**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

- The information of the registered number of students of standard 1 to 5 is given by the graph.
- Standards are shown on the X-axis.
- The registered number of students is shown on the Y-axis.
- The scale on Y-axis is 1 cm = 5 students.
- 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):**