# Mathematics notes for class 10 (KPK)

These notes are according to the syllabus of KPK text book. Other board notes will also be uploaded time to time.

## Mathematics Class 10 Chapters included:

### Mathematics Class 10 Notes (KPK) Chapter # 1

1. The name Quadratic comes from
O Dratic
O Both a & b
O None of these

Explanation:

O Cube
O Cubed root
O Square
O Square root

Explanation:

3. An equation of degree is called quadratic equation.
O 1
O 2
O 3
O 4

Explanation:
The name Quadratic comes from “quad” means square because the highest power of the variable is 2

4. In quadratic equation a x^2+bx+c=0 ,
O a=0
O a \neq 0
O Both a & b
O None of these

Explanation:
a=0 then it becomes linear equation.

5. An equation of degree 2 is called equation.
O Linear
O Cubic
O All of these

Explanation:
a x^2+bx+c=0 is called General or Standard form of Quadratic equation.

6. An equation of degree 1 is called ______ equation.
O Linear
O Cubic
O All of these

Explanation:

7. In quadratic equation a x^2+b c+c=0, \ when \ a=0 then it becomes
O Linear
O Cubic
O All of these

Explanation:
a x^2+b c+c=0
when a=0
0x^2+b c+c=0
b c+c=0
This is linear equation.

8. All those values of the variable for which the given equation is true are called
O Solutions
O Roots
O Both a & b
O None of these

Explanation:

9. The maximum number of roots of quadratic equation are
O One
O Two
O Three
O All of these

Explanation:
The name Quadratic comes from “quad” means square because the highest power of the variable is 2.

10. Which of the following is not a quadratic equation?
O x^2+3 x+9=0
O x^2-16=0
O 9+3 x+x^2=0
O x^2+3 x^3+9=0

Explanation:
For Quadratic equation, the highest power of the variable is 2.

11. There are _______ basic methods to solve Quadratic equation.
O 1
O 2
O 3
O 5

Explanation:
1. Factorization
2. Completing square

12. In factorization method, a quadratic equation can be solved by _________ it in factors.
O Combine
O Separate
O Splitting
O All of these

Explanation:
In factorization method, the middle term of a quadratic equation can be Splitted.

13. To solve quadratic equation, the equation must have in ________ form of quadratic equation.
O Any
O Standard
O Linear form
O All of these

Explanation:
The Standard form of quadratic equation is:
a x^2+b x+c=0

14. In factorization method, the ________ term will be split of a x^2+b x+c=0
O a
O b
O c
O All of these

Explanation:
In factorization, the middle term should be split.

15. In factorization method of a x^2+b x+c=0
O We find the product of a \ (coefficient \ of \ x^2 ) \ and \ c \ (constant \ term) \ i.e. \ a c
O Find two numbers b_1 \ and \ b_2 such that b_1 \pm b_2=b \ and \ also \ b_1 b_2=a \cdot c
O a x^2+b_1 x+b_2 x+c=0 can be factorized into two limear factors.
O All of these

Explanation:
These all are the steps to solve Quadratic equation by Factorization.

16. In factorization method, put all the terms on one side and _____ on other side.
O 0
O 1
O B
O c

Explanation:
because Equate each factor to zero by zero – product property.

17. In factorization method, equate each factor to ______ by zero- product property.
O 0
O 1
O Constant
O Variable

Explanation:
In factorization method, put all the terms on one side and 0 on other side.

18. In factorization method, equate each factor to zero by ________ property.
O Zero-product
O Both a & b
O None of these

Explanation:
In factorization method, put all the terms on one side and 0 on other side.

19. In zero-product, if a b=0 , then either a=0 \ or \ b ______ 0
O =
O \neq
O Both a & b
O None of these

Explanation:
if a b=0 , it means at least one must be zero either a or b.

20. The solution of p^2+p-6=0 is
O p=2,-3
O p=-2,-3
O p=2,3
O p=-2,3

Explanation:
p^2+p-6=0
p^2-2p+3p-6=0
p(p-2)+3(p-2)=0
(p-2)(p+3)=0
p-2=0 \ or \ p+3=0
p=2 \ or \ p=-3

21. Quadratic equation which cannot be solved by factorization, then it will be solved by
O Completing square
O Both a & b
O None of these

Explanation:
By Completing Square & Quadratic Formula, we can solve almost every Quadratic Equation.

22. In completing square method, the co-efficient of x^2 should be
O 0
O 1
O Other than 1
O All of these

Explanation:
It is the first rule to solve Qudratic Equation by Completing Square.

23. In completing square method of a x^2+b x+c=0
O Divide all terms by the co-efficient of x^2 if other than 1
O Shift the constant term to the right side of the equation.
O Multiply the co-efficient of x \ with \ \frac{1}{2} then take Square of it and Add to B.S
O All of these

Explanation:
These all are the steps to solve Quadratic equation by Completing Square.

24. To solve x^2-8 x+9=0 by completing square, then it becomes
O (x-4)^2=7
O \sqrt{x-4}=7
O Both a & b
O None of these

Explanation:
x^2-8 x+9=0
x^2-8 x=-9
x^2-8 x+(4)^2=-9+(4)^2
(x)^2-2(x)(4)+(4)^2=-9+16
(x-4)^2=7

25. By ________ we can solve all types of quadratic equations.
O Factorization method
O Both a & b
O None of these

Explanation:

26. To solve a x^2+b x+c=0 by completing square, we get
O Factors
O None of these

Explanation:
Quadratic Formula is derived by Completing Square

O x=\frac{-b }{2 a} \pm\sqrt{b^2-4 a c}
O x=\frac{-b \pm \sqrt{b^2}}{2 a}-4 a c
O x=-b \pm \frac{\sqrt{b^2-4 a c}}{2 a}
O x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}

Answer: x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}

28. To apply Quadratic formula to 3 x^2-6 x+2=0 then
O a=3, b=6, c=-2
O a=3, b=6, c=2
O a=3, b=-6, c=-2
O a=3, b=-6, c=2

Explanation:
Compare 3 x^2-6 x+2=0 with
a x^2+b x+c=0

29. The solution set of 4 x^2+12 x=0 is
O Solution \ Set =\{3\}
O { S.S }=\{0,3\}
O { S.S }=\{0,-3\}
O { S.S }=\{-3\}

Explanation:
4 x^2+12 x=0
4x(x+3)=0
4x=0 \ or \ x+3=0
x=\frac{0}{4} \ or \ x=-3
x=0 \ or \ x=-3
S.S=\{0, -3\}

30. By ________ we can solve all quadratic equations.
O Factorization method
O Both a & b
O None of these

Explanation:

31. The solution set of x^2+5 x+4=0 is
O Solution \ Set =\{-1\}
O { S.S }=\{1,4\}
O { S.S }=\{-1,-4\}
O { S.S }=\{-4\}

Explanation:
x^2+5 x+4=0
x^2+1x+4x+4=0
x(x+1)+4(x+1)=0
(x+1)(x+4)=0
x+1=0 \ or \ x+4=0
x=-1 \ or \ x=-4
S.S=\{-1, -4\}

32. The solution set of (x-3)^2=4 is
O Solution \ Set =\{1\}
O S.S=\{1,5\}
O S.S=\{-1,-4\}
O S.S =\{-4\}

Explanation:
(x-3)^2=4
Taking Square root on B.S
\sqrt{(x-3)^2}=\pm \sqrt{4}
x-3=\pm 2
x-3=2 \ or \ x-3=-2
x=2+3 \ or \ x=-2+3
x=5 \ or \ x=1
S.S=\{5, 1\}

33. What must be added to x^2+5 x to obtain a perfect square?
O \left(\frac{5}{2}\right)^2
O \frac{5}{2}
O 5
O 2

Explanation:
Multiply the co – efficient of x \ with \ \frac{1}{2} then take Square of it and Add to B.S.

34. What must be added to q^2-4 q to obtain a perfect square?
O (2)^2
O \frac{5}{2}
O 5
O 2

Explanation:
Multiply the co – efficient of x \ with \ \frac{1}{2} then take Square of it and Add to B.S.
4 \times \frac{1}{2}=2
(2)^2

1. Polynomial of degree four is called
O Both a & b
O None of these

Explanation:

2. The equation in the form of a x^4+b x^2+c=0 is called
O Both a & b
O None of these

Explanation:
Here the highest power is 4, that is why it is called Biquadratic

O One
O Two
O Three
O Four

Explanation:
Here the highest power is 4, so there will be 4 roots of Biquadratic equation.

4. The equation a x^4+b x^2+c=0 has solutions.
O One
O Two
O Three
O Four

Explanation:
Here the highest power is 4, so there will be 4 roots of Biquadratic equation.

5. To solve a x^4+b x^2+c=0
O a\left(x^2\right)^2+b x^2+c=0
O Substitute y=x^2
O All of these

Explanation:
These all are the steps of to solve the above equation.

6. The equation a x^4+b x^2+c=0 can be solved by reducing it into
O Both a & b
O None of these

Explanation:
It is to easy to solve Biquadratic equation by reducing it into Quadratic equation.

7. In substitutional it must remember to go back and express the answers in terms of______ the variable.
O New
O Original
O Both a & b
O None of these

Explanation:

8. To solve a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+c=0, we substitute
O x+\frac{1}{x}=y
O x^2+\frac{1}{x^2}=y^2-2
O Both a & b
O None of these

Explanation:
Let x+\frac{1}{x}=y
Taking square root on B.S
\left(x+\frac{1}{x} \right)^2=y^2
x^2+\frac{1}{x^2}+2=y^2
x^2+\frac{1}{x^2}=y^2-2

9. Exponential involving the term a^x is called equations.
O Exponential
O All of these

Explanation:

10. For exponential equation, a^x it must be noted that
O a>0
O a \neq 1
O a=2
O Both a & b

Explanation:
rules to represent the exponential equation.

11. In equation, 4.2^{2 x}-10.2^x+4=0 , substitute
O 2^{2 x}=y
O 2^x=y
O x=y
O All of these

Explanation:
It is the simplest way to convert exponential equation to Quadratic equation.

12. If 2^x=2^3 , then
O 2=2
O x \neq 3
O x=3
O All of these

Explanation:
Here the bases are same and it is called One-to-One Property of Exponential Functions.

13. If b^n=b^m , then
O b=m
O n \neq m
O n=m
O All of these

Explanation:
Here the bases are same and it is called One-to-One Property of Exponential Functions.

14. If b^n=b^m , then n=m is called ________ of exponential functions.
O One-to-one property
O Zero-zeroproperty
O None of these

Explanation:
Here the bases are same and it is called One-to-One Property of Exponential Functions.

15. To solve 4x^2-10 x+4=0 , first we
O Solve by factorization
O Taking 2 common
O Both a & b
O None of these

Explanation:
To make the equation in the simplest way.

16. To solve 4.2^{2 x}-10.2^x+4=0
O 4 .\left(2^x\right)^2-10.2^x+4=0
O 2^x=y
O kAll of these

Explanation:
These all are the steps to solve Exponential equation.

1. An equation in which the variable appear in one or more radicands is called a equation.
O Linear
O All of these

Explanation:

2. In \sqrt{x+2}, \ x+2 is
O All of these

Explanation:

3. \sqrt{x+2}=3 is _______ equation.
O Linear
O All of these

Explanation:

4. Square root is finished by
O Formula
O Squaring
O None of these

Explanation:

5. The solution satisfies the original radical equation is called
O Solution set
O Extraneous
O Squaring

Explanation:

6. The solution that does not satisfy the original radical equation is called
O Solution set
O Extraneous
O Squaring

Explanation:

7. (\sqrt{x+2})^2=
O x^2+4
O x^2+4+2(x)(2)
O x^2
O x+2

Explanation:
(\sqrt{x+2})^2= x+2

O x^2+4+2(x)(2)
O x^2+4+4 x
O x^2+4
O Both a & b

Explanation:
(x+2)^2
=x^2+4+2(x)(2)
=x^2+4+4 x

9. (\sqrt{x+2}+\sqrt{x+7})^2=
O x+2+x+7
O x+2+x+7+\sqrt{(x+2)(x+7)}
O 2 x+9+2 \sqrt{(x+2)(x+7)}
O (\sqrt{(x+2)(x+7)})^2

2 x+9+2 \sqrt{(x+2)(x+7)}
Explanation:
(\sqrt{x+2}+\sqrt{x+7})^2
=(\sqrt{x+2}+\sqrt{x+7})^2
=(\sqrt{x+2})^2+(\sqrt{x+7})^2+2\sqrt{x+2}\sqrt{x+7}
=x+2+x+7+2\sqrt{(x+2)(x+7)}
=2x+9+2\sqrt{(x+2)(x+7)}

O Identical (same)
O All
O Different
O Squaring

Explanation:

11. \sqrt{9}+\sqrt{16}=
O 3+4
O 7
O Both a & b
O None of these

Explanation:
\sqrt{9}+\sqrt{16}
=3+4
=7

12. \sqrt{9+16}=
O \sqrt{25}
O 5
O Both a & b
O None of these

Explanation:
\sqrt{9+16}
=\sqrt{25}
=5

13. \sqrt{9}+\sqrt{16}=
O \sqrt{9+16}
O 7
O Both a & b
O None of these

Explanation:
\sqrt{9}+\sqrt{16}
=3+4
=7

14. \sqrt{9}+\sqrt{16} _______ \sqrt{9+16}
O Equal to
O Not equal to
O None of these

Explanation:
\sqrt{9}+\sqrt{16}
=3+4
=7
And
\sqrt{9+16}
=\sqrt{25}
=5

15. If x^2=9 \ then \ x=
O 3
O -3
O Both a & b
O None of these

Explanation:
x^2=9
\sqrt{x^2}=\pm \sqrt{9}
x=\pm 3
x=3 \ or \ x=-3

16. If x^2=11 \ then \ x=
O \pm \sqrt{11}
O \sqrt{11}
O -\sqrt{11}
O None of these

Explanation:
x^2=11
\sqrt{x^2}=\pm \sqrt{11}

(i). If (x+1)(x-5)=0 then the solutions are
O x=1, -5
O x=1, 5
O x=-1, -5
O x=-1, 5
x=-1, 5
Explanation:
(x+1)(x-5)=0
x+1=o or x-5=0
x=-1 or x=5
Thus x=-1, 5

(ii). if x^2-x-1=0 , then x=
O \frac{-1 \pm \sqrt{5}}{2}
O -1 \pm \frac{\sqrt{5}}{2}
O \frac{1 \pm \sqrt{5}}{2}
O 1 \pm \frac{\sqrt{5}}{2}
\frac{1 \pm \sqrt{5}}{2}
Explanation:
x^2-x-1=0
a=1, b=-1, c=-1
We have
x= \frac{-b \pm \sqrt{b^2-4ac}}{2}
x= \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2}
x= \frac{1 \pm \sqrt{1+4}}{2}
x= \frac{1 \pm \sqrt{5}}{2}

(iii). \frac{-1 \pm \sqrt{5}}{2} in simplified form is
O 1 \pm \sqrt{24}
O 1 \pm \sqrt{6}
O 2 \pm \sqrt{6}
O cannot be simplified
cannot be Simplified
Explanation:

(iv). To apply the quadratic formula to 2x^2-x=3
a=2, b=-1, c=3
a=2, b=1, c=3
a=2, b=-1, c=-3
a=2, b=-1, c=0
a=2, b=-1, c=-3
Explanation:
2x^2-x=3
2x^2-x-3=0
Compare the equation with
ax^2+bx+c=0
a=2, b=-1, c=-3

(v). If x^2-3x-4=0 , then the solutions are
O x=4, -1
O x=-4, 1
O x=4, 1
O x=-4, -1
x=4, -1
Explanation:
x^2-3x-4=0
x^2-4x+1x-4=0
x(x-4)+1(x-4)=0
(x-4)(x+1)=0
x-4=0 or x+1=0
x=4 or x=-1

(vi). If 2x^2+4x-9=0 , then solutions are
O x=\frac{2 \pm \sqrt{22}}{2}
O x=\frac{-2 \pm \sqrt{22}}{2}
O x=2 \pm \frac{\sqrt{22}}{2}
O x=-2 \pm \frac{\sqrt{22}}{2}
x= \frac{-2 \pm \sqrt{22}}{2}
Explanation:
2x^2+4x-9=0
a=2, b=4, c=-9
We have
x= \frac{-b \pm \sqrt{b^2-4ac}}{2}
x= \frac{-4 \pm \sqrt{(4)^2-4(2)(-9)}}{(2)(2)}
x= \frac{-4 \pm \sqrt{16+72}}{4}
x= \frac{-4 \pm \sqrt{88}}{4}
x= \frac{-4 \pm \sqrt{4 \times 22}}{4}
x= \frac{-4 \pm 2\sqrt{22}}{4}
x= \frac{2(-2 \pm \sqrt{22})}{4}
x= \frac{-2 \pm \sqrt{22}}{2}

(vii). x^2 – \frac{1}{4}=0 , then solution are:
O x= \pm \frac{1}{2}
O x= \pm \frac{1}{4}
O x= \pm \frac{1}{8}
O x= \pm \frac{1}{16}
x =\pm \frac{1}{2}
Explanation:

x^2 – \frac{1}{4}=0
x^2 = \frac{1}{4}
\sqrt {x^2} =\pm \sqrt{ \frac{1}{4}}
x =\pm \frac{1}{2}

(viii). What are the solutions of the equation x^2+7x-18=0 ?
O 2 or -9
O -2 or 9
O -2 or -9
O 2 or 9
2 or -9
Explanation:
x^2+7x-18=0
x^2-2x+9x-18=0
x(x-2)+9(x-2)=0
(x-2)(x+9)=0
x-2=0 or x+9=0
x=2 or x=-9

(ix). Which of the following values of x are the roots of the equation x^2-8x+15=0 ?
O x=1 or x=-7
O x=2 or x=4
O x=-2 or x=4
O x=3 or x=5
x=3 or x=5
Explanation:

x^2-8x+15=0
x^2-3x-5x+15=0
x(x-3)-5(x-3)=0
(x-3)(x-5)=0
x-3=0 or x-5=0
x=3 or x=5

### Mathematics Class 10 Notes (KPK) Chapter # 2

1. In quadratic formula, the expression b^{2}-4 a c is called ________ of quadratic equation.
(a) Factorization
(b) Functions
(c) Discriminant
(d) Irrational

Discriminant

2. The value of the discriminant is used to determine the number of solutions of a ________ equation.
(a) Linear
(c) Simultaneous
(d) None of these

3. If b^{2}-4 a c ________ then the roots are real, equal and rational.
(a) =0
(b) <0
(c) >0
(d) None of these

=0

4. If b^{2}-4 a c ________ then the roots are unequal and imaginary.
(a) =0
(b) <0
(c) >0
(d) None of these

<0

5. If b^{2}-4 a c ________ and root is a perfect square, then roots are real, unequal and rational.
(a) =0
(b) <0
(c) >0
(d) None of these

6. If b^{2}-4 a c ________ and root is not a perfect square, then roots are real, unequal and irrational.
(a) =0
(b) <0
(c) >0
(d) None of these

7. If b^{2}-4 a c>0 and roots are perfect square then roots are ________
(a) Rational
(b) Irrational
(c) Equal
(d) Imaginary

Rational

8. If b^{2}-4 a c>0 and roots are not perfect square then roots are ________
(a) Rational
(b) Irrational
(c) Equal
(d) Imaginary

Irrational

9. The discriminant of x^{2}+9 x+2=0
(a) -73
(b) 73
(c) 0
(d) 9 x+2

Explanation:
x^{2}+9 x+2=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=9, c=2
As we have
Discriminant =b^{2}-4 a c
Discriminant =(9)^{2}-4(1)(2)
Discriminant =81-8
Discriminant =73

10. What is the nature of the roots of x^{2}-8 x+16=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Explanation:
x^{2}-8 x+16=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=-8, c=16
As we have
Discriminant =b^{2}-4 a c
Discriminant =(-8)^{2}-4(1)(16)
Discriminant =64-64
Discriminant =0
Thus the roots are real, equal and rational

11. What is the nature of the roots of x^{2}+9 x+2=0
(a) Real, equal and rational

(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Explanation:
x^{2}+9 x+2=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=9, c=2
As we have
Discriminant =b^{2}-4 a c
Discriminant =(9)^{2}-4(1)(2)
Discriminant =81-8
Discriminant =73>0
Thus the roots are real, unequal and irrational

12. What is the nature of the roots of 6 x^{2}-x-15=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Explanation:
6 x^{2}-x-15=0
Compare it with a x^{2}+b x+c=0
Here a=6, b=-1, c=-15
As we have
Discriminant =b^{2}-4 a c
Discriminant =(-1)^{2}-4(6)(-15)
Discriminant =1+360
Discriminant =361
Discriminant =19^{2}>0
Thus the roots are real, unequal and rational

13. What is the nature of the roots of 4 x^{2}+x+1=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Explanation:
4 x^{2}+x+1=0
Compare it with a x^{2}+b x+c=0
Here a=4, b=1, c=1
As we have
Discriminant =b^{2}-4 a c
Discriminant =(1)^{2}-4(4)(1)
Discriminant =1-16
Discriminant =-15 <0
Thus the roots are unequal and imaginary

14. Without solving, determine the nature of the roots of the 3 x^{2}-4 x+6=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Explanation:
3 x^{2}-4 x+6=0
Compare it with a x^{2}+b x+c=0
Here a=3, b=-4, c=6
As we have
Discriminant =b^{2}-4 a c
Discriminant =(-4)^{2}-4(3)(6)
Discriminant =16-72
Discriminant =-56 <0
Thus the roots are unequal and imaginary

15. Determine the value of {k} for which the given quadratic equation have real roots.
k x^{2}+4 x+1=0

(a) k>4
(b) k<4
(c) k \geq 4
(d) k \leq 4

Explanation:
k x^{2}+4 x+1=0
Compare it with a x^{2}+b x+c=0
Here a=k, b=4, c=1
If roots are Real
Discriminant =b^{2}-4 a c \geq 0
b^{2}-4 a c \geq 0
(4)^{2}-4(k)(1) \geq 0
16-4 k \geq 0
16 \geq 4 k
\frac{16}{4} \geq k
4 \geq k
k \leq 4

16. For what value of {k} the roots of the following equation are imaginary 2 x^{2}+3 x+k=0
(a) k>4
(b) k<4
(c) k \leq \frac{9}{8}
(d) k>\frac{9}{8}

Explanation:
2 x^{2}+3 x+k=0
Compare it with a x^{2}+b x+c=0
Here a=2, b=3, c=k
If roots are Imaginary
Discriminant =b^{2}-4 a c <0
b^{2}-4 a c <0
(3)^{2}-4(2)(k) <0
9-8 k <0
9<8k
\frac{9}{8}
k> \frac{9}{8}

17. Quadratic equation $ax^2+bx+c=0$ has equal roots if b^2-4ac=
(a) =0
(b) <0
(c) >0
(d) None of these

=0

1. The cube root of unity are
(a) 1
(b) \frac{-1+i \sqrt{3}}{2}
(c) \frac{-1-i \sqrt{3}}{2}
(d) All of these

All of these

2. The cube root of unity are
(a) 1
(b) \omega
(c) \omega^{2}
(d) All of these

All of these

3. The sum of cube roots of unity is ________
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \alpha \beta

Explanation:
As \omega=\frac{-1+i \sqrt{3}}{2}
And \omega^{2} \frac{-1-i \sqrt{3}}{2}
1+\omega+\omega^{2}
=1+\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2}
=\frac{2+(-1+i \sqrt{3})+(-1-i \sqrt{3})}{2}
=\frac{2-1+i \sqrt{3}-1-i \sqrt{3}}{2}
=\frac{2-1-1+i \sqrt{3}-i \sqrt{3}}{2}
=\frac{1-1}{2}
=\frac{0}{2}
=0

4. 1+\omega+\omega^{2}=
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \alpha \beta

Explanation:
See MCQS No. 3

5. 1+\omega=
(a) \omega
(b) \omega^{2}
(c) -\omega
(d) -\omega^{2}

Explanation:
1+\omega+\omega^{2}=0
1+\omega=\omega^{2}

6. 1+\omega^{2}=
(a) \omega
(b) \omega^{2}
(c) -\omega
(d) -\omega^{2}

Explanation:
1+\omega+\omega^{2}=0
Then
1+\omega^{2}=-\omega

7. \omega+\omega^{2}=
(a) \omega
(b) \omega^{2}
(c) 1
(d) -1

Explanation:
1+\omega+\omega^{2}=0
Then
\omega+\omega^{2}=-1

8. The Product of cube roots of unity is ________
(a) Zero
(b) 1
(c) \alpha+\beta

Explanation:
As \omega=\frac{-1+i \sqrt{3}}{2}
And \omega^{2} \frac{-1-i \sqrt{3}}{2}
1.\omega \cdot \omega^{2}
=1 \cdot\left(\frac{-1+i \sqrt{3}}{2}\right)\left(\frac{-1-i \sqrt{3}}{2}\right)
=\frac{(-1+i \sqrt{3})(-1-i \sqrt{3})}{2 \times 2}
=\frac{(-1)^{2}-(i \sqrt{3})^{2}}{4}
=\frac{1-i^{2}(3)}{4}
=\frac{1-3 i^{2}}{4}
=\frac{1+3}{4}
=\frac{4}{4}
=1

9. 1 . \omega \cdot \omega^{2}=
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \alpha \beta

1

10. \omega^{3}=
(a) \omega
(b) \omega^{2}
(c) 1
(d) -1

1

11. \omega=\frac{1}{\omega^{2}}
(a) 1
(b) \omega^{3}
(c) \frac{1}{\omega}
(d) \frac{1}{\omega^{2}}

Explanation:
1.\omega \cdot \omega^{2}=1
Then
\omega=\frac{1}{\omega^{2}}

12. \omega^{2}=
(a) \omega
(b) \omega^{3}
(c) \frac{1}{\omega}
(d) \frac{1}{\omega^{2}}

Explanation:
1.\omega \cdot \omega^{2}=1
Then
\omega^{2}=\frac{1}{\omega}

13. \omega \cdot \omega^{2}=
(a) 1
(b) \omega^{3}
(c) \frac{1}{\omega}
(d) \frac{1}{\omega^{2}}

1

14. (x+y)(x+\omega y)\left(x+\omega^{2} y\right)=
(a) \omega^{3}-\omega
(b) \omega^{3}
(c) x^{3}+y^{3}
(d) x^{3}-y^{3}

Explanation:
(x+y)(x+\omega y)\left(x+\omega^{2} y\right)
=(x+y)\left(x^{2}+\omega^{2} x y+\omega x y+\omega^{3} y^{2}\right)
=(x+y)\left[x^{2}+x y\left(\omega^{2}+\omega\right)+\omega^{3} y^{2}\right]
As {\omega}^{2}+{\omega}=-1
=(x+y)\left[x^{2}+x y(-1)+(1)^{3} y^{2}\right]
=(x+y)\left(x^{2}-x y+y^{2}\right)
=x^{3}+y^{3}

15. Evaluate \omega^{15}
(a) \omega
(b) \omega^{2}
(c) 1
(d) -1

Explanation:
\omega^{15}=\omega^{3 \times 5}
=\left(\omega^{3}\right)^{5}
=(1)^{5}
=1

16. Let $\omega$ be one of the complex cubed roots of unity, then the value of 3+\omega+\omega^2 is
(a) 0
(b) 1
(c) 2
(d) 3

Explanation:
3+\omega+\omega^2
3+(-1)
3-1
=2

1. Sum of roots =
(a) \frac{-b}{a}
(b) \frac{c}{a}
(c) \omega
(d) \omega^{2}

2. Product of roots =
(a) \frac{-b}{a}
(b) \frac{c}{a}
(c) \omega
(d) \omega^{2}

3. Sum of the roots of 2 x^{2}-3 x-4=0
(a) \frac{3}{2}
(b) -2
(c) \omega
(d) \omega^{2}

Explanation:
2 x^{2}-3 x-4=0
Compare it with a x^{2}+b x+c=0
Here a=2, b=-3, c=-4
Let \alpha \ and \ \beta be the roots of equation
Then sum of roots:
\alpha+\beta=\frac{-b}{a}
\alpha+\beta=\frac{-(-3)}{2}
\alpha+\beta=\frac{3}{2}

4. Products of the roots of 2 x^{2}-3 x-4=0
(a) \frac{3}{2}
(b) -2
(c) \omega
(d) \omega^{2}

Explanation:
2 x^{2}-3 x-4=0
Compare it with a x^{2}+b x+c=0
Here a=2, b=-3, c=-4
Let \alpha \ and \ \beta be the roots of equation
Product of roots:
\alpha \cdot \beta=\frac{c}{a}
\alpha \cdot \beta=\frac{-4}{2}
\alpha \cdot \beta=-2

5. Let \alpha, \beta be the roots of a quadratic equation, then the expressions of the form of \alpha+\beta, \alpha \beta, \alpha^{2}+\beta^{2} are called the _________ of the roots of the quadratic equation.
(a) Solutions
(b) Roots
(c) Functions
(d) None of these

6. By _________ function of the roots of an equation, we mean that the function remains unchanged in values when the roots are interchanged.
(a) Solution
(b) Root
(c) Symmetric
(d) None of these

7. The functions \alpha+\beta, \alpha^{2}+\beta^{2}, \alpha^{3}+\beta^{3} are _________ function of \alpha \ and \ \beta
(a) Solution
(b) Root
(c) Symmetric
(d) None of these

Explanation:
By Symmetric function of the roots of an equation, we mean that the function remains unchanged in values when the roots are interchanged.

8. {\alpha}^{2}+{\beta}^{2}=
(a) (\alpha+\beta)^{2}
(b) (\alpha+\beta)^{2}-2 \alpha \beta
(c) (\alpha-\beta)^{2}-2 \alpha \beta
(d) None of these

Explanation:
As
\alpha^{2}+\beta^{2}+2 \alpha \beta=(\alpha+\beta)^{2}
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta
Then
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta

9. \alpha^{3}+{\beta}^{3}=
(a) (\alpha+\beta)^{3}
(b) (\alpha+\beta)^{3}-3 \alpha \beta
(c) -3 \alpha \beta(\alpha+\beta)
(d) (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)

Explanation:
As
\alpha^{3}+\beta^{3}+3 \alpha\beta(\alpha+\beta)=(\alpha+\beta)^{3}
Then
\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)

10. frac{1}{\alpha}+\frac{1}{\beta}=
(a) \frac{\alpha+\beta}{\alpha \beta}
(b) \beta+\alpha
(c) \beta \alpha
(d) \alpha+\beta

Explanation:
\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}
\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}

11. If S and P be the sum and product of roots of a quadratic equation respectively, then the quadratic equation is
(a) x^2+Sx+P=0
(b) x^2-Sx+P=0
(c) x^2-Sx-P=0
(d) x^2+Sx-P=0

Explanation:
As a x^2+b x+c=0
Divide all terms by $a$
\frac{a x^2}{a}+\frac{b x}{a}+\frac{c}{a}=\frac{0}{a}
x^2+\frac{b x}{a}+\frac{c}{a}=0
Now we can write it as
x^2-\left(-\frac{b}{a}\right) x+\frac{c}{a}=0
As -\frac{b}{a}=\mathrm{S} \ and \ \frac{c}{a}=\mathrm{P}
Then
x^2-\mathrm{S} x+\mathrm{P}=0

12. Form a quadratic equation whose roots are 1+\sqrt{5}, 1-\sqrt{5}
(a) x^{2}-\sqrt{5} x+1=0
(b) x^{2}-S x+P=0
(c) x^{2}-2 x-4=0
(d) None of these

Explanation:
As 1+\sqrt{5} and 1-\sqrt{5} are the roots of required equation
Then sum of roots:
S=1+\sqrt{5}+1-\sqrt{5}
S=1+1+\sqrt{5}-\sqrt{5}
S=2
And product of roots:
P=(1+\sqrt{5})(1-\sqrt{5})
P=(1)^2-(\sqrt{5})^2
P=1-5
P=-4
As required equation is:
x^2-S x+P=0
Now
x^2-2 x+(-4)=0
x^2-2 x-4=0

13. (\alpha-\beta)^{2}=
(a) (\alpha+\beta)^{2}
(b) (\alpha+\beta)^{2}-4 \alpha \beta
(c) 2 \alpha \beta
(d) 4 \alpha \beta

Explanation:
(\alpha-\beta)^{2}
=(\alpha+\beta)^{2}-4 \alpha \beta

14. ___________ division is the process of finding the quotient and remainder with less writing and fewer calculations.
(a) Synthetic
(b) Long
(c) No
(d) None of these

15. ___________ division is the shortcut of long division method and allows one to calculate without writing variables.
(a) Synthetic
(b) Long
(c) No
(d) None of these

16.___________ division can be used only when the divisor is a linear factor.
(a) Synthetic
(b) Long
(c) No
(d) None of these

17. More than one equation which are satisfied by the same values of the variables involved are called ___________ equations.
(b) Linear
(c) Simultaneous
(d) None of these

18. A system of Linear equation consists of two or more ___________ equations in the same variables.
(b) Linear
(c) Function
(d) None of these

19. The sum of roots of an equation x^2-5kx++6k^2=0 is
(a) 5k
(b) -5k
(c) 6k^2
(d) None of these

Explanation:
x^2-5kx++6k^2=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=-5k, c=6k^2
Let \alpha \ and \ \beta be the roots of equation
Then sum of roots:
\alpha+\beta=\frac{-b}{a}
\alpha+\beta=\frac{-(-5k)}{1}
\alpha+\beta=5k

(i). If the sum of roots of
(a+1)x^2+(2a+3)x+(3a+4)=0
is -1 , then roots is

O 0
O 1
O 2
O 3

2
Explanation:
(a+1)x^2+(2a+3)x+(3a+4)=0
a=a+1, b=2a+2, c=3a+4
As
Sum of roots =s=-1
Now
S=\frac{-b}{a}
-1=\frac{-(2a+3)}{a+1}
-1(a+1)=-2a-3
-a-1=-2a-3
-a+2a=-3+1
a=-2
Now
P=\frac{c}{a}
P=\frac{3a+4}{a+1}
Put a=-2
P=\frac{3(-2)+4}{-2+1}
P=\frac{-6+4}{-1}
P=\frac{-2}{-1}
P=2

(ii). The sum of the roots of a quadratic equation is 2 and the sum of the cubes of the roots is 98. The equation is
O x^2-2x-15=0
O x^2-2x+15=0
O x^2-4x+15=0
O None of these

x^2-2x-15=0
Explanation:
Let \alpha, \beta be the roots of quatratic eaquation
As we know that
S=\alpha+ \beta
P=\alpha \beta
Now we have
Sum of roots
S=\alpha+ \beta=2
and sum of cumbes of the roots \alpha^3+ \beta^3 =98
Now
(\alpha+ \beta)^3 -3(\alpha \beta)(\alpha+ \beta)=98
8 -6(\alpha \beta)=98
-6(\alpha \beta)=98-8
-6(\alpha \beta)=90
\alpha \beta=\frac{90}{-6}
\alpha \beta=-15
So
P=\alpha \beta=-15 For
As required equation is
x^2-Sx+P=0
So
x^2-2x+(-15)=0
x^2-2x-15=0

(iii). If a, b, c are positive real number, then both the roots of the equation a^2+bx+c=0 , are always
O real and positive
O real and negative
O rational and unequal
O none of these

None of these

(iv). If a and b are the roots of 4x^2-3x+7=0 then the value of \frac{1}{a}+ \frac{1}{b} is
O -\frac{3}{4}
O \frac{3}{7}
O -\frac{3}{7}
O \frac{4}{7}

\frac{3}{7}
Explanation:
4x^2-3x+7=0
a=4, b=-3, c=7
Let a and b be the roots of equation
Then sum of roots:
a+b=\frac{-b}{a}=\frac{-(-3)}{4}
a+b=\frac{3}{4}
and Product of roots:
ab=\frac{c}{a}=\frac{7}{4}
According to given condition
\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}
\frac{1}{a}+\frac{1}{b}=\frac{3/4}{7/4}
\frac{1}{a}+\frac{1}{b}=\frac{3}{7}

### Mathematics Class 10 Notes (KPK) Chapter # 3

1. The comparison between two quantities of the same kind (same units) is called
(a) Variation
(b) Proportion
(c) Ratio
(d) None of these
Ratio

2. If a and b are two quantities of the same kind then ratio is written as
(a) a: b
(b) \frac{a}{b}
(c) Both a & b
(d) None of these
Both a & b

3. The simplified form of 6 a: 18b
(a) 6 a: 18 b
(b) a: b
(c) a: 3 b
(d) 6: 18
a: 3 b

4. In ratio, a and b are
(a) Imaginary
(b) Integers
(c) Shares
(d) None of these
Integers

5. In simplified form of a: b, there is no common factors other than
(a) 1
(b) 2
(c) 0
(d) None of these
1

6. A ________ is an equation that states that two ratios are equivalent.
(a) Quantity
(b) Proportion
(c) K method
(d) None of these
Proportion

7. __________ is a comparison of the quantity of a part to the quantity of a whole.
(a) Quantity
(b) Proportion
(c) K method
(d) None of these
Proportion

8. In a: b:: c: d, b \ and \ d are _________ to zero.
(a) Equal
(b) Not equal
(c) 1
(d) None of these
Not equal

9. If the value of a quantity remains unchanged under different situations, it is called a
(a) Variable
(b) Constant
(c) Ratio
(d) None of these
Constant

10. If the value of a quantity changes under different situation, it is called a
(a) Variable
(b) Constant
(c) Ratio
(d) None of these
Variable

11. There are ________ types of variation.
(a) Two
(b) Three
(c) Four
(d) Five
Two

12. If one quantity increases, the other also increases is called
(a) Direct variation
(b) Inverse variation
(c) Both a & b
(d) None of these
Direct variation

13. If one quantity decreases, the other also decreases is called
(a) Direct variation
(b) Inverse variation
(c) Both a & b
(d) None of these
Direct variation

14. If y varies directly x then
(a) y \propto x
(b) y \propto \frac{1}{x}
(c) Both a & b
(d) None of these
y \propto x

15. If one quantity increases, the other decreases are called
(a) Direct variation
(b) Inverse variation
(c) Both a & b
(d) None of these
Inverse variation

16. If one quantity decrease, the other increases is called
(a) Direct variation
(b) Inverse variation
(c) Both a & b
(d) None of these
Inverse variation

17. If y varies inversely x then
(a) y \propto x
(b) y \propto \frac{1}{x}
(c) Both a & b
(d) None of these
y \propto \frac{1}{x}

18. If y \propto x then
(a) y=k x
(b) \frac{y}{x}=k
(c) Both a & b
(d) None of these
Both a & b

19. If y \propto \frac{1}{x}
(a) y=k \frac{1}{x}
(b) y=\frac{k}{x}
(c) x y=k
(d) All of these
All of these

20. If y varies directly with x \ and \ y=27 \ when \ x=3 then an equation connecting x \ and \ y is
(a) x \propto y
(b) x=y
(c) x=9 y
(d) y=9 x
y=9 x
Explanation:
As there is direct variation
y \propto x
y=k x \quad \ldots \ldots { equ(i) }
Put x=3 \ and \ y=27 in equ(i)
27=k(3)
\frac{27}{3}=\frac{k(3)}{3}
9=k
k=9
So equ (i) becomes
y=9 x

21. If \mathrm{P} \propto \frac{1}{\mathrm{~V}} \ and \ V=25 \ when \ \mathrm{P}=10 . Find \mathrm{P} \ when \ V=20 .
(d) P=20
Explanation:
P \propto \frac{1}{V}
P=\frac{k}{V} \ldots \ldots equ (\mathrm{i})
Put \mathrm{P}=10 and \mathrm{V}=25 in equ(i)
10=\frac{k}{25}
10 \times 25=k
250=k
k=250
Now
To Find:
P when V=20
P=?, V=20
Put V=20 and k=250 in equ(i)
P=\frac{250}{20}
P=12.5

22. Which is the greater ratio, 5: 7 \ or \ 151: 208 ?
(a) 5: 7
(b) 151: 208
(c) Bothe are equal
(d) None of these
151: 208
Explanation:
As we have
5: 7 \ or \ 151: 208
Now
5: 7=\frac{5}{7}=0.714285
Also 151: 208=\frac{151}{208}=0.725961
Hence 151: 208 is greater ratio.

23. Gold and silver are mixed in the ratio 7: 4. If 36 grams of silver is used. How much gold is used?
(a) 36 grams
(b) 4 grams
(c) 63 grams
(d) 4 grams
63 grams
Explanation:
Let gold used =x
Ratio of Gold and Silver =7: 4
Silver used =36 grams
Now the ratio Gold and Silver
7: 4=x: 36
As we have
Product of mean = Product of extreme
4 \times x=7 \times 36
x=\frac{7 \times 36}{4}
x=63
Thus 63 grams of Gold is used

24. If 11: x-1=22: 27 , find the value of x=
(a) 11
(b) 22
(c) 14.5
(d) 29
14.5
Explanation:
11: x-1=22: 27
As we have
Product of mean = Product of extreme
22(x-1)=11 \times 27
22 x-22=297
22 x-22+22=297+22
22 x=319
Divide B.S by 22
\frac{22 x}{22}=\frac{319}{22}
x=14.5

1. If a, b \ and \ c are in continued proportion then
(a) a^{2}=b c
(b) a=b c
(c) b^{2}=a c
(d) a b c
b^{2}=a c
Explanation:
If a, b \ and \ c are in continued proportion then
a: b:: b: c
Product of mean = Product of extreme
So b^{2}=a c

2. If a, b \ and \ c are in continued proportion then b is called
(a) Geometric mean
(b) Mean proportion
(c) Both a & b
(d) None of these
Both a & b

3. Find the third proportional of a^{2} b^{2} \ and \ a b c
(a) c^{2}
(b) b^{2}
(C) a^{2}
(d) a b c
c^{2}
Explanation:
Let the third proportional =x
So a^{2} b^{2}, a b c, x are in continued proportional
Now we write it
a^{2} b^{2}: a b c=a b c: x
Product of mean = Product of extreme
a b c \times a b c=a^{2} b^{2} \times x
a^{2} b^{2} c^{2}=a^{2} b^{2} \times x
Divide B. S a^{2} b^{2}
\frac{a^{2} b^{2} c^{2}}{a^{2} b^{2}}=\frac{a^{2} b^{2} \times x}{a^{2} b^{2}}
c^{2}=x
x=c^{2}

4. Find the mean proportional of 12,3
(a) 12
(b) 3
(c) 36
(d) 6
6
Explanation:
Let the mean proportional =x
So 12, x, 3 are in continued proportional
Now we write it
12: x=x: 3
Product of mean = Product of extreme
x \times x=12 \times 3
x^{2}=36
Taking square root on B.S
\sqrt{x^{2}}=\sqrt{36}
x=6

5. Is 3,12,39 are in continued proportion.
(a) Yes
(b) No
(c) Same
(d) None of these
No
Explanation:
As 3,12,39 are in continued proportional
So we can write it
3: 12=12: 39
Product of mean = Product of extreme
12 \times 12=3 \times 39
144=117
Thus 4,12,36 are not in continued proportional

6. If 5: 15: x are in continued proportional, find the value of x
(a) x=5
(b) x=15
(c) x=45
(d) x=225
x=45
Explanation:
As 5: 15: x are in continued proportional
So we can write it
5: 15=15: x
Product of mean = Product of extreme
15 \times 15=5 \times x
225=5 x
Divide B.S by 5
\frac{225}{5}=\frac{5 x}{5}
45=x
x=45

7. If a: b=c: d \ then \ a: c=b: d then it is
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Alternendo Property

8. If two ratios are equal, then their inverse are also equal is known as
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Invertendo Property

9. If a: b=c: d \ then \ b: a=d: c then it is
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Invertendo Property

10. If a: b=c: d \ then \ (a+b): b=(c+d): d then it is
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Compnendo Property

11. If a: b=c: d \ then \ \frac{a+b}{b}=\frac{c+d}{d} then it is
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Compnendo Property

12. If a: b=c: d \ then \ (a-b): b=(c-d): d then it is
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Dividendo Property

13. If a: b=c: d \ then \ \frac{a-b}{b}=\frac{c-d}{d} then it is
(a) Invertendo Property
(b) Dividendo Property
(c) Alternendo Property
(d) Compnendo Property
Dividendo Property

14. If a: b=c: d then Componendo – Dividendo Property is:
(a) \frac{a+b}{b}=\frac{c+d}{d}
(b) \frac{a+b}{a-b}=\frac{c+d}{c-d}
(c) \frac{a-b}{b}=\frac{c-d}{d}
(d) None of these
\frac{a+b}{a-b}=\frac{c+d}{c-d}

15. A combination of direct and inverse variation of one or more variables forms ________ variation.
(a) Simple
(b) Joint
(c) No
(d) All of these
Joint

16. If y varies directly as x and inversely as z Then
(a) y \propto x z
(b) y \propto \frac{x}{z}
(c) x y z=k
(d) None of these
y \propto \frac{x}{z}

17. If a: b:: c: d is a proportion, then putting each ratio equal to k is called
(a) Ratio
(b) Alternendo
(c) k method
(d) None of these
k method

18. Volume of gas ‘V’ varies inversely as pressure ‘P’.
P=300 \mathrm{~N} / \mathrm{m}^{2} \ when \ V=4 \mathrm{~m}^{3}.
Find pressure when V=3 m^{3} .
Pressure =400 \frac{\mathrm{N}}{\mathrm{m}^{2}}
Explanation:
As there is Inverse variation
\mathrm{P} \propto \frac{1}{\mathrm{~V}}
P =\frac{k}{V} \qquad \ldots { equ(i) }
Put \mathrm{P}=300 and \mathrm{V}=4 in equ(i)
300=\frac{k}{4}
300 \times 4=k
1200=k
k=1200
Now
To Find:
P when V=3
P=?, V=3
Put V=3 \ and \ k=1200 in equ(i)
P=\frac{1200}{3}
P=400
Thus
Pressure =400 \frac{\mathrm{N}}{\mathrm{m}^{2}}

(i). Direct variation between a and b is expressed as.
O a=b
O a=\frac{1}{b}
O \propto b
O a \propto \frac{1}{b}

(ii). If m \propto \frac{1}{n} then
O m=kn
O n=km
O \frac{m}{n}=k
O mn=k

Expalnation:
m \propto \frac{1}{n}
m = \frac{k}{n}
mn=k

(iii). identify the item that does not have the same ratio as the other thtree.
O \frac{30}{45}
O 4 to 6
O 2 : 3
O 3 to 2

Expalnation:
First Option:
\frac{30}{45}=\frac{6}{9}=\frac{2}{3}
Second Option:
4 to 6 = 2 to 3
Third Option:
Thus the different ration from the three is
3 to 2

(iv). If \frac{a}{b}=\frac{c}{d} then by alternendo property
O \frac{a-b}{b}=\frac{c-d}{d}
O \frac{a}{a+b}=\frac{c}{c+d}
O \frac{a}{c}=\frac{b}{d}
O \frac{b}{a}=\frac{d}{c}

Expalnation:
Alternendo property means that if the second and third term interchange their places, then also the four terms are in proportion.

(v). If 7 : 9 :: x : 27
O x=21
O x=3
O x=7
O x=81

Expalnation:
7 : 9 :: x : 27
Product of mean = Product of extreme
9 \times x=7 \times 27
x=\frac{7 \times 27}{9}
x=7 \times 3
x=21

(vi). The third proportional of x and y is
O xy
O \frac{x}{y}
O \frac{y^2}{x}
O None of these

Expalnation:
Let z be the third proportional
x : y :: y : z
Product of mean = Product of extreme
y \times y=x \times z
\frac{y^2}{x}=z

(vii). If x \propto \frac{1}{y} and y \propto \frac{1}{z} then
O y \propto \frac{1}{z}
O y \propto z
O xy \propto z
O xz \propto y

Expalnation:
x \propto \frac{1}{y} and y \propto \frac{1}{z}
y \propto \frac{1}{z} or z \propto \frac{1}{y}
Then by comparing
x \propto \frac{1}{y} and z \propto \frac{1}{y}
z \propto z

(viii). If 2a+1 : 21 :: 4 : 7 , then
O a=\frac{13}{2}
O a=\frac{11}{2}
O a=10
O a=\frac{9}{2}

Expalnation:
2a+1 : 21 :: 4 : 7
Product of mean = Product of extreme
21 \times 4=7(2a+1)
84=14a+7
84-7=14a
77=14a
\frac{77}{14}=a
\frac{11}{2}=a

(ix). If \frac{a}{b}=\frac{c}{d}=\frac{e}{f} then each fraction is equal to
O \frac{la+mb+ne}{ld+me+nf}
O \frac{la+mc+ne}{lb+md+nf}
O \frac{la+mc+ne}{md+nd+ef}
O \frac{la+mb+nc}{lb+mc+nf}

Expalnation:
See Example No. 15
Book Page No. 63

(x). Which of the following is a situation in which x varies as directly as y ?
O x=\frac{4}{y}
O xy=6
O x=xy
O x=\frac{7}{16}y

Expalnation:
In option a & b, there is inverse varation
Option c:
x=xy or 1=y
Option d:
x=\frac{7}{16}y or x \propto y
This shows direct variaction between x & y

### Mathematics Class 10 Notes (KPK) Chapter # 4

(i). \frac{1}{x^2-1}=
O \frac{1}{x+1} -\frac{1}{x-1}
O \frac{1}{2(x+1)} -\frac{1}{2(x-1)}
O \frac{1}{2(x-1)} -\frac{1}{2(x+1)}
O \frac{1}{x-1} -\frac{1}{2(x+1)}

Explanation:

See Question No. 3
Exercise # 4.1

(ii). If P(x) and Q(x) are two polynomials then \frac{P(x)}{Q(x)}, \neq0 is
O Rational fraction
O Irrational fraction
O Proper fraction
O Improper fraction

Explanation:
Definition of Rational fraction

(iii). \frac{x^2+2}{x^2+2x+2} is
O Proper fraction
O Improper fraction
O Irrational fraction
O None of these

Explanation:
Here the degrees of nemerator and denomenator are equal.
Thus this improper rational fraction.

(iv). What is the quotient when x^3-8x^2+16x-5 is divided by x-5 ?
O x^2-x+5
O x^2-3x+2
O x^2-3x+1
O x^2+13x-49+\frac{240}{(x+5)}

Explanation:
By Division, we get the answer
x^2-3x+1

### Mathematics Class 10 Notes (KPK) Chapter # 5

(i). If A=\{1,2,3\}, B=\{4,5\} and R=\{(1, 4), (2, 5), (3, 4)\} then R is
O A one-one function from A to B
O A function from A to A
O Not a function
O An onto function from A to B

Answer: An onto function from A to B
Explanation:
For function
Dom R=\{1,2,3\}
And there is no repetetion in Domain
Kind of Function:
Range=\{4,5\}
R is an onto function from A to B

(ii). If A has two elements and B has three elements, then number of binary relations in A \times B
O 2 \times 3
O 2^3
O 2^6
O 2^2

Explanation:
A has 2 elements, m=2
B has 3 elements, n=3
Number of elements in A \times B=mn
No. of Binary Relations in A \times B is given by:
2^{m \times n}=2^{2 \times 3}= 2^6

(iii). Which of the following is an example of disjoint sets?
O \{0,1,2,3\} and \{3,2,1,0\}
O \{0,2,4,6\} and \{2,4,6,8\}
O \{0,3,6,9\} and \{9,16,25,36\}
O \{0,4,8,12\} and \{6,10,14,18\}

Explanation:
The intersection of two sets have no any common element is called disjoint set.
Thus there is no common elements in
\{0,4,8,12\} and \{6,10,14,18\}

(iv). If the universal set U=\{x|x is a positive odd integer less than 30 \} , R=\{1,5,7\} and S=\{1,3,7,11,13\} , how many elements in (R \cap S)^/ ?
O 15
O 13
O 7
O 2

Explanation:
U=\{1,3,5,7, \dots,29 \}
R=\{1,5,7\} and S=\{1,3,7,11,13\}
First find R \cap S
R \cap S=\{1,5,7\} \cap \{1,3,7,11,13\}
R \cap S=\{1,7\}
Now
(R \cap S)^/=U-(R \cap S)
(R \cap S)^/=\{1,3,5,7, \dots,29 \}-\{1,7\}
(R \cap S)^/=\{1,5,7,9, \dots,29 \}
As there are 15 odd numbers less than 30.
Through intersection, 13 elements are left in (R \cap S)^/

(v). If f is a function from A to B, then f is onto function if
O Range f= B
O Range f \neq A
O Dom f= A
O second element of all ordered pairs contained in f is not repeated.

Explanation:
f be a function from A to B, then f is onto function if Range f= B.

(vi). If R=\{(0,0), (8,2), (10,3), (14,12) \} , then Dom R =
O \{0,8,10,14\}
O \{0,2,3,12\}
O \{8,10,4\}
O \{0,10\}

Explanation:
The set of all first elements of the ordered pairs in binary relation is called domain of a binary relation.
Thus the Dom R=\{0,8,10,14\}

### Mathematics Class 10 Notes (KPK) Chapter # 6

(i). The difference between upper limit of two consecutive classes in a frequency table is called.
O Class limit
O Class interval
O Class mark
O Range

Explanation:
Definition of Class Interval

(ii). A cummulative frequency histrogram is also called
O Histogram
O Pie Chart
O Ogive
O frequency polygon

(iii). The number of times a value appera on a set of data is called
O frequency
O average
O mode
O median

Explanation:
Definition of Frequency

(iv). The data below represents the number of televisons that 11 students have in their homes. Find the mode of the data.
3, 2, 1, 1, 1, 5, 3, 1, 2, 1, 2

O 1
O 2
O 3
O 4

Explanation:
As the most repeated value in a data is called mode
Here 1 is the most repeated value.

(v). In which data set are the mean, median, mode, and range all the same.
O 1, 2, 3, 3, 2, 1, 2
O 1, 2, 3, 1, 2, 3, 1
O 1, 3, 3, 3, 2, 3, 1
O 2, 2, 1, 2, 3, 2, 3

Answer: 1, 2, 3, 3, 2, 1, 2
Explanation:
Mean=\frac{\sum X}{n}
Mean=\frac{1+2+3+3+2+1+2}{7}
Mean=\frac{14}{7}
Mean=2
Now
For Median, first arrange the data in ascending order
1, 1, 2, 2, 2, 3, 3
As Median is the central exact value of arrange data.
Thus Median=2
As the most repeated value in a data is called mode
Here 2 is the most repeated value.
As Range= largest value smallest value
Range=3-1
Range=2
Thus 2 is same for all.

(vi). The n^{th} root of product of ‘n’ number of values is called.
O arithmetic
O geopmetric mean
O harmonic
O standard deviation

Explanation:
Definition of Geometric Mean

(vii). In a set of data
63, 65, 66, 67, 69, median is

O 63
O 66
O 67
O 69

Explanation:
As Median is the central exact value of arrange data.
Thus Median=6

(viii). In a set of data
41, 43, 47, 51, 57, 52, 59 median is

O 51
O 47
O 47
O none of these

Explanation:
As Median is the central exact value of arrange data.
Thus Median=51

(ix). In the given set of data
5, 7, 7, 5, 3, 7, 2, 8, 2 mode is

O 9
O 5
O 2
O 7

Explanation:
As the most repeated value in a data is called mode
Here 7 is the most repeated value.

(x). In the given set of data,
5, 5, 5, 5, 5, 5, 5 the standard deviation is

O 5
O 0
O 7
O None of these

Explanation:
The standard Deviation will be zero (0) for all same number in a data

(xi). The average pocket money of 30 students is Rs. 20/-. The total amount in the class is
O Rs. 20/-
O Rs. 30/-
O Rs. 300/-
O Rs. 600/-

Explanation:
Average=20
Average=\frac{ Total \ amount}{No. \ of \ Students}
20=\frac{ Total \ amount}{30}
20 \times 30=Total \ amount
600=Total \ amount

(xii). The sum of 30 observations is 1500. Its average will be
O 1500
O 150
O 15
O None of these

Explanation:
Sum \ of \ observation=20
Average=\frac{ Sum \ of \ observation}{No. \ of \ Observations}
Average=\frac{ 1500}{30}
Average=50

(xiii). The difference of the largest and smallest value in the data is called
O Mean
O Mode
O Range
O Standard deviation

(xiv). The formula \frac{\sum x}{n} determines
O Arithmetic Mean
O Median
O Mode
O G. M

(xv). What is the difference between the mean of Set B and the median of Set A?
Set A: \{2,-1,7,-4,11,3\}
Set B: \{12,5,-3,4,7,-7\}

O -0.5
O 0
O 0.5
O 1

Explanation:
Mean of Set B:
Mean=\frac{ 12+5-3+4+7-7}{6}
Mean=\frac{ 18}{6}
Mean=3
Median of Set A:
For Median, first arrange the data in ascending order
-4, -1, 2, 3, 7, 11
Median=\frac{2+3}{2}
Median=\frac{5}{2}
Thus Median=2.5
As difference between Mean and Median is
3-2.5=0.5

(xvi). \frac{\sum f(x-\overline{x})}{\sum f} is called _______
O Range
O Median
O S.D
O Variance

(xvii). The most frequent value in the data is called its
O Mena
O Median
O Mode
O G.M

### Mathematics Class 10 Notes (KPK) Chapter # 7

1. An angle is a union of _________ rays which have a common point called vertex.
(a) Two
(b) Three
(c) Four
(d) No
Explanation:
One of the ray is called “initial side” and other ray is called “terminal side”

2. Sexagesimal is a numeral system with ________ as its base.
(a) Ten
(b) Sixty
(c) Six
(b) Hundred
Explanation:
Sexagesimal is a system of measurement of angles, in which angles are measured in degrees, minutes and seconds.

3. Sexagesimal is a system of measurement of angles, in which angles are measured in ________
(a) Degrees
(b) Minutes
(c) Seconds
(d) All of these
Explanation:
Sexagesimal is a numeral system with Sixty as its base.

4. One complete rotation=
(a) 180^{\circ}
(b) 90^{\circ}
(c) 360^{\circ}
(d) None of these
Explanation:
One complete rotation shows 360^{\circ} which is a circle.

5. \frac{1}{2} of complete rotation =
(a) 180^{\circ}
(b) 90^{\circ}
(c) 360^{\circ}
(d) None of these
Explanation:
\frac{1}{2} \times 360^{\circ}=180^{\circ}

6. 180^{\circ} is called ________ angle.
(a) Straight
(b) Right
(c) Acute
(d) Obtuse
Straight

7. \frac{1}{4} of complete rotation =
(a) 180^{\circ}
(b) 90^{\circ}
(c) 360^{\circ}
(d) None of these
Explanation:
\frac{1}{4} \times 360^{\circ}=90^{\circ}

8. 90^{\circ} is called ________ angle.
(a) Straight
(b) Right
(c) Acute
(d) Obtuse
Right

9. ________ degree is defined as the measure \frac{1}{360} t h of a complete rotation.
(a) One
(b) 360
(c) 90
(d) 180
Explanation:
One degree is one 360th part of a complete rotation (Circle).

10. The measure \frac{1}{360} t h of a complete rotation is called ________
(a) 1^{0}
(b) 180^{\circ}
(b) 90^{\circ}
(c) 360^{\circ}
Explanation:

11. When a degree is divided into 60 equal parts is called ________
(a) Seconds
(b) Minutes
(c) Degree
(d) None of these
Minutes

12. When each minute is divided into 60 equal parts is called ________
(a) Seconds
(b) Minutes
(c) Degree
(d) None of these
Seconds

13. One minute is denoted by
(a) 1^{0}
(b) 1^{\prime}
(c) 1^{\prime \prime}
(d) None of these
1^{\prime}

14. One second is denoted by
(a) 1^{0}
(b) 1^{\prime}
(c) 1^{\prime \prime}
(d) None of these
1^{\prime \prime}

15. One degree is denoted by
(a) 1^{0}
(b) 1^{\prime}
(c) 1^{\prime \prime}
(d) None of these
1^{0}

16. 1^{0}=
(a) 60^{\prime}
(b) 3600^{\prime \prime}
(c) Both a & b
(d) 60^{\prime \prime}
Both a & b

17. 1^{\prime}=
(a) 60^{\prime}
(b) 3600^{\prime \prime}
(c) Both a & b
(d) 60^{\prime \prime}
60^{\prime \prime}

18. 1^{\prime}=
(a) \left(\frac{1}{60}\right)^{0}
(b) \left(\frac{1}{3600}\right)^{0}
(c) \left(\frac{1}{60}\right)^{\prime}
(d) None of these

Explanation:
60^{\prime} =1^{\circ}
1^{\prime} =\left(\frac{1}{60}\right)^{0}

19. 1^{\prime \prime}=
(a) \left(\frac{1}{60}\right)^{0}
(b) \left(\frac{1}{360}\right)^{0}
(c) \left(\frac{1}{60}\right)^{\prime}
(d) None of these

Explanation:
3600^{\prime \prime} =1^{\circ}
1^{\prime \prime} =\left(\frac{1}{3600}\right)^{0}

20. 1^{\prime \prime}=
(a) \left(\frac{1}{60}\right)^{0}
(b) \left(\frac{1}{360}\right)^{0}
(c) \left(\frac{1}{60}\right)^{\prime}
(d) None of these

Explanation:
60^{\prime \prime} =1^{\prime}
1^{\prime \prime} =\left(\frac{1}{60}\right)^{\prime}

21. Circumference of a circle =
(a) 2 \mathrm{r}
(b) 2 \pi
(c) 2 \pi r
(d) \pi r
2 \pi r

22. In circular system unit of measure of angle is _______.
(a) Decimal
(b) Hexa
(d) None of these

(a) 30^{\circ}
(b) 60^{\circ}
(c) 90^{\circ}
(d) 180^{\circ}
Explanation:

(a) 30^{\circ}
(b) 60^{\circ}
(c) 90^{\circ}
(d) 180^{\circ}
Explanation:
\frac{\pi } {2} \ radians =\frac{180^{\circ}}{2}

(a) 30^{\circ}
(b) 60^{\circ}
(c) 90^{\circ}
(d) 180^{\circ}
Explanation:
\frac{\pi } {3} \ radians =\frac{180^{\circ}}{3}

(a) 30^{\circ}
(b) 60^{\circ}
(c) 90^{\circ}
(d) 180^{\circ}
Explanation:
\frac{\pi } {6} \ radians =\frac{180^{\circ}}{6}

(a) \frac{180^{\circ}}{\pi}
(b) \frac{180^{\circ}}{3.14159}
(c) 57.3^{\circ}
(d) All of these
Explanation:

28. 1^{0}=
(d) All of these
Explanation:

29. Convert \frac{4 \pi}{7} radians to degrees.
(a) .84^{\circ}
(b) 102.84^{\circ}
(c) 102^{\circ}
(d) None of these
Explanation:
Now
\frac{4 \pi}{7} \ radians =\frac{4}{7} \times 180^{\circ}
\frac{4 \pi}{7} \ radians =4 \times 25.71^{\circ}

30. Convert 42.25^{\circ} from decimal forms to \mathrm{D}^{\circ} \mathrm{M}^{\prime} S^{\prime \prime}
(a) 42^{\circ}
(b) 15^{\prime}
(c) 42^{\circ} 15^{\prime}
(d) None of these
Explanation:
42.25^{\circ} }
As 1^{\circ}=60^{\prime} \ and \ 1^{\prime}=60^{\prime \prime}
42.25^{\circ}=42^{\circ}+0.25^{\circ}
42.25^{\circ}=42^{\circ}+(0.25 \times 60)^{\prime}
42.25^{\circ}=42^{\circ}+15^{\prime}
42.25^{\circ}=42^{\circ}15^{\prime}

31. Convert \frac{\pi}{6} radians into the measures of degrees.
(a) 30^{\circ}
(b) 60^{\circ}
(c) 90^{\circ}
(d) 180^{\circ}
Explanation:
Now

32. Convert 45^{\circ} in terms of radians.
(b) \frac{180^{\circ}}{\pi}
(c) \frac{\pi}{6}
(d) None of these
Explanation:
45^{\circ}
Now

1. l=
(a) \frac{\pi}{r}
(b) \frac{\theta}{r}
(c) r \ \theta
(d) None of these
Explanation:
l=r \ \theta shows the length of an arc in terms of angle (in radian) subtended at the center of a circle.

2. Area of a circular sector =
(a) \frac{1}{2} r^{2}
(b) \frac{1}{2} \ \theta
(c) \frac{1}{2} r^{2} \ \theta
(d) None of these
Explanation:
This shows the Area of sector of a circle with radius r, whose central angle is \theta radian.

3. Find l \ when \ \theta=\frac{\pi}{6} \ radians, r=6 \mathrm{~cm}
(a) l=\frac{180^{\circ}}{\pi} \mathrm{cm}
(b) l=3.14159 \mathrm{~cm}
(d) None of these
Explanation:
To Find:
l= ?
As we have
l=r \ \theta
Put the values
l=6\left(\frac{\pi}{6}\right)
l=\pi \mathrm{cm}
l=3.14159 \mathrm{~cm}

4. Find \theta \ when \ l=30 \mathrm{~cm}, \mathrm{r}=6 \mathrm{~cm}
(b) \theta=\frac{180^{\circ}}{\pi}
(d) None of these
Explanation:
To Find:
\theta=?
As we have
l=r \ \theta
Put the values
30=6 \ \theta
Divide B.S by 6
\frac{30}{6}=\frac{6 \ \theta}{6}
5=\theta

5. Find the area of sector whose radius is 4 \mathrm{~m} , with central angle 12 radian.
(a) Area of sector =6 \mathrm{~m}^{2}
(b) Area of sector =16 \mathrm{~m}^{2}
(c) Area of sector =24 \mathrm{~m}^{2}
(d) Area of sector =96 \mathrm{~m}^{2}
Answer: Area of sector =96 \mathrm{~m}^{2}
Explanation:
To Find:
Area of sector =A= ?
As we have
Area of sector =\frac{1}{2} r^{2} \ \theta
Put the values
Area of sector =\frac{1}{2}(4)^{2}(12)
Area of sector =\frac{1}{2}(16)(12)
Area of sector =(8)(12)
Area of sector =96 \mathrm{~m}^{2}

1. Angles having the same initial and terminal sides are called _______ angles
(a) Right
(b) Coterminal
(c) Supplementary
(d) Complementary
Coterminal

2. Coterminal angles are differ by a multiple of _______
(b) 360^{\circ}
(c) Both a & b
(d) None of these
Both a & b

3. Coterminal angles are also called _______ angles.
(a) Right
(b) General
(c) Supplementary
(d) Complementary
General

4. Find coterminal angles of 55^{\circ}
(a) 415^{\circ}
(b) -305^{\circ}
(c) Both a & b
(d) None of these
Explanation:
55^{\circ}
As 55^{\circ}+360^{\circ}=415^{\circ}
And 55^{\circ}-360^{\circ}=-305^{\circ}
The coterminal angles of 55^{\circ} \ are \ 415^{\circ} \ and \ -305^{\circ}

4. Find coterminal angles of \frac{\pi}{6}
(a) \frac{13 \pi}{6} \ and \ \frac{-11 \pi}{6}
(b) -305^{\circ}
(c) Both a & b
(d) 55^{\circ}
Answer: \frac{13 \pi}{6} \ and \ \frac{-11 \pi}{6}
Explanation:
\frac{\pi}{6}
As \frac{\pi}{6}+2 \pi=\frac{\pi+12 \pi}{6}
And \frac{\pi}{6}-2 \pi=\frac{\pi-12 \pi}{6}
The coterminal angles of \ \frac{\pi}{6} are \ \frac{13 \pi}{6} \ and \ \frac{-11 \pi}{6}

5. Angles are in _______ position if the vertex of an angle lies at the origin, and initial side lies on positive x – axis.
(a) Imaginary
(b) Negative
(c) Standard
(d) None of these
Standard

6. When an angle is positive than it shows _______ wise direction.
(a) clock
(b) anti – clock
(c) Both Clock and anti – clock
(d) None of these
Anti – clock

7. When an angle is negative then it shows _______ wise direction.
(a) Clock
(b) anti – clock
(c) Both Clock and anti – clock
(d) None of these
Clock

8. An angle of measure 55^{\circ} shows _______ wise direction.
(a) clock
(b) anti – clock
(c) Both Clock and anti – clock
(d) None of these
Anti – clock Explanation:
When an angle is positive than it shows anti – clock wise direction.

9. An angle of measure -55^{\circ} shows _______ wise direction.
(a) clock
(b) anti – clock
(c) Both Clock and anti – clock
(d) None of these
Clock
Explanation:
When an angle is negative then it shows Clock wise direction.

10. Quadrant are obtained when \mathrm{XY} – {plane} is divided into _______ equal parts.
(a) One
(b) Two
(c) Three
(d) Four
Four

11. The Cartesian plan is divided into _______ quadrants.
(a) One
(b) Two
(c) Three
(d) Four
Four

12. 0^{0}, 90^{\circ}, 180^{\circ}, 270^{\circ}, 360^{\circ} are _______ angles.
(b) Supplementary
(c) Complementary
(d) None of these

13. 0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2} and 2 \pi are _______ angles.
(b) Supplementary
(c) Complementary
(d) None of these

14. If \theta lies in _______, then we can write as 0 < \theta < \frac{\pi}{2} \ or \ 0 < \theta < 90^{\circ} .

15. If \theta lies in _______, then we can write as \frac{\pi}{2} < \theta < \pi \ or \ 90^{\circ} < \theta<180^{\circ} .

16. If \theta lies in ________, then we can write as \pi < \theta < \frac{3 \pi}{2} \ or \ 180^{\circ} < \theta<270^{\circ}.

17. If \theta lies in ______, then we can write as \frac{\pi}{2} < \theta < 2 \pi \ or \ 270^{\circ} < \theta < 360^{\circ} .

18. \frac{8 \pi}{5} lies in which quadrant.
(a) I
(b) II
(c) III
(d) IV
Explanation:
\frac{8 \pi}{5}
\frac{8 \pi}{5}=\frac{8 \times 180^{\circ}}{5}
\frac{8 \pi}{5}=8 \times 36^{0}
\frac{8 \pi}{5}=288^{\circ}
Thus \frac{8 \pi}{5} lies in 2nd Quadrant

19. 75^{\circ} lies in which quadrant.
(a) I
(b) II
(c) III
(d) IV
Explanation:
If \theta lies between 0^{\circ} \ and \ 90^{\circ} then it lies in 1st Quadrant.
Thus 75^{\circ} lies in 1st Quadrant

20. -818^{\circ} lies in which quadrant.
(a) I
(b) II
(c) III
(d) IV
Explanation:
As -818^{\circ}=-2\left(360^{\circ}\right)-98^{\circ}
As -818^{\circ} is negative so in anti – clock direction
Thus -98^{\circ} lies in 3rd Quadrant

\begin{array}{|c|c|c|c|} \hline \theta & 30^{\circ} & 45^{\circ} & 60^{\circ} \\ \hline \sin \theta & \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{2} \\ \hline \cos \theta & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} \\ \hline \tan \theta & \frac{1}{\sqrt{3}} & 1 & \sqrt{3} \\ \hline cosec \ \theta & 2 & \sqrt{2} & \frac{2}{\sqrt{3}} \\ \hline \sec \theta & \frac{2}{\sqrt{3}} & \sqrt{2} & 2 \\ \hline \cot \theta & \sqrt{3} & 1 & \frac{1}{\sqrt{3}} \\ \hline \end{array}

1. The side opposite to 90^{\circ} is called
(a) Base
(b) Perpendicular
(c) Hypotenuse
(d) None of these
Explanation:
In right angled triangle, The side opposite to 90^{\circ} is always be called Hypotenuse

2. Side opposite to \theta or angle in consideration is called or _______ opposite side.
(a) Base
(b) Perpendicular
(c) Hypotenuse
(d) None of these
Explanation:
In right angled triangle, the side opposite to angle under consideration or on which we ae working/ calculating is called Perpendicular

3. The side adjacent to \theta or angle in consideration is called or _______ adjacent side.
(a) Base
(b) Perpendicular
(c) Hypotenuse
(d) None of these
Explanation:
In right angled triangle, the side adjacent to angle in consideration is called as Base.

4. \frac{ perp }{ hyp}=
(a) \sin \theta
(b) \cos \theta
(c) \tan \theta
(d) All of these
\sin \theta

5. \frac{h y p}{p e r p}=
(a) \cosec \theta
(b) \sec \theta
(c) \cot \theta
(d) All of these
\cosec \theta

6. \frac{ base }{ hyp }=
(a) \sin \theta
(b) \cos \theta
(c) \tan \theta
(d) All of these
\cos \theta

7. \frac{h y p}{ base }=
(a) \cosec \theta
(b) \sec \theta
(c) \cot \theta
(d) All of these
\sec \theta

8. \frac{ perp }{ base }=
(a) \sin \theta
(b) \cos \theta
(c) \tan \theta
(d) All of these
\tan \theta

9. \frac{\ base }{ perp }=
(a) \cosec \theta
(b) \sec \theta
(c) \cot \theta
(d) All of these
\cot \theta

10. \sin 30^{\circ}=
(a) \frac{1}{2}
(b) \frac{1}{\sqrt{2}}
(c) \frac{\sqrt{3}}{2}
(d) None of these
\frac{1}{2}

11. \sin 45^{\circ}=
(a) \frac{1}{2}
(b) \frac{1}{\sqrt{2}}
(c) \frac{\sqrt{3}}{2}
(d) None of these
\frac{1}{\sqrt{2}}

12. \sin 60^{\circ}=
(a) \frac{1}{2}
(b) \frac{1}{\sqrt{2}}
(c) \frac{\sqrt{3}}{2}
(d) None of these
\frac{\sqrt{3}}{2}

13. \cos 30^{\circ}
(a) \frac{1}{2}
(b) \frac{1}{\sqrt{2}}
(c) \frac{\sqrt{3}}{2}
(d) None of these
\frac{\sqrt{3}}{2}

14. \cos 45^{\circ}=
(a) \frac{1}{2}
(b) \frac{1}{\sqrt{2}}
(c) \frac{\sqrt{3}}{2}
(d) None of these
\frac{1}{\sqrt{2}}

15. \cos 60^{\circ}=
(a) \frac{1}{2}
(b) \frac{1}{\sqrt{2}}
(c) \frac{\sqrt{3}}{2}
(d) None of these
\frac{1}{2}

16. \tan 30^{\circ}=
(a) \frac{1}{\sqrt{3}}
(b) 1
(c) \sqrt{3}
(d) None of these
\frac{1}{\sqrt{3}}

17. \tan 45^{\circ}=
(a) \frac{1}{\sqrt{3}}
(b) 1
(c) \sqrt{3}
(d) None of these
1

18. \tan 60^{\circ}=
(a) \frac{1}{\sqrt{3}}
(b) 1
(c) \sqrt{3}
(d) None of these
\sqrt{3}

19. \cosec 30^{\circ}=
(a) 2
(b) \sqrt{2}
(c) \frac{2}{\sqrt{3}}
(d) None of these
2

20. {cosec} \ 45^{\circ}=
(a) 2
(b) \sqrt{2}
(c) \frac{2}{\sqrt{3}}
(d) None of these
\sqrt{2}

21. \cosec \ 60^{\circ}=
(a) 2
(b) \sqrt{2}
(c) \frac{2}{\sqrt{3}}
(d) None of these
\frac{2}{\sqrt{3}}

22. \sec 30^{\circ}=
(a) 2
(b) \sqrt{2}
(c) \frac{2}{\sqrt{3}}
(d) None of these
\frac{2}{\sqrt{3}}

23. \sec 45^{\circ}=
(a) 2
(b) \sqrt{2}
(c) \frac{2}{\sqrt{3}}
(d) None of these
\sqrt{2}

24. \sec 60^{\circ}=
(a) 2
(b) \sqrt{2}
(c) \frac{2}{\sqrt{3}}
(d) None of these
2

25. \cot 30^{\circ}=
(a) \frac{1}{\sqrt{3}}
(b) 1
(c) \sqrt{3}
(d) None of these
\sqrt{3}

26. \cot 45^{\circ}=
(a) \frac{1}{\sqrt{3}}
(b) 1
(c) \sqrt{3}
(d) None of these
1

27. \cot 60^{\circ}=
(a) \frac{1}{\sqrt{3}}
(b) 1
(c) \sqrt{3}
(d) None of these
\frac{1}{\sqrt{3}}

28. In ________ Quadrant, all Trigonometric ratios are positive.
(a) 1^{st}
(b) 2^{nd }
(c) 3^{rd }
(d) 4^{th }
1^{st}

29. In ________ Quadrant, \sin \theta \ and \ \cosec \theta are positive.
(a) 1^{st}
(b) 2^{nd }
(c) 3^{rd }
(d) 4^{th }
2^{nd }

30. In ________ Quadrant, \tan \theta \ and \ \cot \theta are positive.
(a) 1^{st}
(b) 2^{nd }
(c) 3^{rd }
(d) 4^{th }
3^{rd }

31. In ________ Quadrant, \cos \theta \ and \ \sec \theta are positive.
(a) 1^{st}
(b) 2^{nd }
(c) 3^{rd }
(d) 4^{th }
4^{th }

32. Find the signs of \sin 105^{\circ} and tell in which quadrant it lie?
Explanation:
\sin 105^{\circ}
Since 105^{\circ} lies in 2^{ {nd }} Quadrant.
As \sin \theta is positive in 2^{ {nd }} Quadrant.
So \sin 105^{\circ} is positive in 2^{ {nd }} Quadrant

33. Find the signs of \cot 710^{\circ} and tell in which quadrant it lie?
Explanation:
\cot 710^{\circ}
As 710^{\circ}=360^{\circ}+350^{\circ}
Since 350^{\circ} lies in 4^{ {th }} Quadrant.
As \tan \theta \ and \ \cot \theta are negative in 4^{ {th }} Quadrant.

34. Find the value of 2 \sin 45^{\circ} \cos 45^{\circ}
(a) 1
(b) 2
(c) \sin 45^{\circ}
(d) \cos 45^{\circ}
Explanation:
2 \sin 45^{\circ} \cos 45^{\circ}
As \sin 45^{\circ}=\frac{1}{\sqrt{2}}
And \cos 45^{\circ}=\frac{1}{\sqrt{2}}
Now
2 \sin 45^{\circ} \cos 45^{\circ}
=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)
=2\left(\frac{1}{\sqrt{2 \times 2}}\right)
=2\left(\frac{1}{2}\right)
=1

35. In which quadrant \theta lies when \sin \theta>0, \tan \theta>0
(a) First
(b) Second
(c) Third
(d) Fourth
Explanation:
As \sin \theta \ and \ \tan \theta are positive in first quadrant.
Thus \theta lies in first quadrant.

36. In which quadrant \theta lies when \sin \theta<0, \cot \theta>0
(a) First
(b) Second
(c) Third
(d) Fourth
Explanation:
As \cot \theta is positive and \sin \theta is negative in 3^{ {rd }} quadrant.
Thus \theta \ lies \ in \ 3^{ {rd }} quadrant.

1. \sin \theta=
(a) \frac{1}{\cosec \theta}
(b) \frac{1}{\sec \theta}
(c) \frac{1}{\cot \theta}
(d) None of these
Explanation:
\sin \theta \ and \ \cosec \theta are the reciprocal of each other.

2. \cos \theta =
(a) \frac{1}{\cosec \theta}
(b) \frac{1}{\sec \theta}
(c) \frac{1}{\cot \theta}
(d) None of these
Explanation:
\cos \theta \ and \ \sec \theta are the reciprocal of each other.

3. \tan \theta=
(a) \frac{1}{\cosec \theta}
(b) \frac{1}{\sec \theta}
(c) \frac{1}{\cot \theta}
(d) None of these
Explanation:
\tan \theta \ and \ \cot \theta are the reciprocal of each other.

4. {cosec} \theta=
(a) \frac{1}{\sin \theta}
(b) \frac{1}{\cos \theta}
(c) \frac{1}{\tan \theta}
(d) None of these
Explanation:
\sin \theta \ and \ \cosec \theta are the reciprocal of each other.

5. \sec \theta=
(a) \frac{1}{\sin \theta}
(b) \frac{1}{\cos \theta}
(c) \frac{1}{\tan \theta}
(d) None of these
Explanation:
\cos \theta \ and \ \sec \theta are the reciprocal of each other.

6. \cot \theta=
(a) \frac{1}{\sin \theta}
(b) \frac{1}{\cos \theta}
(c) \frac{1}{\tan \theta}
(d) None of these
Explanation:
\tan \theta \ and \ \cot \theta are the reciprocal of each other.

7. \tan \theta=
(a) \frac{\sin \theta}{\cos \theta}
(b) \frac{\cos \theta}{\sin \theta}
(c) Both a & b
(d) None of these
\frac{\sin \theta}{\cos \theta}

8. \cot \theta=
(a) \frac{\sin \theta }{\cos \theta}
(b) \frac{\cos \theta}{\sin \theta}
(c) Both a & b
(d) None of these
\frac{\cos \theta}{\sin \theta}

9. \sin ^{2} \theta+\cos ^{2} \theta=
(a) 1
(b) 1-\sin ^{2} \theta
(c) 1-\cos ^{2} \theta
(d) None of these
Explanation:
\sin ^{2} \theta+\cos ^{2} \theta=1

10. \cos ^{2} \theta=
(a) 1
(b) 1-\sin ^{2} \theta
(c) 1-\cos ^{2} \theta
(d) None of these
Explanation:
\sin ^{2} \theta+\cos ^{2} \theta=1
\cos ^{2} \theta=1-\sin ^{2} \theta

11. \sin ^{2} \theta=
(a) 1
(b) 1-\sin ^{2} \theta
(c) 1-\cos ^{2} \theta
(d) None of these
Explanation:
\sin ^{2} \theta+\cos ^{2} \theta=1
\sin ^{2} \theta=1-\cos ^{2} \theta

12. 1+\tan ^{2} \theta=
(a) \sec ^{2} \theta
(b) \sec ^{2} \theta-1
(c) 1
(d) None of these
Explanation:
1+\tan ^{2} \theta = \sec ^{2} \theta

13. \tan ^{2} \theta=
(a) \sec ^{2} \theta
(b) \sec ^{2} \theta-1
(c) 1
(d) None of these
Answer: 1+\tan ^{2} \theta = \sec ^{2} \theta
Explanation:
1+\tan ^{2} \theta = \sec ^{2} \theta
\tan ^{2} \theta= \sec ^{2} \theta-1

14. 1+\cot ^{2} \theta=
(a) \cosec^{2} \theta
(b) \cosec^{2} \theta-1
(c) 1
(d) None of these
Explanation:
1+\cot ^{2} \theta= \cosec^{2} \theta

15. \cot ^{2} \theta=
(a) \cosec^{2} \theta
(b) \cosec^{2} \theta-1
(c) 1
(d) None of these
Explanation:
1+\cot ^{2} \theta= \cosec^{2} \theta
\cot ^{2} \theta= \cosec^{2}-1 \theta

16. \sqrt{1-\cos ^{2} \theta}=
(a) \cosec^{2} \theta
(b) \cosec^{2} \theta-1
(c) \sin \theta
(d) None of these
Explanation:
\sqrt{1-\cos ^{2} \theta}
As 1-\cos ^{2} \theta=\sin ^{2} \theta
=\sqrt{\sin ^{2} \theta}
=\sin \theta

17. If an object is above the level of observer, then the angle formed between the horizontal and observer’s line of sight is called
(a) Angle of depression
(b) Angle of elevation
(c) Obtuse angle
(d) None of the above
Angle of elevation

18. If an object is below the level of observer, then the angle formed between the horizontal and observer’s line of sight is called
(a) Angle of depression
(b) Angle of elevation
(c) Obtuse angle
(d) None of the above
Angle of depression

19. A building that is 21 meters tall casts a shadow 25 meters long. Find the angle of elevation of the sun to the nearest degree.
(a) 45^{\circ}
(b) 90^{\circ}
(c) 40.03^{\circ}
(d) None of these
Explanation:
See Question # 1
Ex # 7.6

20. A tree is 50 \mathrm{~m} high. Find the angle of elevation of its top to a point on the ground 100 m away from the foot of a tree.
(a) 45^{\circ}
(b) 26.56^{\circ}
(c) 40.03^{\circ}
(d) None of these
Explanation:
See Question # 3
Ex # 7.6

1. If an object is above the level of observer, then the angle formed between the horizontal and observer’s line of sight is called
(a) Angle of depression
(b) Angle of elevation
(c) Obtuse angle
(d) None of the above
Angle of elevation

2. \cot \theta=
(a) \frac{\sin \theta}{\cos \theta}
(b) \frac{1}{\cos \theta}
(c) \frac{\cos \theta}{\sin \theta}
(d) \frac{1}{\sin \theta}
\frac{\cos \theta}{\sin \theta}

3. 1+\tan ^{2} \theta=
(a) \sin ^{2} \theta
(b) \cos ^{2} \theta
(c) \cosec^{2} \theta
(d) \sec ^{2} \theta
\sec ^{2} \theta

4. If \tan \theta=1 \ then \ \sin \theta= ______ when \theta lies in 3^{rd } quadrant.
(a) \frac{1}{2}
(b) \frac{-1}{2}
(c) -\frac{1}{\sqrt{2}}
(d) \frac{1}{\sqrt{2}}
Explanation:
Let x=base, r=hyp, y=perp
As \theta lies in 3^{rd} Quadrant
Then x \ and \ y are negative.
Now
\tan \theta = \frac{perp}{bas} = \frac{y}{x} = \frac{-1}{-1}
So x=-1, y=-1
By Pythagoras theorem
\left(r \right)^2=\left(x \right)^2+\left(y \right)^2
\left(r \right)^2=\left(-1 \right)^2+\left(-1 \right)^2
r^2 = 1+1
r^2 = 2
r= \sqrt{2}
Now
\sin \theta= \frac{y}{r}
\sin \theta= \frac{-1}{\sqrt{2}}
\sin \theta= -\frac{1}{\sqrt{2}}

5. \sin \left(-350^{\circ}\right) lies in
(c) 3^ {rd } \ quadrant
Explanation:
As \theta is negative which shows clock wise direction.
Since \left(-350 \right)^0 lies in 1^{st} Quadrant.
Thus \sin \left(-350 \right)^0 lies in 1^{st} Quadrant.

(a) \frac{\pi}{3}
(b) \frac{\pi}{4}
(c) \frac{\pi}{6}
(d) \frac{\pi}{2}
Explanation:
45^{\circ}
Now

7. If the measure of the hypotenuse of a right triangle is 5 feet and m \angle B=58^{0} , what’s the measure of the leg adjacent to \angle B ?
(a) 4.3402
(b) 8.0017
(c) 0.10060
(d) 2.6496
Explanation:
As hyp=5 \ feet
m \angle B=58^{0}
To find adjacent to \angle B
So Base= ?
As
\cos \theta=\frac{Base}{Hyp}
\cos 58^{0}=\frac{Base}{5}
0.5299=\frac{Base}{5}
5 \times 0.5299=Base
2.6495=Base

8. Find the value of \tan P to the nearest tenth.
(a) 2.6
(b) 0.5
(c) 0.4
(d) 0.1
Explanation:
Here perp=8 \ and \ base=21
Now
\tan P= \frac{Perp}{Base}
\tan P= \frac{8}{21}
\tan P= 0.38
Thus the nearest tenth is 0.4

9. Find RT in the given figure.
(a) 2 \sqrt{6}
(b) 2 \sqrt{3}
(c) 4 \sqrt{3}
(d) 2 \sqrt{2}
Explanation:
By Pythagoras theorem
\left(QR \right)^2=\left(QT \right)^2+\left(RT \right)^2
\left(4 \right)^2=\left(2 \right)^2+\left(RT \right)^2
16=4+\left(RT \right)^2
16-4=\left(RT \right)^2
12=\left(RT \right)^2
\sqrt{12}=\sqrt{\left(RT \right)^2}
\sqrt{4 \times 3}=RT
2 \sqrt{3}=RT

10. RT is equal to TS. What is \angle S ?
(a) 25^{0}
(b) 30^{\circ}
(c) 45^{0}
(d) 60^{\circ}
Explanation:
By Pythagoras theorem
\left(QR \right)^2=\left(QT \right)^2+\left(RT \right)^2
\left(4 \right)^2=\left(2 \right)^2+\left(RT \right)^2
16=4+\left(RT \right)^2
16-4=\left(RT \right)^2
12=\left(RT \right)^2
\sqrt{12}=\sqrt{\left(RT \right)^2}
\sqrt{4 \times 3}=RT
2 \sqrt{3}=RT
As RT=TS=2\sqrt{3}
Now
\tan \theta=\frac{Perp}{Base}
\tan \angle S=\frac{2\sqrt{3}}{2\sqrt{3}}
\tan \angle S=1
\angle S= \tan^{-1}1
\angle S= 45^0

### 38 thoughts on “Mathematics notes for class 10 (KPK)”

1. Thanks sir bahot acha tareeka hai yarr ye Maine pehli Barr nikala ye chrome per istara ke subject Jo hamara pora sabaq hai 10 class maths ka to sir plz our subject is Mai milega Kai to mihrabani kr ke moje Bata de

1. Sir but aghy k chaptr kaha hain .4 k bad nhe mel rhy or bio k 2 chaptr h baki nhe mel rhy hain