# NCERT Solutions Class 6 Maths Chapter 11

## NCERT Solutions for Class 6 Maths Chapter 11 – Algebra

Algebra is one of the broad areas of maths that is the study of mathematical symbols and the rules for manipulating these symbols through formulas. It is a unifying thread of almost all maths. Apart from this, it also focuses on various regulations to use different symbols mathematically.

NCERT Solutions for Class 6 Maths Chapter 11 provided by Extramarks will make your understanding of algebra a lot easier and more efficient. These are carefully curated solutions to textbook exercises by subject-matter experts while adhering to the latest CBSE syllabus and guidelines. Students will find them very useful for their exam preparations and revisions.

### Access NCERT Solutions for Class 6 Maths Chapter 11 – Algebra

The concepts that you will be studying in Class 6 play an important role in strengthening the foundation of the subject. Getting an in-depth understanding of the concepts and getting your doubts cleared will ensure that there aren’t any gaps in your learning. NCERT Solutions for Class 6 Maths Chapter 11 Algebra included below for your access. You can even cross-check your answers from them while solving the exercise questions from the chapter.

### SOLUTIONS

#### Exercise 11.1

NCERT Solutions of Exercise 11.1 of  Class 6 Chapter 11 discusses answers to the questions related to the concepts of matchstick patterns and the idea of variables.

#### Exercise 11.2

NCERT Solutions of Exercise 11.2 of Chapter 11 Algebra discusses answers to the questions related to the concepts of rules of variables in common rules: geometric and arithmetic rules. Under arithmetic rules, it deals with solutions related to the perimeter of squares and rectangles and various other properties in arithmetics.

#### Exercise 11.3

NCERT Solutions of Exercise 11.3 of this chapter discusses the method of solving problems based on the expressions in variables. The main aim of this exercise is to help students gain expertise in the topic and to improve their problem-solving speed.

#### Exercise 11.4

NCERT Solutions of Exercise 11.4 of Class 6 Maths Chapter 11 provides students with the basic idea of how to practically use expressions in our day to day life. It is one of the fundamental concepts that are important from the exam point of view.

#### Exercise 11.5

Exercise 11.5 discusses the answers to questions based on equations and the steps involved in solving those algebraic equations. The equation mainly has both LHS and RHS and students will learn how to solve such questions by referring to Extramarks’ NCERT solutions.

### NCERT Solutions for Class 6 Maths Chapter 11 Exercises are given in the table below:

 Chapter 11 – Algebra Exercises Exercise 11.1 11 Questions & Solutions Exercise 11.2 5 Questions & Solutions Exercise 11.3 6 Questions & Solutions Exercise 11.4 3 Questions & Solutions Exercise 11.5 5 Questions & Solutions

### NCERT Solutions for Class 6 Maths Chapter 11 – Algebra

Algebra is a branch of Maths in which the students understand the method of using letters in the form of equations. The solutions for Algebra Class 6 NCERT discusses the solutions related to questions based on equations involved in solving algebraic equations. Solving Algebra problems will become easier for you by referring to Extramarks’ NCERT Solutions Class 6 for Chapter 11 prepared by subject experts. The Maths faculty possessing a vast knowledge of the concepts have covered every exercise of Chapter 11 Algebra.

### NCERT Solutions for Class 6 Maths  Chapter 11 Algebra

NCERT Solutions for Class 6 Maths contains answers to all exercise questions of algebra, which is one of the most important parts of maths. A distinct feature of NCERT Class 6 Maths  Chapter 11 is that it uses letters to deduce mathematical analogies. This will give students of Class 6 an ample amount of understanding about using algebraic values and their applications.

NCERT Solutions of Maths Class 6 Chapter 11 are easily accessible and these study materials are manageable and convenient. An expert team of teachers having valuable knowledge and experience in teaching students for several years have prepared them.

### NCERT Solutions for Class 6 Maths  Chapter 11

Algebra Class 6 NCERT Solutions teaches students how to solve different problems of Algebra using systematic methods by laying down the solutions to exercise questions. They further contain illustrated representations, and graphs to make Algebra Class 6 student-friendly. This will interest Class 6 students and will help them develop fundamental knowledge about algebra by going through the solved problems of the solutions.

