Kerala Syllabus Class 8 Maths Model Question Paper Set 2

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Class 8 Maths Kerala Syllabus Model Question Paper Set 2

Times: 1½ hr.
Score: 40

Instructions

• Use the first 15 minutes to read the questions and think about answers.
• There are 16 questions, split in to 4 parts A, B, C, D
• Answer all questions; but in questions of the type A or B you need answer only one of those.
• You can answer the questions in any order, writing the correct question number.
• Answers must be explained, whenever necessary.

Section – A
This section has 4 questions of score 1 each. Select the correct answer.

Question 1.
A table showing classes and frequencies is called ________
(a) Pie chart
(b) Bar graph
(c) Histogram
(d) Frequency table
Answer:
(d) Frequency table

Question 2.
Which statement is TRUE?
(a) Equal chords are at unequal distances from the centre
(b) Chords at equal distances from the centre are equal
(c) Perpendicular bisector of a chord never passes through the centre
(d) Radius is always perpendicular to the chord
Answer:
(b) Chords at equal distances from the centre are equal

Question 3.
Two statements are given below:
(i) The diagonals of all parallelograms bisects each other perpendicularly.
(ii) The diagonal of a rhombus bisects each other perpendicularly.
(a) (i) and (ii) are true
(b) Only (i) is true
(c) Only (ii) is true
(d) (i) and (ii) is not true
Answer:
(c) Only (ii) is true

Question 4.
Two sides of a triangle are equal. Small angle is 30°. What is the one of the equal angles?
(a) 100°
(b) 75°
(c) 70°
(d) 50°
Answer:
(b) 75°

Angle sum 180
Smallest is 30°. Sum of the equal angles is 150°
Equal angle is 75°

Section – B
This section has 4 questions of score 2 each.

Question 5.
Calculate \(\frac{1}{7}-\frac{3}{7}+\frac{5}{7}-\frac{7}{7}+\cdots+\frac{21}{7}\)
Answer:
\(\frac{1-3+5-7+9-11+13-15+17-19+21}{7}\)
= \(\frac{-10+21}{7}\)
= \(\frac{11}{7}\)

Question 6.
When \(\frac{2}{3}\) of a number is subtracted from a number we get 13 . What is the number?
Answer:
\(\frac{1}{3}\) of the number is 13.
Number is 3 × 13 = 39

Question 7.
The angles of a triangle are in the ratio 4:5:9.
(a) Write the angles as multiples of x.
Answer:
Angles are 4x, 5x, 9x

(b) Find the angles.
Answer:
Sum of angles =180
⇒ 4x + 5x + 9x = 180
⇒ 18x = 180
⇒ x = 10
So, angles are 40°, 50°, 90°

Question 8.
Draw the following trapezium.

Answer:

Section – C
This section has 4 questions of socre 3 each.

Question 9.
Sum of two consecutive natural numbers is 63
(a) If the smaller is x then what is the other number.
Answer:
x, x + 1 are the numbers

(b) Write the equations and find the numbers.
Answer:
2x + 1 = 63, 2x = 62, x = 31
Numbers are 31, 32

Question 10.
In a class of 52 students, number of boys is \(\frac{6}{7}\) of number of girls. How many are boys ?
Answer:
Number of boys x, number of girls 52 — x 6
x = \(\frac{6}{7}\) × (52 – x),7x = 6(52 – x)
7x = 312 – 6x, 13x = 312, x – 24
Boys 24 girls 28

Question 11.
[A] This is the section of a regular polygon.

(a) What are the angles of triangle ?
Answer:
18°, 18°, 144°

(b) What is an outer angle ?
Answer:
180 – 144 = 36°

(c) How many sides does this polygon have ?
Answer:
\(\frac{360}{36}\) = 10

OR

[B] The diagram shows an equilateral triangle, a square, and one diagonal of the square. What is the value of ?

Answer:
Equilateral triangle has one inner angle 60°
Square has inner angle 90°
A diagonal divides it into 45° each
x = 60 + 45 = 105°

Question 12.
[A] In the figure ABCD is a parallelogram. AB = 6 cm. The distance between AB and CD is 3 cm and the distance between AD and BC is 2 cm. Find the perimeter of the parallelogram.

Answer:
Area of the parallelogram ABCD = 6 × 3 = 18 cm²
Also, area – BC × 2 = 18
⇒ BC = 18 + 2 = 9 cm
AB = CD = 6 cm
AD = BC = 9 cm
∴ Perimeter = 2(6 + 9) = 30 cm

OR

[B] The area of a rhombus is 120 cm² and the length of one of its diagonals is 24 cm.
(a) Find the length of its second diagonal.
Answer:
Area =120 cm²
⇒ \(\frac{1}{2}\) × d₁ × d₂ = 120
⇒ \(\frac{1}{2}\) × 24 × d₂ = 120
⇒ d₂ = 120 ÷ 12 = 10 cm

(b) Find the perimeter.
Answer:
Length of side of the rhombus
= \(\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}\) = 13 cm
∴ Perimeter = 4 × 13 = 52 cm

Section – D
This section has 4 questions of score 4 each.

Question 13.
Check the steps given below:
\(\frac{1}{3}=\frac{1}{10} \times \frac{10}{3}\)
= \(\frac{1}{10}\left(3+\frac{1}{3}\right)\)
= \(\frac{3}{10}+\frac{1}{30}\)
\(\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

(a) Write two more steps and convert \(\frac{1}{3}\) into decimal form.
Answer:
Step 1:

Decimal form of \(\frac{1}{3}\) is 0.3333….

(b) Write the decimal form of \(\frac{1}{9}\) and \(\frac{2}{9}\)
Answer:
\(\frac{1}{9}\) = 0.111 …..
\(\frac{2}{9}\) = 0.222……

(c) Calculate \(\sqrt{0.444} \ldots\)
Answer:
\(\sqrt{0.444} \ldots\) = \(\sqrt{\frac{4}{9}}=\frac{2}{3}=\frac{6}{9}\) = 0.666 ………..

Question 14.
One side of a triangle has a length of 6 cm, and the angles at the ends of that side are 40° and 70°. Draw the triangle and construct its circumcircle.
Answer:

Question 15.
[A] (i) 11² + 10² = 121 +100 = 221 By using this , write 442 as the sum of the two perfect squares.
(ii) If the square of a number ending in 5 is of the form 110pq. Then p × q = ________.
(iii) (a) 7.6² = ________
(b) 75.4² = ________
Answer:
(i) 442 = 2 × 221
= 2(11² + 10²)
= (11 + 10)² + (11 – 10)²
442 = 21² + 1²

(ii) p × q = 2×5 = 10

(iii) (a) 7.6² = 72 + 2 × 7 × 0.6 + (0.6)²
= 49 + 8.4 + 0.36
= 57.76

(b) 75.4² = 75² + 2 × 75 × 0.4 + (0.4)²
= 5625 + 60 + 0.16
= 5685.16

OR

[B] The perimeter of a rectangle is 48 centimetres. The ratio between the length and breadth is 5:3,
(a) Length + breadth =
(b) Calcualte the length and breadth.
(c) If the breadth is increased by 1 centimetre, then find the ratio between length and breadth
Answer:
a) Length + breadth = \(\frac{48}{2}\) = 24 cm

b) Length : breadth = 5:3
∴ Length = 24 × \(\frac{5}{8}\) = 15 cm
Breadth = 24 × \(\frac{3}{8}\) = 9 cm 8

c) If the breadth is increased by 1 cm
Means
New breadth = 9 + 1 = 10 cm
∴ Length : breadth = 15:10 = 3:2

Question 16.
[A] The marks obtained by 25 students in an examination are given below. Prepare a grouped frequency table and answer the following questions, (maximum mark: 40)
25, 30, 33, 18, 36
12, 25, 30, 15, 25
19, 25, 30, 33, 20
35, 37, 29, 31, 30
31, 28, 33, 31, 30
(i) How many students scored less than 15 marks?
(ii) How many students scored 30 marks or more?
(iii) How many students scored more than 25 marks?
(iv) How many students scored between 30 and 40 marks?
Answer:

Class interval	Number of students
10-15	1
15-20	3
20-25	1
25-30	6
30-35	11
35-40	3
Total	25

i. 1 student
ii. 14 students
iii. 20 students
iv. 14 students

OR

[B] Some natural numbers can be, written as the difference of two perfect squares. For example,
8 = 4 × 2 × 1 = 3² – 1²
12 = 4 × 3 × 1 = 4² – 2²
16 = 4 × 4 × 1 = 5² – 3² ……..
(a) Write 20 as the difference of two perfect squares.
Answer:
20 = 4 × 5 × 1 = 6² – 4²

(b) Explain using algebra, the method of writng all multiples of 4, starting with 8 as the difference of two perfect squares.
Answer:
4 × x × 1 = (x + 1)² – (x – 1)²
(x + 1)² – (x – 1)²
= (x + 1 + x – 1)(x + 1 – [x – 1])
= 2x × 2 = 4 × x × 1