Kerala Syllabus Class 8 Maths Model Question Paper Set 5

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Class 8 Maths Kerala Syllabus Model Question Paper Set 5

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 J each. Select the correct answer.

Question 1.
305² = 93025
(30.5)² = ………….
(a) 93.025
(b) 9.3025
(c) 930.25
(d) 9302.5
Answer:
(c) 930.25

Question 2.
One third of a number is 10 less than the number. Which of the following is the number?
(a) 30
(b) 15
(c) 10
(d) 27
Answer:
\(\frac{2}{3}\) of the number is 10
Number is \(\frac{3}{2}\) times 10.
Number = 10 × \(\frac{3}{2}\) = 15

Question 3.
Read the given statements.
Statement I: A mixture of milk and water has a ratio of 3:1. If 5 liters of water are added, the new ratio of milk to water is 3:2. The quantity of milk in the mixture is 15 liters.
Statement II: The addition of water does not change the quantity of milk in the mixture.
A) Statement I is true, Statement II is false
B) Statement I is false, Statement II is true
C) Statement I is true, Statement II is the reason for Statement I
D) Statement I is true, Statement II is not the reason for Statement I
Answer:
C) Statement I is true, Statement II is the reason for Statement I

Milk = 3x
Water = x
If 5 litres of water are added,
Milk = 3x
Water = x + 5
The new ratio of milk to water is 3:2.
That is milk is \(\frac{3}{2}\) of water.
⇒ 3x = \(\frac{3}{2}\) × (x + 5)
⇒ 6x = 3x + 15
⇒ 3x = 15
⇒ x = 5
Milk = 3 × 5 = 15 litres
So, Statement I is true.
Statement II says that adding water to a mixture of milk and water does not change the quantity of milk. This is a fundamental concept of mixture problems where only one component is added or removed. The amount of the other component remains constant. Therefore, Statement II is true.
Hence, Statement I is true, Statement II is the reason for Statement I

Question 4.
If x = 0.333 …, then x² + x =
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{9}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{9}\)
Answer:
(b) \(\frac{4}{9}\)

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

Question 5.
In the figure AB = AC, BD = CD
(a) Name two equal triangles in the figure
(b) If ∠BAC = 40°, ∠ABD = 30° then what are the angles of triangle ADB ?

Answer:
(a) AB = AC, BD = CD, AD is common
Triangle ABD and triangle ADC are equal.

(b) In triangle ADB, ∠A = 20°, ∠B = 30°, ∠D = 130°

Question 6.
A polygon has inner angle sum is two times the outer angle sum.
(a) What is its inner angle sum ?
Answer:
2 × 360 = 720°

(b) Find the number of sides of the polygon
Answer:
180 × (n – 2) = 720
n – 2 = 4, n = 6

Question 7.
A chord that is 8 cm long lies 3 cm away from the centre of a circle. Calculate the diameter of the circle.
Answer:
The perpendicular drawn from the centre to the chord bisects the chord.

OB² = 3² + 4² = 25
OB = 5 cm
Therefore the diameter = 2 × 5 = 10 cm

Question 8.
In the figure ABCD is a parallelogram. The area of the parallelogram is 36 cm². The distance between parallel sides is 3 cm. Find the length of the side AB.

Answer:
Area = 36 cm²
3 × base = 36
So, base = 36 ÷ 3 = 12 cm
∴ AB = 12 cm

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

Question 9.
Given below is an unfinished magic square made up of positive numbers and negative numbers. The sum of the numbers in rows, columns and diagonals are equal.

Find x + y?
Answer:
Consider the number that can be written in the top right side of the column be A.
—7 + 6 + A = x + y + A
⇒ -7 + 6 = x + y
x + y = 6 – 7 = – 1

Question 10.
Draw a square with perimeter is 23 centimeter.
Answer:
Draw a line of length 11.5 centimeter.
Draw the perpendicular bisector of the line. Mark the half length of the line on the perpendicular bisector. Complete the square.
One side = \(\frac{11.5}{2}\) cm, Perimeter = 4 x \(\frac{11.5}{2}\) = 23cm

Question 11.
[A] The perimeter of a rectangle is 28 cm. The ratio between length and breadth is 5:2.
(a) What is the sum of length and breadth?
(b) Find length and breadth.
(c) Find the area of the rectangle.
Answer:
Length = 5x
Breadth = 2x

