Exemplar Questions
Mathematics EOYE 2022
Class 8
Topic: Introduction to Set Theory
1. If S = {1, 2, 3} are the elements of set S then;
A. 1 belongs to set S
B. 3 does not belong to set S
C. 2 does not belong to set S
(1 mark)
Solution: A [1 belongs to set S]
1, 2 and 3 are the elements of set S and therefore A is the correct answer.
The statements in options B and C are false.
2.
Write down the elements of the following descriptive set in numerical form: (1 mark)
a) set of all integers between 10 and 15
Solution: {11, 12, 13, 14}
Integers between 10 and 15 are: 11, 12, 13 and 14.
3.
30 students were asked if they had a bicycle (B), a mobile phone (M) and a computer
(C). The results are shown in the Venn diagram:
a) Find the value of x
(1 mark)
b) Shade M ∩ C on the Venn Diagram below:
(1 mark)
c) Shade B U C on the Venn Diagram below:
(1 mark)
d) List the number of students who own a bicycle, a mobile phone and a computer
(1 mark)
Solution:
a) Total number of students = 30
X = 30 – [2 + 4 + 1 + 7 + 6 + 3 + 2] = 30 – 25
X=5
b) M ∩ C
Intersection of sets is the set of elements, which are common to both the given sets. In set
theory, for any two sets A and B, the intersection is defined as the set of all the elements in
set A that are also present in set B. We use the symbol '∩' that denotes 'intersection of'
The overlapping region of two circles represents the intersection of the two sets.
Therefore, the overlapping region of circles M and C represents the intersection of the M
and C
c) B U C
The union of two sets A and B is defined as the set of all the elements, which lie in set A
and set B, or both the elements in A and B altogether. The union of the set is denoted by
the symbol '∪'.
The entire region of two circles represents the union of the two sets. Therefore, the entire
region of circles B and C represents the union of the B and C
d) 7 students
The number of students who own a bicycle, a mobile phone and a computer is indicated by
the shaded region in the Venn diagram below:
Therefore, the respective number of students are 7
Topic: Graphical Representation of Simultaneous Linear Inequalities
1. The graph below represents which of the following linear inequality:
A. Y < 3
B. Y > 3
C. Y ≥ 3
(1 mark)
Solution: B [Y > 3]
The area above the line Y = 3 is shaded in the graph above.
In addition, because the line is dotted, it means that the line Y = 3 is not included in the
shaded region. Therefore, the shaded region represents the inequality Y > 3
2.
Shade the region satisfies x ≤ -1 in the graph below:
(2 marks)
Solution:
The shaded region on the graph below indicates the inequality x ≤ -1
All of the area to the left of the line x = -1 is to be shaded.
In addition to this, the line x = -1 has to be solid because it is included in the shaded region.
[1 mark for shading the correct region. 1 mark for making the line x = -1 solid]
Topic: Area and Volume (Last 3 ATs)
1. The volume of this rectangular prism is:
A. 24 cm3
B. 60 cm
C. 20 cm3
D. 240 cm3
(1 mark)
Solution: D [240 cm3]
Volume of rectangular prism or cuboid = L x W x H = 10 x 6 x 4 = 240 cm3
2.
The volume of a cube of length 4 cm is:
A. 8 cm3
B. 64 cm3
C. 36 cm3
D. 16 cm3
(1 mark)
Solution: B [64 cm3]
Volume of cube = L x L x L = 4 x 4 x 4 = 64 cm3
3.
The height and radius of a closed cylinder are 23 cm and 5 cm. Find its volume and
total surface area.
(6 marks)
Solution: (2 marks for correct volume formula and entering values correctly. 1 mark for
correct volume answer) (2 marks for correct surface area formula and entering values
correctly. 1 mark for correct surface area answer)
Volume of cylinder = πr2h= 3.142 x 52 x 23 = 1806.65 cm3
Total Surface Area of cylinder = 2πrh + 2πr2 = (2 x 3.142 x 5 x 23) + (2 x 3.142 x 52) = 879.76
cm2
4.
Find the surface area of the given cuboid.
(3 marks)
Solution: (2 marks for correct formula and entering values correctly. 1 mark for correct
answer)
Surface area of cuboid = 2(L x W) + 2(L x H) + 2(W x H) = 2(7 x 4) + 2(7 x 3) + 2(4 x 3) = 122
cm2
5.
Calculate the area of the shaded region
(5 marks)
Solution: (2 marks for correctly calculating the area of triangle. 2 marks for correctly
calculating the area of circle. 1 mark for correctly calculating the area of the shaded region)
Area of Circle = π r2 = 3.142 x 22 = 12.57 m2
Area of Triangle = ½ x b x h = ½ x 6 x 6 = 18 m2
Area of shaded region = 18 – 12.57 = 5.43 m2
6.
Calculate the perimeter of the given shape:
(5 marks)
Solution: (1 mark for correctly calculating the value of the unknown side. 3 marks for
correctly adding the lengths of all sides to find the total perimeter)
Length of unknown side = 12 – 3 – 3 = 6 cm
Perimeter of the given shape = 10 + 12 + 10 + 3 + 8 + 3 + 8 + 6 = 60 cm
7.
