Oasis Math 6

Exercise 8.6

1. Find the volumes of the following cuboids and cubes by counting the unit cubes
of volume 1 cu. cm.

(a) (b) (c)

2. Find the volume of the given cuboids. (c)
(a) (b)

4 cm 6 cm 5 cm 8 cm

6 cm 5 cm 6 cm 3 cm

5 cm

3. Find the volume of the following cubes.

(a) length of edge is 4 cm (b) length of edge is 6 cm.

(c) length of edge is 3 cm (d) length of edge is 5 m.

4. Find the volume of the following cuboids.

(a) length = 6 cm (b) length = 8 cm (c) length = 10 cm
breath = 8 cm
breadth = 5 cm breadth = 6 cm height = 5 cm

height = 4 cm height = 4 cm

5. Solve the following.

(a) Find the volume of a room whose length, breadth and height are 6 m, 4 m and 2
m respectively.

(b) Find the volume of a cubical box of length 2 cm.
(c) length and breadth of a cubical box are equal and each equal to 6cm, if its height

is 5cm, find its volume.

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6. Solve the following.

(a) The volume of a cube is 27cm³. Find the length of its edge.
(b) The volume of a cuboid is 60 cm³ and its length and breadth are 5 cm and 4 cm

respectively. Find its height.
(c) Volume of a cuboid is 90 cubic cm. If the breadth and height of the cuboid are

6cm and 3cm respectively, find its length.

7. (a) A rectangular water tank is 50 cm long, 30 cm broad and 20 cm high. Find the
capacity of the tank.

(b) Capacity of a rectangular tank is 6m3. If its length and breadth are 5m and 2m
respectively. Find its height.

(c) Capacity of a rectangular tank is 12000cm3. If its breadth and height are 4m and
1m respectively, find its length.

8. (a) How many cubical blocks of side 1 cm can be put in a box of 8 cm long, 6 cm
wide and 4 cm high?

(b) How many unit cubes can be put in the box of 12cm long, 9cm wide and 6cm
high?

Answers (b) 12 cu. cm (c) 24cu.cm 2. (a) 144cu.cm (b) 125cu.cm

1. (a) 3 cu.cm 3. (a) 64cu.cm (b) 216 cu.cm (c) 27cu.cm (d) 125cu.m.
(c) 144cu.cm
4. (a) 120cu.cm (b) 192cu.cm (c) 400cu.cm 5. (a) 48cu.m (b) 8cu.cm
(c) 180cu.cm
7. (a) 30,000 cm3 6. (a) 3cm (b) 3cm (c) 5cm

(b) 60cm (c) 0.3cm 8. (a) 192 (b) 648

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Objective Questions

Choose the correct alternative.

1. Which of the following relation is not true?

(i) 1 ft = 30 cm (ii) 1 ft = 30.48 cm (iii) 1 ft = 12 inches

2. How many cm is equal to 1 Inch?

(i) 2.50 cm (ii) 2.54 cm (iii) 2 cm

3. Which one of the following relation is true?

(i) 1 ft = 3.28 m (ii) 1m = 3.28 ft. (iii) 1 m = 12 Inches.

4. Among the given length of three lines, which one is the shortest line?

(i) 1 Inch (ii) 1 cm (iii) 1 ft.

5. The length of four sides of the outer boundary of a field are 35m, 45m, 40m and 50m,
then its perimeter is:

(i) 180m (ii) 160m (iii) 170cm

6. Perimeter of a rhombus is 60cm, then the length of each side is:

(i) 120cm (ii) 240 cm (iii) 15cm

7. Area of the shaded part in the given figure is : 12 cm
(i) 10cm² (ii) 120cm² (iii) 110cm² 5 cm

8. Area of a square is 49cm², then its perimeter is : 10cm
2 cm
(i) 28cm (ii) 49cm (iii) 7cm

9. Length of a rectangular field is 15m if its perimeter is 50m, then its area is :

(i) 150m² (ii) 100 m² (iii) 50m²

10. Area of shaded part in the given figure is 20 cm
8 cm
(i) 300cm² (ii) 252cm² (iii) 48cm² 15cm
6 cm

11. Volume of a cube is 216cm³, then the length of each side is:

(i) 12cm (ii) 3cm (iii) 6cm

12. Volume of a cubical tank is 8000 cm3, then the length of each side is:

(i) 2m (ii) 2cm (iii) 20cm

13. How many cubical blocks of side 1cm can be put in a box having length, breadth and
height 8cm, 6cm and 4cm respectively?

(i) 192 (ii) 96 (iii) 48

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14. Volume of a cube is 729 cm³, then the length of each side of the cube is :

(i) 4.5 cm (ii) 18 cm (iii) 9 cm

15. Approximate area of given irregular shape is

(i) 14 sq. units (ii) 13 sq. units (iii) 15 sq. units

16. Perimeter and area of given figure are 5cm

(i) area = 50cm2, perimeter = 40 cm 10 cm 5cm
5cm10cm
(ii) area = 75 cm2, perimeter = 40 cm.
5cm
(ii) area = 10 cm2, perimeter = 30 cm.

Assessment Test Paper

Attempt all the questions Full marks : 30

Group 'A' 5 × 1 = 5

1. (a) How many Inches is equal to 1 m?
(b) Write the formula to calculate the perimeter of rectangle.
(c) Write the formula to calculate the perimeter of square.
(d) How many feet is equal to 1 m?

(e) What is area of shaded part?

Group 'B' 5 × 2 = 10 2cm
2. (a) Convert 2 Inches into cm.
(b) Find the perimeter of given figure .

6cm

(c) What is the area of a rectangle whose length is 15cm and the breadth is 10cm?
(d) Find the length of each side of a square whose area if 49cm².

148 Oasis School Mathematics – 6

(e) The length and breadth of a cuboid are 8cm and 5cm respectively. If its volume
is 200cm³, find its height.

Group 'C' 5 × 3 = 15

3. (a) Find the approximate area of given figure:

(b) Perimeter of a rectangular field is 60m, if its length is 25m, find its breadth and

area.
8m

(c) Find the area of the given figure. 10m 30m 12m
10m
20m20m
10cm10m
(d) Find the area of given field.
10m

40m

(e) Length, breadth and height of a rectangular tank are 30 ft., 20 ft. and 15 ft.
respectively. Find its volume.

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Algebra

30Estimated Teaching Hours

Contents
• Concept of Indices
• Addition and Subtraction of Algebraic terms and

Expressions
• Multiplication and Division of Algebraic Expressions
• Equation, Inequality and Graph.
Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:
• To understand the concept of indices
• To add, subtract, multiply and divide algebraic

expressions
• To identify whether the given statement are true, false

or open.
• To solve the equation using the principle of balance
• To apply the trichotomy rules on inequality
• To represent the inequality in the number line
Teaching Materials
• A4 size paper, graph sheet, chart paper, scissors, etc.

150 Oasis School Mathematics – 6

Unit

9 Indices

9.1 Introduction

Lets see some examples:
3+3+3+3
Here 3 is added 4 times.
It is repeated addition. It is written as 4 × 3 .
Again, 3 × 3 × 3 × 3
Here same number 3 is multiplied repeatedly with itself.
It is written as 34. It is read as "3 to the power 4".
In 34,

3 is called base and 5 is power or exponent.

The power 2 and 3 are also called by special name squared and cubed respectively.

52 can be read as 5 squared.

72 can be read as 7 squared.

Again,

43 can be read as 4 cubed.
113 can be read as 11 cubed.

Remember!
• 4+4+4+4+4=5×4
• 4 × 4 × 4 × 4 × 4 = 45
• (–2)3 = (–2) × (–2) × (–2)




Exercise 9.1

1. Identify the base and the power of following :

a) 35 b) (–2)4 c) 58 d) 47 e) x3
e) x7
2. Write the following in the expanded form:

a) 43 b) 56 c) 62 d) (–3)7

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3. Write the following in the short form:

a) 5+5+5

b) 3×3×3×3×3×3

c) 4×4×4

d) (–2) × (–2) × (–2) × (–2)

e) (–7) × (–7) × (–7) × (–7) × (–7)

f) x × x × x × x × x × x × x × x

9.2 Multiplication of Indices

Lets be clear with the following examples;

43 × 42 = (4 × 4 × 4) × (4 × 4)

= 4×4×4×4×4

= 45

T­ he same result can be obtained

43 × 42 = 43+2 = 45.

Again, 54 × 53 = (5 × 5 × 5 × 5 × 5) × (5×5×5)

= 5×5×5×5×5×5×5

= 57

The same result can be obtained

54 × 53 = 54+3 = 57.

Again, x4 × x2 = (x × x × x × x) × (x × x)

= x×x×x×x×x×x

= x6.

The same result can be obtained

x4 × x2 = x4+2 = x6.

From above examples, we can conclude that if two numbers with different power are

multiplied, the result is same base with the sum of powers. i.e. xa × xb = xa+b

Division of Indices

Lets be clear with the following examples;

25 ÷ 22 = 25 = 2×2×2×2×2 = 2 × 2 × 2 = 23.
22 2×2

The same result can be obtained.

25 ÷ 22 = 25–2 = 23.

Again,

47 ÷ 42 = 47 = 4×4×4×4×4×4×4 = 4 × 4 × 4 × 4 = 45.
42 4×4

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The same result can be obtained.

47 ÷ 42 = 47–2 = 45.

Again, x4 ÷ x2 = x4 = x×x×x×x×x = x2.
x2 x×x

The same result can be obtained.

x4 ÷ x2 = x4–2 = x2.

From the above examples, we conclude that if a number is divided by ta number with the

same base but different power the result is the same base with difference of powers

i.e. xa ÷ xb = xa–b.

Remember!

xa × xb = xa+b
xa ÷ xb = xa–b

Exercise 9.2

1. Simplify the following using the law of indices.

a) 24 × 23 b) 32 × 36

c) 42 × 48 d) (–3)7 × (–3)3

e) (–4)5 × (–4)3 f) x5 × x8 g) y7 × y2

2. Simplify the following using the law of indices.

a) 24 ÷ 22 b) 37 ÷ 33

c) 410 ÷ 43 d) 56 ÷ 52

e) (–3)8 ÷ (–3)5 f) (–5)10 ÷ (–5)6

g) x10 ÷ x3 h) y12 ÷ y8

Answers
Consult your teacher.

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Unit

10 Algebraic Expressions

10.1 Algebraic Terms and Expressions

x, 2y, 3z etc are the algebraic terms.

Again, look at some examples,

• Bishal has x copies, his father gave him 2 more.
• How many copies he has now?
• How to write this statement in Mathematical form?

Now, Bishal has (x + 2) copies.
Again,
Seema has y pencils, she lost 4 pencils.
Now she has (y–4) pencils.

Ram has x sweets. Sita has It is easy! Sita
twice as much as Ram has. has 2x sweets.
How many sweets Sita has?

Here (x + 2), (y - 4), 2x, etc. are algebraic expressions.

Hence an expression made of algebraic terms connected by arithmetic operations is
called an algebraic expression.

x + 2 is an algebraic expression having 2 terms.

2x is an algebraic expression having single term.

Formation of an algebraic expression

Arithmetic operations can be used to connect variables and constants. See some
examples here and get the idea to form an algebraic expression.

5 more than x is x + 5. Use '+' sign in the case of 'more than' 'is
y is added to z is y + z added to' and 'increased by'.
Increase z by 6 is z + 6

Again, Use '–' sign in the case of 'less than' 'is
subtracted from' and 'decreased by'.
7 less than x is x – 7
9 is subtracted from y is y – 9

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z is decreased by 4 is z – 4. Use '×' sign in the case of 'times',
Again, 'product of' and 'multiplied by'.

5 times x is 5x. Use ÷ sign in the case of 'quotient of'
The product of 6 and y is 6y and 'is divided by'.
9 is multiplied by z is 9z
Again,

Quotient of x by 5 is x ÷ 5
y divided by 7 is y ÷7

Terms of an expression:

4x + 5y is an algebraic expression.
4x and 5y are the terms of the algebraic expression 4x + 5y
Again,

3x – 2y is an algebraic expression.
3x and 2y are the terms of the algebraic expression 3x – 2y .

Terms are separated by '+'
or '–' only.

Types of algebraic expression

Monomial:
An algebraic expression consisting only one term is called a monomial.
4x, 3x2y, 5a2b, etc. are monomials.
Binomial:
An algebraic expression having two terms is called a binomial.
2x + 3y, 3x2y - 2xy­2, a + b are binomials.
Trinomial:
An algebraic expression consisting of three terms is called a trinomial.
2x + 3y - 2z, 4x2 - 5xy + 2yz are trinomials.
Polynomials:
An expression consisting of several terms is called a polynomial. It is also called
multinomial.

Factors and Co-efficients

Factors: When we multiply two or more numbers or literal numbers, then each of
them is called the factor of the product.

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Factor Product
(numerical factor) Factor
(literal factor)

Co-efficients: In the product of numbers and literal numbers, any of the factor is

called the coefficient of other factors.
In 50ab,
co-efficient of b is 50a
co-efficient of a is 50b
co-efficient of ab is 50

Note:

If the coefficient is a number then it is a numerical coefficient. If the
co-efficient is a letter, then it is a literal co-efficient.

Coefficient, base and power of algebraic terms

Recall that when a number is multiplied by itself, letters can be raised to different
powers.

Like 5 × 5 = 5² , x × x is x².
x² is read 'x squared' or 'x raised to power 2'.
Similarly, x × x × x × x × x × x is written as x6 and read as x raised to power 6.
In x6, x is the base and 6 is the exponent or index or power.
In 4 × y³ × 3 = 12 y3, 12 is coefficient, y is base and 3 is exponent or index or power.

Worked Out Examples Power
Base
Co-efficient

Example: 1

Write the following using numbers, literal numbers and signs of fundamental
operations.

(a) the sum of x and z (b) 3 more than x (c) x decreased by 6.

(d) x divided by b. (e) the product of 5 and y. (f) m increased by twice n.
Solution:

(a) the sum of x and z = x + z (b) 3 more than x = x + 3
x
(c) x decreased by 6 = x – 6 (d) x divided by b = b .

(e) the product of 5 and y = 5y. (f) m increased by twice n = m + 2n.

156 Oasis School Mathematics – 6

Example: 2

Write the coefficient of x in each of the following expressions:

(a) – 4xy (b) 16xyz (c) – 5xy² (d) –x
Solution:

(a) In – 4xy, coefficient of x is –4y. (b) In 16xyz, coefficient of x is 16yz.
(c) In – 5xy², coefficient of x is –5y². (d) In –x, coefficient of x is – 1.

Example: 3

Write down the numerical coefficient of each term of:

(a) x in – 6xy (b) xyz in – xyz
Solution:

(a) The numerical coefficient of x in –6xy is –6.

(b) The numerical coefficient of xyz in –xyz is –1.

Example: 4

Write down the following in product form:

(a) 3b5 (b) 7a3b2 (c) -2p3q2r (d) 5xy2z
Solution:

(a) 3b5 = 3 × b × b × b × b × b (b) 7a3b2 = 7× a×a×a× b×b

(c) -2p3q2r = -2 × p × p × p × q × q × r (d) 5xy²z = 5 × x × y × y × z

Example: 5

Write in the exponential form:

(a) 2 × p × p × p (b) 6 × a × a × b × b × c × c × c
Solution:

(a) 2 × p × p × p = 2p³
(b) 6 × a × a × b × b × c × c × c = 6a²b²c³.

Exercise 10.1

1. Write the terms of the given algebraic expressions:

(a) 3a + b (b) 3x - 5y + z (c) 4x2y + 5xy2 - 3z3

2. Write the following using numbers, literal numbers and signs of fundamental

operations:

Statements Using Symbol (Algebraic form)

(a) the sum of a and b .......................................
(b) 4 more than x .......................................
(c) x decreased by 6 .......................................
(d) 2 minus c .......................................
(e) the product of 8 and y .......................................

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(f) x divided by b .......................................

(g) 4 less than the quotient x divided by y .......................................

3. Form the algebraic expression for the following:

(a) Two times of x is increased by 7. (b) Three times of x is decreased by 2.
(c) Three times of x is increased by y. (d) Five times of x is decreased by y.
(e) Three times of y is subtracted from two times of x.
(f) Quotient of x by 7 is added with 2.
(g) Quotient of x by y is added to the product of x and y.

4. Express the following problems in the mathematical form:

(a) Anasuya had 15 chocolates, she ate x chocolates, how many chocolates she has now?

(b) Abdullah had Rs. 50, he gave Rs. x to his daughter. How much money is left with

him?

(c) A man had to travel 20 km. If he travelled y km, how much distance is left to

travel?

5. Write the following algebraic expressions in words: x
x y
(a) x + 7 (b) y - 10 (c) 2x + 15 (d) 3x - 20 (e) y (f) − 10

6. Identify the monomials, binomials, trinomials and polynomials in the following

expressions:

(a) 4x (b) 2a + 5b + c (c) 6x + 7y – 3z + 6

(d) 3x + 5y (e) px + qy + r (f) x³ – y³

7. Write the coefficient of x in each of the following expressions:
2
(a) – 6xy (b) 12xyz (c) –7xy² (d) –xab (e) 3 xy (f) 3xy³z4

8. Write down the numerical coefficient of each term of:

(a) y in – 6xy (b) x in 5xyz (c) abc in – abc (d) z in xyz

(e) w in wxy (f) x in –2xy²

9. Study the table and fill in the blanks:

S.N. Term Numerical Coefficient Literal Coefficient
(a) 8xyz 8 xyz
(b) 5x²y³z
(c) 2xy
(d) 3abx
(e) x²yz
(f) 7x²y

10. Write down the following in product form:

(a) 5 b5 (b) x7 (c) 3a2b3c4 (d) 7p3q4r2 (e) -2x2y3z (f) 11 x3y4

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11. Write in the exponential form: (b) 7 × a × a × a × a × b × b × b
(a) 5 × x × x × y × y (d) 11 × x × x × y × y × y × z × z × z
(c) -3 × -p × -p × q × r × r (f) -b × -b × -b × -b × -b × -b × -b
(e) x × -y × -y × -z × -z

Answers

1. Consult your teacher. 2. Consult your teacher. x x
7 y
3. (a) 2x + 7 (b) 3x – 2 (c) 3x + y (d) 5x – y (e) 2x – 3y (f) +2 (g) + xy
4. (a) 15 – x
(b) Rs. (50 – x) (c) (20 – y)km

5. (a) 7 is added to x . (b) 10 is subtracted from y.

