GSEB Solutions for Class 6 Mathematics – Revision 1

GSEB Solutions for Class 6 Mathematics – Revision 1 (Sem – I)

GSEB SolutionsMathsScience
Revision – 1

Solution 1:

Registered number of students of standard 1 to 5

1.  The information of the registered number of students of standard 1 to 5 is given by the graph.
2. Standards are shown on the X-axis.
3. The registered number of students is shown on the Y-axis.
4. The scale on Y-axis is 1 cm = 5 students.
5. There are maximum students in standard 2.

Solution 2:

Weight of 6 students in my class

Solution 3:

The trees found in my school garden and surroundings:

Solution 4:

(1)1 sq. m = 10,000 sq. cm
∴ 8 sq. m = 80,000 sq.cm
(2) 10,000 sq cm = 1 sq. m
(3) Perimeter of a square = 4× length
Perimeter of the given square = 4× 10 cm = 40 cm.
Perimeter of a square, whose length of the side 10 cm is 40 cm.
(4) Perimeter of a rectangle = 2 (length + breadth)
Perimeter of the given rectangle = 2 (6 + 4) = 2 (10) = 20 m
Perimeter of a rectangle whose length is 6 m and breadth is 4 m is 20 m.
(5) 0.18 = 18%

Solution 5:

Solution 6(1):

Let us first find the area of the floor of the classroom first.
Area of a rectangle = length × breadth
∴ Area of the rectangular classroom = 10 m × 8 m = 80 sq.m
The cost of polishing 1 sq.m tiles = Rs.12
∴ The cost of polishing 80 sq.m tiles = (12 × 80) = Rs.960
Thus, the cost of polishing the floor of the classroom is Rs.960.

Solution 6(2):

Let us first find the area of the field.
Area of a square= length × length
∴ Area of the square field = 80 m × 80 m = 6400 sq.m
The cost of preparing after tiling of 1 sq m field = Rs.3
The cost of preparing after tiling field = (3 × 6400) = Rs.19200
Thus, the cost of preparing after tiling the field is Rs.19200.

Solution 6(3):

Solution 6(4):

Solution 6(5):