Author Archives: Swati

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.1

Question 1Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically and graphically.Solution 1

Question 2Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’ this interesting?) Represent this situation algebraically and graphically.Solution 2Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago,
Age of Aftab = x – 7
Age of his daughter = y – 7

According to the given condition,

Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3

According to the given condition,

Thus, the given conditions can be algebraically represented as:
x – 7y = -42
x – 3y = 6

Three solutions of this equation can be written in a table as follows:

x	-7	0	7
y	5	6	7

Three solutions of this equation can be written in a table as follows:

x	6	3	0
y	0	-1	-2

The graphical representation is as follows:

Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter’s present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words ‘seven years ago’ means we have to subtract 7 from their present ages, and ‘three years from now’ or ‘three years hence’ means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.Solution 7Let the cost of 1 kg of apples and 1 kg grapes be Rsx and Rsy.
The given conditions can be algebraically represented as:

Three solutions of this equation can be written in a table as follows:

x	50	60	70
y	60	40	20

Three solutions of this equation can be written in a table as follows:

x	70	80	75
y	10	-10	0

The graphical representation is as follows:

Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale.

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Since, the graph of the two lines coincide, the given system of equations have infinitely many solutions.Question 13

Solution 13

Question 14

Solution 14

Question 15Show graphically that each one of the following systems of equations is in-coinsistent (i.e. has no solution):

3x – 5y = 20

6x – 10y = – 40Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19(i)

Solution 19(i)

Question 19(ii)

Solution 19(ii)

Question 20

Solution 20

Question 21(i)

Solution 21(i)

Question 21(ii)

Solution 21(ii)

Question 22(i)

Solution 22(i)

Question 22(ii)

Solution 22(ii)

Question 22(iii)Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

2x + y – 11 = 0

x – y – 1 = 0Solution 22(iii)

Question 22(iv)Solve graphically each of the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of y.

x + 2y – 7 = 0

2x – y – 4 = 0Solution 22(iv)

Question 22(v)

Solution 22(v)

Question 22(vi)

Solution 22(vi)

Question 23(i)

Solution 23(i)

Question 23(ii)

Solution 23(ii)

Question 23(iii)

Solution 23(iii)

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28(i)

Solution 28(i)

Question 28(ii)

Solution 28(ii)

Question 28(iii)

Solution 28(iii)

Question 28(iv)

Solution 28(iv)

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis. Calculate the area of the triangle so formed.Solution 32

Three solutions of this equation can be written in a table as follows:

x	0	1	2
y	-5	0	5

x	0	1	2
y	-3	0	3

The graphical representation of the two lines will be as follows:

It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

Concept insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

Question 33

Form the pair of linear equations in the following problems, and find their solutions graphically.

(i)  10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii)  5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

(iii) Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa a bought.Solution 33

(i) Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x – y = 4
x + y = 10  x = 10 – y
Three solutions of this equation can be written in a table as follows:

x	4	5	6
y	6	5	4

x – y = 4  x = 4 + y
Three solutions of this equation can be written in a table as follows:

x	5	4	3
y	1	0	-1

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So, x = 7 and y = 3.

Thus, the number of girls and boys in the class are 7 and 3 respectively.


(ii)    Let the cost of one pencil and one pen be Rs x and Rs y respectively.

According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46

Three solutions of this equation can be written in a table as follows:

x	3	10	-4
y	5	0	10

Three solutions of this equation can be written in a table as follows:

x	8	3	-2
y	-2	5	12

The graphical representation is as follows:


From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So, x = 3 and y = 5.

Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.



(iii)

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:

y = 2x – 2 …(1)

and y = 4x – 4 …(2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.

They are given in Table

x	2	0
y = 2x – 2	2	-2

x	0	1
y = 4x – 4	-4	0

Plot the points and draw the lines passing through them to represent the equations, as shown in fig.,

The two lines intersect at the point (1,0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Concept insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.Question 34(i)

Solution 34(i)

Question 34(ii)

Solution 34(ii)

Question 35

Solution 35

Question 36

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) parallel lines

(iii) Coincident linesSolution 36

(i) For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0, to be intersecting, we must have

So, the other linear equation can be 5x + 6y – 16 = 0
 
(ii)  For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0, to be parallel, we must have

So, the other linear equation can be 6x + 9y + 24 = 0,

(iii)  For the two lines a₁x + b₁x + c₁ = 0 and a₂x + b₂x + c₂ = 0 to be coincident, we must have

So, the other linear equation can be 8x + 12y – 32 = 0,


Concept insight: In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.Question 37(i)

Solution 37(i)

Question 37(ii)

Solution 37(ii)

Question 38

Graphically, solve the following pair of equations:

2x + y = 6

2x – y + 2 = 0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.Solution 38

The lines AB and CD intersect at point R(1, 4). Hence, the solution of the given pair of linear equations is x = 1, y = 4.

From R, draw RM ⊥ X-axis and RN ⊥ Y-axis.

Then, from graph, we have

RM = 4 units, RN = 1 unit, AP = 4 units, BQ = 4 units

Question 39

Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8.Solution 39

From the graph, the vertices of the triangle AOP formed by the given lines are A(4, 4), O(0, 0) and P(6, 2).Question 40

Draw the graph of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also, find the area of the quadrilateral formed by the lines and the x-axis.Solution 40

The graph of x = 3 is a straight line parallel to Y-axis at a distance of 3 units to the right of Y-axis.

The graph of x = 5 is a straight line parallel to Y-axis at a distance of 5 units to the right of Y-axis.

Question 41

Draw the graphs of the lines x = -2, and y = 3. Write the vertices of the figure formed by these lines, the x-axis and the y-axis. Also, find the area of the figure.Solution 41

The graph of x = -2 is a straight line parallel to Y-axis at a distance of 2 units to the left of Y-axis.

The graph of y = 3 is a straight line parallel to X-axis at a distance of 3 units above X-axis.

Question 42

Draw the graphs of the pair of linear equations x – y + 2 = 0 and 4x – y – 4 = 0. Calculate the area of the triangle formed by the lines so drawn and the x-axis.Solution 42

Question 5

Solve the following equations graphically:

x – y + 1 = 0

3x + 2y – 12 = 0Solution 5

Given equations are:

x – y + 1 = 0 … (i)

3x + 2y – 12 = 0 … (ii)

From (i) we get, x = y – 1

When x = 0, y = 1

When x = -1, y = 0

When x = 1, y = 2

We have the following table:

x	0	-1	1
y	1	0	2

From (ii) we get,  

When x = 0, y = 6

When x = 4, y = 0

When x = 2, y = 3

We have the following table:

x	0	4	2
y	6	0	3

Graph of the given equations is:

As the two lines intersect at (2, 3). 

Hence, x = 2, y = 3 is the solution of the given equations.

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solve the following systems of equation:

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solve the pair of equations:

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solve the following systems of equation:

21x + 47y = 110

47x + 21y = 162Solution 48

Question 49

If x + 1 is a factor of 2x³ + ax² + 2bx + 1, the find the values of a and b given that 2a – 3b = 4.Solution 49

Question 50

Find the solution of the pair of equations  and . Hence, find λ, if y = λx + 5.Solution 50

Question 51

Find the values of x and y in the following rectangle.

Solution 51

Question 52

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x – y = 1. How many such lines can we find?Solution 52

Question 53

Write a pair of linear equations which has the unique solution x = -1, y = 3. How many such pairs can you write?Solution 53

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.5

Question 29

Find c if the system of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions.Solution 29

The given system of equations will have infinite number of solutions if

Question 1

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x – 3y = 3

3x – 9y = 2Solution 1

Question 2

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

2x + y = 5

4x + 2y = 10Solution 2

Question 3

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

3x – 5y = 20

6x – 10y = 40Solution 3

Question 4

Determine whether the following system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

x – 2y = 8

5x – 10y = 10Solution 4

Question 5

Find the value of k for which the following system of equations has a unique solution:

kx + 2y = 5

3x + y = 1Solution 5

Question 6Find the value of k for which the following system of equations has a unique solution:

4x + ky + 8 = 0

2x + 2y + 2 = 0Solution 6

Question 7

Find the value of k for which the following system of equations has a unique solution:

4x – 5y = k

2x – 3y = 12Solution 7

Question 8

Find the value of k for which the following system of equations has a unique solution:

x + 2y = 3

5x + ky + 7 = 0Solution 8

Question 9

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y – 5 = 0

6x + ky – 15 = 0Solution 9

Question 10

Find the value of k for which the following systems of equations have infinitely many solutions:

4x + 5y = 3

kx + 15y = 9Solution 10

Question 11

Find the value of k for which the following systems of equations have infinitely many solutions:

kx – 2y + 6 = 0

4x – 3y + 9 = 0Solution 11

Question 12

Find the value of k for which the following systems of equations have infinitely many solutions:

8x + 5y = 9

kx + 10y = 18Solution 12

Question 13

Find the value of k for which the following systems of equations have infinitely many solutions:

2x – 3y = 7

(k + 2)x – (2k + 1)y = 3(2k – 1)Solution 13

Question 14

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k – 1)Solution 14

Question 15

Find the value of k for which the following systems of equations have infinitely many solutions:

x + (k + 1)y = 4

(k + 1)x + 9y = (5k + 2)Solution 15

Question 16

Find the value of k for which the following systems of equations have infinitely many solutions:

kx + 3y = 2k + 1

2(k + 1)x + 9y = 7k + 1Solution 16

Question 17

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + (k – 2)y = k

6x + (2k – 1)y = 2k + 5Solution 17

Question 18

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = 7

(k + 1)x + (2k – 1)y = 4k + 1Solution 18

Question 19

Find the value of k for which the following systems of equations have infinitely many solutions:

2x + 3y = k

(k – 1)x + (k + 2)y = 3kSolution 19

Question 20Find the value of k for which the following system of equations has no solution:

kx – 5y = 2

6x + 2y = 7Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Find he value of k for which of the following system of equation has no solution:

kx + 3y = k – 3

12x + ky = 6Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 30

Solution 30

Question 31For what value of k, the following system of equations will represent the coincident lines?

x + 2y + 7 = 0

2x + ky + 14 = 0Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36 (i)

Solution 36 (i)

Question 36 (ii)

Solution 36 (ii)

Question 36 (iii)

Solution 36 (iii)

Question 36 (iv)

Solution 36 (iv)

Question 36 (v)

Find the values of a and b for which the following system of equations has infinitely many solutions:

2x + 3y = 7

(a – b) x + (a + b)y = 3a + b – 2Solution 36 (v)

Question 36 (vi)

Solution 36 (vi)

Question 36 (vii)

Solution 36 (vii)

Question 36(viii)

Find the values of a and b for which the following system of equations has infinitely many solutions:

x + 2y = 1

(a – b)x + (a + b)y = a + b – 2Solution 36(viii)

Question 36(ix)

Find the values of a and b for which the following system of equations has infinitely many solutions:

2x + 3y = 7

2ax + ay = 28 – bySolution 36(ix)

Question 37(i)

For which value(s) of λ, do the pair of linear equations λx + y = λ² and x + λy = 1 have no solution?Solution 37(i)

Question 37(ii)

For which value(s) of λ, do the pair of linear equations λx + y = λ² and x + λy = 1 have infinitely many solutions?Solution 37(ii)

Question 37(iii)

For which value(s) of λ, do the pair of linear equations λx + y = λ² and x + λy = 1 have a unique solution?Solution 37(iii)

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.8

Question 9

A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.Solution 9

Let the fraction be 

According to the given conditions, we have

Subtracting (ii) from (i), we get x = 7

Substituting the value of x in (ii), we get

y = 15Question 1

Solution 1

Question 2A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.Solution 2

Question 3

Solution 3

Question 4If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction?Solution 4

Question 5

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9Old

Solution 9Old

Question 10

Solution 10

Question 11

Solution 11

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solve each of the following systems of equations by the method of cross-multiplication:

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Jamila sold a table and a chair for Rs.1050, thereby making a profit of 10% on a table and 25% on the chair. If she had taken profit of 25% on the table and 10% on the chair she would have got Rs.1065. Find the cost price of each.Solution 7

Question 8

Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs.1860 as annual interest. However, had she interchanged the amount of investment in the two schemes, she would have received Rs.20 more as annual interest. How much money did she invest in each scheme?Solution 8

Question 9

The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.Solution 9

Question 10

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.Solution 10

Question 11

The cost of 4 pens and 4 pencils boxes is Rs.100. Three times the cost of a pen is Rs.15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.Solution 11

Question 12

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?Solution 12

Question 13

A and B each have a certain number of mangoes. A says to B, “if you give 30 of your mangoes, I will have twice as many as left with you. “B replies, “if you give me 10, I will have thrice as many as left with you. “How many mangoes does each have?Solution 13

Question 14

Vijay had some bananas, and he divided them into two lots A and B. He sold first lot at the rate of Rs.2 for 3 bananas and the second lot at the rate of Rs.1 per banana and got a total of Rs.400. If he had sold the first lot at the rate of Rs.1 per banana and the second lot at the rate of Rs.4 per five bananas, his total collection would have been Rs.460. Find the total number of bananas he had.Solution 14

Question 15

Solution 15

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5The sum of two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14The sum of digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.Solution 14

Question 15

Solution 15

Question 16

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.Solution 16

Question 17

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.Solution 17

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6The present age of a father is three more than three times the age of the son. Three years hence father’s age will be 10 years more than twice the age of the son. Determine their present ages.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju.Solution 11

The difference between the ages of Ani and Biju is given as 3 years. So, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2 × x = 2x years
Case I: Ani is older than Biju by 3 years
x – y = 3        … (1)

4x – y = 60     ….(2)
Subtracting (1) from (2), we obtain: 3x = 60 – 3 = 57

Age of Ani = 19 years
Age of Biju = 19 – 3 = 16 years

Case II: Biju is older than Ani by 3 years
y – x = 3        … (3)

4x – y = 60        … (4)

Adding (3) and (4), we obtain:
3x = 63
x = 21

Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Concept Insight: In this problem, ages of Ani and Biju are the unknown quantities. So, we represent them by variables x and y. Now, note that here it is given that the ages of Ani and Biju differ by 3 years. So, it is not mentioned that which one is older. So, the most important point in this question is to consider both cases  Ani is older than Biju and  Biju is older than Ani. For second condition the relation  on the ages of Dharam and Cathy can be implemented . Pair of linear equations can be solved using a suitable algebraic method.

Question 12

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?Solution 12

Question 13

The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.Solution 13

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.10

Question 1

Solution 1

Question 2

Solution 2

Question 3The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of the boat in still water.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

A person rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of stream.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.Solution 12

Question 13

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.Solution 13

Question 14

Solution 14

Question 15A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time.  Find the distance covered by the train.Solution 15Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Or, d = xt        … (1)

According to the question,

By using equation (1), we obtain:
3x – 10t = 30        … (3)

Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) x (50) + 10t = 20
-100 + 10t = 20
10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 x 12 = 600

Thus, the distance covered by the train is 600 km.

Concept insight: To solve this problem, it is very important to remember the relation . Now, all these three quantities are unknown. So, we will represent these

by three different
variables. By using the given conditions, a pair of equations will be obtained. Mind one thing that the equations obtained will not be linear. But they can be reduced to linear form by using the fact that . Then two linear equations can be formed which can

be solved easily by elimination method.

Question 16Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hours. What are the speeds of two cars?Solution 16

Question 17

Solution 17

Question 18

Solution 18

Chapter 3 Pairs of Linear Equations in Two Variables Exercise Ex. 3.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

ABCD is a cyclic quadrilateral such that A = (4y + 20)^o, B = (3y – 5)^o, C = (-4x)^o and D = (7x + 5)^o. Find the four angles.Solution 6

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

A + C = 180
4y + 20 – 4x = 180
-4x + 4y = 160
x – y = -40            … (1)
Also, B + D = 180
3y – 5 – 7x + 5 = 180
-7x + 3y = 180        … (2)
Multiplying equation (1) by 3, we obtain:

3x – 3y = -120        … (3)
Adding equations (2) and (3), we obtain:
-4x = 60
x = -15
Substituting the value of x in equation (1), we obtain:
-15 – y = -40
y = -15 + 40 = 25

A = 4y + 20 = 4(25) + 20 = 120^o
B = 3y – 5 = 3(25) – 5 = 70^o
C = -4x = -4(-15) = 60^o
D = -7x + 5 = -7(-15) + 5 = 110^o

Concept insight: The most important idea to solve this problem is by using the fact that the sum of the measures of opposite angles in a cyclic quadrilateral is 180^o. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12The car hire charges in a city comprise of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is Rs 89 and for a journey of 20 km, the charge paid is Rs. 145. What will a person have to pay for travelling a distance of 30 km?Solution 12

Question 13A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, he has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per day.Solution 13

Question 14

Solution 14

Question 15The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.Solution 15

Question 162 Women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroideery, and that taken by 1 man alone.Solution 16

Question 17

Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes Rs 50 and Rs 100 she received.Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find number of students in the class.Solution 21

Question 22

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?Solution 22

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = -300        … (1)

6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y = 70            … (2)

Multiplying equation (2) by 2, we obtain:
12x – 2y = 140        … (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

Concept insight: This problem talks about the amount of capital with two friends. So, we will represent them by variables x and y respectively. Now, using the given conditions, a pair of linear equations can be formed which can then be solved easily using elimination method.

Question 23

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of Rs.1008. If she had sold the saree at 10% profit and sweater at 8% discount, she would have got Rs.1028. Find the cost price of the saree and the list price (price before discount) of the sweater.Solution 23

Question 24

In a competitive examination, one mark is awarded for each correct answer while ½ mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?Solution 24

Question 25

A shopkeeper gives book on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs.22 for a book kept for 6 days, while Rs.16 for the book kept for four days. Find the fixed charges and charge for each extraday.Solution 25

Chapter 3 Pairs of Linear Equations in Two Variables Exercise 3.114

Question 1

Solution 1

So, the correct option is (b).Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 3 Pairs of Linear Equations in Two Variables Exercise 3.115

Question 6

Solution 6

Question 7

If   am ≠ bl, then the system of equations

ax + by = c

lx + my = n

(a)  has a unique solution        

(b) has no solution

(c) has infinite many solution    

(d) may or may not have a solutionSolution 7

So, the correct option is (a).Question 8

Solution 8

So, the correct option is (b).Question 9

Solution 9

So, the correct option is (a).Question 10

If 2x-3y=7 and (a+b)x – (a+b-3)y = 4a+b represent coincident lines than a and b satisfy the equation

(a) a+5b=0    (b) 51+b=0    (c) a-5b=0     (d) 5a-b=0Solution 10

So, the correct option is (c).Question 11

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are

(a) Intersecting             (b) parallel

(c) always coincident       (d) intersecting or coincidentSolution 11

Consistent solution means either linear equations have unique solutions or infinite solutions.

⇒ In case of unique solution; lines are intersecting

⇒ If solutions are infinite, lines are coincident.

So, lines are either intersecting or coincident

So, the correct option is (d).Question 12

Solution 12

So, the correct option is (c).Question 13

The area of the triangle formed by the lines

y=x,  x=6, and  y=0  is

(a) 36 sq. units      

(b) 18 sq. units

(c) 9 sq. units         

(d) 72 sq. unitsSolution 13

So, the correct option is (b).Question 14

If the system of equations 2x + 3y=5,  4x + ky =10 has infinitely many solutions, then k=

(a) 1     

(b)     

(c) 3    

(d) 6Solution 14

So, the correct option is (d).Question 15

If the system of equations  kx – 5y = 2,  6x +2y=7 has no solution, then k=

(a) -10     

(b) -5      

(c) -6     

(d)-15Solution 15

So, the correct option is (d).Question 16

The area of the triangle formed by the lines

x = 3, y = 4 and x = y is

(a)  sq. unit    

(b) 1 sq. unit

(c) 2 sq. unit     

(d) None of theseSolution 16

So, the correct option is (a).

