Practicing with 6th Standard Maths Question Paper with Answers Kerala Syllabus and 6th Standard Maths First Term Model Question Paper will help students prepare effectively for their upcoming exams.
Class 6 Maths First Term Model Question Paper Kerala Syllabus
Time : 2 Hours
Total Score : 60
Activity – I
a) Using set squares make 15° angle.
Answer:
∠BAD = 45°
∠BAC = 30°
∠CAD = 15°
b) Name the angle measuring device.
What is the measure of right angle.
Answer:
Protractor
The measure of right angle is 90°
c) Make a quadrilateral using the given measurements.
Answer:
Activity – II
a) The average age of 30 students of a class is 9 years. If teacher’s age is also added, average is increased by 1. What is the age of teacher ?
Answer:
Average age of 30 students = 9
Total age of 30 students = 9 × 30 = 270
Average after teacher’s age is added = 9 + 1 = 10
Total age for 31 people = 10 × 31 = 310
Age of teacher = 310 – 270 = 40 yrs
b) How can we calculate the average of a group of values?
Answer:
Average of a group of values = \(\frac{\text { Sum of the values }}{\text { Number of values }}\)
c) The given table indicates the library books distributed to class VI students. Complete the table.
Answer:
| Class | No. of students | No. of books | Average |
| 6A | 30 | 120 | 4 |
| 6B | 50 | 150 | 3 |
| 6C | 30 | 210 | 7 |
| 6D | 25 | 75 | 3 |
Activity – III
a) 1) \(\frac{1}{3}\) of \(\frac{1}{5}\)
2) 1 millilitre = ____ litre
3) 2 times of \(\frac{1}{3}\)
4) \(\frac{13}{5}\) =
5) 4\(\frac{1}{8}\) = ____
Answer:
1) \(\frac{1}{5} \times \frac{1}{3}=\frac{1}{15}\)
2) 1 millilitre = \(\frac{1}{1000}\) litre
3) \(\frac{1}{3}\) × 2 = \(\frac{2}{3}\)
4) \(\frac{13}{5}\) = 2 \(\frac{3}{5}\)
5) 4\(\frac{1}{8}\) = \(\frac{33}{8}\)
i) What is the area of the figure ?
ii) It is divided into two. Then what is the area of one part ?
iii) One part is divided into 3 pieces. Find the area of one piece.
Answer:
1) Area of the figure = Length × Breadth
Length = 15 cm
Breadth = 1 cm
Area = 15 × 1 = 15 cm²
ii) When divided into two
Area of one part = \(\frac{15}{2}\) = 7\(\frac{1}{2}\) cm²
iii) When one part is divided into three
Area of one piece = 7\(\frac{1}{2}\) × \(\frac{1}{3}\)
= \(\frac{15}{2} \times \frac{1}{3}\)
= \(\frac{15}{6}\) = \(\frac{5}{2}\)
= 2\(\frac{1}{2}\) cm²
Activity – IV
a) What is the volume of a pillar of 4 m length, 1 m breadth and 25 cm height ?
Answer:
Volume of pillar = l × b × h
= 4 × 1 × \(\frac{1}{4}\) = 1 m3
b) 1 cubic metre = ______ litre
1 litre = ____ cubic centimeter
Answer:
1 m3 = 1000 l
1 l = 1000 cm3
c) What is the formula for volume calculation?
Answer:
Volume l capaicty of a container
= length × breadth × height
= l × b × h
Activity – V
a) What is the angle between the hands of a clock showing 10 O’clock. Explain with figure?
Answer:
Angle within a circle = 360°
There are 12 equal divisions.
Angle within a division = \(\frac{360^{\circ}}{12}\) =30°
Two divisions are there.
∴ Angle between the hands = 2 × 30 = 60°
b) How many angles are there in the figure? Name them.
Answer:
There are 9 angles.
∠APC, ∠CPB, ∠BPD, ∠DPE, ∠EPA, ∠APD, ∠EPB, ∠EPC, ∠DPC
c) Categorize them under right angle, angle less than 90°, angle greater than 90°.
Answer:
| Right Angle | Angle Less than 90° | Angle greater than 90° |
| ∠APE | ∠EPD | ∠APD |
| ∠EPB | ∠BPC | ∠CPD |
| ∠BPD | ∠APC | |
| ∠EPC |
Activity – VI
\(\frac{1}{5}\) portion of a courtyard is converted into a play ground. Half of the rest is changed into a car shed. How much put in the shed out of the whole courtyard. Remaining part is divided and \(\frac{1}{3}\) is used for garden. The remaining is how much part out of the whole?
Answer:
Area of play ground = \(\frac{1}{5}\)th
Remaining portion = 1 – \(\frac{1}{5}\)
= \(\frac{5-1}{5}\)
= \(\frac{4}{5}\)
Area of part of car shed = \(\frac{1}{2}\) of \(\frac{4}{5}\)
= \(\frac{4}{5} \times \frac{1}{2}\)
= \(\frac{2}{5}\)
Remaining portion = 1 – \(\left(\frac{1}{5}+\frac{2}{5}\right)\)
= 1 – \(\frac{3}{5}=\frac{2}{5}\)
Garden = \(\frac{1}{3}\) of \(\frac{2}{5}=\frac{2}{15}\)
Remaining portion = 1 – \(\left(\frac{1}{5}+\frac{2}{5}+\frac{2}{15}\right)\)
= 1 – \(\left(\frac{3}{15}+\frac{6}{15}+\frac{2}{15}\right)\)
= 1 – \(\frac{11}{15}=\frac{4}{15}\)
Activity – VII
The three division of 6th standard having equal number of students. Totally 60 students for the 6th standard. The average body weight is given below.
| A | B | C |
| 46 | 44 | 48 |
a) What is the number of students in each division?
Answer:
Total number of students = 60
Number of students in each division = \(\frac{60}{3}\) = 20
b) A new student is added to each division then the average of A remains the same, B increased by 1 and C decreased by 1. What is the weight of new comer ?
Answer:
i) Average weight of students in A division = 46 kg
Average after the arrival of new student
Weight of new student = 46 kg = 46 kg
ii) Average of B division = 44 kg
Total weight = 44 × 20 = 880 kg
Average after the arrival of new student
Total weight = 45 kg = 45 × 21
Weight of new student = 945 kg
= 945 – 880
= 65 kg
iii) Average of C division = 65 kg = 48 kg
Total weight = 48 × 20
= 960 kg
Average after the arrival of new student
= 47 kg
Total weight of 21 students = 47 × 21
= 987 kg
Weight of new student = 987 – 960
= 27 kg
Activity – VIII
a) A rectangular shaped container having inner side measures 20 cm, 10 cm and 5 cm can hold how much of water ?
Answer:
Length = 20 cm
Breadth = 10 cm
Height = 5 cm
Volume of container = 20 × 10 × 5
= 1000 cm3
b) This water is transferred to a square shaped container which is completely filled, then what is the measure of the side of the container ?
Answer:
Volume of square container = 1000 cm3
One side of the container = 10 cm
c) If the water is filled in another container of length 20 cm and breadth 15 cm at what height is the water ?
Answer:
Length of new container = 20 cm
Breadth = 15 cm
Height of water level = \(\frac{\text { Volume }}{\text { length } \times \text { breadth }}\)