Also, NCERT Solutions for Class 6 Maths  Chapter 11 Algebra contain solutions to problems in an uncomplicated and straightforward language.

### The Topics That Are Covered in Algebra For Class 6 Are

The following topics are covered in Class 6 Algebra:

• Introduction to Algebra
• Matchstick Patterns
• The idea of a variable
• Use of variables in the common rule: Rules from Geometry and Rules from Arithmetic
• Expressions with variables
• Practical applications of expressions
• What is the equation?
• A solution to an equation

### Benefits of NCERT Solutions for Class 6 Maths  Chapter 11 Algebra

The following are a few benefits of referring to NCERT Solutions for Class 6 Maths  Chapter 11 Algebra:

• NCERT Solutions for Class 6 Algebra include step-by-step solutions.
• NCERT Solutions help in understanding the solutions and the methodology of writing the answers in the exams.
• It is described in an easy language, keeping in mind the needs of the Class 6 students.

Q.1 Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.

$\begin{array}{l}\left(a\right)\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{T}\mathrm{as}\overline{\text{\hspace{0.17em}}|\text{\hspace{0.17em}}}\\ \left(b\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{Z}\mathrm{as}\overline{\underset{¯}{\text{\hspace{0.17em}}\overline{)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}}\text{\hspace{0.17em}}}\\ \left(c\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{U}\mathrm{as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|}\text{\hspace{0.17em}}\\ \left(d\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{V}\mathrm{as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}/\\ \left(e\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{E}\mathrm{as}\text{\hspace{0.17em}}\frac{|\text{\hspace{0.17em}}\overline{\text{​}\text{​}\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}}{|\text{\hspace{0.17em}}\underset{¯}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\\ \left(f\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{S}\mathrm{as}\frac{|\text{\hspace{0.17em}}\overline{\text{​}\text{​}\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}}{\underset{¯}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|}\\ \left(g\right)\text{\hspace{0.17em}}\mathrm{A}\mathrm{pattern}\mathrm{of}\mathrm{letter}\mathrm{A}\mathrm{as}\frac{|\text{\hspace{0.17em}}\overline{\text{​}\text{​}\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}}{|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\end{array}$

Ans-

(a) In the formation of pattern of letter T, we can see that it will require two matchsticks.

 Number of T 1 2 3 4 5 6 – Required matchsticks 2 4 6 8 10 12 –

Therefore, the pattern is 2n.

(b) In the formation of pattern of letter Z, we can see that it will require three matchsticks.

 Number of Z 1 2 3 4 5 6 – Required matchsticks 3 6 9 12 15 18 –

Therefore, the pattern is 3n.

(c) In the formation of pattern of letter U, we can see that it will require three matchsticks.

 Number of U 1 2 3 4 5 6 – Required matchsticks 3 6 9 12 15 18 –

Therefore, the pattern is 3n.

(d) In the formation of pattern of letter V, we cansee that it will require two matchsticks.

 Number of V 1 2 3 4 5 6 – Required matchsticks 2 4 6 8 10 12 –

Therefore, the pattern is 2n.

(e) In the formation of pattern of letter E, we can see that it will require five matchsticks.

 Number of E 1 2 3 4 5 6 – Required matchsticks 5 10 15 20 25 30 –

Therefore, the pattern is 5n.
(f) In the formation of pattern of letter S, we can see that it will require five matchsticks.

 Number of S 1 2 3 4 5 6 – Required matchsticks 5 10 15 20 25 30 –

Therefore, the pattern is 5n.

(g) In the formation of pattern of letter A, we can see that it will require six matchsticks.

 Number of A 1 2 3 4 5 6 – Required matchsticks 6 12 18 24 30 36 –

Therefore, the pattern is 6n.

Q.2 We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?

Ans-

In the formation of pattern of letter L, we can see that it will require two matchsticks.