Chapter 3 – Pairs of Linear Equations in Two Variables Exercise 3.116

Question 17

The area of the triangle formed by the lines

2x + 3y = 12,     x – y – 1= 0  and x = 0

(a) 7 sq. units          

(b) 7.5 sq. units

(c) 6.5 sq. units        

(d) 6 sq. units   Solution 17

So, the correct option is (b).Question 18

The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is

1. 25
2. 72
3. 63
4. 36

Solution 18

Question 19

If x = a, y = b is the solution of the system of equations x – y = 2 and x + y = 4, then the values of a and b are, respectively

1. 3 and 1
2. 3 and 5
3. 5 and 3
4. -1 and -3

Solution 19

Since x = a and y = b is the solution of given system of equations x – y = 2 and x + y = 4, we have

a – b = 2 ….(i)

a + b = 4 ….(ii)

Adding (i) and (ii), we have

2a = 6 ⇒ a = 3

⇒ b = 4 – 3 = 1

Hence, correct option is (a).Question 20

For what value k, do the equations 3x – y + 8 = 0 and 6x – ky + 16 = 0 represent coincident lines?

1. 
2. 
3. 2
4. -2

Solution 20

Question 21

Aruna has only Rs.1 and Rs.2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs.75, then the number of Rs.1 and Rs.2 coins are, respectively

1. 35 and 15
2. 35 and 20
3. 15 and 35
4. 25 and 25

Solution 21

Chapter 2 Polynomials Exercise Ex. 2.1

Question 1(i)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

f(x) = x² – 2x – 8 Solution 1(i)

x² – 2x – 8 = x² – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4)(x + 2) The zeroes of the quadratic equation are 4 and -2. Let ∝ = 4 and β = -2 Consider f(x) = x² – 2x – 8 Sum of the zeroes = …(i) Also, ∝ + β = 4 – 2 = 2 …(ii) Product of the zeroes = …(iii) Also, ∝ β = -8 …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients isverified.Question 1(ii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

g(s) = 4s² – 4s + 1 Solution 1(ii)

4s² – 4s + 1 = 4s² – 2s – 2s + 1 = 2s(2s – 1) – (2s – 1) = (2s – 1)(2s – 1) The zeroes of the quadratic equation are and . Let ∝ =  and β =  Consider4s² – 4s + 1 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(iii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

h (t) = t² – 15 Solution 1(iii)

h (t) = t² – 15 = (t + √15)(t – √15)  The zeroes of the quadratic equation areand . Let ∝ =  and β =  Considert² – 15 = t² – 0t – 15 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(iv)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

f(s) = 6x² – 3 – 7x Solution 1(iv)

f(s) = 0 6x² – 3 – 7x =0 6x² – 9x + 2x – 3 = 0 3x (2x – 3) + (2x – 3) = 0 (3x + 1) (2x – 3) = 0 The zeroes of a quadratic equation are  and . Let ∝ =  and β =  Consider6x² – 7x – 3 = 0 Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(v)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(v)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(vi)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(vi)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider Sum of the zeroes = …(i) Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(vii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(vii)

 The zeroes of a quadratic equation are  and 1. Let ∝ =  and β = 1 Consider Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(viii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(viii)

 The zeroes of a quadratic equation are a and. Let ∝ = a and β = Consider Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(ix)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(ix)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(x)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(x)

 The zeroes of a quadratic equation are and . Let ∝ =  and β =  Consider=0 Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(xi)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(xi)

 The zeroes of a quadratic equation are  and. Let ∝ =  and β = Consider=0 Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 1(xii)

Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Solution 1(xii)

 The zeroes of a quadratic equation are  and. Let ∝ =  and β = Consider=0 Sum of the zeroes = …(i)  Also, ∝ + β = …(ii) Product of the zeroes = …(iii) Also, ∝ β = …(iv) Hence, from (i), (ii), (iii) and (iv),the relationship between the zeroes and their coefficients is verified.Question 2(i)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(i)

Question 2(ii)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(ii)

Question 2(iii)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(iii)

Question 2(iv)

For the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution 2(iv)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

(i)



(ii)



(iii)



(iv)



(v)



(vi)



(vii)



(viii)

 

Chapter 2 Polynomials Exercise Ex. 2.2

Question 1Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:

Solution 1

On comparing the given polynomial with the polynomial ax³ + bx² + cx + d, we obtain a = 2, b = 1, c = -5, d = 2

Thus, the relationship between the zeroes and the coefficients is verified.

On comparing the given polynomial with the polynomial ax^₃ + bx^₂ + cx + d, we obtain a = 1, b = -4, c = 5, d = -2.

Thus, the relationship between the zeroes and the coefficients is verified.

Concept insight: The zero of a polynomial is that value of the variable which makes the polynomial 0. Remember that there are three  relationships between the zeroes of a cubic polynomial and its coefficients which involve the sum of zeroes, product of all zeroes and the product of zeroes taken two at a time.

Question 2

Solution 2

Question 3

Find all zeroes of the polynomial 3x³ + 10x² – 9x – 4, if one of its zeroes is 1.Solution 3

Let f(x) = 3x³ + 10x² – 9x – 4 

As 1 is one of the zeroes of the polynomial, so (x – 1) becomes the factor of f(x).

Dividing f(x) by (x – 1), we have

Hence, the zeroes are  Question 4

If 4 is a zero of the cubic polynomial x³ – 3x² – 10x + 24, find its other two zeroes.Solution 4

Let f(x) = x³ – 3x² – 10x + 24 

As 4 is one of the zeroes of the polynomial, so (x – 4) becomes the factor of f(x).

Dividing f(x) by (x – 4), we have

Hence, the zeroes are 4, -3 and 2.Question 5

Solution 5

Question 6

Solution 6

Chapter 2 Polynomials Exercise Ex. 2.3

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Find all zeros of the polynomial 2x⁴ – 9x³ + 5x² + 3x – 1, if two of its zeros are  Solution 11

Let f(x) = 2x⁴ – 9x³ + 5x² + 3x – 1 

As  are two of the zeroes of the polynomial, so  becomes the factor of f(x).

Dividing f(x) by  we have

Hence, the other zeros  Question 12

For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5?Solution 12

Let f(x) = 3x⁴ – 9x³ + x² + 15x + k 

As f(x) is completely divisible by 3x² – 5, it becomes one of the factors of f(x).

Dividing f(x) by 3x² – 5, we have

As (3x² – 5) is one of the factors, the remainder will be 0.

Therefore, k + 10 = 0

Thus, k = -10.Question 13

Solution 13

Question 14

Solution 14

Chapter 2 Polynomials Exercise 2.61

Question 1

(a) 1

(b) -1

(c) 0

(d) None of theseSolution 1

Question 2

Solution 2

So, the correct option is (d).Question 3

If one zero of the polynomial f(x) = (k² + 4)x² + 13x + 4k is reciprocal of the other, then k =

(a) 2

(b) -2

(c) 1

(d) -1Solution 3

Chapter 2 Polynomials Exercise 2.62

Question 4

If the sum of the zeros of the polynomial f(x) = 2x³ – 3kx² + 4x – 5 is 6, then value of k is

(a) 2

(b) 4

(c) -2

(d) -4Solution 4

So, the correct option is (b).Question 5

(a) x² + qx + p

(b) x² – px + q

(c) qx² + px + 1

(d) px² + qx + 1Solution 5

Question 6

If α, β are the zeros of polynomial f(x) = x² – p(x + 1) – c, then (α + 1) (β + 1) =

(a) c – 1

(b) 1 – c

(c) c

(d) 1 + cSolution 6

Question 7

If α, β are the zeros of the polynomial f(x) = x² – p(x + 1) – c such that (α + 1) (β + 1) = 0 then c =

(a) 1

(b) 0

(c) -1

(d) 2Solution 7

Given that (α + 1) (β + 1) = 0

So, the correct option is (a).Question 8

If f(x) = ax² + bx + c has no real zeros and a + b + c < 0, then

(a) c = 0

(b) c > 0

(c) c < 0

(d) None of theseSolution 8

We know that, if the quadratic equation ax² + bx + c = 0 has no real zeros

then

Case 1:

a > 0, the graph of quadratic equation should not intersect x – axis

It must be of the type

Case 2 :

a < 0, the graph will not intersect x – axis and it must be of type

According to the question,

a + b + c < 0

This means,

f(1) = a + b + c

f(1) < 0

Hence, f(0) < 0 [as Case 2 will be applicable]

 So, the correct option is (c).Question 9

If the diagram in figure show the graph of the polynomial f(x) = ax² + bx + c then

(a) a < 0, b < 0 and c > 0

(b) a < 0, b < 0 and c < 0

(c) a < 0, b > 0 and c > 0

(d) a < 0, b > 0 and c < 0Solution 9

Question 10

Figure shows the graph of the polynomial f(x) = ax² + bx + c for which

(a) a < 0, b > 0 and c > 0

(b) a > 0, b < 0 and c > 0

(c) a < 0, b < 0 and c < 0

(d) a > 0, b > 0 and c < 0Solution 10

Question 11

Solution 11

Question 12

If zeros of the polynomial f(x) = x³ – 3px² + qx – r are in A.P, then

(a) 2p³ = pq – r

(b) 2p³ = pq + r

(c) p³ = pq – r

(d) None of theseSolution 12

Chapter 2 Polynomials Exercise 2.63

Question 13

Solution 13

So, the correct option is (b).Question 14

Solution 14

Question 15

In Q. No. 14, c =

(a) b

(b) 2b

(c) 2b²

(d) -2bSolution 15

So, the correct option is (c).Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

If two of the zeros of the cubic polynomial ax³  + bx² + cx + d are each equal to zero then the third zero is

Solution 21

Question 22

Solution 22

Question 23

The product of the zeros of x³ + 4x² + x – 6 is 

(a) -4

(b) 4

(c) 6

(d) -6Solution 23

Chapter 2 Polynomials Exercise 2.64

Question 24

What should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the resulting polynomial ?

