Practicing with 6th Standard Maths Question Paper with Answers Kerala Syllabus and 6th Standard Maths First Term Question Paper 2022-23 will help students prepare effectively for their upcoming exams.
Class 6 Maths First Term Question Paper 2022-23 Kerala Syllabus
Time : 2 Hours
Total Score : 60
Activity – 1
Let’s Draw & Measure
A) Write the names of all angles in the given figure.
Answer:
1) ∠ ABC
2) ∠ ACB
3) ∠ CAB
B) Draw the triangle with given measures.
Answer:
Activity – 2
Sale of Mangoes
The table given below shows the price of mangoes sold by a vendor from Monday to Friday.
| Days | Price of Mangoes (Rs) |
| Monday | 680 |
| Tuesday | 700 |
| Wednesday | 640 |
| Thursday | 660 |
| Friday | 620 |
A) Find the total price of mangoes sold.
Answer:
Total price of mangoes sold
= 680 + 700 + 640 + 660 + 620
= Rs. 3300
B) What is the average price of mangoes sold per day?
Answer:
Average price of mangoes sold per day
= \(\frac{\text { Total price }}{\text { Total no. of days }}\)
= \(\frac{3300}{5}\)
= Rs. 660
C) In which days the total price of mangoes is greater than the average?
Answer:
Monday = 680
Tuesday = 700
D) If the prices are written in ascending order, which is the middle number?
Answer:
Price in ascending order 620, 640, 660, 680, 700
Middle number is 660.
Activity – 3
Angle and time in a clock
A) Find the angle between the hands of the clock?
Answer:
Angle within a circle = 360° There are 12 equal divisions.
Angle with each division = \(\frac{360^{\circ}}{12}\) = 30°
At 10 O’clock
No. of divisions within the hands = 2
Angle between the hands = 2 × 30° = 60°
B) Write the angle between the hands when the time is 4 O’ clock.
Answer:
At 4 O’clock
No. of division between the handles = 4
Angle between the handles = 4 × 30° = 120°
C) Draw a clock with the hands showing 3 O’clock and write the angle between the hands.
Answer:
At 3 O’ clock
Equal division between the hands = 3
Angle between the hands = 3 × 30° = 90°
D) Write the angle between the hands when time is 6:30.
Answer:
At 6.30
Angle with in the clock = 360°
There are 12 equal divisions.
Angle within equal division = \(\frac{360^0}{12}\) = 30°
Each of these division again divided into 4 equal parts
Angle with small division = \(\frac{30}{4}\) = 7\(\frac{1°}{2}\)
At 6.30, there are 2 smaller divisions
∴ Angle = 2 × 7½ = 150
Activity – 4
Average Quantity of milk
The table shows the quantity of milk gained in Aji’s farm from September to December.
| Month | No. of days | Total quantity of milk (in litre) | Average quantity of milk (in litre) |
| September | 30 | 360 | |
| October | 31 | 11 | |
| November | 30 | 10 | |
| December | 31 | 403 |
A) Find the values in the missing column of corresponding months.
Answer:
| Month | No. of days | Total quantity of milk (in litre) | Average quantity of milk (in litre) |
| September | 30 | 360 | 12 |
| October | 31 | 341 | 11 |
| November | 30 | 300 | 10 |
| December | 31 | 403 | 13 |
B) In which month the production of milk is highest?
Answer:
In December the production is highest
Activity – 5
Part and Times
A) Find the area of the rectangle in the above figure.
Answer:
Area of the rectangle = Length × Breadth
Length = 15cm
Breadth 1 cm
Area = 15 × 1=15 cm²
B) If the rectangle is divided into 4 rectangles of equal area, find the area of one small rectangle.
Answer:
The rectangle is divided into 4 equal parts.
Area of each part = \(\frac{1}{4}\) of the area
= \(\frac{1}{4}\) of 15 = \(\frac{15}{4}\)
= 3\(\frac{3}{4}\) cm²
C) If 12 such small rectangles are put together, find the total area.
Answer:
No. of small rectangles part together = 12
Area of 1 small rectangle = \(\frac{15}{4}\) cm²
Total area of 12 small rectangles = 12 × \(\frac{15}{4}\)
= 3 × 15 = 45 cm²
Activity – 6
Tug of War Competition
A & B are two tug of war teams of a school with 8 members each. The average weight of one member in team A is 54 kg and that of team B is 55 kg.
A) When one member left and another one joined in team Athere is no change in average weight. Find the weight of the new member.
Answer:
No. of members = 8
Average weight of team A = 54 kg
Average weight of team B = 55 kg
Change in average after replacing one member in team A = 0 (no change)
Weight of new member = Average weight = 54 kg
B) Total weight of 8 members in a team should not exceed 435 kg. Can team A and team B participate in the competition? Why?
Answer:
Average of team A = 54 kg
No. of players = 8
Total weight of team A = 54 × 8 = 432 kg
432 kg is less than 435 kg.
∴ Team A can participate.
Average of team B = 55 kg
No. of players = 8
Total weight of Team B = 55 × 8 = 440 kg
440 kg is greater than 435.
∴ Team B is not eligible.
C) A member of weight 60 kg is excluded from the team B having more total weight. Find the maximum possible weight of the new member.
Answer:
Total weight of Team B = 440 kg
Weight of excluded member = 60 kg
Total weight of remaining 7 member
= 440 – 60
= 380 kg
Possible weight for new member = 435 – 380
= 55 kg
Activity – 7
Circle and Part
A) What part of the whole circle is shaded?
Answer:
Angle of shaded portion = 45°
Angle within a circle = 360°
Shaded portion = \(\frac{45}{360}\) part
= \(\frac{1}{8}\) part
B) What part of the whole circle is unshaded?
Answer:
Unshaded part = 1 – \(\frac{1}{8}=\frac{8}{8}-\frac{1}{8}\)
= \(\frac{7}{8}\) part
C) Draw a circle and shade \(\frac{8}{12}\) part?
Answer:
\(\frac{8}{12}\) part shaded
Activity – 8
Orange Juice
Three litres of orange juice in a vessel is equally poured in to 4 bottles.
A) How many litres of orange juice are in a bottle?
Answer:
Total quantity of orange juice = 3 litre
No. of bottles = 4
Juice in one bottle = \(\frac{1}{4}\) of 3 litre
= \(\frac{1}{4}\) × 3 = \(\frac{1}{4}\) litre
B) Orange juice in the bottle is poured equally into 3 cups. Write the quantity of orange juice in one cup.
Answer:
No. of cups = 3
Quantity of juice in one cup = \(\frac{1}{3}\) of \(\frac{3}{4}\) litre
= \(\frac{1}{3} \times \frac{3}{4}=\frac{1}{4}\) litre
C) What fraction of the orange juice in the vessel does each cup contain?
Answer:
Fraction of juice in each cup
= \(\frac{1}{3}\) × (\(\frac{1}{4}\) of 3)
= \(\left(\frac{1}{3} \times \frac{1}{4}\right)\) × 3
= \(\frac{1}{12}\) of 3 litre
Cup contains \(\frac{1}{12}\) fraction of juice in vessel.