region correct to two decimal places (Take Pie= ²²⁄₇) (4mks)
Solve the equation 2x²+ 3x=5 by completing the square method (3mks)
Simplify the expression
3x² – 4xy²+ y (3mks)
9x²—y²
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)
Given that Cos (x – 20)⁰ = Sin (2x + 32)⁰ and x is an acute angle, Find tan (x – 4)⁰ (3 Marks)
In the figure below PQRS is a trapezium with QR parallel to PS. QR = 6cm, RS = 4cm, QS = 9cm and PS = 10cm.

Calculate
1. The size of angle SQR (2 Marks)
2. The area of triangle PQS (2 Marks)
Wanza sold a bag of potatoes for Sh. 420 and made a profit. If she sold it at Sh. 320, she could have made a loss. Given that the profit is thrice the loss, how much did she pay for the bag of potatoes? (3 Marks)
The sum of digits formed in a two digit number is 16. When the number is subtracted from the number formed by reversing the digits, the difference is 18. Find the number (3 Marks)
Make P the subject of the formula XY^P = Q^PX(3 marks)
The length and breadth of a rectangular floor garden were measured and found to be 4.1m and 2.2m respectively. Find the percentage error in its area. (3 marks)

SECTION B (CHOOSE 3 QUESTIONS)

The figure below represents a towel model. The base PQR is an equilateral triangle of side 9cm. The length VP=VQ=VR=20.5cm. Point M is the mid-point of PQ and VM =20cm. Point N is on the base and vertically below V.

Calculate
1. 1. Length of RM (2mks)
  2. Height of the model (2mks)
  3. Volume of the model (3mks)
2. The model is made up of a material whose density = 2700kg/m³. Find the mass of the model. (3mks)
The table below shows distribution of marks scored in a test by standard eight pupils in a Mathematics test.

Marks

30-34

35-39

40-44

45-49

50-54

55-59

60-64

65-69

70-74

75-79

No. of pupils

1

5

10

10

19

20

8

8

4

3

Using assumed mean of mark of 57 calculate.
1. The actual Mean of the grouped data. (3mks)
2. The standard deviation. (7mks)
OABC is a parallegram with O(0,0), A(2,0), B(3,2) C(1,2). O¹A¹B¹C¹ is the image of OABC under transformation -2 0
0 -2
1. 1. Find the co-ordinates of O¹A¹B¹C¹ (2mks)
  2. On the grid provided draw OABC and O¹A¹B¹C¹ (2mks)
2. 1. Find O¹¹A¹¹B¹¹C¹¹ the image of O¹A¹B¹C¹ under the transformation matrix (10 -20). (2mks)
  2. Draw O¹¹A¹¹B¹¹C¹¹
3. Find a single matrix that maps O¹¹A¹¹B¹¹C¹¹ onto OABC. (3mks)
Using trapezium rule find the area bounded by the curve (4mks)
1. Y=-X²+6X+1 X=0 and X=6
2. Calculate the exact area under the curve. (4mks)
3. Find the percentage error introduced by the trapezium rule. (2mks)
1. Complete the table below, giving the rules correct to 1 decimal place.
  