(a) 1

(b) 2

(c) 4

(d) 5Solution 24

We know that, if α and β are roots of ax² + bx + c = 0 then they must satisfy the equation.

According to the question, the equation is

x² – 5x + 4 = 0

If 3 is the root of equation it must satisfy equation.

x² – 5x + 4 = 0

but f(3) = 3² – 5(3) + 4 = -2

so, 2 has to be added in the equation.

So, the correct option is (b).Question 25

What should be subtracted to the polynomial x² – 16x + 30, so that 15 is the zero of resulting polynomial?

(a) 30

(b) 14

(c) 15

(d) 16Solution 25

We know that, if α and β are roots of ax²  + bx + c = 0, then α and β must satisfy the equation.

According to the question, the equation is

x² – 16x + 30 = 0

If 15 is a root, then it must satisfy the equation x² – 16x + 30 = 0, 

But f(15) = 15² – 16(15) + 30 = 225 – 240 + 30 = 15

and so 15 should be subtracted from the equation.

So, the correct option is (c).Question 26

A quadratic polynomial, the sum of whose zeros is 0 and one zero is 3, is 

(a) x² – 9

(b) x² + 9

(c) x²  + 3

(d) x² – 3Solution 26

Question 27

If two zeros of the polynomial x³  + x² – 9x – 9 are 3 and -3, then its third zero is

(a) -1

(b) 1

(c) -9

(d) 9Solution 27

Question 28

Solution 28

Question 29

If x + 2 is a factor of x²  + ax + 2b and a + b = 4, then

(a) a = 1, b = 3

(b) a = 3, b = 1

(c) a = -1, b = 5

(d) a = 5, b = -1Solution 29

Question 30

The polynomial which when divided by -x² + x – 1 gives a quotient x – 2 and remainder is 3, is

(a) x³ – 3x² + 3x – 5

(b) -x³ – 3x² – 3x – 5

(c) -x³ + 3x²  – 3x + 5

(d) x³ – 3x² – 3x + 5Solution 30

We know that 

Dividend = Divisor × quotient  + remainder

Then according to question,

Required polynomial

= (-x² + x – 1) (x – 2) + 3

= -x³ + 2x² + x² -2x – x + 2 + 3

= -x³ + 3x² – 3x + 5

So, the correct option is (c).Question 31

The number of polynomials having zeroes -2 and 5 is

a. 1

b. 2

c. 3

d. more than 3Solution 31

Correct option: (d)

The polynomials having -2 and 5 as the zeroes can be written in the form 

k(x + 2)(x – 5), where k is a constant. 

Thus, number of polynomials with roots -2 and 5 are infinitely many, since k can take infinitely many values.Question 32

If one of the zeroes of the quadratic polynomial (k – 1)x² + kx + 1 is – 3, then the value of k is

a. 

b. 

c. 

d. Solution 32

Question 33

The zeroes of the quadratic polynomial x² + 99x + 127 are

a. Both positive

b. Both negative

c. both equal

d. One positive and one negativeSolution 33

The zeroes of the quadratic polynomial x² + 99x + 127 are both negative since all terms are positive.

Hence, correct option is (b).Question 34

If the zeroes of the quadratic polynomial x² + (a + 1)x + b are 2 and -3, then

a. a = -7, b = -1

b. a = 5, b = -1

c. a = 2, b = -6

d. a = 0, b = -6Solution 34

Question 35

Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is

a. 

b. 

c. 0

d. Solution 35

Question 36

The zeroes of the quadratic polynomial x² + ax + a, a ≠ 0,

a. cannot both be positive

b. cannot both be negative

c. are always unequal

d. are always equalSolution 36

Question 37

If one of the zeros of the cubic polynomial x³ + ax² + bx + x is -1, then the product of other two zeroes is

a. b – a + 1

b. b – a – 1

c. a – b + 1

d. a – b – 1Solution 37

Chapter 2 Polynomials Exercise 2.65

Question 38

Given that two of the zeros of the cubic polynomial ax³ + bx² + cx + d are 0, the third zero is

a. 

b. 

c. 

d. Solution 38

Question 39

If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is

a. 10

b. -10

c. 5

d. -5Solution 39

Question 40

If the zeros of the quadratic polynomial ax² + bx + c, c ≠ 0 are equal , then

a. c and a have opposite signs

b. c and b have opposite signs

c. c and a have the same sign

d. c and b have the same signSolution 40

It is given that the zeros of the quadratic polynomial ax² + bx + c, c ≠ 0 are equal.

⇒ Discriminant = 0

⇒ b² – 4ac = 0

⇒ b² = 4ac

Now, b² can never be negative,

Hence, 4ac also can never be negative.

⇒ a and c should have same sign.

Hence, correct option is (c).Question 41

If one of the zeros of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it

a. has no linear term and constant term is negative.

b. has no linear term and the constant term is positive

c. can have a linear term but the constant term is negative

d. can have a linear term but the constant term is positiveSolution 41

Question 42

Which of the following is not the graph of a quadratic polynomial?

Solution 42

The graph of a quadratic polynomial crosses X-axis at atmost two points.

Hence, correct option is (d).

Chapter 1 Real Numbers Exercise Ex. 1.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

For any positive integer n, prove that n³ – n divisible by 6.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.Solution 9

Question 10

Solution 10

Question 11

Show that any positive odd integer is of the form  6q + 1, or  6q + 3, or 6q + 5, where q is some integer.Solution 11

Let a be any odd positive integer we need to prove that a is of the form 6q+1 , or 6q+3 , or 6q+5 , where q is some integer.

Since a is an integer consider b = 6 another integer applying Euclid’s division lemma  we get
a = 6q + r f or some integer q  0, and r = 0, 1, 2, 3, 4, 5  since
0  r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4
(since all these are divisible by 2)

Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 +  1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.

Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, or 6q + 5 where q is some integer

Concept Insight:  In order to solve such problems  Euclid’s division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must  be of the form 6q + 1, 6q + 3, 6q + 5.

Basic definition of even (divisible by 2) and odd numbers (not divisible by 2) and the fact that addition and  multiplication of integers is always an integer are applicable here.Question 12

Prove that one of every three consecutive positive integers is divisible by 3.Solution 12

Let the three consecutive positive integers be m, (m + 1) and (m + 2).

By Euclid’s division lemma, when m is divided by 3, we have

m = 3q + r for some integer q ≥ 0 and r = 0, 1, 2

Case 1: When m = 3q

In this case, clearly, m is divisible by 3.

But, (m + 1) and (m + 2) are not divisible by 3.

Case 2: When m = 3q + 1

In this case, m + 2 = 3(q + 1), which is divisible by 3.

But, m and (m + 1) are not divisible by 3.

Case 3: When m = 3q + 2

In this case, m + 1 = 3(q + 1), which is divisible by 3.

But, m and (m + 2) are not divisible by 3.

Hence, one of m, (m + 1) and (m + 2) is always divisible by 3.Question 13

Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

[NCERT EXEMPLER]Solution 13

Let x be any positive integer.

When we divide x by 6, the remainder is either 0 or 1 or 2 or 3 or 4 or 5.

So, x can be written as

x = 6a or x = 6a + 1 or x = 6a + 2 or x = 6a + 3 or x = 6a + 4 or x = 6a + 5.

Thus, we have the following cases:

CASE I:

When x = 6a,

x² = 36a² = 6(6a²) = 6m, where m = 6a²

CASE II:

When x = 6a + 1,

x² = (6a + 1)² = 36a² + 12a + 1 = 6(6a² + 2a) + 1 = 6m + 1, where m = 6a² + 2a

CASE III:

When x = 6a + 2,

x² = (6a + 2)² = 36a² + 24a + 4 = 6(6a² + 4a) + 4 = 6m + 4, where m = 6a² + 4a

CASE IV:

When x = 6a + 3,

x² = (6a + 3)² = 36a² + 36a + 9 = (36a² + 36a + 6) + 3 = 6(6a² + 6a + 1) + 3 = 6m + 3, where m = 6a² + 6a + 1

CASE V:

When x = 6a + 4,

x² = (6a + 4)² = 36a² + 48a + 16 = (36a² + 48a + 6) + 10 = 6(6a² + 8a + 1) + 10 = 6m + 10, where m = 6a² + 8a + 10

CASE VI:

When x = 6a + 5,

x² = (6a + 5)² = 36a² + 60a + 25 = (36a² + 60a + 6) + 19 = 6(6a² + 10a + 1) + 19 = 6m + 19, where m = 6a² + 10a + 19

Here, x is of the form 6m or 6m + 1 or 6m + 3 or 6m + 4 or 6m + 10 or 6mn + 19.

So, it cannot be of the form 6m + 2 or 6m + 5.Question 14

Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.Solution 14

Let x be any positive integer.

Then, it is of the form 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4 or 6n + 5.