Find the volume of the given triangular prism
(3 marks)
Solution: (2 marks for correct formula and entering values correctly. 1 mark for correct
answer)
1
Volume of triangular prism = (𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎)(ℎ𝑒𝑖𝑔ℎ𝑡) = (2 𝑥 3 𝑥 4) (15) = 90 cm3
8. Find the perimeter of the following semi-circle:
(Take π = 3.142)
(3 marks)
Solution: (2 marks for correct formula and entering values correctly. 1 mark for correct
answer)
Perimeter of semi-circle = (𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑠𝑒𝑚𝑖 − 𝑐𝑖𝑟𝑐𝑙𝑒) + (𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑠𝑒𝑚𝑖 − 𝑐𝑖𝑟𝑐𝑙𝑒)
1
= (2 𝑥 2 𝑥 𝜋 𝑥 𝑟) + (10)
1
= (2 𝑥 2 𝑥 3.142 𝑥 5) + (10)
= (15.71) + (10)
= 25.71 𝑐𝑚
Topic: Probability
1. Write down the sample space when a six-sided dice is rolled
(3 marks)
Solution: (1 mark for two correct numbers. For example, if a writes 2 and 3 only, he/she will
get 1 mark. If a student writes 3, 4 and 5, he/she will get 1.5 marks. If a student writes 1, 2,
3, 4, 5 and 6, he/she will get full marks)
Sample Space - A sample space is a collection or a set of ALL possible outcomes of an
experiment
Therefore, the sample space for the roll of a six-sided die is: 1, 2, 3, 4, 5 and 6
2. A fair six-sided die is rolled once. Find the probability of:
(6 marks – 2 marks each)
a) Getting a 4
b) Getting an even number
c) Getting a prime number
Solution:
a) 1/6
The sample space for the roll of a six-sided die is: 1, 2, 3, 4, 5 and 6
Number of all Possible outcomes is 6
Number of Target outcomes is 1 (The number: 4)
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐓𝐚𝐫𝐠𝐞𝐭 𝐨𝐮𝐭𝐜𝐨𝐦𝐞
So, Probability of getting a 4 is: 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐥𝐥 𝐏𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 = 1/6
[1 mark to be deducted if either of number of favorable outcomes or number of all possible
outcomes is incorrect]
b) 3/6 or ½
The sample space for the roll of a six-sided die is: 1, 2, 3, 4, 5 and 6
Number of all Possible outcomes is 6
Number of Target outcomes is 3 (Even numbers: 2, 4 and 6)
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐓𝐚𝐫𝐠𝐞𝐭 𝐨𝐮𝐭𝐜𝐨𝐦𝐞
So, Probability of getting an even number is: 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐥𝐥 𝐏𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 = 3/6 = ½
[1 mark to be deducted if either of number of favorable outcomes or number of all possible
outcomes is incorrect]
c) 3/6 or ½
The sample space for the roll of a six-sided die is: 1, 2, 3, 4, 5 and 6
Number of all Possible outcomes is 6
Number of Target outcomes is 3 (Prime numbers: 2, 3 and 5)
𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐓𝐚𝐫𝐠𝐞𝐭 𝐨𝐮𝐭𝐜𝐨𝐦𝐞
So, Probability of getting a prime number is: 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐚𝐥𝐥 𝐏𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐨𝐮𝐭𝐜𝐨𝐦𝐞𝐬 = 3/6 = ½
[1 mark to be deducted if either of number of favorable outcomes or number of all possible
outcomes is incorrect]
Topic: Coordinate Geometry
1. Find the gradient and the y-intercept of the line given by the following equation:
(2 marks)
𝑦 = 3 − 2𝑥
Solution:
In the equation y = mx + c the value of m is called the slope, (or gradient), of the line and
the value of c is called the y-intercept of the line
So, re-arranging the above equation in the standard form:
𝑦 = −2𝑥 + 3
m = -2 (Gradient)
c = 3 (y-intercept)
2. A is the point with coordinates (4, 11). B is the point with coordinates (8, 3).
a) Work out the coordinates of the midpoint of AB.
(3 marks)
b) Calculate the gradient of the line AB
(3 marks)
Solution:
a) Mid-point is of a line can be calculated by using the formula below:
𝑀=
𝑀=
𝑥1 + 𝑥2 𝑦1 + 𝑦2
+
2
2
4 + 8 11 + 3
+
2
2
𝑀=
12 14
+
2
2
𝑴 = (𝟔, 𝟕)
(2 marks for correct formula and entering values correctly. 1 mark for correct answer)
b) Gradient of a line can be calculated by using the formula below:
𝑚=
𝑚=
𝑦2 − 𝑦1
𝑥2 − 𝑥2
3 − 11
8−4
𝑚=
−8
4
𝒎 = −𝟐
(2 marks for correct formula and entering values correctly. 1 mark for correct answer)
Topic: Transformation
1. Draw the reflection of Polygons in the given line, on the graphs below:
a) X-axis
(3 marks)
b) Y-axis
(3 marks)
Solution:
a) When you reflect a point across the x-axis, the x-coordinate remains the same, but
the sign of the y-coordinate reverses. The reflection of point (x, y) across the x-axis
is (x, -y).
(1 mark for each of the three points correctly reflected)
b) When you reflect a point across the y-axis, the y-coordinate remains the same,
but the sign of the x-coordinate reverses. The reflection of point (x, y) across the
x-axis is (-x, y). The point at the y-axis remains the same when reflected across
y-axis
(1 mark for each of the three points correctly reflected)