  X^O
  
  0
  
  40
  
  80
  
  120
  
  160
  
  200
  
  240
  
  2SIN(X+20)²
  
  0.7
  
  -
  
  2.0
  
  -
  
  00
  
  -
  
  -2.0
  
  [3 COS X
  
  1.7
  
  1.3
  
  0.9
  
  -1.6
2. Using the same scale drawing on X and Y axis draw the graphs of 2sin(x+20) and [3 cos x for 0≤x≤240.
3. Using the graphs above determine the values for X which 2sin(x+20)² =[cosX.
4. Determine the difference in amplitude between the two graphs.
Three business partners, Bela Joan and Trinity contributed Kshs 112,000, Ksh,128,000 and ksh,210,000 respectively to start a business. They agreed to share their profit as follows:
30% to be shared equally
30% to be shared in the ratio of their contributions
40% to be retained for running the business.
If at the end of the year, the business realized a profit of ksh 1.35 Million. Calculate:
1. The amount of money retained for the running of the business at the end of the year. (1mk)
2. The difference between the amounts received by Trinity and Bela (6mks)
3. Express Joan’s share as a percentage of the total amount of money shared between the three partners. (3mks)
A bag contains 5 red, 4 white and 3 blue beads . two beads are selected at random one after another.
1. Draw a tree diagram and show the probability space. (2mks)
2. From the tree diagram, find the probability that;
  1. The last bead selected is red (3mks)
  2. The beads selected were of the same colour (2mks)
  3. At least one of the selected beads is blue. 3(mks)
A surveyor recorded the following information in his field book after taking measurement in metres of a plot.
1. Sketch the layout of the plot. 4 mks.
2. Calculate the area of the plot in hectares. 6mks

MARKING SCHEME

SECTION A

The equation of a circle is given by X² + Y² – 10Y + 16 = 0.
Find the radius of a circle and its Centre. (4mks)
- x² + y² = 10y + 25 = -16 + 25
  x² + y2 - 10y + 25 = 9
  √(x²) √y² - 1xy + 25
  = √9
  
  Radius = 3 units
  x = 0
  y = 5
  
  centre -(0,5)
  
  Radius = 3 units
The point (5,2) undergoes transformation 3 2 followed by translation -6. Determine the Co-ordinates of the image. (3mks)
Using a ruler and a pair of compasses only construct a parallelogram.
1. PQRS in which PQ=6 QR=4 and angle SPQ=75º (3mks)
2. Determine the perpendicular distance between PQ and SR. (1mk)
  - 3.8 ± 0.1 cm
1. Find the values between Oº and 360º which satisfy the equation 2 Ө = -0.5. (4mks)
  - σsin 20 = 0.5
    2σ = 30
    
    = 210/2 and 330/2
    = 105º and 165º
2. Determine the amplitude, period and phase angle of the following equation. (3mks)
  Y = 6 Sin (x/2 – 90)º
  - period ³⁶⁰/_¹/2 = 720
    Amplitude = 6
    Phase angle = ⁹⁰/_¹/2= 180º
The first term of AP is 2 and the sum of 10 terms of the AP is 650. Find the value of d. (2mks)
- Ap = 2
  N/2 (2G + (N - 1)d = 650
  10/2 (2 x 2 + (10 - 1)d) = 650
  5(4 + 9d) = 650
  
  20 + 45d = 650
  45d/45 = 630/45
  d = 14
Area of the Rhombus = 92 Sin 60° = 70.1481
Area of sector = ⁶⁰/₃₆₀ × ²²/₇ × 92 = 42.4286
Shaded area = 70.1481 − 42.4286
Shaded area = 277195
≅ 27.72 (2dp)
2x² + 3x = 5
x² + ³/₂x + (¾)² = ⁵/₂ + (¾)²
(x+¾⁾² = ⁴⁹/₁₆
x + ¾ = ± √(⁴⁹/₁₆)
x = − ¾ ±⁷/₄
x = − 2.5 or x = 1
3x² – 4xy²+ y
9x²—y²
3x² – 3xy – xy + y²
(3x – y) (3x + y)
= (3x – y) (x – y)
(3x – y) (3x + y)
= x – y
3x + y)
Let the ratio be x:y in kg respectively.
Cost of the mixture = Sh. (42x + 467y)
Total mass of mixture = (x+y)kg
Cost per kg of the mixture = Total cost of the mixture
Total mass
Sh. 46 = 42x + 47y
x + y
46x + 46y = 42x + 47y
4x = y
^x/_y = ¼ ⇒ x:y = 1:4
Ratio 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 hr
Work done in 1 hour =⁸/₁₅ × 1 = ⁸/₁₅ of the volume
Volume still empty = 1 − ⁸/₁₅
= ⁷/₁₅Rate 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
Cos(x − 20)° = Sin (2x + 32)°
x − 20 + 2x + 32 = 90
3x + 12 = 90
3x +78
x = 26°
Tan(x − 4) = Tan (26 − 4)
= Tan 22
= 0.4040
1. 4° = 9° + 6° − 2 × 6 × 9 Cos Q
  108Cos Q = 101
  Cos Q = 0.9352
  Q = Cos−1 0.9352
  Q = 20.74°
2. ∠PSQ = 20.740
  Area of triangle PQS = ½ × 9 × 10 Sin 20.74° = 15.94cm²
HFC 0f 240, 320 and 380

