# Abhinay Sir Maths Class Notes And Book PDF 2022

## Abhinay Sir Maths Class Notes And Book PDF 2022

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### Abhinay Sir Maths Class Notes Questions Answers

1. 121 Divided by 11 is

1. 11

2. 10

3. 19

4. 18

2. 60 Times of 8 Equals to

1. 480

2. 300

3. 250

4. 400

3. Find the Missing Term in Multiples of 6 : 6, 12, 18, 24, _, 36, 42, _ 54, 60.

1. 32, 45

2. 30, 48

3. 24, 40

4. 25, 49

4. What is the Next Prime Number after 7 ?

1. 13

2. 12

3. 14

4. 11

5. The Product of 131 × 0 × 300 × 4

1. 11

2. 0

3. 46

4. 45

6. Solve 3 + 6 × ( 5 + 4) ÷ 3 – 7

1. 11

2. 16

3. 14

4. 15

7.  Solve 23 + 3 ÷ 3

1. 24

2. 25

3. 26

4. 27

8. What is 6% Equals to

1. 0.06

2. 0.6

3. 0.006

4. 0.0006

9. How Many Years are there in a Decade?

1. 5 years

2. 10 years

3. 15 years

4. 20 years

10. How Many Months Make a Century?

1. 12

2. 120

3. 1200

4. 12000

Answer: c, 1200 months. As a century has a total of 100 years and each year has a total of 12 months, then a century will have a sum of 1200 months in it (100 × 12 = 1200 months).

### Abhinay Sir Maths Class Notes Questions

1. Which is greater than 4?

(a) 5,

(b) -5,

(c) -1/2,

(d) -25.

Solution:5 greater than 4.

2. Which is the smallest?

(a) -1,

(b) -1/2,

(c) 0,

(d) 3.

Solution:The smallest number is -1.

3. Combine terms: 12a + 26b -4b – 16a.

(a) 4a + 22b,

(b) -28a + 30b,

(c) -4a + 22b,

(d) 28a + 30b.

Solution:12a + 26b -4b – 16a.

= 12a – 16a + 26b – 4b.

= -4a + 22b.

4. Simplify: (4 – 5) – (13 – 18 + 2).

(a) -1,

(b) –2,

(c) 1,

(d) 2.

Solution:(4 – 5) – (13 – 18 + 2).

= -1-(13+2-18).

= -1-(15-18).

= -1-(-3).

= -1+3.

= 2.

5. What is |-26|?

(a) -26,

(b) 26,

(c) 0,

(d) 1

Solution:|-26|= 26.

6. Multiply: (x – 4)(x + 5)

(a) x2 + 5x – 20,

(b) x2 – 4x – 20,

(c) x2 – x – 20,

(d) x2 + x – 20.

Solution:(x – 4)(x + 5).

= x(x + 5) -4(x + 5).

= x2 + 5x – 4x – 20.

= x2 + x – 20.

7. Factor: 5x2 – 15x – 20.

(a) 5(x-4)(x+1),

(b) -2(x-4)(x+5),

(c) -5(x+4)(x-1),

(d) 5(x+4)(x+1).

Solution:5x2 – 15x – 20.

= 5(x2 – 3x – 4).

= 5(x2 – 4x + x – 4).

= 5{x(x – 4) +1(x – 4)}.

= 5(x-4)(x+1).

8. Factor: 3y(x – 3) -2(x – 3).

(a) (x – 3)(x – 3),

(b) (x – 3)2,

(c) (x – 3)(3y – 2),

(d) 3y(x – 3).

Solution:3y(x – 3) -2(x – 3).

= (x – 3)(3y – 2).

9. Solve for x: 2x – y = (3/4)x + 6.

(a) (y + 6)/5,

(b) 4(y + 6)/5,

(c) (y + 6),

(d) 4(y – 6)/5.

Solution:2x – y = (3/4)x + 6.

or, 2x – (3/4)x = y + 6.

or, (8x -3x)/4 = y + 6.

or, 5x/4 = y + 6.

or, 5x = 4(y + 6).

or, 5x = 4y + 24.

or, x = (4y + 24)/5.

Therefore, x = 4(y + 6)/5.

10. Simplify:(4x2 – 2x) – (-5x2 – 8x).

Solution:(4x2 – 2x) – (-5x2 – 8x)

= 4x2 – 2x + 5x2 + 8x.

= 4x2 + 5x2 – 2x + 8x.

= 9x2 + 6x.

= 3x(3x + 2).

11. Find the value of 3 + 2 • (8 – 3)

(a) 25,

(b) 13,

(c) 17,

(d) 24,

(e) 15.

Solution:3 + 2 • (8 – 3)

= 3 + 2 (5)

= 3 + 2 × 5

= 3 + 10

= 13

12. Rice weighing 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?

Solution:33/4 ÷ 4 pounds.

= (4 × 3 + 3)/4 ÷ 4 pounds.

= 15/4 ÷ 4 pounds.

= 15/4 × 1/4 pounds.

= 15/16 pounds.

Now we know that, 1 pound = 16 ounces.

Therefore, 15/16 pounds = 15/16 × 16 ounces.

= 15 ounces.

13. Factor: 16w3 – u4w3
Solution:16w3 – u4w3.

= w3(16 – u4).

= w3(42 – ((u2)2).

= w3(4 + u2)(4 – u2).

= w3(4 + u2)(22 – u2).

= w3(4 + u2)(2 + u)(2 – u).

Answer: w3(4 + u2)(2 + u)(2 – u).

14. Factor: 3x4y3 – 48y3.
Solution:3x4y3– 48y3.

= 3y3(x4 – 16).

= 3y3[(x2)2 – 42].

= 3y3(x2 + 4)(x2 – 4).

= 3y3(x2 + 4)(x2 – 22).

= 3y3(x2 + 4)(x + 2)(x -2).

Answer: 3y3(x2 + 4)(x + 2)(x -2)

15. What is the radius of a circle that has a circumference of 3.14 meters?

Solution:
Circumference of a circle = 2πr.

Given, circumference = 3.14 meters.

Therefore,

2πr = Circumference of a circle

or, 2πr = 3.14.

or, 2 × 3.14r = 3.14,[Putting the value of pi (π) = 3.14].

or, 6.28r = 3.14.

or, r = 3.14/6.28.

or, r = 0.5.

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