So, we have the following cases:

CASE I:

When x = 6n,

x³ = (6n)³ = 216n³ = 6(36n³) = 6q, where q = 36n³

CASE II:

When x = 6n + 1,

x³ = (6n + 1)³ = 216n³ + 108n² + 18n + 1 = 6(36n³ + 18n² + 3n) + 1 = 6q + 1, where q = 36n³ + 18n² + 3n

CASE III:

When x = 6n + 2,

x³ = (6n + 2)³ = 216n³ + 216n² + 72n + 8 = 216n³ + 216n² + 72n + 6 + 2 =6(36n³ + 36n² + 12n + 1) + 2 = 6q + 2, where q = 36n³ + 54n² + 12n + 1

CASE IV:

When x = 6n + 3,

x³ = (6n + 3)³ = 216n³ + 324n² + 162n + 27 = 216n³ + 324n² + 162n + 24 + 3 =6(36n³ + 54n² + 27n + 4) + 3 = 6q + 3, where q = 36n³ + 54n² + 27n + 4

CASE V:

When x = 6n + 4,

x³ = (6n + 4)³ = 216n³ + 432n² + 288n² + 64 = 216n³ + 432n² + 288n + 60 + 4 =6(36n³ + 72n² + 48n + 10) + 4 = 6q + 4, where q = 36n³ + 72n² + 48n + 10

CASE VI:

When x = 6n + 5,

x³ = (6n + 5)³ = 216n³ + 540n² + 450n² + 125 = 216n³ + 540n² + 450n + 120 + 5 =6(36n³ + 90n² + 75n + 20) + 5 = 6q + 5, where q = 36n³ + 72n² + 48n + 10

Thus, the cube of any positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.Question 15

Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.Solution 15

Given numbers are n, n + 4, n + 8, n + 12 and n + 16.

Let n = 5q + r, where 0 ≤ r < 5

n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4 for any natural number q.

So, we have the following cases:

CASE I:

When n = 5q

n = 5q is divisible by 5

n + 4 = 5q + 4 is not divisible by 5

n + 8 = 5q + 8 = 5q + 5 + 3 = 5(q + 1) + 3 is not divisible by 5

n + 12 = 5q + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5

n + 16 = 5q + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5

CASE II:

When n = 5q + 1

n = 5q + 1 is not divisible by 5

n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) is divisible by 5

n + 8 = 5q + 1 + 8 = 5q + 9 = 5q + 5 + 4 = 5(q + 1) + 4 is not divisible by 5

n + 12 = 5q + 1 + 12 = 5q + 13 = 5q + 10 + 3 = 5(q + 2) + 3 is not divisible by 5

n + 16 = 5q + 1 + 16 = 5q + 17 = 5q + 15 + 2 = 5(q + 3) + 2 is not divisible by 5

CASE III:

When n = 5q + 2

n = 5q + 2 is not divisible by 5

n + 4 = 5q + 2 + 4 = 5q + 6 = 5q + 5 + 1 = 5(q + 1) + 1 is not divisible by 5

n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) is divisible by 5

n + 12 = 5q + 2 + 12 = 5q + 14 = 5q + 10 + 4 = 5(q + 2) + 4 is not divisible by 5

n + 16 = 5q + 2 + 16 = 5q + 18 = 5q + 15 + 3 = 5(q + 3) + 3 is not divisible by 5

CASE IV:

When n = 5q + 3

n = 5q + 3 is not divisible by 5

n + 4 = 5q + 3 + 4 = 5q + 7 = 5q + 5 + 2 = 5(q + 1) + 2 is not divisible by 5

n + 8 = 5q + 3 + 8 = 5q + 11 = 5(q + 2) + 1 is not divisible by 5

n + 12 = 5q + 3 + 12 = 5q + 15 = 5(q + 3) is divisible by 5

n + 16 = 5q + 3 + 16 = 5q + 19 = 5q + 15 + 4 = 5(q + 3) + 4 is not divisible by 5

CASE V:

When n = 5q + 4

n = 5q + 4 is not divisible by 5

n + 4 = 5q + 4 + 4 = 5q + 8 = 5q + 5 + 3 = 5(q + 1) + 3 is not divisible by 5

n + 8 = 5q + 4 + 8 = 5q + 12 = 5(q + 2) + 2 is not divisible by 5

n + 12 = 5q + 4 + 12 = 5q + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5

n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4) is divisible by 5

Hence, in each case, one and only one out of n, n + 4, , n + 8, n + 12 and n + 16 is divisible by 5.Question 16

Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.Solution 16

We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5 for some integer m.

Thus, an odd positive integer x can be of the form 6m + 1, 6m + 3 or 6m + 5.

Thus, we have

CASE I:

x = 6m + 1

x² = (6m + 1)² = 36m² + 12m + 1 = 6(6m² + 2m) + 1 = 6q + 1, where q = 6m² + 2m

CASE II:

x = 6m + 3

x² = (6m + 3)² = 36m² + 36m + 9 = 36m² + 36m + 6 + 3 = 6(6m² + 6m + 1) + 3 = 6q + 3, where q = 6m² + 6m + 1

CASE III:

x = 6m + 5

x² = (6m + 5)² = 36m² + 60m + 25 = 36m² + 60m + 24 + 1 = 6(6m² + 10m + 4) + 1 = 6q + 1, where q = 6m² + 10m + 4

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.Question 17

A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m? Justify your answer.Solution 17

By Euclid’s Lemma,

a = bq + r, 0 ≤ r < b

Here, a is any positive integer and b = 3,

a = 3q + r

So, this must be in the form 3q, 3q + 1 or 3q + 2.

Now,

(3q)² = 9q² = 3m

(3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1

(3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1

Thus, square of a positive integer of the form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.Question 18

Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.Solution 18

Let x be any positive integer.

So, x can be written as

x = 3a or x = 3a + 1 or x = 3a + 2

Thus, we have the following cases:

CASE I:

When x = 3a,

x² = 9a² = 3(3a²) = 3m, where m = 3a²

CASE II:

When x = 3a + 1,

x² = (3a + 1)² = 9a² + 6a + 1 = 3(3a² + 2a) + 1 = 3m + 1, where m = 3a² + 2a

CASE III:

When x = 3a + 2,

x² = (3a + 2)² = 9a² + 12a + 4 = 9a² + 12a + 3 + 1 = 3(3a² + 4a + 1) + 1 = 3m + 1, where m = 3a² + 4a + 1

Thus, the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

Chapter 1 Real Numbers Exercise Ex. 1.2

Question 1(i)Find H.C.F. of 32 and 54
Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)Find H.C.F. of 100 and 190Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 2(i)

Use Euclid’s division algorithm to find the HCF of: 

135 and 225 

Solution 2(i)

 135 and 225

Step 1: Since 225 > 135, apply Euclid’s division lemma, to a =225 and b=135 to find q and r such that 225 = 135q+r, 0 r

On dividing 225 by 135 we get quotient as 1 and remainder as 90
i.e 225 = 135 x 1 + 90

Step 2: Remainder r which is 90 0, we apply Euclid’s division lemma to a = 135 and b = 90 to find whole numbers q and r such that
135 = 90 x q + r 0 r<90
On dividing 135 by 90 we get quotient as 1 and remainder as 45
i.e 135 = 90 x 1 + 45

Step 3: Again remainder r = 45 0 so we apply Euclid’s division lemma to a = 90 and b = 45 to find q and r such that
90 = 45 x q + r 0 r<45
On dividing 90 by 45 we get quotient as 2 and remainder as 0
i.e 90 = 2 x 45 + 0

Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (135, 225).

Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.Question 2(iv)

Use Euclid’s division algorithm to find the HCF of

184, 230 and 276Solution 2(iv)

Question 2(v)

Use Euclid’s division algorithm to find the HCF

136, 170 and 255Solution 2(v)

Question 2(vi)

Use Euclid’s division algorithm to find the HCF of

1260 and 7344Solution 2(vi)

Step 1: Since 7344 > 1260, apply Euclid’s division lemma, to a = 7344 and b = 1260 to find q and r such that 7344 = 1260q + r, 0 ≤ r < 1260

On dividing 7344 by 1260 we get quotient as 5 and remainder as

i.e. 7344 = 1260 x 5 + 1044

Step 2: Remainder r which is 1044, we apply Euclid’s division lemma to a = 1260 and b = 1044 to find whole numbers q and r such that

1260 = 1044 x q + r, 0 ≤ r < 1044

On dividing 1260 by 1044 we get quotient as 1 and remainder as 216

i.e. 1260 = 1044 x 1 + 216

Step 3: Again remainder r = 216, so we apply Euclid’s division lemma to a = 1044 and b = 216 to find q and r such that

1044 = 216 x q + r, 0 ≤ r < 216

On dividing 1044 by 216 we get quotient as 4 and remainder as 180

i.e. 1044 = 216 x 4 + 180

Step 4: Again remainder r = 180, so we apply Euclid’s division lemma to a = 216 and b = 180 to find q and r such that

216 = 180 x q + r, 0 ≤ r < 180

On dividing 216 by 180 we get quotient as 1 and remainder as 36

i.e. 216 = 180 x 1 + 36

Step 5: Again remainder r = 36, so we apply Euclid’s division lemma to a = 180 and b = 36 to find q and r such that

180 = 36 x q + r, 0 ≤ r < 36

On dividing 180 by 36 we get quotient as 5 and remainder as 0

i.e. 180 = 36 x 5 + 0

Step 6: Since the remainder is zero, the divisor at this stage will be HCF of (7344, 1260).

Since the divisor at this stage is 36, therefore, the HCF of 7344 and 1260 is 36.Question 2(vii)

Use Euclid’s division algorithm to find the HCF of

2048 and 960Solution 2(vii)

Step 1: Since 2048 > 960, apply Euclid’s division lemma, to a = 2048 and b = 960 to find q and r such that 2048 = 960q + r, 0 ≤ r < 960

On dividing 2048 by 960 we get quotient as 5 and remainder as

i.e. 2048 = 960 x 2 + 128

Step 2: Remainder r which is 128, we apply Euclid’s division lemma to a = 960 and b = 128 to find whole numbers q and r such that

960 = 128 x q + r, 0 ≤ r < 128

On dividing 960 by 128 we get quotient as 7 and remainder as 216

i.e. 960 = 128 x 7 + 64

Step 3: Again remainder r = 64, so we apply Euclid’s division lemma to a = 128 and b = 64 to find q and r such that

128 = 64 x q + r, 0 ≤ r < 64

On dividing 128 by 64 we get quotient as 2 and remainder as 0

i.e. 128 = 64 x 2 + 0

Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (2048, 960).

Since the divisor at this stage is 64, therefore, the HCF of 2048 and 960 is 64.Question 3(i)

Solution 3(i)

Question 3(ii)Find H.C.F. of 592 and 252 and express it as a linear combination of them.Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Find the largest number which divides 615 and 963 leaving remainder 6 in each case.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 2(ii)

Use Euclid’s division algorithm to find the HCF of: 
 

196 and 38220 

Solution 2(ii)

196 and 38220

Step 1: Since 38220 > 196, apply Euclid’s division lemma
to a =38220 and b=196 to find whole numbers q and r such that
38220 = 196 q + r, 0 r < 196
On dividing 38220 we get quotient as 195 and remainder r as 0
i.e 38220 = 196 x 195 + 0
Since the remainder is zero, divisor at this stage will be HCF
Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.

NOTE: HCF( a,b) = a if a is a factor of b. Here, 196 is a factor of 38220 so HCF is 196.

Question 2(iii)

Use Euclid’s division algorithm to find the HCF of: 

867 and 255Solution 2(iii)

867 and 255

Step 1: Since 867 > 255, apply Euclid’s division lemma, to a =867 and b=255 to find q and r such that 867 = 255q + r, 0 r<255
On dividing 867 by 255 we get quotient as 3 and remainder as 102
i.e 867 = 255 x 3 + 102

Step 2: Since remainder 102 0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that
255 = 102q + r where 0 r<102
On dividing 255 by 102 we get quotient as 2 and remainder as 51
i.e 255 = 102 x 2 + 51

Step 3: Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51 to find whole numbers q and r such that
102 = 51 q + r where 0 r < 51

On dividing 102 by 51 quotient is 2 and remainder is 0
i.e 102 = 51 x 2 + 0
Since the remainder is zero, the divisor at this stage is the HCF
Since the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51.

Concept Insight: To crack such problem remember to apply the Euclid’s division Lemma which states that “Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0 r < b” in the correct order.

Here, a > b.

Euclid’s algorithm works since Dividing ‘a’ by ‘b’, replacing ‘b’ by ‘r’ and ‘a’ by ‘b’ and repeating the process of division till remainder 0 is reached, gives a number which divides a and b exactly.

i.e HCF(a,b) =HCF(b,r)

Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps.

lemma to a=135 and b=

Chapter 1 Real Numbers Exercise Ex. 1.3

Question 1

Solution 1

Question 2

Solution 2

Question 3Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.Solution 3Numbers are of two types – prime and composite. Prime numbers has only two factors namely 1 and the number itself  whereas composite numbers have factors other than 1 and itself.

It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6

The given expression has 6 and 13  as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x  4 x 3 x 2 x 1 + 1)
 = 5 x (1008 + 1)
 = 5 x 1009

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.

Question 4Check whether 6ⁿ can end with the digit 0 for any natural number n.Solution 4If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorisation must include primes 2 and  5 both
Prime factorisation of 6_ⁿ = (2 x 3)_ⁿ

By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6^_n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6^_n cannot end with the digit 0 for any natural number n.

Concept Insight: In order solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorisation of a number is unique.

Question 5

Explain why 3 × 5 × 7 + 7 is a composite number.Solution 5

Numbers are of two types – prime and composite. Prime numbers has only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself. It can be observed that 3 × 5 × 7+ 7 = 7 × (3 × 5 + 1) = 7 × (15 + 1) = 7 × 16

The given expression has 7 and 16 as its factors. Therefore, it is a composite number. 

Chapter 1 Real Numbers Exercise Ex. 1.4

Question 1

Solution 1

Question 1(iv)

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers: 

404 and 96Solution 1(iv)

404 = 2 × 2 × 3 × 37 = 2² × 101

96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

H.C.F. = 2² = 4

L.C.M. = 2⁵ × 3 × 101 = 9696

Now, H.C.F. × L.C.M. = 4 × 9696 = 38784

Product of numbers = 404 × 96 = 38784

Hence, H.C.F. × L.C.M. = Product of numbers.Question 2Find the LCM and HCF of the following integers by applying the prime factorisation method:

(i) 12,15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(iv) 40, 36 and 126

(v) 84, 90 and 120

(vi) 24, 15 and 36.Solution 2

Concept Insight: HCF is the product of common prime factors of all three numbers  raised to least power, while LCM is product of prime factors of all here  raised to highest power.  Use the fact that HCF is always a factor of the LCM to verify the answer. Note HCF of (a,b,c)  can also be calculated by taking two numbers at a time i.e HCF (a,b) and then HCF (b,c) .Question 3

Given that HCF (306, 657) = 9, find LCM (306, 657).Solution 3

Concept Insight: This problem must be solved using product of two numbers = HCF x LCM rather than prime factorisationQuestion 3(ii)

Write the smallest number which is divisible by both 306 and 657.Solution 3(ii)

The smallest number divisible by both 306 and 657 is the LCM if these numbers.

306 = 2 × 3 × 3 × 17 = 2 × 3² × 17

657 = 3 × 3 × 73 = 3² × 73

L.C.M. = 2 × 3² × 17 × 73 = 22338

Hence, the smallest number which is divisible by both 306 and 657 is 22338. Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?Solution 17

Required minimum distance each should walk so that they can cover the distance in complete step is the L.C.M. of 30 cm, 36 cm and 40 cm.

30 = 2 × 3 × 5;

36 = 2²× 3²;

40 = 2³× 5

∴ LCM (30, 36, 40) = 2³× 3²× 5

∴ LCM = 2³× 3²× 5 = 360 cm.Question 18

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.Solution 18

Clearly, the required number is the H.C.F of the numbers

 1251 – 1 = 1250, 9377 – 2 = 9375, and 15628 – 3 = 15625.

First we’ll find the H.C.F of 1250 and 9375 by Euclid’s algorithm as given below:

9375 = 1250 × 7 + 625

1250 = 625 × 2 + 0

Clearly, H.C.F of 1250 and 9375 is 625.

Let us now find the H.C.F of 625 and the third number 15625 by Euclid’s algorithm:

15625 = 625 × 25 + 0

Hence, the required number is 625.

Chapter 1 Real Numbers Exercise Ex. 1.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Prove that  is an irrational number.Solution 10

Question 11

Given that   is irrational, prove that   is an irrational number.Solution 11

Let us assume that   is a rational number.

So, there exist co-prime positive integers a and b such that

As  is rational, so   is rational.

Then,   is also rational.

Thus,   is rational.

But this is a contradiction to the fact that   is an irrational number.

Hence,   is an irrational number.Question 12

Prove that   is an irrational number, given that   is an irrational number.Solution 12

Let us assume that   is a rational number.

So, there exist co-prime positive integers a and b such that

As  is rational, so   is rational.

Then,   is also rational.

Thus,   is rational.

But this is a contradiction to the fact that   is an irrational number.

Hence,   is an irrational number.Question 13

Prove that   is an irrational number, given that   is an irrational number.Solution 13

Let us assume that   is a rational number.

So, there exist co-prime positive integers a and b such that

As  is rational, so   is rational.

Then,   is also rational.

Thus,   is rational.

But this is a contradiction to the fact that   is an irrational number.

Hence,   is an irrational number.Question 14

Solution 14

Chapter 1 Real Numbers Exercise Ex. 1.6

Question 1(i)

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution 1(i)

Question 1(ii)

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution 1(ii)

Question 1(iii)

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution 1(iii)

Question 1(iv)

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution 1(iv)

Question 1(v)

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution 1(v)

Question 1(vi)

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution 1(vi)

Question 2

Solution 2

Question 3

Write the denominator of the rational number  in the form 2^m × 5ⁿ, where m, n are non-negative integers. Hence, write the decimal expansion, without actual division.Solution 3

Question 4

Solution 4

Question 5

A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form? Give reasons.Solution 5

Chapter 1 Real Numbers Exercise 1.59

Question 1

The exponent of 2 in the prime factorisation of 144, is

(a) 4

(b) 5

(c) 6

(d) 3Solution 1

Factorisation of 144 can be done as follows

so 144 = 2 × 2 × 2 × 2 × 3 × 3

          = 2⁴ × 3²

Exponent of 2 in the prime factorisation of 144 is 4.

So, the correct option is (a).Question 2

The LCM of two numbers is 1200. Which of the following cannot be their HCF ?

(a) 600

(b) 500

(c) 400

(d) 200Solution 2

We know that LCM of two numbers is divisible of HCF of these two numbers.

Hence

(a) 1200 is divisible by 600. So 600 can be the HCF.

(b) 500 cannot be the HCF because 1200 is not divisible by 500.

(c) 400 can be the HCF because 1200 is divisible by 400.

(d) 200 can be the HCF because 1200 is divisible by 200.

So, the correct option is (b).Question 3

If n = 2³ × 3⁴ × 5⁴ × 7, then the number of consecutive zeros in n, where n is a natural number, is

(a) 2

(b) 3

(c) 4

(d) 7Solution 3

n can also be written as

3⁴ × 2³ × 5³ × 5 × 7

3⁴ × (2 × 5)³ × 5 × 7

3⁴ × 5 × 7 × 10³

exponent of 10 in n is 3.

Hence number of consecutive zeros in n is 3.

So, the correct option is (b).Question 4

The sum of the exponents of the prime factors in the prime factorisation of 196, is

(a) 1

(b) 2

(c) 4

(c) 6Solution 4

Factorisation of 196 is

so 196 = 2 × 2 × 7 × 7

           = 2² × 7²

exponent of 2 is 2

exponent of 7 is 2

Hence sum of exponents is 4.

So, the correct option is (c).Question 5

(a) 1

(b) 2

(c) 3

(d) 4Solution 5

So, the correct option is (b).Question 6

(a) an even number

(b) an odd number

(c) an odd prime number

(d) a prime numberSolution 6

So, the correct option is (a).Question 7

If two positive integers a and b are expressible in the form a = pq² and b = p³q ; p, q being prime numbers,

then LCM (a, b) is

(a) pq

(b) 

(c) 

(d) Solution 7

LCM (a, b) is

LCM (a, b) = p × q × q × p²

               = p³q²

So, the correct option is (c).Question 8

In Q. no. 7, HCF (a, b) is

(a) pq

(b) 

(c) 

(d) Solution 8

HCF (a, b) is

No further common division is possible

Hence HCF (a, b) = p × q

                         = pq

So, the correct option is (a).