HCF = 2 × 2 × 5 = 20cm
Area = 20² = 400cm²
Let the digits be x and y
x + y = 16.......(i)
original no. = 10x + y
Reversed no. = 10y + x
(10y+x) − (10x+y) = 18ii
x + y = 16
9y − 9x = 18
9y + 9x = 144
9y − 9x = 18
18x = 126
x = 7
y = 9
The no. is 79
log x + log_yP = log_QPX
log x + p log y = px log Q
Log x = px log Q − p Log y
Log x = p(x log Q − Log y)
Log x = p
x log Q − log y
Maximum area 4015 × 2.25 = 9.3375
Actual area 4.1 × 2.2 = 9.02
Minimum area 4.05 × 2.15 = 8.7075
A.e = 9.3375 − 8.7075
2
= 0.315
% error = (0.315 × 100)%
9.02
= 3.492%

SECTION B (CHOOSE 3 QUESTIONS)

The figure below represents a towel model. The base PQR is an equilateral triangle of side 9cm. The length VP=VQ=VR=20.5cm. Point M is the mid-point of PQ and VM =20cm. Point N is on the base and vertically below V.

Calculate
1. 1. Length of RM (2mks)
    - 7.794 cm
  2. Height of the model (2mks)
    - 19.83 cm
  3. Volume of the model (3mks)
    - Volme = Base area x height
      1/3 x 9 x 7.794 x 18.82 x 1/2 = 231.8177 cm³
2. The model is made up of a material whose density = 2700kg/m³. Find the mass of the model. (3mks)
  - mass = volume x density
    m = 231.8197 x 2.7 g/cm³
    = 625.91 grams or 0.62591 kgs
The table below shows distribution of marks scored in a test by standard eight pupils in a Mathematics test.

Marks

30-34

35-39

40-44

45-49

50-54

55-59

60-64

65-69

70-74

75-79

No. of pupils

1

5

10

10

19

20

8

8

4

3

Using assumed mean of mark of 57 calculate.
1. The actual Mean of the grouped data. (3mks)
  - 54.386
2. The standard deviation. (7mks)
  - 10.002
OABC is a parallegram with O(0,0), A(2,0), B(3,2) C(1,2). O¹A¹B¹C¹ is the image of OABC under transformation -2 0
0 -2
1. 1. Find the co-ordinates of O¹A¹B¹C¹ (2mks)
    - O¹= (0,0)
    - A¹ = (4,0)
    - B¹ = -6,-4)
    - C¹ = (-2, -4)
  2. On the grid provided draw OABC and O¹A¹B¹C¹ (2mks)
2. 1. Find O¹¹A¹¹B¹¹C¹¹ the image of O¹A¹B¹C¹ under the transformation matrix (10 -20). (2mks)
    - O¹¹= (0,0)
    - A¹¹ = (4,0)
    - B¹¹ = -6,-8)
    - C¹¹ = (-2, 8)
  2. Draw O¹¹A¹¹B¹¹C¹¹
3. Find a single matrix that maps O¹¹A¹¹B¹¹C¹¹ onto OABC. (3mks)
  - =(0.5 0)
    (0 0. 25)
Using trapezium rule find the area bounded by the curve (4mks)

x 0 1 2 3 4 5 6

y 1 6 9 10 9 6 1
1. Y=-X²+6X+1 X=0 and X=6
  - 1/2n (1 + 1 + 2(6 + 9 + 104 + 9 + 6 + m)
    1/2 (2 + 2(40) = 41 square units
2. Calculate the exact area under the curve. (4mks)
  - 42 square units
3. Find the percentage error introduced by the trapezium rule. (2mks)
  - 42 - 41
    42
    1/42 x 100
    = 2.38%
1. Complete the table below, giving the rules correct to 1 decimal place.
  
  X^O
  
  0
  
  40
  
  80
  
  120
  
  160
  
  200
  
  240
  
  2SIN(X+20)²
  
  0.7
  
  1.73
  
  2.0
  
  1.3
  
  00
  
  -1.3
  
  -2.0
  
  √3 COS X
  
  1.7
  
  1.3
  0.3
  0.9
  -1.627
  -1.6
  -0.9
2. Using the same scale drawing on X and Y axis draw the graphs of 2sin(x+20) and [3 cos x for 0≤x≤240.
3. Using the graphs above determine the values for X which 2sin(x+20)² =[cosX.
  - 27º and 208º
4. Determine the difference in amplitude between the two graphs.
  - 2 - 1.7 = 0.3
1. ⁴⁰/₁₀₀ × 1,350,000 = 540,000
2. Shared equally = ³⁰/₁₀₀ × 1350000 × ¹/₃ = 135000
  Ratio 112:128:210
  Trinity shared in the ratio
  ²¹⁰/₄₅₀ × ³⁰/₁₀₀ × 1,350,000 = 189000
  Total Trinity's amount
  135,000 + 189,000 = 324,000
  Bela's ratio share
  ¹¹²/₄₅₀× ³⁰/₁₀₀ × 1350000 = 100800
  Total Bela's amount
  135000 + 100800 = 235,800
  Difference = 324000 − 235800 = 88,200
3. Joan's
  135000 + ¹²⁸/₄₅₀ ×³⁰/₁₀₀× 1350000 = 250,20
  250200 × 100
  (⁶⁰/₁₀₀) × 1350000
  250,200 × 100
  810,000
  = 30⁸/₉%
2. 1. P(RR) or p(WR) or P(BR)
    (⁵/₁₂ × ⁴/₁₁) + (⁴/₁₂ × ³/₁₁) + (³/₁₂ × ²/₁₁)
    ⁵/₃₃ + ⁵/₃₃ + ⁵/₄₄ = 20+20+15
    132
    = ⁵⁵/₁₃₂ =⁵/₁₂
  2. P(RR) or p(WW) or P(BB)
    (⁵/₁₂ × ⁴/₁₁) + (⁴/₁₂ × ⁵/₁₁) + (³/₁₂ × ⁵/₁₁)
    20+12+6 = 38 = 19
    132 132 66
  3. P(RB) or p(WB) or P(BB) or P(BW) or P(BR)
    (⁵/₁₂ × ³/₁₁) + (⁴/₁₂ × ³/₁₁) + (³/₁₂ × ²/₁₁) + (³/₁₂ × ⁴/₁₁) + (³/₁₂ × ⁵/₁₁)
    15+12+6+12+15
    132
    = ⁶⁰/₁₃₂ =⁵/₁₁
1. 1cm rep 100m
2. Area 1 = ½ × 400 × 240 = 48000m²
  Area 2 = ½ × 240 × (240+720) = 115200m²
  Area 3 = ½ × 360 × 720 = 129 600m²
  Area 4 = ½ × 120 × 320= 19200m²
  Area 5 = ½ × 400 × (600+320) = 184000m²
  Area 6 = ½ × 280× (400+600) = 140000m²
  Area 7 = ½ × 200 × 400 = 40000m²
  Total area = 48000+ 115200 + 129600 + 19200 + 184000 + 140000 + 40000 = 676,000m²
  = 67.6ha