QUESTIONS
- Use logarithms to 4 decimal places to evaluate: (4 marks)
(0.7841 x √0.1356)1/3
Log 84.92 - A globe representing the earth has a radius of 0.5m. point A(0º, 10ºW), B (0º, 35ºE), P(60ºN, 110ºE) and Q(60ºN, 120ºW) are marked on the globe.
Find the length of arc AB, leaving your answer in term of (3mks) - A circle centre is the point C(2,3) passes through a point P(a,b). A point M(-2 , -5/2) is the mid-point of the line CP.
- Calculate the coordinates of P. (1mk)
- Determine the equation of the circle in the form x2 +y2 +ax + by + c = 0 (3mks)
- Make a the subject of the formula:
x = y +√(x2+ a2) (3marks) - Given that Sin (2/3 x+ 20º ) - Cos (5/6x + 10º) = 0. Without using a mathematical table or a calculator, determine tan (x+ 20º). (3 marks)
- Two fair dice one a regular tetrahedron (4 faces) and the other a cube are thrown. The scores are added together. Complete the table below to show all possible outcomes. (2 mark)
Find the probability that:- The sum is 6. (1 mark)
- The sum is 6 or 9. (2 marks)
- A particle moves along a straight line such that its displacement s metres from a given point is s = t3 – 5t2 + 3t + 4 where t is time in seconds. Find:
- The displacement of the particle at t = 8. ( 2 marks )
- The velocity of the particle when t = 10. (3 marks )
- A classroom measures (x + 2) m by (x – 5)m. If the area of the classroom is 60m2.
Find its length. (3 marks )
SECTION B
- Lengths of 100 mango leaves from a certain mango tree were measured t the nearest centimeter and recorded as per the table below,
Length in cm No. of leaves
10 to 12 3
13 to 15 16
16 to 18 36
19 to 21 31
22 to 24 14- On the grid provided draw a cumulative frequency graph to represent this data. (5mks)
- Use your graph to estimate
- The median length of the leaves (2mks)
- The number of leaves whose lengths lie between 13cm and 17cm. (3mks)
-
- Draw ΔPQR whose vertices are P (1, 1), Q (-3, 2) and R (0, 3) on the grid provided.(1mk)
Find and draw the image of ΔPQR under the transformation whose matrix (31 01) is and label the image P¹Q¹R¹. (2 marks) - P¹Q¹R¹ is then transformed into P¹¹Q¹¹R¹¹ by the transformation with the matrix (-11 03) .Find the co-ordinates of P¹¹Q¹¹R¹¹ and draw P¹¹Q¹¹R¹¹. (3 marks)
- Describe fully the single transformation which maps PQR onto P¹¹Q¹¹R¹¹ find the matrix of this transformation. (3 marks)
- Describe fully the single transformation which maps PQR onto P¹¹Q¹¹R¹¹ find the matrix of this transformation. (3 marks)
- Draw ΔPQR whose vertices are P (1, 1), Q (-3, 2) and R (0, 3) on the grid provided.(1mk)
MARKING SCHEME
No Log 0.7841
0.1356½
Log 84.92 = 1.929
0.53101.8944
1.1323 = 1.5662
2 14606
1.2853
1.1753
3/3 + 2.1753 = 1.7251
3
1.7251- Difference in longitude = 10 + 35 = 45º
length of arc AB = 45/360 x 2 x π x 0.5
= 1/8π = 0.125π -
- 2 + a = 2 = 2 + a = -4
2
a = -6
3 + b = -2.5 = 3 + b = -5
2
b = -8 - (a, b) = (-6,8)
r = √(2 --6)2 + (3--8)2 = √185
(x-2)2 + (y-3)2 = 185
x2 - 4x + 4 + y2 - 6y + 9 = 185
x2 + y2 - 4x - 6y - 172 = 0
- 2 + a = 2 = 2 + a = -4
- x = y + √(x2+ a2)
x – y = √(x2+ a2)
(x – y)2 x2 + a2
(x – y)2 x2 = a2
± √((x-y+x)(x-y-x) = a
± √(2x-y)(-y) = a
± √(y2- 2xy) = a - (2/3 x+20)+5/6 x+10º=90
9/6 x+30=90
9/6 x=60º
x = 40º
=Tan (x+20)
=Tan 60 M1 - CUBE
1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 - P(6) = 4/24 = 1/6
- P(6 or 9) = ¼
P(6) = 4/24
P(9) = 2/24
4/24 + 2/24
6/24 = 1/4
-
- s = 83 – 5 x 82 + 3 x 8 + 4
512 – 320 + 24 + 4
= 220m - V = ds/dt = 3t2 – 10t + 3
= 3 x 102 – 10 x 10 + 3
300 – 100 + 3 = 197m
- s = 83 – 5 x 82 + 3 x 8 + 4
- (x + 2) (x – 5 ) = 60
x2 – 5x + 2x – 10 = 60
x2 – 3x – 70 = 0
x2 – 10x + 7x – 70 = 0
x (x – 10) + 7(x – 10) = 0
(x – 10) (x +7) = 0
x = 10
x = -7
Length 10 + 2 = 12m - Upper class limits 12.5, 15.5, 18.5, 21.5 24.5
Cumulative frequency 3, 19 55, 86,100 -
- Median = 50th 18.2 ± 0.15
- Leaves below 13 =4 leaves
leaves below 17 =35 leaves
leaves between 13 and 17 = 35-4=31
-
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