**GSEB Solutions for Class 6 Mathematics – Revision 2 (Sem – I)**

GSEB SolutionsMathsScience

**Revision 2**

**Solution 1(A):**

**Solution 1(B):**

(1) Given, C.P. = Rs. 300, S.P. = Rs. 350

S.P. > C.P.

Hence, there is a profit

Profit = S.P. – C.P. = Rs. 350 – Rs. 300 = Rs. 50

Thus, there is a profit of Rs. 50.

(2) Value of 2 × 2 × 3 × 3 × 3 × 3 = 4 × 81 = 324.

Exponent form of 2 × 2 × 3 × 3 × 3 × 3 = 22 × 34

(3) Length of the carrom board = 110 cm.

A Carrom board is always in the shape of a square.

Hence, perimeter of square = 4 × length

Perimeter of the carrom board = (4 × 110) cm = 440 cm

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

The given polynomials can be classified into monomial, binomial and trinomial as follows:

**Solution 6:**

**Solution 7:**

(1)y2 – xy + y

= (y × y) – (x × y) + y

= (3 × 3) – (1 × 3) + 3

= 9 – 3 + 3

= 9

(2) x3 + ay + 10

= (x × x × x) + (a × y) + 10

= (1 × 1 × 1) – (2 × 3) + 10

= 1 + 6 + 10

= 17

(3) 2×2 – x + 2

= 2(x × x) – x + 2

= 2(1 × 1) – 1 + 2

= 2 – 1 + 2

= 3

(4) x3 + y2

= (x × x × x) + (y × y)

= (1 × 1 × 1) + (3 × 3)

= 1 + 9

= 10

(5) 3×4 – ax3+ 5x – 3

= 3(x × x × x × x) – a (x × x × x) + 5x – 3

= 3(1 × 1 × 1 × 1) – 2 ((1 × 1 × 1) + (5 × 1) -3

= 3 – 2 + 5 – 3

= 8 – 5

= 3

**Solution 8:**

The three lines perpendicular to line l can be shown symbolically as:

**Solution 9:**

**Solution 10:**

**Solution 11:**

Length of the ground = 20 m…. (Given)

Breadth of the ground = 15 m…. (Given)

Area of the rectangle = length × breadth

∴ Area of the rectangular ground = 20 × 15 = 300 sq. m

Cost to prepare 1 sq m lawn in the ground = Rs. 5

∴ Cost to prepare 300 sq. m lawn = (5 × 300) = Rs. 1500

Thus, the cost of preparing lawn in the ground is Rs. 1500.

**Solution 12:**