Practicing with 6th Standard Maths Question Paper with Answers Kerala Syllabus and 6th Standard Maths Annual Exam Question Paper 2023-24 will help students prepare effectively for their upcoming exams.
Class 6 Maths Annual Exam Question Paper 2023-24 Kerala Syllabus
Time : 2 Hours
Total Score : 60
Activity – 1
Angle of the shaded portion in the figure is 40°.
A) What is the angle of the unshaded portion?
Answer:
360° – 40° = 320°
B) What part of the circle is shaded?
Answer:
\(\frac{40}{360}=\frac{1}{9}\)
C) What part of the circle is unshaded?
Answer:
\(\frac{320}{360}=\frac{8}{9}\)
or
1 – \(\frac{1}{9}=\frac{8}{9}\)
Activity – 2
In the figure ∠DOC is twice of ∠BOC.
A) What is the measure of DOC?
Answer:
∠DOC = 2 × ∠ BOC
= 2 × 36°
= 72°
B) What is the measure of ∠AOE?
Answer:
∠AOE = ∠BOD [Opposite angles are equal]
= 72 + 36
= 108°
C) Find the measure of ∠AOD.
Answer:
∠AOD + ∠BOD = 180° [Linear pair]
∠AOD = 180 – ∠BOD
= 180 – 108
= 72°
D) Name any two linear pairs in the figure.
Answer:
∠BOD, ∠AOD
∠BOE, ∠AOE
Activity – 3
The total number of students in Manachadikkunnu UP school in 2021 -22 was 900.60% them were girls and the rest were boys.
A) What percentage of the total are boys?
Answer:
100 – 60 = 40%
B) What is the number of girls?
Answer:
Total number of students = 900
Percentage of girls = 60%
Number of girls = 900 × \(\frac{60}{100}\) = 540
C) The number of students increased by 20% in 2022 – 23 than in the previous year. What is the total number of students in 2022-23?
Answer:
Total number of students in 2022-23
= \(\frac{120}{100}\) 900
= 1080
×
D) On 2023-24 the number of children decreased by 10% than the previous year. What is the percentage of the present number of children compared to the previous year?
Answer:
Percentage of students in 2023 – 24 = 90%
Activity – 4
Write the fraction in the table as fractions with denominators 10,100,1000 and write its decimal form.
Answer:
Activity – 5
A) Which of the following numbers has odd numbers of factors?
(18, 8, 16, 13)
Answer:
16 [∵ Square numbers have odd number of factors]
B) What is the number of factors of 12?
Answer:
12 = 2 × 2 × 3
= 2² × 3
∴ Number of factors
= (2 + 1) (1 + 1)
= 3 × 2
= 6
C) Write 20 as the product of prime numbers?
Answer:
20 = 2 × 2 × 50
D) Write 60 as product of prime factors and write all its factors.
Answer:
60 = 2 × 2 × 3 × 5
∴ Factors = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Activity – 6
It was decided to construct a house for a classmate in Seena’s school given below is the bar diagram showing the amount received from the UP section for this purpose.
A) Which class collected the least amount?
Answer:
5 A
B) Which are the classes that raised equal amount?
Answer:
6 A and 7 B
C) What is the total amount obtained from 5 A and 5B classes?
Answer:
Amount raised by 5 A = 4500
Amount raised by 5 B = 5000
Total amount raised by them together
= 4500 + 5000 = 9500
D) How much more amount is received from 7B than 6B?
Answer:
Amount received from 7 B = 7000
Amount received from 6 B = 6500
More amount received = 7000 – 6500 = 500
E) Prepare a question based on the diagram.
Answer:
Which class collected more amount?
Activity – 7
The table below shows the information about the amount brought by some students to invest in School Savings Account.
| Name |
20 Rs (Numbers) |
10 Rs (Numbers) |
Total Amount |
| Neenu |
2 |
5 |
(2 × 20) + (5 × 10)= 90 |
| Meenu |
3 |
2 |
(3 × 20) + (2 × 10) = 80 |
| Ramla |
4 |
4 |
……………….. |
| Meera |
1 |
8 |
……………….. |
A) Complete the table.
Answer:
| Name |
20 rupees (Numbers) |
10 rupees (Numbers) |
Total Amount |
| Neenu |
2 |
5 |
(2 × 20) + (5 × 10) = 90 |
| Meenu |
3 |
2 |
(3 × 20) + (2 × 10) = 80 |
| Ramla |
4 |
4 |
(4 × 20) + (4 × 10) = 120 |
| Meera |
1 |
8 |
(1 × 20) + (8 × 10) = 100 |
B) If the number of 20 rupee notes are denoted by ‘a’ the number of 10 rupee notes by ‘b’ and the total amount is denoted by‘t’ then how can you write the relation among them?
Answer:
20 rupee notes amounts to 20a rupees
10 rupee notes amounts to 10b rupees
∴ 20 a + 10 b = t
Activity – 8
A) Find the volume of the box shown in the figure.
Answer:
Volume = l × b × h
= 80 × 40 × 10
= 32000 cm3
B) What is the maximum number of boxes of length 10 centimeters, breadth 4 centimetres and height 2 centimetres that can be stacked in the this box?
Answer:
Volume of the small box with length 10 cm, breadth 4 cm and height 2cm = 10 × 4 × 2
= 80 cm3
Number of small boxes that can be stacked = \(\frac{32000}{80}\) = 400
C) If the length, breadth and height of the box in the picture are doubled, what will be the volume? How many times this volume will be the volume ofthe box given in the picture?
Answer:
If length, breadth and height of the box doubled,
l = 160 cm, b = 80 cm, h = 20 cm
Volume = 160 × 80 × 20
= 256000 cm3
The volume becomes \(\frac{256000}{32000}\) = 8 times