 Number of T 1 2 3 4 5 6 – Required matchsticks 2 4 6 8 10 12 –

Therefore, the pattern is 2n. V and T are the letters in Q.1, which give us the same rule as that given by L. This is happened because in the formation of letters L, T and V, we require only two matchsticks in each.

Q.3 Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)

Ans-

Let number of rows be n.
Number of cadets in each row = 5
Number of cadets in n rows = 5n
The general rule to get the number of cadets in n rows is 5n.

Q.4 If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)

Ans-

Let number of boxes b.
Number of mangoes in each box = 50
Number of mangoes in b boxes = 50b
The general rule to get the number of mangoes in n boxes is 5n.

Q.5 The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)

Ans-

Let number of students be s.
Each student got pencils = 5
s students got pencils = 5s
Therefore, 5s pencils are needed to distribute in s students.

Q.6 A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)

Ans-

Let the flying time of a bird be t minutes.
Distance covered by the bird in one minute = 1 km
Distance covered by the bird in t minutes = t km
Therefore, the distance covered by a bird in t minutes is t km.

Q.7 Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows?
How many dots are there if there are 8 rows?
If there are 10 rows?

Ans-

Number of dots in a row = 9
Number of rows = r
Number of dots in r rows = 9r
Number of dots in 8 rows = 9 x 8
= 72
Number of dots in 10 rows = 9 x 10 = 90

Q.8 Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.

Ans-

Let age of Radha = x years
Difference between the age of Leela and Radha = 4 years
Therefore, age of Leela = Age of Radha – 4
= (x – 4) years

Q.9 Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?

Ans-

Number of laddus gave away by mother = l
Remaining laddus with mother = 5

Q.10 Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?

Ans-

Number of oranges in smaller box = x
Number of boxes filled by oranges = 2
Remaining oranges after filling 2 smaller boxes = 10
Total oranges in a big box = 2x + 10

Q.11 (a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (b) Fig 11.7 gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles. Ans-

(a) We see that number of matchsticks in given pattern are 4, 7, 10 and 13, which are 1 more than thrice of the number of square. Therefore, the pattern is (3n + 1), where n is the number of squares in pattern.

(b) We see that number of matchsticks in given pattern are 3, 5, 7 and 9, which are 1 more than twice of the number of triangles. Therefore, the pattern is (2n + 1), where n is the number of triangles in pattern.

Q.12 The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.

Ans-

Length of side of equilateral triangle = l
Perimeter of equilateral triangle = l + l + l = 3 l

Q.13 The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l. Ans-

Length of each side of regular hexagon = l
Perimeter of regular hexagon = l + l + l + l + l+ l

= 6 l

Q.14 A cube is a three-dimensional figure as shown in Fig 11.11. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube. Ans-

Length of each side of cube = l
Number of edges in cube = 12
Total length of the edges of cube = Number of edges × Length of one edge

= 12 l

Q.15 The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.12) AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius(r). Ans-

Diameter of a circle is double of its radius.
So, diameter (d) of circle = 2r

Q.16 To find sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54.
Thus, (14 + 27) + 13 = 14 + (27 + 13)This can be done for any three numbers.
This property is known as the associativity of addition of numbers.
Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.

Ans-

For three whole numbers a, b and c. We, have (a + b) + c = a + (b + c)

Q.17 Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.

Ans-

By using addition, subtraction and multiplication, we can make many expressions by using 5, 7 and 8. Some of these are as given below:
(5 + 7) – 8
(5 – 7) + 8
(8 – 7) + 5
(8 – 5) + 7
(7 – 5) + 7
(5 – 7) × 8
(5 – 8) × 7
(8 – 7) × 5
(5 + 7) × 8
(8 + 7) × 5
(5 + 8) × 7; etc.

Q.18 Which out of the following are expressions with numbers only?
(a) y + 3 (b) (7 × 20) – 8z
(c) 5(21 – 7) + 7 × 2 (d) 5
(e) 3x (f) 5 – 5n
(g) (7 × 20) –(5 × 10) – 45 + p

Ans-

Expressions (c) and (d) have only numbers.

Q.19 Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.