(c) Two times of x is added to 15. (d) 3 times of x is decreased by 20.

(e) x is divided by y. (f) Quotient of x and y is decreased by 10.

6. (a) Monomial. (b) Trinomial (c) Polynomials (d) Binomial (e) Trinomial (f) Binomial.

7. (a) – 6y (b) 12yz (c) – 7y² (d) – ab (e) 2y (f) 3y³z4 8. (a) – 6 (b) 5 (c) – 1
(d) 1 (e) 1 3
(f) – 2. 9. (b) 5, x²y³z (c) 2, xy (d) 3, abx (e) 1, x²yz (f) 7, x²y.

10. (a) 5×b×b×b×b×b (b) x×x×x×x×x×x×x (c) 3×a×a×b×b×b×c×c×c×c

(d) 7×p×p×p×q×q×q×q×r×r (e) –2×x×x×y×y×y×z (f) 11×x×x×x×y×y×y×y

11. (a) 5x²y² (b) 7a4b³ (c) – 3 (– p)²qr² (d) 11x²y³z³ (e) x(–y)²(–z)² (f) (–b)7

Let's find the number Oh! it's
interesting!
Think a number x Every time the
result is 74.
Add 50 on it x + 50

Double it 2(x + 50)

Add 48 on it 2x + 100 + 48

Divide the result by 2 x + 74

Take away the number you thought of 74

10.2 Addition and Subtraction of Algebraic Terms

Like and unlike terms

x2, 3x2, 7x2 are like terms.
4xy, 7xy, 12xy are like terms.
12x2y, x2y, 20x2y are like terms.
In the first example, variable part x2 is the same.
In the second example, variable part xy is the same.
In the third example, variable part x2y is the same.
Hence, the terms with the same variables with the same power or exponent are

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called like terms.
Again,

x2, 3y2, 4z2 are unlike terms
5x2y, 6xy2­, 10xy are unlike terms.
In the first example, variable part of each term is different.
In the second example, variables are the same, but their power in each term is
different.
Hence, the terms which have different variables or same variables with different
power or exponent are unlike terms.

Addition and subtraction of like terms

Let's see the following example and get the idea of addition of like terms.

First basket contains 5 mangoes.
Second contains 3 mangoes.
How many mangoes are there altogether?
8 mangoes

\ 5 mangoes + 3 mangoes = 8 mangoes

Similarly, 5x + 3x = 8x

Again, from the plate of 5 mangoes, 3 mangoes are taken out, how many mangoes
left on the plate?

5 mangoes – 3 mangoes = 2 mangoes.

Similarly, 5x – 3x = 2x.

Hence the like terms can be added by adding the co-efficient of like terms and
writing the variable only once.
5y + 7y = (5 + 7)y = 12 y

5ab + 2ab + 6ab = (5 + 2 + 6) ab = 13 ab

Similarly,

Like terms can be subtracted by subtracting the co-efficient of like terms and writing

the variable only once.

12y - 5y Subtract the co-efficient
= (12 - 5)y
= 7y

160 Oasis School Mathematics – 6

Addition and Subtraction of unlike terms

Let's see the given example and get the idea about the addition of unlike terms.

First basket contains 5 mangoes.

Second basket contains 3 apples.
Altogether there are 5 mangoes + 3 apples.
\ 5 mangoes + 3 apples= 5 mangoes + 3 apples
Similarly,
5x + 3y = 5x + 3y (no change)
In the case of subtraction also,
5x - 3y = 5x - 3y (no change)

10x - 7y = 10x - 7y itself.
No more simplification is possible.

Exercise 10.2

1. Identify whether the following are like or unlike terms.

(a) 5a, 6a, 7a (b) 7x, 3y, 4z (c) 2ab, 5ab, 9ab (d) 4x2y, - 7x2y, 9x2y
(h) 4b2c, -5b2c, 6b2c
(e) x2y, 2xy2, y3 (f) 4abc, -2ab, 3bc (g) 3ab, 5bc, 6ac

2. Pick out the like terms from each group.

(a) x2y, 5xy, -6x2y, 2x3, 4x2y (b) 5a, 6b, -7a, 2a, 6c
(c) a2x, 2ax, 3ax2, 4a2x (d) 4xy, 3yz, 2yx
(e) 2x, 2y, 5x, 10z
3. Add or subtract:

(a) 5x + 7x (b) 3x + 8x (c) 6m - 2m (d) 13p - 10 p
(e) 8y - 3y (f) 10z - 2z (g) 4x - 9x (h) -12x + 3x

4. Simplify the following expressions by adding or subtracting the like terms.

(a) 5a + 7a - 2b (b) 8x - 7x + y
(c) 8y + 3y + 2z + 3z (d) 4a + b - 2c
(e) 8p + 2p - 3q (f) 5m - 2n + 3m + 4n

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5. Find the perimeter of the given figures.

(a) (b) 3y 4y

3x 5x

2y
5y

4x

Answers

Consult your teacher.

10.3 Addition and Subtraction of Algebraic Expressions

Addition of algebraic expressions

Let's understand the method of addition Steps:
of algebraic expressions with the help of
given example: W rite each expression in separate
Add: 3x + 4y + 5z and 3x + 2y - 2z rows in such a way that like terms
are arranged one below the other
3x + 4y + 5z in column.

3x + 2y - 2z     A dd the like terms in each column.

6x + 6y + 3z

Alternatively,

Let's see an example and get the idea of addition of algebraic expression by
alternative method,

Add: (5x - 2y + 6z) and (2x + 5y - 2z) Steps:

Here, W rite the given expressions
(5x - 2y + 6z) + (2x + 5y - 2z) horizontally.
= 5x - 2y + 6z + 2x + 5y - 2z
Collect the like terms and

= 7x + 3y + 4z simplicity.

Let's see an example and get Steps:
the idea of subtraction of two R ewrite the expressions in two lines
algebraic expressions with the
such that lower line is the expression to
help of given example.
be subtracted and the like terms of both
Subtract: expressions are one below the other.
3x - 4y + 2z from 5x - 6y - 4z C hange the sign of each term of lower

Here, line

3x - 4y + 2z Combine the terms column wise.

5x - 6y - 4z

(-) (+) (+)    

-2x + 2y + 6z

162 Oasis School Mathematics – 6

Alternatively, Steps:

Let's see an example and get the Enclose the expressions to be subtracted
idea of subtraction of two algebraic in the bracket with a minus (–) sign.
expressions using alternative
method. Open the bracket with (–) sign.
Subtract: Combine the like terms.

(3a + 4b - 3c) from (5a - 2b + 8c)

Here,
(5a - 2b + 8c) - (3a + 4b - 3c)
= 5a - 2b + 8c - 3a - 4b + 3c

= 2a - 6b + 11c

Worked Out Examples

Example 1 4x2 – 2y2 + 6z2
5x2 – 3y2 – 5z2
Add the following algebraic expressions. (-) (+) (+)   
4x2 - 2y2 + 6z2 and 5x2 - 3y2- 2z2 9x2 – 5y2 + 4z2
Solution:
Here, (4x2 - 2y2 + 6z2) + (5x2 - 3y2 - 2z2)

= 4x2 - 2y2 + 6z2 + 5x2 - 3y2 - 2z2

= 9x2 - 5y2 + 4z2

Example 2

Subtract 3x2 - 2xy - 5y2 from 8x2 + 3xy - 2y2 8x2 + 3xy - y2
Solution: 3x2 - 2xy - 5y2
Here, (8x2 + 3xy - y2) - (3x2 - 2xy - 5y2) (-) (+) (+)   
5x2 + 5xy + 4y2
= 8x2 + 3xy - y2­ - 3x2 + 2xy + 5y2

= 5x2 + 5xy + 4y2

Exercise 3

From what 5a2 - 2ab + b2 be subtracted to get 4a2 - 3ab + 2b2 ?
Solution:

Here, Let, from 'K' 5a2–2ab+b2 be subtracted to get
5a2 - 2ab + b2 4a2–3ab+2b2.
+ 4a2 - 3ab + 2b2   K–(5a2–2ab+b2) = 4a2–3ab+2b2)
K–5a2 + 2ab–b2 = 4a2 – 3ab + 2b2
9a2 - 5ab + 3b2 K = 4a2–3ab+2b2 + 5a2–2ab+b2

= 9a2 – 6ab + 3b2

\ From 9a2 - 5ab + 3b2, (5a2 - 2ab + b2) be subtracted to get 4a2 - 3ab + 2b2

Oasis School Mathematics – 6 163

Why Think! From what 5 should be
addition ? subtracted to get 20?
Your answer is 25:

It is obtained by adding 5 and 20.

Example 4

What should be added to 2m3 - 3m2n + 4n3 to get 4m3 + 2m2n - 2n3?

Solution:

Here, Why
4m3 + 2m2n - 2n3­ subtraction?

2m3 - 3m2n + 4n3

(-) (+) (-)   

2m3 + 5m2n - 6n3

What should be added to 10 to get
15?

Your answer is 5.
It is obtained subtracting the former

from the later (15 - 10)

Let 'X' should be added to 2m3–3m2n+4n3 to get 4m3+2m2n–2n3
then,

X + 2m3 – 3m2n + 4n3 = 4m3 + 2m2n–2n3
X = 4m3 + 2m2n–2n3–2m3+3m2n–4n3

= 2m3 + 5m2n – 6n3
Hence, 2m3 + 5m2n–6n3 should be added to 2m3–3m2n+4n3 to get

4m3+2m2n–2n3

Example 5

What should be subtracted from 3a2 - 4ab + b2 to get a2 + ab - b2?

Solution:

Here, What should be subtracted from
3a2 - 4ab + b2 6 to get 2? It is 4.
a2 + ab - b2
(-) (-) (+)     It can be obtained by
­2a2 - 5ab + 2b2 subtracting the latter from the

former.

164 Oasis School Mathematics – 6

Let 'Y' should be subtracted from 3a2–4ab+b2 to get a2+ab–b2.
Now, 3a2–4ab+b2 – Y = a2 + ab + b2 = Y 2a2 – 5ab + 2b2 = Y
Hence, 2a2–5ab+2b2 should be subtracted from 3a2 – 4ab + b2 to get a2+ab–b2,

Exercise 10.3

1. Add the following algebraic expressions:

(a) 2x + 5y and 3x + 2y
(b) 3a - 2b and 5a - 4b
(c) 2x2 + 5xy + 3y2 and x2 - 2xy + 2y2
(d) x + 3y - 4z and 2x - y + 5z
(e) 4p2 - 3pq + q2 and 3p2 + 5pq - 2q2
(f) 2m2n - 3mn2 - n3 and -3n3 + 5m2n

2. Subtract:

(a) 2x - y from 3x + 2y
(b) 2a + 5b - c from 3a - 2b + c
(c) 5m2 - 4mn - 6n2 from 2m2 + 3mn + n2
(d) 4m2 + 6mn - n2 from 2m2 + 3mn - n2
(e) 4x3 - 2x2- xy from x3 + 4x2

3. Simplify:

(a) (4x + 2y - z) + (3x - 5y)
(b) (4a - 2b + c) - (a + b + c)
(c) (x2 - xy + 3y2) + (2x2 - 5xy + y2)

(d) (4m2 + n2 - 6mn) - (m2 + 2mn - n2)

(e) (a2 - 6ab + 2b2) - (2a2 + ab - b2)

4. If a = 3x - y + z, b = 2x - 3y + 2z and c = 4x - y + 3z, find the value of

(a) a + b + c (b) a - b +c (c) -a + b + c (d) a + b - c

5. Find the perimeter of the given figures: m+n+2p
(a) 3x–2y+z (b) m+n+p
4x–2y+5z 4p–m–n
x–5y+2z 2m+3n–p

Oasis School Mathematics – 6 165

6. (a) From what a + b + 2c be subtracted to get 3a + b + 2c?
(b) From what x² – xy + 2y² be subtracted to get 3x² + 2xy + y²?

7. (a) What should be added to 3m² – 2mn + n² to get 2m² + 4mn – 2n²?
(b) What should be added to 8x – 5y + 6z so that the sum may be 8x + 9y + 6z?

8. (a) What should be subtracted from 2x – 3y + 4z to get x + 2y + z?
(b) What should be subtracted from m² – 2mn + 3n² to get 2m² – n²?

9. (a) By how much x² + xy + y² is less than 2x² – xy?
(b) By how much 2m² – 3n² + 4mn is greater than m² – 2mn + n²?
(c) Take a²b – 3c²a from 4a²b – 5b²c + 2c²a.

Answers

1. (a) 5x + 7y (b) 8a – 6b (c) 3x² + 3xy + 5y² (d) 3x + 2y + z (e) 7p² + 2pq – q² (f) 7m²n – 3mn² – 4n³

2. (a) x + 3y (b) a – 7b + 2c (c) –3m² + 7mn + 7n² (d) –2m² – 3mn (e) –3x³ + 6x² + xy

3. (a) 7x – 3y – z (b) 3a – 3b (c) 3x² – 6xy + 4y² (d) 3m² + 2n² – 8mn (e) –a² – 7ab + 3b²

4. (a) 9x – 5y + 6z (b) 5x + y + 2z

5. (a) 8x – 9y + 8z (b) 3m + 4n + 6p

6. (a) 4a + 2b + 4c (b) 4x² + xy+3y² 7. (a) – m² + 6mn – 3n² (b) 14y

8. (a) x – 5y + 3z (b) – m² – 2mn + 4n²

9. (a) x² – 2xy – y² (b) m² + 6mn – 4n² (c) 3a²b – 5b²c + 5c²a

10.4 Multiplication of Algebraic Expressions

We know that a5 × a6 = a5 + 6 = a11

x × x = x²
x × x × x = x3

Again, Power of the same base is
x² × x³ = x × x × x × x × x added after multiplication
x² × x³ = x5 = x2 + 3
\ xm × xn = xm + n

Again, we know that, (+) × (+) = +
(+) × (–) = –
(+ a) × (+b) = (+ab) (–) × (+) = –
(+a) × (– b) = –ab (–) × (–) = +
(– a) × (+b) = –ab
(– a) × (– b) = + ab



166 Oasis School Mathematics – 6

Multiplication of a monomial by a monomial

Let's find the product of 5a²b³ × 6a³b7. • Multiply the coefficients.
Here, • 5 × 6 = 30
5a²b³ × 6a³b7 • Add the powers of like factors.
= 5 × 6 × a² × a³ × b³ × b7 • a² × a³ = a2 + 3 = a5
= 30 a2 + 3 . b3+ 7 • b³ × b7 = b3 + 7 = b10
= 30 a5b10

Again,

Let's find the product of 3mn and 2ab.

3mn × 2ab. • Multiply the co-efficients.
= 3 × 2 mnab • Write all the unlike factors as
= 6 mnab.
it is.

Hence,

While multiplying a monomial by a monomial

• Multiply the Co-efficients.
• Add the power of like factors.
• Write unlike factor as it is.

Worked Out Examples

Example 1:

Multiply : 2x²y³ and – 7x5y²

Solution :

2x²y³ × (– 7x5y²) (+) × (–) = –
= 2 × (–7) . x². x5y³y²

= – 14x2 + 5 . y3 + 2 = – 14x7y5

Example 2:

Find the product of:

− 1 x³y²z³, − 2 x 2 yz and 5xyz.
2 5
Solution :
( ) ( ) − 1 2
2 × − 5 × 5 × x³ × x² × x × y² × y × y × z³ × z × z (–) × (–) = +

= 1 × 5 × x 3+ 2 +1.y 2 +1+1.z 3+1+1 = x6y4z5.
5

Oasis School Mathematics – 6 167

Example 3: b = 2x

Applying the formula area of l= 1 x2y3
rectangle (A) = l × b, find the area 2
of given rectangle:

Solution :

Here, 1
2
l = x²y³ , b = 2x .

Using the formula,

A = l × b
1
= 12 x²y³ × 2x x× y3
= 2 × 2 × x² ×

= x2 + 1y³ = x³y³

Example 4:

If x = 1 and y = 4, find the area of l = 6xy cm
given rectangle.

Solution : b = 2y cm

Here, l = 6xy cm, b = 2ycm

We have,

Area of rectangle = l × b

= 6xy × 2y cm2

= 12xy2 cm2

= 12 × 1 × 42 cm2

= 192 cm2

Example 5: z = 3z cm

Ifx = 1, y = 2, z = 3, find the l = 3xyz cm b = 2x2 cm
volume of given cuboid.
Solution :

Here, l = 3xyz cm, b = 2x2 cm, h = 3z cm

We have,

Volume of cuboid (V) = l × b × h

= 3xyz × 2x2 × 3z cm3

= 18 x3yz2

= 18 × 13 × 2 × 32 cm3

= 18 × 1 × 2 × 9 = 324 cm3

168 Oasis School Mathematics – 6

Exercise 10.4

1. Multiply:

(a) x × y (b) 2a × 3b (c) (– a) × 2b (d) a × (–2b)
(g) y³ × y² (h) z5 × z³
(e) (– 3m) × (–2n) (f) x² × x (k) a8 × a10

(i) 2b × 3c × 2a (j) y8 × y6

2. Find the product of: (b) x³y5 and xy³
(a) 2x²y and 3xy²

(c) – 4a²b³c and 5a5bc6 (d) 3m³n² and 2m4n³

(e) –3a³b², –3a²c³ and 4a²b6c³ (f) 2x³y, 3y4z² and 4z²x²

(g) –4 x6y6, 5y²z³ and –6z4x² (h) 2 m²n³p4, 3 m6n²p³ and 6p²
3 2
3 2 10
(i) − 4 x6y4z², − 5 x²y4z6 and 3 x²y³z4

3. Using the formula, area of a rectangle = l × b, find the area of given rectangles.

(a) l = 3x5 (b) l = 2x2y3

b = 4x3 b = 5xy

4. If a = 2, b = 1, c = 3, find the area of given rectangle.

(a) l = 4a2b c m (b) l = 2a2b cm

b = ac cm b = 3ac cm

5. Using the formula, volume of the cuboid = l × b × h, find the volume of the
given cuboid.

(a) (b) h = 4ab

h = 4xy

b = 3xy l = 2ab b = 3ab

l = 2xy

6. If x = 1, y = 2, z = 3, find the volume of given cuboid.

(a) (b) h = 2yz

h = xz

b = xy b = x2z

l = 2xyz l = xyz

Oasis School Mathematics – 6 169

Answers (b) 6ab (c) – 2ab (d) – 2ab (e) 6mn (f) x³
(j) y14 (k) a18 (f) 24x5y5z4
1. (a) xy (h) z8 (i) 12abc (d) 6m7n5 (e) 36a7b8c6 (b) 648 cm3
3. (a) 12x8 (b) 10x³y4
(g) y5 (b) x4y8 (c) –20a7b4c7 (b) 24ab 6. (a) 72 cm3
2. (a) 6x³y³
(h) 6m8n5p6 (i) x10y11z12
(g) 120x6y10z7
4. (a) 96cm2 (b) 144cm2 5. (a) 24xy

10.5 Multiplication of a Binomial by a Monomial

Let's see an example and get the idea about the multiplication of a binomial by a
monomial.
Multiply:

3ab × (4a²b + 2b²) Multiply each term of binomial
= 3ab × 4a²b + 3ab × 2b² by a monomial.

= 12 a1 + 2 b1+ 1 + 6ab1 + 2

= 12a³b² + 6ab³

Alternatively method: Note:

4a2b + 2b2 Method of multiplication of a trinomial by a
× 3ab monomial is the same as the multiplication of a
binomial by a monomial.
4a2b × 3ab + 2b2 × 3ab

= 12a³b² + 6ab³



Multiplication of binomial by a binomial

Let's see an example and get the idea of the multiplication of binomial by a binomial.
Multiply: (x + 5) and (2x + 3)

Here, Multiply each term of one
(x + 5) (2x + 3) expression by the other expression.
= x (2x + 3) + 5 (2x + 3) Add or subtract the like terms.

= 2x2 + 3x + 10x + 15

= 2x² + 13x + 15

Alternatively method: Multiplying (x + 5) by 2x
Multiplying (x + 5) by 3
x+5
× 2x + 3

2x2 + 10x
+ 3x + 15

2x2 + 13x + 15

170 Oasis School Mathematics – 6

Worked Out Examples

Example 1: = 6x² – xy – 2y²

Multiply: (3x – 2y) by (2x + y)
Solution :

(3x – 2y) (2x + y)
= 3x (2x + y) –2y (2x + y)
= 6x² + 3xy – 4xy – 2y²

Example 2:

Using the formula, area of a rectangle = l × b, find the area of given rectangle.

Solution: l = (x+2y+z) units b = (x+y) units
Here, l = (x + 2y) units

b = (x + y) units.

We have,

Area of a rectangle = l × b
A = (x + 2y) × (x + y)
= x (x + y) + 2y (x + y)
= x² + xy + 2xy + 2y²
= (x² + 3xy + 2y²) sq. units.

Exercise 10.5

1. Multiply:

(a) (a + b) × a (b) (2a – 3b) × b² (c) 3xy (2x + 5y)
(d) a²b (a² + b²) (e) pq² (p² – q²) (f) 2mn(m³ – n³)

2. Multiply the following algebraic expressions:

(a) (x + 2) and (x + 3) (b) (x + y) and (3x – 2y)
(c) (x + y) and (2x – 3y) (d) (4x + y) and (x – 3y)
(e) (m + n) and (m + n) (f) (a + b) and (a – b)

Oasis School Mathematics – 6 171

3. Using the formula of area of a square = l², find the area of given squares.
(a) 2(b)(3x+2y) units

(2x+3) units

4. Using the formula of area of rectangle = l × b, find the area of given rectangles.
(a) (b)

b = (3x+y) units b = (x+y) units

l = (2x+3y) units l = (3x+2y) units

Answers (b) 2ab² – 3b³

1. (a) a² + ab (d) a4b + a²b3 (e) p3q² – pq4 (f) 2m4n – 2mn4
(c) 6x²y + 15xy²
2. (a) x² + 5x + 6 (b) 3x² + xy –2y² (c) 2x² – xy – 3y²
(d) 4x² – 11xy – 3y²
3. (a) 4x² + 12x + 9 (e) m² + 2mn + n² (f) a² – b ²
4. (a) (6x2 + 11xy + 3y2)
(b) 9x² + 12xy + 4y²

(b) 3x2 + 5xy + 2y2) sq. units

10.6 Division of Algebraic Expressions

Division is the reverse process of multiplication. Let's understand how to divide an

algebraic term by another with the help of given example.

x5 ÷ x² = x5 = x × x ×x × x × x = x × x × x = x³
x² x× x
xm ÷xn = xm – n
\ x5 ÷ x² = x5 – 2 = x³

Note:

+x = x , +−xy++xy= =−yxxy , +x = −x −x = x a−−nxyd =+−xyxy a=nd−yx+−xy = −x
+y y −y y −y y y


Division of a monomial by a monomial

Let's get an idea of division of monomial by a monomial with the help of given
example.
Divide : 15x6 ÷5x4

Here, 15x6 = 3 × 5x6
5x 4 5x 4

172 Oasis School Mathematics – 6

= 3x6 – 4 = 3x²

Again,

Divide: – 30a4b² by – 10ab • Divide the co-efficient of dividend by
the co-efficient of divisor 15÷5 = 3
Here, −2 × 3 × 5a4b²
−−3100aa4bb² −2 × 5ab • Divide the literal numbers using
=

= 3a4 – 1.b2 – 1 = 3a³b. xm÷xn = xm–n

Exercise 10.6

1. Divide the following:

(a) a6 ÷a4 (b) x6 ÷x2 (c) y8 ÷ y5 (d) p9 ÷ p3

(e) x3y2 ÷ xy2 (f) a5b3c6 ÷ a2b3c5

2. Divide:

(a) 10p3q2 ÷ 2pq (b) 90a6b5c4 ÷ 10a3b2c3
(c) –25x4y3z2 ÷ 5xyz (d) (–20a3b2c2) ÷ (–10ac2)

3. Divide:

(a) (16a+20) ÷ 2 (b) (70p–10) ÷ (-5)

(c) (15a5b – 27ab4) ÷ 3ab (d) (6x4y2 – 4x2­y3 + 2x2y) ÷ 2xy

(e) (21a³b² – 70a³b³ + 7a²b²) ÷ (-7a²b)

5. Find the length of unknown side of given rectangle.

(a) (b) (c)

b = 7x units. l = (a+b) units.

A = 21x2–7x sq. units. A = 56a7b8c3 + 24a8b3c2 A = (a2–b2) sq. units.
sq. units.

l = 8a3b2c units.

6. (a) If x = 6a³b² and y = 2ab, find the value of x .
y
p× r
(b) If p = 2a²b², q = 3a³b³ and r = 3a4b4 ,find the value of q .

Answers

1. (a) a² (b) x4 (c) y³ (d) p6 (e) x² (f) a³c

2. (a) 5p²q (b) 9a³b³c (c) –5x³y²z (d) 2a²b²

3. (a) 8a + 10 (b) – 14p + 2 (c) 5a4 – 9b³ (d) 3x³y – 2xy² +x (e) – 3ab + 10ab² – b
6. (a) 3a²b (b) 2a³b³
4. (a) (3x – 1) units (b) (7a4b6c² + 3a5bc) units

Oasis School Mathematics – 6 173

Unit Equation, Inequality

11 and Graphs

11.1 Mathematical Statements

The statements which contain the fundamental operations addition (+), subtraction
(–), multiplication (×) and division (÷) along with variables or constants are called
mathematical statements.

Observe the following statements:

S.N. Mathematical Statements Symbolic Form
(i) The sum of 6 and 4 is 10 6 + 4 = 10
(ii) The difference between 13 and 6 is 7. 13 – 6 = 7
(iii) The product of 6 and 2 is 12. 6 × 2 = 12
(iv) 5 is greater than 2 5>2

All the above statements are numerically true. A mathematical statement may be
true or false.
For example: The sum of 6 and 4 is 10, which is a true statement.
But the sum of 6 and 4 is 12, which is a false statement.

Open Statements

Now consider the mathematical statements:
4 added to x is equal to 6.
We can write the above statements as follows:
4 + x = 6
If x is replaced by 1, 4 + 1 = 6, which is false.
If x is replaced by 2, 4 + 2 = 6, which is true.
The above statements are true or false according as x is replaced by a suitable
number or non–suitable number. Simply, we can say it is an open statement.

Open Statement True Statement False Statement

+ 6 = 10 4 + 6 = 10 4 + 6 = 12
× 5 = 15 3 × 5 = 15 3 × 5 = 10
z < 5 2 < 5 6<5

174 Oasis School Mathematics – 6

Worked Out Examples

Example: 1

Identify whether the given statements are true, false or open statements.

(a) 2 + 10 = 12 (b) 5 × 2 = 15 (c) x + 3 = 4

Solution:

(a) 2 + 10 = 12
L.H.S. = 2 + 10 = 12 = R.H.S.
Hence, it is a true statement
(b) 5 × 2 = 15
R.H.S. = 5 × 2 = 10 ≠ 15
Which is a false statement
(c) x + 3 = y
Which is open statement.

Example: 2

What values of the letters make the following open statements true?

(a) x + 6 = 10, x = ... (b) y – 5 = 12, y = ...

Solution:

(a) We have x + 6 = 10, x = 4 because the sum of 4 + 6 = 10.
(b) We have y – 5 = 12, y = 17 because the difference between 17 and 5 is 12.

Exercise 11.1

1. Write T for true, F for false and statements of the following:

(a) 2 + 3 = 5 (b) 6 × 5 = 10 × 3 (c) 7 < 2

(d) 14 ÷ 2 = 12 (e) x ≥ 2 (f) x + 3 = 10

2. Identify whether the given statements are true, false or open:
(a) 2 is a prime number (b) 4 is a square number (c) 6 is a composite number.

(d) 24 is a prime number (e) 20 is an even number (f) 11 is an odd number.

(g) Sum of 5 and 7 is 35 (h) 7 is a prime number (i) 3x = 18

3. What value of a letter makes the following open statements true.

(a) z + 2 = 6, z = ... (b) 2x = 10, x = ... (c) y – 6 = 14, y = ...

4. Determine a replacement for each of the symbols or letters given below so that

such replacement changes the open statement into a true statement.

(a) ∆ – 5 = 2, ∆ = ... (b) 2 ×  = 14,  = ...

(c)  is one–fourth of 32,  = ... (d) The sum of z and 7 is 10, z = ...

Answers

Consult your teacher

Oasis School Mathematics – 6 175

11.2 Equation

Study the given picture carefully. It will help us to understand

the simple equation.

We observe that the two weights in the pans balance each

other. x+5 8

This observation, we may write mathematically as follows:

x + 5 = 8,

The symbol '=' stands for 'equal to'.

Clearly, x must be 3 for the two pans to balance.
Thus, x + 5 = 8 is called an equation, and x = 3 is the solution of the above equation.

Solving the equation using the principle of balance

Let us see in this balance. If one ball is added
on the left side this balance, then it will not be
balanced. To balance it one ball should be added
on right side also. If one ball is removed from
one side of the balance, then one ball should also
be removed from the other side to balance it.

Again, if the weight on one side is doubled
then the weight on the other side should also be
doubled.

Solving the equation

Finding the roots of the equation is called the solving of the equation.

Properties

Let us consider an equation x + 2 = 5

Let us take the number {1, 2, 3, 4, 5, …}

When x = 1, 1 + 2 = 5 (not true)

When x = 2, 2 + 2 = 5 ( not true)

when x = 3, 3 + 2 = 5 (true)

\ x = 3 is the solution of this equation. This method solving equation is called "Trial
and error" method.
Hence, the following facts are to be used in solving equation:

176 Oasis School Mathematics – 6

Fact: 1 Fact: 2

If same or equal quantity is added to both If the same number or quantity is
sides of an equation, the equation remains subtracted from the both sides of an
true. equation, the equation remains true.

For example: For example:
Solve x – 6 = 12
Solution: Solve: y + 9 = 14

Solution:

Here, y + 9 = 14

Here, x – 6 = 12 or, y + 9 – 9 = 14 – 9
[∵Subtracting 9 from both sides]
or, x – 6 + 6 = 12 + 6 ∴ y = 5

[∵ adding 6 in both sides]

∴ x = 18

Fact: 4

Fact: 3 If the same number or quantity divides
both sides of an equation (except zero)
If the same number or quantity the equation remains true.
For example:
is multiplied on both sides of an
Solve: 15x = 30
equation, the equation remains
Solution:
true.

For example:
z
Solve: 2 = 5 Here, 15x = 30

Solution: or, 15x = 30
z 15 15
Here, 2 = 5 [∵ dividing both sides by 15]

or, z ×2 = 5 × 2 ∴ x = 2
2
[∵ multiplying both sides by 2]

∴ z = 10

Note: It is noted that division by zero is not possible.

Worked Out Examples

Example: 1

Solve: 2x – 20 = 12

Solution:

Here, 2x – 20 = 12 Dividing both sides by 2

Add 20 to both sides. 2x 32
2 2
2x – 20 + 20 = 12 + 20 =

or, 2x = 32 ∴ x = 16

Oasis School Mathematics – 6 177

Example: 2 Example: 3

Solve: 2x – 15 = 5 Solve x = 4
3 3 6
Solution:
Solution:
Here, 23x – 15 = 5 x 4
[Adding 15 on both sides] Here, 3 = 6

2x 6x = 12
3
or, = 20 6x = 12 [Dividing both sides by 6]
6 6
or, 2x × 3 = 20 × 3
3 \x = 2
or, 6x = 60

or, 6x = 60
6 6

\ x = 10

Exercise 11.2 (b) (c) z + 3 = 16

1. Solve the following:
(a)

x+4 10 y+2 12 (f) x + 3 1 = 9
2 2
(d) y + 4 = -9 (e) x + 0.5 = -2.5
(c) x – 1.4 = – 6
2. Solve: (b)
(a)

x–5 6 y–2 6

(d) p – 2.4 = 3.6 (e) x − 2 1 = 2 1 (f) z + 4 1 = 2 1
3. Solve: 2 2 2 2

(a) (b) (c) 5z = 20

2x 20 4y 20

(d) 3x = 45 (e) 5x = 5.5 (f) 1.1p = 3.3

178 Oasis School Mathematics – 6

4. Solve: (b) (c) x = -23
(a) 5

x 30 y 3
2 3

(d) x = 3 1 (e) y = 2 1 (f) m = − 20
2 2 6 3 9 3

5. Solve: (b) 2y – 3 = 7 (c) 11x – 4 = 40
(a) 2x + 12 = 20
(e) m – 5 = 10 (f) 5x = 5 + 2x
(d) 2x – 6 = 6 3 2

6. Solve:
3x
(a) 5 + 3 = 6 (b) 18 – y = 3y – 2 (c) 5y – 9 = 6 + 2y

(d) x−5 = 3x − 1 (e) 5(12x – 2) – 8(3x + 2) = 10
3 2

Answers (b) 10 (c) 13 (d) –13 (e) –3 (f) 1

1. (a) 6 (b) 8 (c) –4.6 (d) 6 (e) 5 (f) –2
2. (a) 11 (b) 5 (c) 4 (d) 15 (e) 1.1 (f) 3
3. (a) 10 (b) 9 (c) –115 (d) 7 (e) 14 (f) –60
4. (a) 60 (b) 5 (c) 4 (d) 6 (e) 45 (f) 10
5. (a) 4 (b) 5 (c) 5
6. (a) 5

11.3 Word Problems on Equation

If the relations involving unknown quantities are stated in words, it can be converted
into the equation. Such equation can be solved to get the value of unknown quantities.

Worked Out Examples

Example: 1

If 16 is added to a number, the result becomes 36, find the number.

Solution:

Let the required number be x.

From the given condition,

x + 16 = 36

or, x = 36 – 16
∴ x = 20

Hence, the required number = 20

Oasis School Mathematics – 6 179

Example: 2

The sum of two consecutive numbers is 71, find the numbers.
Solution:

Let two consecutive numbers be x and (x + 1)

From the given condition,

x + x + 1 = 71

2x + 1 = 71

or, 2x = 71 – 1

or, 2x = 70

or, x = 70
2

or, x = 35

\ Two consecutive numbers are,

x =35, and x + 1 = 35 + 1 = 36.

Exercise 11.3

1. Form the equation and solve:

(a) Sum of x and 7 is 18. (b) Difference between x and 13 is 7.

(c) A number increased by 18 is 30. (d) One third of a certain number is 20.

(e) Twice a number added to 10 gives 32.

(f) 7 added to thrice a number gives 43.

(g) Twice a number decreased by 9 gives 17.

2. Form the equation and solve.
(a) If the sum of two consecutive numbers is 25, find the numbers.
(b) If the sum of two consecutive odd numbers is 48, find the numbers.
(c) If the sum of two consecutive even numbers is 102, find the numbers.

3. From the given figure make the equation and find the value of x.

(a) 7 cm (b) 2x cm 6cm

x cm 20 cm
15 cm

(c) 3 x cm 3 2 cm 8cm (d) 2x+1 2x+4

3x

(e) 2x+3

x+7

x+4

180 Oasis School Mathematics – 6

Answers (b) 20 (c) 12 (d) 60 (e) 11 (f) 12 (g) 13
(b) 23, 25 (c) 50, 52 (d) 3cm (e) 2 cm
1. (a) 11 (b) 7cm (c) 8cm
2. (a) 12,13
3. (a) 8cm,

11.4 Inequation or Inequality

If the mathematical statements are connected by either < or > or ≤ or ≥, it is called
inequality or inequation. For example, 2x > 3, 3x < 5, etc. are inequalities.

Consider the following open statements.

+ 2 > 6, – 1 < 7, 2x ≥ 4, y ≤3
6

The above open statements are called the inequations or inequalities.

Trichotomy Law:

Law of Trichotomy states that the comparison between any two real numbers 'a' and
'b', one and only one of the following is true.

a > b or, a < b or, a = b

The property of above relations i.e. '>' (greater than), '<' (less than) and '='

(equal to) are called the trichotomy symbols.
For example, let's consider any two numbers 5 and 6.

Then, the comparison between 5 and 6 is as follows.

5 < 6 or 5 > 6 or 5 = 6

But in this relation 5 > 6 and 5 = 6 are false and 5 < 6 is a true statement.

Thus, out of these three relations, only one is true is 5 < 6.

Trichotomy Rules

Addition (or subtraction) rule

If the same number (non-zero) is added or subtracted on either sides of an inequality,
then the sign of inequality does not change.
Let a, b and c are three real numbers,
(a) If a < b then a + c < b + c and a – c < b – c
For example: 3 < 5 ⇒ 3 + 2 < 5 + 2 and 3 – 2 < 5 – 2
(b) If a > b, then a + c > b + c and a – c > b – c
For examples: 7 > 4 ⇒ 7 + 2 > 4 + 2 and 7 – 2 > 4 – 2

Multiplication rule

If the same number (non–zero) is multiplied on both sides of an inequality, then the
sign of inequality does not change, that is:

Oasis School Mathematics – 6 181

(a) If a > b and c > 0, then ac > bc [∵ Multiplying both sides by c]
For example: 4 > 2, ⇒ 4 × 3 > 2 × 3
(b) If a < b and c > 0, then ac < bc
For example: 3 < 6, ⇒ 3 × 2 < 6 × 2
I f an inequality is multiplied by a negative number on both sides, then the sign of
the inequality reverses i.e.
(a) If a < b and c < 0, then ac > bc
For example: 3 < 7, ⇒ 3 (-2) > 7(-2)
(b) If a < b and c < 0, then ac > bc.

For example: 2<5, ⇒ 2 (–4) > 5(–4)

Division rule

If an inequality is divided by the same positive number the sign of inequality does

not change; that is:

(a) If a > b and c ≠ 0, then a > b
c c
6 4
For example: 6 > 4, ⇒ 2 > 2 i.e. 3 > 2

(b) If a < b and c ≠ 0, then a > b
c c
6 9
For example: 6 < 9 ⇒ 3 < 3 i.e. 2 < 3

I f an inequality is divided by the same negative number inequality is reversed that is,

(a) If a > b and c ≠ 0, then [∵ c < 0]
6 4
For example: 6 > 4, ⇒ ac−2> < −2 i.e. –3 < – 2
(b) If a < b and c ≠ 0, then b
c [∵ c < 0]

For example: 6 < 8, ⇒ 6 > 8 i.e. –3 > – 4
−2 −2
Comparison rule

(a) If a > b, b > c, then a > c. For example: 7>5, 5>3 ⇒ 7>3.

(b) If a < b, b < c, then a < c. For example: 5 < 8, 8 < 9 ⇒ 5 < 9.

Graphical Representation of Inequalities on a real Number Line

A real number line can be used to represent the solution set of an equality. Remember
the following statements.

(i) implies that the number is not included in the solution. Thus, '' (a

hollow circle) marks the end of a range with inequality < or >.

(ii) implies that the number is included in the solution. Thus '' (a darkened

circle) marks the end of a range involving an equality as well as inequality

(i.e. ≥ or ≤)

182 Oasis School Mathematics – 6

Worked Out Examples

Example: 1

Write the negation statement from the following statements.

(a) 2 is a prime number (b) 4 is a factor of 16.
Solution:

(a) 2 is not a prime number. (b) 4 is not a factor of 16.

Example: 2

Show the given in equations in the real number line.

(a) x ≥ 4 (b) x > 4 (c) x < 4 (d) x ≤ 4

Solution: x≥4
(a)

–3 –2 –1 0 1 2 3 4 5 6

(b) x > 4

–3 –2 –1 0 1 2 3 4 5 6

(c) x < 4

–3 –2 –1 0 1 2 3 4 5 6

x<4

(d) –3 –2 –1 0 1 2 3 4 5 6

Example: 3

From the given number line form an inequation.

(a)

–3 –2 –1 0 1 2 3 4

(b)

Solution: –3 –2 –1 0 1 2 3 4 5

(a) x ≥ – 2 (b) x < 4

Example: 4

Solve 2x – 4 < 12

Solution:

Here, 2x – 4 < 12

or, 2x – 4 + 4 < 12 + 4 (adding 4 on both sides)

or 2x < 16

or, 2x < 16 (dividing both sides by 2)
∴ 2 2

x < 8.

Oasis School Mathematics – 6 183

Exercise 11.4

1. Write the Trichotomy symbol (>, < or =) in the box provided.

(a) 6  5 (b) –2  1 (c) –8  –3

(d) 12  –10 (e) 2  (–6+3) (f) –1  (3–4)

2. Identify whether the following Trichotomy is true or false.

(a) 8 > 6 (b) –1 < 3 (c) –1 > –14 (d) (6–9) > (12–5)

3. Show the following inequation on the number line.

(a) x ≥ 3 (b) x ≤ – 2 (c) x < 2 (d) 3x + 1 > 4

4. From each of the given number lines form an inequation.

(a)

–3 –2 –1 0 1 2 3 4

(b) –3 –2 –1 0 1 2 3 4

5. Solve the following inequations:

(a) 4x–2 > 6 (b) 2z+5 < 9 (c) 3y–7 ≥ 14 (d) 7x + 2 ≤ 9

Answers 2. Consult your teacher.
(b)
1. Consult your teacher
3. (a)

(c) (d)

4. (a) x ≥ –1 (b) x < 2 5. (a) x > 2 (b) z < 2 (c) y ≥ 7 (d) x ≤ 1

184 Oasis School Mathematics – 6

Objective Questions

Choose the correct alternatives.

1. Algebraic expression of the statement "5 times of x is subtracted from 3 times of y" is:

(i) 5x – 3y (ii) 3y–5x (iii) 3x + 5y

2. What should be added to 4x+3y–2z to make x+y+z?

(i) –3x–2y+3z (ii) –3x+2y–3z (iii) 3x+2y–2z

3. x8 ÷x4 is equal to:

(i) x2 (ii) x12 (iii) x4

4. -a is equal to:
+b
a a -a
(i) - b (ii) b (iii) -b

5. If 4x–3 = 17, then the value of x is:

(i) 5 (ii) 7 (iii) 2
2

6. Sum of two consecutive even numbers is 22, then its equation is:

(i) x+(x+1) = 22 (ii) x + (x+2) = 10 (iii) x + (x+2) = 22

7. ≤ represents. (i) Less than (ii) Less and equal to (iii) Less or equal to

8. If 5 is added to twice a number, the result is 15, then the number is

(i) 20 (ii) 4 (iii) 5

9. Inequality represented by the number line is:

01 2 3 4 5 6 7

(i) x<3 (ii) x≤3 (iii) x≥3

10. Area of rectangle having length and breadth (x+3) units and 2x units is

(i) 2x2 + 6x (ii) 2x2 – 3x (iii) (3x2–2x) sq. units

11. If 3x – 1 = 8, then the value of x is

(i) 4 (ii) 3 (iii) 2

12. If l = 2x3y4, b = x7y2 and h 1 xy5, then the volume of the cuboid is
2

(i) x10y10 (b) x10y11 (c) x11y11

Oasis School Mathematics – 6 185

Assessment Test Paper

Group - A [6×1 = 6]
1. (a) What is the sum of the terms 2x2y and 3x2y?
(b) Simplify: 3xy – 2xy + 4xy
2. (a) What is the product of –4a3b5 and 5a2b7?
(b) Divide 4a3b2 –2a2b3 by 2ab.
3. (a) '2 is a prime number. Identify whether this statement is true or false.
(b) Write the inequality of given number line

Group - B [7×2=14]

4. (a) Multiply: 3x2y × 6x10y7
(b)
5. (a) If a = 1, b = 2 and c = 0, find the value of 3a2b – 4a2b – 6ac
(b)
6. (a) Simplify : 8y–3y+5x+2x
(b)
(c) What should be subtracted from 4a+3b–5c to get 2a+5b–3c?
x
If x = 8a3b2 and y = 2ab, find y .

Solve : 3x – 5= 10

Solve : 2x ≥ –6

Group–C [ 5×4 = 20]

7. If x = 1, y = 2 and z = 3, find the perimeter of given figures?

A P

x-y+3z 3x–y+2z
2x+y
2x+y+z

B x+y+2z C Q x+y+z R

8. If the length, breadth and height of a cuboid are (2x+y)cm, (x+2y)cm and 3y cm respec-
tively, find its volume.

9. Divide : (i) 16x4y3z ÷ 4x2y2 (ii) 8x3y2 + 4xy ÷ 2xy

10. Solve : From what 2x2–5x+6 be subtracted to get x2 + 10x + 13?
11. Make the equation from the given figure and find the value of x.

x cm (x+1)cm 6 cm
27 cm

186 Oasis School Mathematics – 6

Geometry

Contents 38Estimated Teaching Hours

• Line and line segment:

– Parallel lines and intersecting lines

– Perpendicular lines

– Construction of parallel lines and perpendicular lines

– Construction of perpendicular bisector of given line segment

• Angles:

– Measurement of angles

– Classification of angles and measurement

– Construction of angles

• Triangle and Quadrilateral:
– Types of triangle
– Properties of triangle
– Types of quadrilaterals and their properties
– Circle and its different parts

• Circle
– Circle and its different parts
– Introduction of cylinder, sphere and cone

• Solid Figures:
– Vertices, edges and faces of solid figure

– Construction of models and skeleton of solid figure.

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following competencies:
• To identify whether the given lines are parallel, perpendicular or intersecting

• To construct parallel lines and perpendicular lines

• To draw the line segment of given length and to measure the length of line
segment

• To measure the given angle

• To construct the angle of given measurement

• To categorize the triangle and to apply their properties to solve the problems
related to them

• To categorize the quadrilateral and to apply their properties
• To identify the different parts of the circle
• To identify the vertices, edges and faces of solid objects

• To make the skeleton of solid objects.

Teaching Materials

• geometrical instruments like ruler, protractor, set square, flash cards, scissors,
glue, models of solid objects, etc.

Oasis School Mathematics – 6 187

Unit Line and Line
12 Segment

12.1 Line and Line Segment - Review (Point A)

Point: In the given figure, A is a point. It has no length, AB
breadth or thickness. It is denoted by a capital letter. (A line or a straight line)
A point has position only.
CD
Line: A line has length but it does not have thickness. (A curved line)
We can produce a line in both directions without E 5cm F
ever ending. It will be either curved or straight such
as:
AB is a straight line and CD is a curved line.

Line segment: A line which has particular
measurement is called a line segment.
Here, EF ( = 5 cm) is a line segment. EF and FE both
represent the same line segment.

Ray: A ray is a part of a line which is closed on one side. A Ray B
AB is a ray.

Remember!
• Infinitely many lines can be drawn from the given point
• Only one straight line can be drawn through two given points.
• Infinitely many rays can be drawn from the given point.
• A point has position only.
• A line can be produced in both direction without ever ending.
• A line having fixed measurement is the line segment.
• A ray is a line which is closed in one end.

188 Oasis School Mathematics – 6

Exercise 12.1

1. Study the given figures and write three points and three line segments.

(a) C (b) P

BQ

AR

2. Identify whether the given figures represent straight lines, line segment, ray or
curved line in the following figures:

(a) (b) (c) N (d)

A BA B M CD

3. Name the rays in the given figures:

(a) B (b) A (c)

OA OB P QR
C

4. Locate the following points in your copy and join the set of the following points

P, Q, R,... in order and come back to the starting point you have drawn.

(a) (b) R (c) P Q

PR Q P QS R

5. Name the points that represent the vertices of given figures. (Recall the concept

of vertex).

(a) A (b) Q (c) D

P HE

BC SR G F
A
6. Name the line segments which form the given figures: B
C
(a) D (b) M (c)
D
N PF

EF E
O

Oasis School Mathematics – 6 189

7. Answer the following questions:
(a) How many lines can be drawn from the given point A?
• A
(b) How many line segments can be drawn by joining P and Q?
P • • Q
(c) How many rays can be drawn from the given point M?
• M

8. Fill in the blanks:
(a) A point has .......... only.
(b) A point has no ........., ........... and ............ .
(c) .......... lines can be drawn from the given point.
(d) .......... line can be drawn from the given two points.
(e) A ray has ............ end point.

9. Name 5 line segments in the given figure.

B
Q
P

A

Answers
Consult your teacher.

12.2 Parallel lines and Intersecting lines A
BO
Let's observe the given figures:
Figure (i)
In the given figure (i), two hands of the clock A
meet at the point O. So two hands of clock OA
and OB are intersecting rays. C
B
In the given figure (ii), side AB, BC, CD and DA
are four edges of the table. D
Can AB and DC meet each other at a point? Figure (ii)
Can AB and BC meet each other at a point?

190 Oasis School Mathematics – 6

AB and DC do not AB and DC are parallel
meet at a point. lines. AB and BC are
intersecting lines.
AB and BC meet at
a point.

Hence, when two lines meet at a point, they are intersecting lines and when two lines

do not meet when produced on either sides, they are parallel lines.
Let's observe one more figure,
C

Here,

Line AB and CD are intersecting lines. AB
What is the value of angle between them?

Angle between them is 90°.

So, they are perpendicular lines. D

Hence, if the angle between two intersecting lines is Recall the concept
90º, they are perpendicular lines. of measurement of
an angle.

Note

• If AB and CD are parallel lines, it is denoted by A B
AB||CD. C
D
C

• If CD is perpendicular to AB, it is denoted by A D B
CD ⊥ AB. AE G IB
JD
• If AB//CD and EF⊥CD, GH⊥CD and IJ⊥CD CF H
then EF = GH = IJ.

Activity-I
I. Observe the given objects in the class and determine whether the following

are parallel or perpendicular.
(a) opposite edges of white/black board.
(b) length and breadth of your classroom.
(c) two edges of a set square.
(d) opposite edges of a ruler.
(e) the length and breadth of the windows.

Oasis School Mathematics – 6 191

Activity-II
. • Take a rectangular sheet of paper.


• Fold it once as shown in the diagram.

• Again fold it as shown in the figure

• Open the fold, give the name on 4
corners and on the points where the
crease meets its edges.

• Name the pair of parallel lines and the perpendicular lines.

Exercise 12.2

1. Identify whether the following lines are parallel lines or perpendicular lines in
the following figures:
(a) (b) (c) (d)

2. Name two pairs of intersecting lines from the given figures:

(a) A D (b) C A (c) E C G

EF EF AB
CB BD HF
D

3. Name a pair of parallel lines in each of the given figures:

(a) A (b) P Q (c) DE H

A

DE S R B CF G
BC

192 Oasis School Mathematics – 6

4. Name a pair of perpendicular lines in each of the given figures:

(a) A (b) P Q (c) A

CBD S R B DC

5. Study the given figure and answer the questions given below: A P B
(a) Which pairs of lines are perpendicular lines? C D
(b) Which pair of lines are parallel lines? R
S

Q

6. Write the geometrical notation which represents the relation between two lines:

(a) B (b) P

D

A
C R QS

7. Name the letters which are formed from two intersecting lines.



8. Name the letters which are formed from perpendicular lines.

9. Name the letters which contain two parallel lines.



10. Identify whether the following statements are true or false.

(a) Distance between two parallel lines is always the same.

(b) Angle between two perpendicular lines is not a right angle.

(c) Parallel lines do not meet when produced on either sides.

(d) Angle between the two perpendicular lines is 900.

(e) If the distance between two lines is always the same, the lines are parallel.

11. (a) How many lines parallel to AB can be drawn from the P B
Q
given point P? A

(b) How many lines perpendicular to PQ can be drawn from A

the point A? P

Oasis School Mathematics – 6 193

12. Observe the following figures and separate the pair of parallel and perpendicular
lines.

AB

C D
F E

13. Observe the given figure and answer the questions given below:

(a) Which two lines are perpendicular to AB? A B
(b) Which two lines are perpendicular to GH? C DE F

(c) Which four lines are parallel to AB?

(d) What is the relation between DG and EH?

GH

Answers
Consult your teacher.

12.3 Measurement and construction of given line segment:

I. Measurement of length of line segment
To measure the length of a given line segment AB.

AB


Steps:

• Place a ruler with the centimeter mark along the line segment AB in such a way that the
zero mark on the ruler is at A.

• Read the mark on the ruler which corresponds to point B. Thus, reading gives the length
of the line.

• In the example above, the length of the line AB is 6 cm.

II. Construction of a line segment of a given length
Lets draw a line segment of length 6 cm.

194 Oasis School Mathematics – 6