Chapter 1 Real Numbers Exercise 1.60

Question 9

If two positive numbers m and n are expressible in the form m = pq³ and n = p³q², where p, q are prime numbers,

then HCF (m, n) = 

(a) pq

(b) pq²

(c) p³q³

(d) p²q³Solution 9

HCF of m, n is

No further division is possible

Hence HCF is p × q² = pq²

So, the correct option is (b).Question 10

If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =

(a) 2

(b) 3

(c) 4

(d) 1Solution 10

We know,

LCM (a, b) × HCF (a, b) = a × b

So LCM (a, 18) × HCF (a, 18) = a × 18

36 × 2 = a × 18

a = 4

So, the correct option is (c).Question 11

The HCF of 95 and 152, is 

(a) 57

(b) 1

(c) 19

(d) 38Solution 11

HCF (95, 152)

No further common division is possible.

Hence HCF (95, 152) is 19.

So, the correct option is (c).Question 12

If HCF (26, 169) = 13, then LCM (26, 169) =

(a) 26

(b) 52

(c) 338

(d) 13Solution 12

We know

LCM (a, b) × HCF (a, b) = a × b

so LCM (26, 169) × HCF (26, 169) = 26 × 169

So, the correct option is (c).Question 13

If a = 2³ × 3, b = 2 × 3 × 5, c = 3ⁿ × 5 and LCM (a, b, c) = 2³ × 3² × 5 then n =

(a) 1

(b) 2

(c) 3

(d) 4Solution 13

LCM (a, b, c) is

LCM (a, b, c) = 3 × 2 × 5 × 2² × 3^{n – 1}

                  = 2³ × 3ⁿ × 5            ……..(1)

given that 

LCM (a, b, c) = 2³ × 3² × 5             ……..(2)

from (1) & (2)

n = 2

So, the correct option is (b).Question 14

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) four decimal placesSolution 14

So, the correct option is (d).Question 15

If p and q are co – prime numbers, then p² and q² are

(a) coprime

(b) not coprime

(c) even

(d) oddSolution 15

If p and q are co-prime numbers then

HCF (p, q) = 1

After squaring the numbers, we get p² and q² 

If two numbers have HCF = 1 then after squaring the numbers their HCF remains equal to 1.

Hence HCF (p² , q²) = 1

so p² and q² are co – prime numbers.

Ex : 2 and 3 are co – prime numbers.

HCF (2, 3) = 1

after squaring

HCF (4, 9) = 1

Hence, 4, 9 are also co – prime.

So, squares of two co – prime numbers are also co – prime.

So, the correct option is (a).Question 16

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (i) and (iii)

(d) (i) and (iv)Solution 16

So, the correct option is (d).

*Note: Since the book has a typo error, the question has been modified.Question 17

If 3 is the least prime factor of number a and 7 is the least prime factor of number b,

then least prime factor of a + b is 

(a) 2

(b) 3

(c) 5

(d) 10Solution 17

It is given that 3 is the least prime factor of number a so a can be 3 (least possible value)

It is given that 7 is the least prime factor of number b so least possible value of b is 7.

Hence a + b = 10 (least possible value)

prime factors of 10 are 2 and 5

Hence the least prime factor of a + b is 2.

So, the correct option is (a).Question 18

(a) an integer

(b) a rational number

(c) a natural number

(d) an irrational numberSolution 18

We know that decimal expansion of a rational number is either terminating or non-terminating and recurring.

So the correct option is (b).Question 19

(a) 

(b) 

(c) 

(d) 3Solution 19

Question 20

Solution 20

Question 21

If n is a natural number, then 9²ⁿ – 4²ⁿ is always divisible by

(a) 5

(b) 13

(c) both 5 and 13

(d) None of theseSolution 21

We know a²ⁿ  – b²ⁿ is always divisible by a – b and a + b

On comparing with 9²ⁿ – 4²ⁿ, we get a = 9 & b = 4

Hence 9²ⁿ – 4²ⁿ  is divisible by 9 – 4 & 9 + 4

                                             = 5 & 13

So, the correct option is (c).

Chapter 1 Real Numbers Exercise 1.61

Question 22

If n is any natural number, then 6ⁿ  –  5ⁿ  always ends with

(a) 1

(b) 3

(c) 5

(d) 7Solution 22

6ⁿ always ends with 6

5ⁿ always ends with 5

Hence 6ⁿ – 5ⁿ always with 6 – 5 = 1

So, the correct option is (a).Question 23

The LCM and HCF of two rational numbers are equal, then the numbers must be 

(a) prime

(b) co-prime

(c) composite

(d) equalSolution 23

(a) If two numbers are prime then their HCF must be 1 but LCM can’t be 1

      Example: 2, 3

      LCM (2, 3) = 6

      HCF (2, 3) = 1

(b) If two numbers are co – prime then their HCF must be 1 but LCM can’t be 1.

(c) If two numbers are composite then their LCM and HCF can only be equal if the two numbers are same.

(d) If the numbers are equal.

     Example: 6, 6

      LCM (6, 6) = 6

      HCF (6, 6) = 6

      LCM = HCF

So, the correct option is (d).Question 24

If sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is

(a) 203400

(b) 194400

(c) 198400

(d) 205400Solution 24

Let numbers be a, b

It is given that LCM (a, b) + HCF (a, b) = 1260           ……….(1)

                    LCM (a, b) – HCF (a, b) = 900             ……….(2)

Adding equations (1) and (2), we get 2LCM (a, b) = 2160 

Subtracting equations (1) and (2), we get 2HCF (a, b) = 360 

So, LCM (a, b) = 1080 and

HCF (a, b) = 180

We know LCM (a, b) × HCF (a, b) = ab

ab = 1080 × 180

     = 194400

So, the correct option is (b).Question 25

The remainder when the square of any prime number greater than 3 is divided by 6, is 

(a) 1

(b) 3

(c) 2

(d) 4Solution 25

Question 26

For some integer m, every even integer is of the form

1. m
2. m + 1
3. 2m
4. 2m + 1

Solution 26

m is an integer.

⇒ m = ….., -2, -1, 0, 1, 2, …..

⇒ 2m = ……., -4, -2, 0, 2, 4, ……

Hence, correct option is (c).Question 27

For some integer q, every odd integer is of the form

1. q
2. q + 1
3. 2q
4. 2q + 1

Solution 27

q is an integer.

⇒ q = ….., -2, -1, 0, 1, 2, …..

⇒ 2q + 1 = ……., -3, -1, 0, 3, 5, ……

Hence, correct option is (d).Question 28

n² – 1 is divisible by 8, if n is

1. an integer
2. a natural number
3. an odd integer
4. an even integer

Solution 28

Let a = n² – 1

Now, when n is odd, i.e. n = 2k + 1, we have

a = (2k + 1)² – 1 =4k² + 4k + 1 – 1 = 4k(k + 1)

At k = -1, we get

a = 4(-1)(-1 + 1) = 0, which is divisible by 8.

At k = 0, we get

a = 4(0)(0 + 1) = 0, which is divisible by 8.

At k = 1, we get

A = 4(1)(1 + 1) = 4(2) = 8, which is divisible by 8.

Hence, correct option is (c).Question 29

The decimal expansion of the rational number  will terminate after

1. one decimal place
2. two decimal places
3. three decimal places
4. more than 3 decimal places

Solution 29

Question 30

If two positive integers a and b are written as a = x³y² and b = xy³, x, y are prime numbers, then HCF (a, b) is

1. xy
2. xy²
3. x³y³
4. x²y²

Solution 30

Question 31

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

1. 13
2. 65
3. 875
4. 1750

Solution 31

Question 34

Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy

1. 1 < r < b
2. 0 < r ≤ b
3. 0 ≤ r < b
4. 0 < r < b

Solution 34

Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy 0 ≤ r < b.

Hence, correct option is (c).Question 32

The product of a non-zero rational number and an irrational number is

(a) always irrational

(b) always rational

(c) rational or irrational

(d) oneSolution 32

The product of a non-zero rational number and an irrational number is always irrational. Question 33

The HCF and LCM of 12, 21, 15 respectively are

(a) 3, 140

(b) 12, 420

(c) 3, 420

(d) 420, 3Solution 33

Here, 12 = 2²× 3, 21 = 3 × 7 and 15 = 3 × 5

HCF = 3

LCM = 2²× 3 × 5 × 7 = 420

Chapter 26 Ellipse Exercise Ex. 26.1

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 5(v)

Solution 5(v)

Question 5(vi)

Solution 5(vi)

Question 5(vii)

Solution 5(vii)

Question 5(viii)

Solution 5(viii)

Question 5(ix)

Solution 5(ix)

Question 5(x)

Solution 5(x)

Question 5(xi)

Solution 5(xi)

Question 5(xii)

Solution 5(xii)

Question 5(xiii)

Solution 5(xiii)

Question 6

Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 10(iii)

Solution 10(iii)

Question 10(iv)

Solution 10(iv)

Question 10(v)

Solution 10(v)

Question 10(vi)

Solution 10(vi)

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 3(v)

Find the eccentricity, coordinates of foci, length of the latus-rectum of the ellipse 9x² + 25y² = 225Solution 3(v)

Question 18

A rod of length 12m moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the, which is 3cm from the end in contract with x-axis.Solution 18

Question 19

Find the equation of the set of all points whose distances from (0, 4) are of their distances from the line y = 9.Solution 19

Question 20

Find the equation of the set of all points whose distances from (0, 4) are 2/3 of their distances from the line y = 9.Solution 20

Chapter 13 Complex Numbers Exercise Ex. 13.1

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 2

Solution 2

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Question 3 (v)

Solution 3 (v)

Question 3 (vi)

Solution 3 (vi)

Question 3(vii)

Find the values of the following expressions:

(1 + i)⁶ + (1 – i)³Solution 3(vii)

Chapter 13 Complex Numbers Exercise Ex. 13.2

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

What is the smallest positive integer n for which

(1+i)²ⁿ = (1-i)²ⁿ?Solution 24

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Solution 1 (ix)

Question 1 (x)

Solution 1 (x)

Question 1 (xi)

Solution 1 (xi)

Question 1 (xii)

Solution 1 (xii)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Question 3 (v)

Solution 3 (v)

Question 3 (vi)

Solution 3 (vi)

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 16 (i)

Solution 16 (i)

Question 16 (ii)

Solution 16 (ii)

Question 16 (iii)

Solution 16 (iii)

Question 16 (iv)

Solution 16 (iv)

Question 16(v)

Evaluate the following :

Solution 16(v)

Question 25

If z₁, z₂, z₃ are complex numbers such that |z₁| = |z₂| = |z₃| =  = 1, then find the value of |z₁ + z₂ + z₃|.Solution 25

Question 26

Find the number of solutions of z² + |z|² = 0Solution 26

Chapter 13 Complex Numbers Exercise Ex. 13.3

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Find the square root of the following complex numbers: -iSolution 1 (ix)

Chapter 13 Complex Numbers Exercise Ex. 13.4

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 2

Write (i²⁵)³ in polar form.Solution 2

Question 3(i)

Express the following complex numbers in the form

 r(cos q + I sin q): 1 + i tan a Solution 3(i)

Question 3(ii)

Express the following complex numbers in the form

 r(cos q + I sin q): tan a – iSolution 3(ii)

Question 3(iii)

Express the following complex numbers in the form

r (cos q + I sin q): 1 – sin a + i cos a Solution 3(iii)

Question 3(iv)

Express the following complex numbers in the form

 r (cos q + I sin q):Solution 3(iv)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 33 Probability Exercise Ex. 33.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

An experiment consists of tossing a coin and then tossing it second time if head occurs. If a tail occurs on the first toss, then a die is tossed once. Find the sample space.Solution 10

In this experiment, a coin is tossed and if the outcome is tail then a die is tossed once.

Otherwise, the coin is tossed again.

The possible outcome for coin is either head or tail.

The possible outcome for die is 1,2,3,4,5,6.

If the outcome for the coin is tail then sample space is S1={(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

If the outcome is head then the sample space is S2={(H,H),(H,T)}

Therefore the required sample space is S={(T,1),(T,2),(T,3),(T,4),(T,5),(T,6),(H,H),(H,T)}Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

In a random sampling, three items are selected so it could be any of the following:

a) All defective or

b) All non-defective or

c) Combination of defective and non defective.

Sample space associated with this experiment is

S={DDD, NDN, DND, DNN, NDD, DDN, NND, NNN}Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

In this experiment, a die is rolled. If the outcome is 6 then experiment is over. Otherwise, die will be rolled again and again.

Chapter 33 Probability Exercise Ex. 33.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

A card is picked up from a deck of 52 playing cards.

(i)  What is the sample space of the experiment?

(ii) What is the event that the chosen card is back faced card?Solution 9

Chapter 33 Probability Exercise Ex. 33.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

\

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 9

Solution 9

Question 10

Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 8

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:

(i) All the three balls are white

(ii) All the three balls are red

(iii) One ball is red and two balls are white.Solution 8

Chapter 33 Probability Exercise Ex. 33.4

Question 1(a)

Solution 1(a)

Question 1(b)

Solution 1(b)

Question 1(c)

Solution 1(c)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.Solution 25

Question 26

In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either any one or both kinds of sets?Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 32 Statistics Exercise Ex. 32.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Calculate the mean deviation about the mean of the following data:

(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59Solution 2(v)

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 6

Solution 6

Question 3

Calculate the mean deviation on the following income groups of five and seven members from their medians:

I Income in Rs.	II income in Rs.
4000	3800
4200	4000
4400	4200
4600	4400
4800	4600
	4800
	5800

Solution 3

Note: Answer given in the book is incorrect.

Chapter 32 Statistics Exercise Ex. 32.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 4(v)

Find the mean deviation from the mean for the following data:

Size:	1 3 5 7 9 11 13 15
Frequency:	3 3 4 14 7 4 3 4

Solution 4(v)

Question 5(i)

Find the mean deviation from the median for the following data:

xi	15 21 27 30
fi	3 5 6 7

Solution 5(i)

Note: Answer given in the book is incorrect.

Chapter 32 Statistics Exercise Ex. 32.3

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 7

Solution 7

Question 8

Solution 8

Question 6

Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age:	16-20	21-25	26-30	31-35	36-40	41-45	45-50	50-55
Number of persons	5	6	12	14	26	12	16	9

Solution 6

Chapter 32 – Statistics Exercise Ex. 32.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Find the mean, variance and standard deviation for the data:

6, 7, 10, 12, 13, 4, 8, 12.Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Show that the two formulae for the standard deviation of ungrouped data

Solution 11

Chapter 32 Statistics Exercise Ex. 32.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

(i)

(ii)

Solution 3

Question 4

(ii)

Solution 4

Chapter 32 Statistics Exercise Ex. 32.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Calculate the Mean, median and Standard Deviation of the following distribution:
Solution 5

Question 7

Solution 7

Question 8

Mean and standard deviation of 100 observation were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.Solution 8

Question 9

While calculating the mean and variance of 10 readings , a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.Solution 9

Question 6

Find the mean and variance of frequency distribution given below:

xi:	1x <3	3 x<5	5x<7	7x<9
Fi:	6	4	5	1

Solution 6

Note: Answer given in the book is incorrect. Question 10

Calculate mean, variance and standard deviation of the following frequency distribution:

Class:	0-10	10-20	20-30	30-40	40-50	50-60
Frequency:	11	29	18	4	5	3

Solution 10

Chapter 32 Statistics Exercise Ex. 32.7

Question 1

Solution 1

We observe that the average monthly wages in both firms is same i.e. Rs. 2500. Therefore the plant with greater variance will have greater variability. Thus plant B has greater variability in individual wages. Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Life of bulbs product by two factories A and B are given below:

Length of life(in hours):Factory A:	550-650 650-750 750-850 850-950 950-1050
(Number of bulbs)Factor B:	10   22 52 20 16
(Number of bulbs)	8  60 24 15 12

The bulbs of which factory are more consistent from the point of view of length of life?Solution 11

Question 12

Following are mark obtained, out of 100, by two students Ravi and Hashina in 10 tests:

Ravi:	25 50 45 30 70 42 36 48 35 60
Hashina:	10 70 50 20 95 55 42 60 48 80

Who is more intelligent and who is more consistent?Solution 12

Chapter 31 Mathematical Reasoning Exercise Ex. 31.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

It is not a statement.

The sentence “This sentence is a statement.” cannot  be assigned a truth value of either true or false, because either assignment contradicts the sense of the sentence.Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 1(xiii)

Solution 1(xiii)

Question 1(xiv)

Solution 1(xiv)

Question 1(xv)

Solution 1(xv)

Question 1(xvi)

Solution 1(xvi)

Question 2

Solution 2

Chapter 31 Mathematical Reasoning Exercise Ex. 31.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 31 Mathematical Reasoning Exercise Ex. 31.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 31 Mathematical Reasoning Exercise Ex. 31.4

Question 1

Solution 1

Question 2

Solution 2

Chapter 31 Mathematical Reasoning Exercise Ex. 31.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 31 Mathematical Reasoning Exercise Ex. 31.6

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 30 Derivatives Exercise Ex. 30.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 7 (iii)

Solution 7 (iii)

Question 7(iv)

Solution 7(iv)

Chapter 30 Derivatives Exercise Ex. 30.2

Question 1(i)

Solution 1(i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Solution 1 (ix)

Question 1 (x)

Solution 1 (x)

Question 1 (xi)

Solution 1 (xi)

Question 1 (xii)

Solution 1 (xii)

Question 1 (xiii)

Solution 1 (xiii)

Question 1 (xiv)

Solution 1 (xiv)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 3 (vii)

Solution 3 (vii)

Question 3 (viii)

Solution 3 (viii)

Question 3 (ix)

Solution 3 (ix)

Question 3 (x)

Solution 3 (x)

Question 3 (xi)

Solution 3 (xi)

Question 3 (xii)

Solution 3 (xii)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Question 3 (v)

Solution 3 (v)

Question 3 (vi)

Solution 3 (vi)

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5(i)

Solution 5(i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5(iv)

Solution 5(iv)

Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 1(xv)

Solution 1(xv)

Question 2(v)

Differentiate -x using first principles.Solution 2(v)

Question 2(vi)

Differentiate (-x)^-1 using first principles.Solution 2(vi)

Question 2(vii)

Differentiate sin(x + 1) using first principles.Solution 2(vii)

Question 2(viii)

Differentiate cos using first principles.Solution 2(viii)

Chapter 30 Derivatives Exercise Ex. 30.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Defferentiate f (x) = Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Chapter 30 Derivatives Exercise Ex. 30.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Differentiate the following functions with respect to x:

Solution 25

Question 26

Differentiate the following functions with respect to x:

(ax + b)ⁿ (cx + d)^mSolution 26

Question 27

Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answer are the same.Solution 27

Question 28(i)

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.

(3x² + 2)²Solution 28(i)

Question 28(ii)

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.

(x + 2)(x + 3)Solution 28(ii)

Question 28(iii)

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.

(3 sec x – 4 cosec x) (-2 sin x + 5 cos x)Solution 28(iii)

Chapter 30 Derivatives Exercise Ex. 30.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Differentiate the following functions with respect to x:

Solution 29

Question 30

Differentiate the following functions with respect to x:

Solution 30