$\begin{array}{l}\left(a\right)z+1,z-1,y+17,y–17\text{}\left(b\right)17y,\frac{y}{17},5z\\ \left(c\right)2y+17,2y-17\left(d\right)7m,-7\text{\hspace{0.17em}}m+3,-7\text{\hspace{0.17em}}m-3\end{array}$

Ans-

(a) z + 1, here is addition because 1 is added to z.
z – 1, here is subtraction because 1 is subtracted from z.
y + 17, here is addition because 17 is added to y.
y – 17, here is subtraction because 17 is subtracted from y.

(b) 17y, here is multiplication because y is multiplied by 17.
(y/17), here is division because y is divided by 17.
5z, here is multiplication because z is multiplied by 5.

(c) 2y + 17, here is multiplication and addition because 17 is added to twice of y.
2y – 17, here is multiplication and subtraction because 17 is subtracted from twice of y.

(d) 7m, here is multiplication because m is multiplied by 7.
–7m + 3, here is multiplication and addition because 3 is added to –7 times of m.
– 7m – 3, here is multiplication and subtraction because 3 is subtracted from – 7 times of m.

Q.20 Give expressions for the following cases.
(a) 7 added to p (b) 7 subtracted from p
(c) p multiplied by 7 (d) p divided by 7
(e) 7 subtracted from – m (f) – p multiplied by 5
(g) – p divided by 5 (h) p multiplied by – 5

Ans-

(a) 7 added to p = p + 7
(b) 7 subtracted from p = p – 7
(c) p multiplied by 7 = 7p
(d) p divided by 7 = p/7
(e) 7 subtracted from – m =– m – 7
(f) – p multiplied by 5 = 5(– p) = – 5p
(g) – p divided by 5 = (– p/ 5)
(h) p multiplied by – 5 = – 5p

Q.21 Give expressions in the following cases.
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to16.

Ans-

(a)11 added to 2m = 2m + 11
(b) 11 subtracted from 2m = 2m – 11
(c)5 times y to which 3 is added = 5y + 3
(d) 5 times y from which 3 is subtracted = 5y – 3
(e) y is multiplied by – 8 = – 8y
(f) y is multiplied by – 8 and then 5 is added to the result = – 8y + 5
(g) y is multiplied by 5 and the result is subtracted from 16 = 16 – 5y
(h) y is multiplied by – 5 and the result is added to 16 = 16 + (– 5y) = 16 – 5y

Q.22 (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.

Ans-

(a) The expressions formed by using t and 4 are: 4 + t, 4 – t, t – 4, 4t, (4/t), (t/4)
(b) The different expressions formed by using y, 2 and 7 are: 2y + 7, 2y – 7, 7y + 2, 7y – 2, …

(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?

(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.

(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.

(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.

Ans-

(a) Present age of Sarita = y years
(i) Age of Sarita after 5 years = (y + 5) years
(ii) Age of Sarita 3 years back = (y – 5) years
(iii) Age of Sarita’s grandfather = 6 times of y = 6y years
(iv)The Grandmother is 2 years younger than grandfather. So, Age of the grandmother = (6x – 2) years
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. So, father’s age = (3y + 5) years

(b) The breadth of rectangular hall = b m The length of rectangular hall = (3b – 4) m

(c) Height of the box = h cm
Length of the box = 5h cm
Breadth of the box = (h – 10) cm

(d) Meena is at steps = s
Since, Beena is 8 steps ahead. Then Beena is at steps = (s + 8)
Since, Leena is 7 steps behind to Meena. So, Leena is at steps = (s – 7)
Total number of steps = (4s – 10)

e) Speed of bus = v km/hr
Distance covered by bus in 5 hours = time x speed
= 5v
Remaining distance = 20 km
Total distance between Daspur to Beespur = (5v + 20) km

Q.24 Change the following statements using expressions into statements in ordinary language.
(For example, Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)
(a) A notebook costs ₹ p. A book costs ₹ 3p.
(b) Tony puts q marbles on the table. He has 8q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.

Ans-

(a) The cost of a book is three times the cost of a notebook.
(b) The box of tony contains 8 times the number of marbles on the table.
(c) The total students in the school is 20 times of the students of our class.
(d) Jaggu’s uncle is 4 times older than Jaggu. Jaggu’s unty is 3 years younger than his uncle.
(e) Number of rows = r
Number of dots in a row = 5
Total number of dots = 5r
Therefore, total number of dots is 5 times of the total number of rows.

Q.25 (a) Given Munnu’s age to be x years, can you guess what (x – 2) may show? Can you guess what (x + 4) may show? What (3 x + 7) may show?

$\begin{array}{l}\text{(b) Given Sara’s age today to be y years. Think of her}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}age in the future or in the past.}\\ \text{What will the following expression indicate?}\\ y+7,y–3,y+4\frac{1}{2},\text{ }y–2\frac{1}{2}.\\ \text{(c) Given n students in the class like football, what}\\ \text{may 2n show? What may }\frac{\text{n}}{\text{2}}\text{ show?}\end{array}$

Ans-

(a) The present age of Mannu is x years. Age of any person is 2 years less than the age of Mannu i.e., (x – 2) years.
Age of any person is 4 years more than the age of Mannu i.e., (x + 4) years.
Age of any person is 7 years more than thrice of the age of Mannu i.e., (3x + 7) years.
(b) The present age of Sara = y years
(y + 7) is the age of Sara after 7 years.
(y – 3) is the age of Sara before 3 years.

$\begin{array}{l}\left(y+4\frac{1}{2}\right)\text{is the age of Sara after 4}\frac{1}{2}\text{years}\text{.}\\ \left(y-2\frac{1}{2}\right)\text{is the age of Sara before 2}\frac{1}{2}\text{years}\text{.}\end{array}$

(c) n students like football, but 2n students may like other game or football. Similarly, (n/2) students may like any other game out of the number of students who like football.

Q.26 State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}17=x+7\left(\mathrm{b}\right)\left(t-7\right)>5\left(c\right)\frac{4}{2}=2\\ \left(d\right)\left(7×3\right)-19=8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×4-8=2x\left(f\right)x-2=0\\ \left(g\right)\text{\hspace{0.17em}}2m<30\left(h\right)\text{\hspace{0.17em}}2n+1=11\left(i\right)\text{\hspace{0.17em}}7=\left(11×5\right)-\left(12×4\right)\\ \left(j\right)\text{\hspace{0.17em}}7=\left(11×2\right)+p\left(k\right)\text{\hspace{0.17em}}20=5y\left(l\right)\text{\hspace{0.17em}}\frac{3q}{2}<5\\ \left(m\right)\text{z}+\text{12>24}\left(n\right)20-\left(10-5\right)=3×5\\ \left(0\right)\text{\hspace{0.17em}}7-x=5\end{array}$

Ans-

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}17=x+7\text{, it is an equation in variable x because equal to sign is there}\text{.}\\ \left(b\right)\left(t-7\right)>5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is an not equation in variable t because equal}\text{\hspace{0.17em}}\text{to sign is not there}\text{.}\\ \left(c\right)\frac{4}{2}=2,\text{it is numerical equation because there is no variable}\text{.}\\ \left(d\right)\left(7×3\right)-19=8,\text{it is numerical equation because there is}\text{\hspace{0.17em}}\text{no variable}\text{.}\\ \left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×4-8=2x,\text{It is an equation in variable x because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(f\right)x-2=0,\text{\hspace{0.17em}}\text{It is an equation in variable x because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(g\right)\text{\hspace{0.17em}}2m<30,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(h\right)\text{\hspace{0.17em}}2n+1=11,\text{\hspace{0.17em}}\text{It is an equation in variable n because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(i\right)\text{\hspace{0.17em}}7=\left(11×5\right)-\left(12×4\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is numerical equation because there is no variable}\text{.}\\ \left(j\right)\text{\hspace{0.17em}}7=\left(11×2\right)+p,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable p because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(k\right)\text{\hspace{0.17em}}20=5y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable y because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(l\right)\text{\hspace{0.17em}}\frac{3q}{2}<5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(m\right)\text{z}+\text{12>24,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(n\right)20-\left(10-5\right)=3×5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is numerical equation because there is no variable}\text{.}\\ \left(o\right)\text{\hspace{0.17em}}7-x=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable x because equal to sign is there}\text{.}\end{array}$

Q.27 Complete the entries in the third column of the table.

