Practicing with 6th Standard Maths Question Paper with Answers Kerala Syllabus and 6th Standard Maths Second Term Question Paper 2022-23 will help students prepare effectively for their upcoming exams.
Class 6 Maths Second Term Question Paper 2022-23 Kerala Syllabus
Time : 2 Hours
Total Score : 60
Activity – 1
Let’s complete the table
Fill in the blanks in the table.
| Number | Product of prime | Number of factors |
| 12 | 2 × 2 × 3 | 3 × 2 = 6 |
| 80 | 2 × 2 × 2 × 2 × 5 | (i) ………. |
| (iii) …… | 2 × 2 × 3 × 5 | (ii) ……….. |
| 225 | (iv) …… | (v) ……….. |
Answer:
| Number | Product of prime | Number of factors |
| 12 | 2 × 2 × 3 | 3 × 2 = 6 |
| 80 | 2 × 2 × 2 × 2 × 5 | (i) 5 × 2 = 10 |
| (ii) 60 | 2 × 2 × 3 × 5 | (ii) 3 × 2 × 2 = 12 |
| 225 | (iv) 3 x 3 x 5 x 5 | (v) 3 × 3 = 9 |
Activity – 2
Let’s arrange
Ammu has many rectangular blocks of 5cm length, 3 cm width and 1cm height.
a) What is the volume of a rectangular block?
Answer:
Volume of the rectangular block
V = l × b × h
l = 5 cm
b = 3 cm
h = 1 cm
V = 5 × 3 × 1
= 15 cm3
b) A big box has inner length 20cm, breadth 12cm and height 4cm.
Answer:
Inner volume of the big box
V = l × b × h
l = 20 cm
b = 12 cm
h = 4 cm
V = 20 × 12 × 4
= 960 cm3
c) How many small rectangular blocks are needed to fill the big box?
Answer:
Number of small rectangular blocks
No. of blocks = \(\frac{\text { Inner volume of big block }}{\text { Volume of small block }}\)
= \(\frac{960 \mathrm{~cm}^3}{15 \mathrm{~cm}^3}\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5}{3 \times 5}\)
= 2× 2 × 2 × 2 × 2 × 2
= 64
∴ 64 small rectangular blocks are needed to fill the big box.
Activity – 3
Number Cards
Here are the number cards distributed by a teacher for a game.
a) Which is the largest number?
Answer:
15.823
b) What is the difference between the largest and the smallest numbers in these cards?
Answer:
Largest number = 15.823
Smallest number = 14.028
Difference = 15.823 – 14.028
= 1.795
c) Write these numbers in ascending order.
Answer:
14.028, 14.12, 14.623, 14.732, 15.283, 15.823
Activity – 4
How many Bangles?
Anu has two bangles of 4½ grams each.
a) What is the total weight of Anu’s bangles (in grames)?
Answer:
Weight of one bangle = 4½ gm
Number of bangles = 2
Total weight = Weight of one bangle × Number of bangles
= 4½ × 2
= 4 + \(\frac{1}{2}\)
= \(\frac{(4 \times 2)+1}{2}\)
= \(\frac{9}{2}\) × 2
= 9 grams
b) How many such bangles are needed to make it 63 grams?
Answer:
Total weight of gold = 63 gm
Weight of one bangle = 4½ g.
Number ofbangles = 63 + 4½ g.
= 63 ÷ \(\frac{9}{2}\)
= \(\frac{63}{1} \times \frac{2}{9}\)
= \(\frac{3 \times 3 \times 7}{1} \times \frac{1 \times 2}{3 \times 3}\)
= 7 × 2
= 14 bangles
Activity – 5
A Garden
The picture of a rectangular shape garden is given below.
a) Find the perimeter of the garden.
Answer:
Perimeter of the garden
= 2 (Length + Breadth)
Length = 5.75 m,
Breadth = 3.65 m
Perimeter = 2 (5.75 + 3.65)
= 2(9.40)
= 18.8 m
b) What is the difference between the length and the breadth of the garden?
Answer:
Length = 5.75 m
Breadth = 3.65 m
Difference = 5.75 – 3.65
= 2.10 m
c) Find out the fraction that shows the breadth of the garden?
a) 3\(\frac{65}{10}\)
b) 3\(\frac{65}{100}\)
c) 3\(\frac{100}{65}\)
d) 3\(\frac{65}{1000}\)
Answer:
Breath = 3.65 = 3\(\frac{65}{100}\)
Option b) 3\(\frac{65}{100}\)
Activity – 6
Let’s Find the Capacity
The length, breadth and height of a cubical vessel is 10 cm each.
a) What is the capacity of this vessel in cubic centimetres? Convert this capacity in litres.
Answer:
Capacity of the Vessel = Volume of the vessel
Length = Breadth = Height = 10 cm
Inner volume of cubic vessel = (side) × (side) × (side)
= 10 × 10 × 10
= 1000 cm3
1 l = 1000 cm3
Capacity of cube vessel = 1 l
b) Another vessel of the same capacity has length 20 cm and breadth 10 cm. What is its height?
Answer:
Capacity of rectangular vessel = 1 l
= 1000 cm3
Length = 20 cm,
Breadth = 10 cm
Height = \(\frac{\text { Volume }}{\text { Length } \times \text { Breadth }}\)
= \(\frac{1000}{20 \times 10}\)
= 5 cm
c) If the measurements of all the sides of a cube are doubled. How many times the volume will be?
Answer:
The measurements of sides are doubled.
One side = 10 × 2 = 20 cm
Volume = 20 × 20 × 20
= 8 × 1000 cm3
Capacity increased by 8 times
Activity – 7
Study Room
Ayisha’s study room is rectangular in shape. It has 4.75 meter length and 3.84 meter breadth.
a) What is the area of the study room in square metres?
Answer:
Length of study room = 4.75 m
Breadth = 3.84 m
Area = Length × Breadth
= 4.75 × 3.84
= 18.24 m2 2
b) The cost of tiling one square metre of floor is Rs. 350. What will be the total cost for tiling the floor.
Answer:
Cost for tiling per square metre
= Rs.350
Total cost for tiling the floor
= 18.24 × 350
= Rs. 6384
c) 184 × 525 = 96600. What is 18.4 × 525?
Answer:
18.4 × 525 = Rs. 9660.00
Activity – 8
Which is Next?
Some numbers and their factors are given below.
Fill in the blanks
a) Numbers with two factors.
Answer:
b) Numbers with three factors.
Answer:
c) Numbers with four factors.
Answer:
d) Factorize and write all the factors of 216.
Answer:
216 = 2 × 2 × 2 × 3 × 3 × 3
Factors of 216 are :
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216