**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):**

y^{8} × y^{5} ÷ y^{10} (y ≠ 0)

= y^{8 + 5} ÷ y^{10}

= y^{13} ÷ y^{10}

= y^{13 – 10}

= y^{3}

**Solution 4(2):**

c^{8} ÷ c^{5} × c^{4}

= c^{8 – 5 }× c^{4}

= c^{3} × c^{4}

= c^{3 + 4}

= c^{7}

**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):**

4m^{2} – 3[m – 2{3m – (m^{2} – 5)}]

= 4m^{2} – 3[m – 2{3m – m^{2} + 5}]

= 4m^{2} – 3[m – 6m + 2m^{2} – 10]

= 4m^{2} – 3[-5m + 2m^{2} – 10]

= 4m^{2} + 15m – 6m^{2} + 30

= -2m^{2} + 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):**

2^{5} × (2^{3})^{2} ÷ 2^{10}

= 2^{5} × 2^{3 × 2} ÷ 2^{10}

= 2^{5} × 2^{6} ÷ 2^{10}

= 2^{5 + 6} ÷ 2^{10}

= 2^{11} ÷ 2^{10}

= 2^{11-10}

= 2^{1}

= 2

**Solution 8(2):**