 S.No. Equation Value of variable Equation satisfied Yes/No (a) 10y = 80 y = 10 (b) 10y = 80 y = 8 (c) 10y = 80 y = 5 (d) 4l = 20 l = 20 (e) 4l = 20 l = 80 (f) 4l = 20 l = 5 (g) b + 5 = 9 b = 5 (h) b + 5 = 9 b = 9 (i) b + 5 = 9 b = 4 (j) h – 8 = 5 h = 13 (k) h – 8 = 5 h = 8 (l) h – 8 = 5 h = 0 (m) p + 3 = 1 p = 3 (n) p + 3 = 1 p = 1 (o) p + 3 = 1 p = 0 (p) p + 3 = 1 p = – 1 (q) p + 3 = 1 p = – 2

Ans-

 S.No. Equation Value of variable Equation satisfied Yes/No (a) 10y = 80 y = 10 No (b) 10y = 80 y = 8 yes (c) 10y = 80 y = 5 No (d) 4l = 20 l = 20 No (e) 4l = 20 l = 80 No (f) 4l = 20 l = 5 Yes (g) b + 5 = 9 b = 5 No (h) b + 5 = 9 b = 9 No (i) b + 5 = 9 b = 4 Yes (j) h – 8 = 5 h = 13 Yes (k) h – 8 = 5 h = 8 No (l) h – 8 = 5 h = 0 No (m) p + 3 = 1 p = 3 No (n) p + 3 = 1 p = 1 No (o) p + 3 = 1 p = 0 No (p) p + 3 = 1 p = – 1 No (q) p + 3 = 1 p = – 2 Yes

Q.28 Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.

$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}5\mathrm{m}=60\left(10,5,12,15\right)\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\mathrm{n}+12=20\left(12,8,20,0\right)\\ \left(\mathrm{c}\right)\mathrm{p}-5=5\left(0,10,5,-5\right)\\ \left(\mathrm{d}\right)\frac{\mathrm{q}}{2}=7\left(7,2,10,14\right)\\ \left(\mathrm{e}\right)\mathrm{r}-4=0\left(4,-4,8,0\right)\\ \left(\mathrm{f}\right)\mathrm{x}+4=2\left(-2,0,2,4\right)\end{array}$

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{We have},\text{5m}=\text{6}0\dots \left(\text{i}\right)\\ \text{Putting m}=\text{1}0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(\text{1}0\right)\\ \mathrm{}=\text{5}0\ne 60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{10 is not the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(5\right)\\ =\text{}25\ne 60\\ ⇒\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{5 is not the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\text{\hspace{0.17em}Putting m}=12\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(12\right)\\ =60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{12 is the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=12\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \mathrm{}\text{5m}=\text{5}\left(12\right)\\ =\text{}60\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{12 is the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=15\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \mathrm{}\text{5m}=\text{5}\left(15\right)\\ =\text{}75\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{15 is not the solution of equation 5m}=\text{6}0.\\ \left(\text{b}\right)\text{We have},\text{n}+\text{12}=\text{2}0\dots \left(\text{i}\right)\\ \text{Putting n}=\text{1}2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{12}+\text{12}\\ =24\ne 20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{12 is not the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=8\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{8}+\text{12}\\ =20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{8 is the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=20\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{20}+\text{12}\\ =32\ne 20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{20 is not the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{0}+\text{12}\\ =12\ne 20\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{0 is not the solution of equation n}+\text{12}=\text{2}0.\\ \left(\text{c}\right)\text{We have},\text{p}-\text{5}=5\dots \left(\text{i}\right)\\ \text{Putting p}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=\text{0}-\text{5}\\ =-5\ne 5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{0 is not the solution of equation p}-\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=10\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=1\text{0}-\text{5}\\ =5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=1\text{0 is the solution of equation p}-\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=\text{5}-\text{5}\\ =0\ne 5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{5 is not the solution of equation p}-\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=-\text{\hspace{0.17em}}5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}-\text{5}=-\text{5}-\text{5}\\ =-10\ne 5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=-\text{\hspace{0.17em}5 is not the solution of equation p}-\text{5}=5.\\ \left(\text{d}\right)\text{We have},\text{}\frac{\mathrm{q}}{2}=7\dots \left(\text{i}\right)\\ \text{Putting q}=7\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{7}{2}\\ =3.5\ne 7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{7 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{2}{2}\\ =1\ne 7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{2 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=10\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{10}{2}\\ =5\ne 7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{10 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=14\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{14}{2}\\ =7\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{14 is the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \left(\text{e}\right)\text{We have},\text{}\mathrm{r}-4=0\dots \left(\text{i}\right)\\ \text{Putting r}=4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=4-4\\ =0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{4 is the solution of equation}\mathrm{r}-4=0.\\ \mathrm{Again},\\ \text{Putting r}=-\text{\hspace{0.17em}}4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=-\text{\hspace{0.17em}}4-4\\ =-\text{\hspace{0.17em}}8\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=-\text{\hspace{0.17em}4 is not the solution of equation}\mathrm{r}-4=0.\\ \mathrm{Again},\\ \text{Putting r}=8\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=8-4\\ =4\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{8 is not the solution of equation}\mathrm{r}-4=0.\\ \mathrm{Again},\\ \text{Putting r}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}-4=0-4\\ =-\text{\hspace{0.17em}}4\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{r}=\text{0 is not the solution of equation}\mathrm{r}-4=0.\\ \left(\text{f}\right)\text{We have},\text{}\mathrm{x}+4=2\dots \left(\text{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=-\text{\hspace{0.17em}}2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=-\text{\hspace{0.17em}}2+4\\ =2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=-\text{\hspace{0.17em}}2\text{is the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=0+4\\ =4\ne 2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=0\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=2+4\\ =6\ne 2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=2\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=4+4\\ =8\ne 2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{x}=4\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\end{array}$

Q.29 (a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.

 m 1 2 3 4 5 6 7 8 9 10 – – – m+10

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.

 t 3 4 5 6 7 8 9 10 11 – – – – 5t

(c) Complete the table and find the solution of the equation z/3 =4 using the table.

 z 8 9 10 11 12 13 14 15 16 – – – z/3 $2\frac{2}{3}$ 3 $3\frac{1}{3}$

(d) Complete the table and find the solution to the equation m – 7 = 3.

 m 5 6 7 8 9 10 11 12 13 – – – m-7

Ans-

(a) Table having solution of m+10 = 16, is given below

 m m+10 1 1+10 = 11 2 2 + 10 = 12 3 3 + 10 = 13 4 4 + 10 = 14 5 5 + 10 = 15 6 6 + 10 = 16 7 7 + 10 = 17 8 8 + 10 = 18 9 9 + 10 = 19 10 10 + 10 = 20

By inspection of the table, the solution to the equation m + 10 = 16 is m = 6.

(b) Table having solution of 5t = 35, is given below:

 t 5t 3 5(3) = 15 4 5(4) = 20 5 5(5) = 25 6 5(6) = 30 7 5(7) = 35 8 5(8) = 40 9 5(9) = 45 10 5(10) = 50 11 5(11) = 55 12 5(12) = 60

By inspection of the table, the solution to the equation 5t = 35 is t = 7.

(c) Table having solution of z/3 = 4, is given below:

 z z/3 8 $\frac{\text{8}}{\text{3}}=2\frac{2}{3}$ 9 9/3 = 3 10 $\frac{\text{10}}{\text{3}}=3\frac{1}{3}$ 11 $\frac{\text{11}}{\text{3}}=3\frac{2}{3}$ 12 12/3 = 4 13 $\frac{\text{13}}{\text{3}}=4\frac{1}{3}$ 14 $\frac{\text{14}}{\text{3}}=4\frac{2}{3}$ 15 15/3 = 5 16 $\frac{\text{16}}{\text{3}}=5\frac{1}{3}$ 17 $\frac{\text{17}}{\text{3}}=5\frac{2}{3}$

By inspection of the table, the solution to the equation z/3 = 4 is z = 12.