(a) Perimeter = 28
⇒ 2(length + breadth) = 28
⇒ length + breadth = 14

(b) 5x + 2x = 14
⇒ 7x = 14
⇒ x = 2
Length = 10 cm Breadth = 4 cm

(c) Area = 10 × 4 = 40 sq.cm

OR

[B] Sides of a triangular plot are in the ratio 2:4:6. Perimeter of the plot is 360 metres. Find the length of the sides?
Answer:
Let the sides be 2x, 4x, 6x.
Perimeter = 360 metres
2x + 4x + 6x = 360
⇒ 12x = 360
⇒ x = 30
So, the length of the sides are:
2 × 30 = 60 metre
4 × 30 = 120 metre
6 × 30 = 180 metre

Question 12.
[A] Find the value of (1\(\frac{1}{2}\))² + (2\(\frac{1}{2}\))² + (3\(\frac{1}{2}\))² + (4\(\frac{1}{2}\))²
Answer:

OR

[B] (a) 3² + 2² = 9 + 4 = 13. By using this , write 26 as the sum of the two perfect squares
Answer:
26 = 2 × 13
= 2 (3² + 2²)
= (3 + 2)² + (3 – 2)²
26 = 5² + 1²

(b) (99\(\frac{1}{2}\))² + (100\(\frac{1}{2}\))² = _______________
Answer:
= 99 × 100 + \(\frac{1}{4}\) + 100 × 101 + \(\frac{1}{4}\)
= 9900 + \(\frac{1}{4}\) + 10100 + \(\frac{1}{4}\)
= 20000\(\frac{1}{2}\)
= 20000.5

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

Question 13.
In the figure D is the midpoint of BC, DL is perpendicular to AB, DM is perpendicular to AC, DL = DM

(a) Prove that triangle BLD and triangle CMD have same side and shape.
Answer:
BD = CD = x, DL = DM = y
Triangle BLD and triangle CMD are right triangles
BD² = BL² + DL²,
CD² = DM² + CM²
Since D is the mid point, BD = CD
BL² + DL² = DM² + CM²,
BL² = CM² ⇒ BL = CM
Three sides of triangle BLD are equal to three sides of triangle CMD. So these triangles are equal

(b) Is ∠B = ∠C?
Answer:
Sides opposite to equal angles are equal, zB = zC

(c) If AB = 12 cm then what is AC?
Answer:
Since ∠B = ∠C we can write AB = AC
⇒ AB = AC = 12 cm

Question 14.
The measure of an outer angle of a regular polygon is 2x, and the measure of an inner angle is 4x.
(a) Use the relationship between inner and outer angles to find x.
Answer:
Sum of the inner angle and outer angle at; corner is 180°
6x = 180, x = 30

(b) Find the measure of one inner and outer angle.
Answer:
Angles are 60°, 120°

(c) Find the number of sides in the polygon and the type of polygon.
Answer:
Regular hexagon

Question 15.
[A] Check the steps given below:

(a) Write two more steps and express \(\frac{1}{9}\) in decimal form.
Answer:

(b) Write the decimal form of \(\frac{1}{3}\) and \(\frac{2}{9}\).
Answer:
\(\frac{1}{3}=\frac{3}{9}\) = 0.333 ………..
\(\frac{2}{9}\) = 0.222 ………..

(c) Write the decimal 0.999… in whole number form.
Answer:
0.999.. = 1

OR

[B] Look at the pattern given below:
1 + 2 + 3 = 2 × 3 = 6
1 + 2 + 3 + 4 + 5 = 3 × 5 = 15
1 + 2 + 3 + 4 + 5 + 6 + 7 = 4 × 7 = 28
(a) Write the sum of the first 9 natural numbers?
Answer:
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
= 5 × 9
= 45

(b) Write the sum of the first 25 natural numbers?
Answer:
13 × 25 = 325

(c) What is – 1 – 2 – 3 …. – 49 ?
Answer:
-(1 + 2 + 3 + .. .+49) = -(25 × 49)
= -1225

Question 16.
[A] The height of students in a class is given below. Prepare a frequency table and draw a histogram. (Heights in cm)

Answer:

Height (cm)	Number of students
148-152	6
152-156	7
156-160	6
160-164	4
164-168	3
168-172	4
Total	30

OR

[B] Detail of rainfall in June and July are given in the table below. Draw a histogram.

Rain fall (mm)	Days
15-25	3
25-35	6
35-45	9
45 – 55	12
55-65	14
65-75	3
75 – 85	2
85 – 95	1
95-105	9

Answer: