GSEB Solutions for Class 7 Mathematics – Revision 2 (Sem – II)

GSEB Solutions for Class 7 Mathematics – Revision 2 (Sem – II) (English Medium)

GSEB SolutionsMathsScience
Revision 2

Solution 1:

Solution 2:

Solution 3:

1. [2 × 3{4 – 1(4 – 2)}]
= [2 × 3{4 – 1(2)}]
= [2 × 3{4 – 2}]
= [2 × 3{2}]
= [2 × 6]
= 12
2. [5 + 3{4 × 2(5 – 2)}]
= [5 + 3{4 × 2(3)}]
= [5 + 3{4 × 6}]
= [5 + 3{24} ]
= 5 + 72
= 77
3. [4 × 3{3 – 1(2 + 5)}]
= [4 × 3{3 – 1(7)}]
= [4 × 3{3 – 7}]
= [4 × 3{-4}]
= [2 × (- 12)]
= -24
4. [2 + 5{3 – 1(2 × 5)}]
[2 + 5{3 – 1(10)}]
= [2 + 5{3 – 10}]
= [2 + 5{-7}]
= [2 + (-35)]
= [2 – 35]
= -33
5. [3 × 2{5 – 1(4 + 2)}]
= [3 × 2{5 – 1(6)}]
= [3 × 2{5 – 6}]
= [2 × 3{-1}]
= [2 × (-3)]
= -6

Solution 4(1):

y⁸ × y⁵ ÷ y¹⁰ (y ≠ 0)
= y^{8 + 5} ÷ y¹⁰
= y¹³ ÷ y¹⁰
= y^{13 – 10}
= y³

Solution 4(2):

c⁸ ÷ c⁵ × c⁴
= c^{8 – 5} × c⁴
= c³ × c⁴
= c^{3 + 4}
= c⁷

Solution 4(3):

Solution 4(4):

x – [x – {5 – (2x + 3)} + 15]
= x – [x – {5 – 2x – 3} + 15]
= x – [x – 5 + 2x + 3 + 15]
= x – [3x + 13]
= x – 3x – 13
= -2x – 13

Solution 4(5):

a – 2[a – 3{a – (5 – 2a)} – 10]
= a – 2[a – 3{a – 5 + 2a} – 10]
= a – 2[a – 3{3a – 5} – 10]
= a – 2[a – 9a + 15 – 10]
= a – 2[-8a + 5]
= a + 16a – 10
= 17a – 10

Solution 4(6):

4m² – 3[m – 2{3m – (m² – 5)}]
= 4m² – 3[m – 2{3m – m² + 5}]
= 4m² – 3[m – 6m + 2m² – 10]
= 4m² – 3[-5m + 2m² – 10]
= 4m² + 15m – 6m² + 30
= -2m² + 15m + 30

Solution 4(7):

4y – [y – 2(y – 8) + 7]
= 4y – [y – 2y + 16 + 7]
= 4y – [-y + 23]
= 4y + y – 23
= 5y – 23

Solution 5A(1):

Solution 5A(2):

Solution 5A(3):

Solution 5A(4):

Solution 5A(5):

Solution 5A(6):

Solution 5A(7):

Solution 5A(8):

Solution 5A(9):

Solution 5B(1):

Solution 5B(2):

Solution 5B(3):

Solution 5B(4):

Solution 5B(5):

Solution 5B(6):

Solution 6(2):

Solution 6(3):

Let x be the breadth of the rectangle.
Hence, its length = 3x
Perimeter of rectangle = 2(length + breadth)
According to the given information, we have
2(x + 3x) = 32 cm
∴ 2x + 6x = 32 cm
∴ 8x = 32 cm
∴ x = 4
Hence, breadth of the rectangle = 4 cm
Length of the rectangle = 3x = 3 × 4 = 12 cm

Solution 6(4):

Let Magan have Rs. x.
Then, amount of money with Raman = Rs. (x + 25)
According to the given information, we have
x + (x + 25) = 125
∴ x + x + 25 = 125
∴ 2x = 125 – 25
∴ 2x = 100
∴ x = 50
Hence, amount of money with Magan = Rs. 50
And, amount of money with Raman = Rs. (50 + 25) = Rs. 75

Solution 6(5):

Let x be the smallest number.
Hence, the three consecutive numbers would be
x, x + 1 and x + 2.
According to the given information, we have
x + (x + 1) + (x + 2) = 42
∴ x + x + 1 + x + 2 = 42
∴ 3x + 3 = 42
∴ 3x = 42 – 3
∴ 3x = 39
∴ x = 13
∴ x + 1 = 13 + 1 = 14
∴ x + 2 = 13 + 2 = 15
Hence, the three consecutive numbers are 13, 14 and 15.

Solution 6(6):

Let the present age of Naresh be x years
Then, age of Dashrath will be (x + 2) years
Now, sum of their ages is 56.
x + (x + 2) = 56
∴ x + x + 2 = 56
∴ 2x + 2 = 56
∴ 2x = 56 – 2
∴ 2x = 54
∴ x = 27
∴ x + 2 = 27 + 2 = 29
Thus, the present age of Naresh is 27 years and that of Dashrath is 29 years.

Solution 6(7):

Length of a cubical box = 4 cm
Volume of cube = length × length× length
= 4 × 4 × 4
= 64 cubic cm
Hence, volume of the cube is 64 cubic cm.

Solution 6(8):

Length of a cube = 1.5 m
Now, 1 m = 100 cm
∴ 1.5 metre = 150 cm
Volume of cube = length × length× length
= 150 × 150 × 150
= 33,75,000 cubic cm
Hence, the volume of the cube is 33,75,000 cubic cm.

Solution 6(9):

Solution 6(10):

Solution 6(11):

Length of the cuboidal tank = 3 m
Breadth of the cuboidal tank = 2 m
Height of the cuboidal tank = 2 m
Volume of a cuboid = Length × Breadth × Height
∴ Volume of the cuboidal tank
= (3 × 2 × 2) m3
= 12 m3
Now, 1 m3 = 1000 litres
∴ 12 m3 = (12 × 1000) litres = 12000 litres
Hence, 12,000 litres of water can be stored in the tank.

Solution 7(1):

Solution 7(2):

Solution 7(3):

Solution 7(4):

Solution 8(1):

2⁵ × (2³)² ÷ 2¹⁰
= 2⁵ × 2^{3 × 2} ÷ 2¹⁰
= 2⁵ × 2⁶ ÷ 2¹⁰
= 2^{5 + 6} ÷ 2¹⁰
= 2¹¹ ÷ 2¹⁰
= 2^11-10
= 2¹
= 2

Solution 8(2):