(d) Table having solution of m – 7 = 3, is given below:

 m m-7 5 5 – 7 = – 2 6 6 – 7 = – 1 7 7 – 7 = 0 8 8 – 7 = 1 9 9 – 7 = 2 10 10 – 7 = 3 11 11 – 7 = 4 12 12 – 7 = 5 13 13 – 7 = 6

By inspection of the table, the solution to the equation m – 7 = 3 is m = 10.

Q.30 Solve the following riddles, you may yourself construct such riddles. Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!

Ans-

(i) Let value of me be x.
Number of corners in a square = 4
Thrice of 4 = 12
Then, x + 12 = 24
x = 24 – 12
= 22
Therefore, the value of me is 22.

(ii) Let value of me be x.
Upcount on Sunday from me = x + 1
Upcount on Monday from me = x + 2
Upcount on Tuesday from me = x + 3
Upcount on Wednesday from me = x + 4
Upcount on Thursday from me = x + 5
Upcount on Friday from me = x + 6
Upcount on Saturday from me = x + 7
In this way, x + 7 = 23
x = 16
Therefore, the value of me is 16.

(iii) Let the special number be x
Number of player in cricket team
= 11
Subtracted number = 6
Then, according to condition,
x – 6 = 11

(iv) Let the value of me be x.
Then, according to condition,
22 – x = x
22 = 2x
x = 11
Therefore, the special number is 11.

## 1. Is Chapter 11 of Class 6 Maths easy?

Algebra is a branch of maths that is based on varied rules and assumptions. The rules that govern the manipulation of symbols in algebra will help the students in understanding the basics of algebra in Class 6 through multiple examples. Since it is an interesting topic, it develops curiosity among students, who as a result will tend to pay more attention, making the concepts easier for them.

## 2. How can one score good marks in Chapter 11 of Class 6 Maths by going through NCERT Solutions?

Chapter 11 is one of the important chapters of Class 6 Maths. Once students understand its basics, it will become easier for them to solve the problems exercise-wise. Solving the problems of this chapter will further improve their analytical and logical thinking skills. All of this is simplified and made easier by choosing the right study material. The NCERT Solutions of Extramarks prepared by experts are to help students analyse the concepts in which they are lagging in and work on them for scoring better marks in exams.

## 3. How to solve Chapter 11 Algebra in Class 6?

Solving an algebraic equation requires a few basic steps, although the steps further have their own other smaller steps depending on the type of question and given algebraic expression. The basic steps that are generally followed while solving questions based on Algebra are:

• Understanding what is to be found or proved by solving the question
• Grouping the similar terms and bringing them to one side
• Solving both the sides simultaneously
• Being careful with the changes in signs while shifting the numbers and variables from one side to another

## 4. What are Variables?

Variables are the signs or symbols which, we are yet to find a valuation. They are mostly denoted as X or Y and are known as variables as they are changeable. An example of a variable can be; Y+3= 12, where Y is a variable. When the variables are applied, they are not just one value. They can be easily represented in numerical forms and are productive in graphical illustrations. Speaking of their types, they Are mainly of three types- dependent variables, controlled variables and independent variables.

## 5. What are the topics covered in Algebra for Class 6?

The following topics are covered in Class 6 Algebra:

• Introduction to Algebra
• Matchstick Patterns
• The idea of a variable
• Use of variables in the common rule: Rules from Geometry and Rules from Arithmetic
• Expressions with variables
• Practical applications of expressions
• What is the equation?
• A solution to an equation

## 6. Explain terms of an expression as per Chapter 11 of Class 6 Maths.

The parts of an expression that are formed separately first and later subtracted or added, are called the terms of an expression. In the given example, the terms 7x and 10 are added to form the expression (7x+10).