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:
