# NCERT Solutions for Class 6 Maths Chapter 11

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

Algebra is a part of mathematics that focuses on different symbols and various regulations to use those symbols mathematically. NCERT solutions for Class 6 Maths Chapter 11 Algebra focus on one of the chapters of class 6, Algebra.

Algebra uses different symbols that are generally of Latin and Greek origin, giving specific utility. The most frequently used letters are X and Y that are variables. Students of Class 6 Maths Algebra learn to find value and answers of differential equations using these two variables. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. You can download NCERT Solution for Class 6 Science to score more marks on your examinations.

The knowledge about these variables help students develop analytical skills and prepare them for higher studies like engineering, computer science and so on.

### Important Topics under NCERT Solutions for Class 6 Maths Chapter 11

NCERT Solutions Class 6 Maths Chapter 11 is on Algebra. This chapter is an essential one for the 6th-grade students since it covers all essential points and sections that are needed to grow a strong understanding of numbers and equations in Algebra, which will go on to help the students greatly in higher mathematics. This chapter is covered in the Class 6 maths syllabus and has 8 major sections or topics. Students must go through each topic and concept covered in the chapter very carefully to be able to solve and practise the sums on Algebra.

• Introduction to Algebra

• Matchstick Problems

• The Idea of a Variable

• Use of Variables in Common Rules

1. Rules from Geometry

2. Rules from Arithmetic

• Expressions with Variables

• Practical Applications of Expressions

• What is an Equation?

• Solution to an Equation

Students should learn each and every topic carefully and then try to attempt and practise the NCERT Solutions for Class 6 Maths Chapter 11 Algebra.

Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)
Book your Free Demo session
Get a flavour of LIVE classes here at Vedantu

## Exercise 11.1

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

(a). A pattern of letter T

Ans. It is observed that number of matchsticks required is $2$. Hence, the pattern of letter T is $2n$.

(b) A pattern of letter Z

Ans. It is observed that number of matchsticks required is $3$. Hence, the pattern of letter Z is $3n$.

(c) A pattern of letter U

Ans. It is observed that number of matchsticks required is $3$. Hence, the pattern of letter V is $3n$.

(d) A pattern of letter V

Ans. It is observed that number of matchsticks required is $2$. Hence, the pattern of letter V is $2n$.

(e) A pattern of letter E

Ans. It is observed that number of matchsticks required is $5$. Hence, the pattern of letter E is $5n$.

(f) A pattern of letter S

Ans. It is observed that number of matchsticks required is $5$. Hence, the pattern of letter S is $5n$.

(g) A pattern of letter A

Ans. It is observed that number of matchsticks required is $6$. Hence, the pattern of letter A is $6n$.

2. We already know the rule for the pattern of letter 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. We Observe that letter L requires $2$ matchsticks. Hence, it has pattern $2n$. Letters ‘T’ and ‘V’ similarly require $2$ matchsticks and hence has pattern $2n$.

3. Cadets are marching in a parade. There are $\mathbf{5}$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. Number of cadets= Number of cadets in each row × Number of rows

Number of rows= $n$

Number of cadets in each row=$5$ (∵Given)

∴ Total number of cadets= $5\times n=5n$

4. If there are $\mathbf{50}$ 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. Number of mangoes= Number of mangoes in each box × Number of boxes

Number of boxes= b

Number of mangoes in each box= $50$ (∵Given)

∴ Total number of mangoes= $50\times b=50b$

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

Ans. Number of pencils needed= Number of pencils distributed per student × Number of students

Number of students= s

Number of pencils distributed per student= $5$ (∵Given)

∴ Total number of pencils needed= $5\times s=5s$

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. Distance covered by birds (in km) = speed × time

Time (in minutes) = t

Speed (in km/min) =$1$ km/min (∵Given)

∴ Total distance = $1\times t=t$km

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

Ans. Number of dots = Number of dots in each row × Number of rows

Number of rows= r

Number of dots in each row= $8$ (∵Given)

∴ Total number of dots= $8\times r=8r$……(Equation 1)

For $8$ rows,

$r=8$

Total number of dots= $8\times r=8\times 8=64$……(∵ From Equation 1)

For $10$ rows,

$r=10$

Total number of dots= $8\times r=8\times 10=80$……(∵ From Equation 1)

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 Radha’s age be x years

Given that, Leela is 4 years younger than Radha.

Hence, Leela’s age = $\left( x-4 \right)$ years

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?

Laddus given to guests and family = l

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 smaller boxes = 2

Number of oranges in 1 small box = $x$

Oranges remaining outside = 10

Total number of oranges in larger box = (number of smaller boxes) × (number of oranges in 1 smaller box) + oranges remaining outside

∴ Total oranges in larger box =$2x+10$

11. (a) Look at the following matchstick pattern of squares. 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. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)

Ans. If we remove the last vertical stick, we are left with a pattern of C.

Letter C has 3 matchsticks which gives us pattern $3n$.

Now, add the removed vertical stick. This gives us the required equation$3n+1$.

(b) Figs. Below 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. We can see that the figures have 1 matchstick more than twice the number of triangles in the pattern.

Hence, required equation is $2n+1$. (where, n is the number of triangles)

## Exercise 11.2

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

Ans. Side of equilateral triangle= l

Number of sides of triangle = 3

Perimeter of equilateral triangle=$3\times l=3l$

Hence the perimeter of the given equilateral triangle is 3l.

2. The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l.

Ans. Side of hexagon = l

Number of sides in a hexagon = 6

Perimeter of hexagon =$6\times l=6l$

Hence, the perimeter of given regular hexagon is 6l.

3. A cube is a three- dimensional figure. 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 edged of cube.

Ans. Length of one edge of a cube= l

Number of edges in a cube = 12

Total length of edges of the given cube =$12\times l=12l$

Hence, the total length of edges in a cube is 12l.

4. 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 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. From figure, we can see that AC=BC=PC=r

Also, diameter AB (d)= AC+BC =$r+r=2r$

Hence, $d=2r$

5. 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 $\left( \mathbf{14}\text{ }+\text{ }\mathbf{27} \right)\text{ }+\text{ }\mathbf{13}\text{ }=\text{ }\mathbf{14}\text{ }+\text{ }\left( \mathbf{27}\text{ }+\text{ }\mathbf{13} \right)$

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. Let a, b and c be three variables.

Property Associativity of addition of Whole Numbers can be expressed as –

$\left( a+b \right)+c=a+\left( b+c \right)$

## Exercise 11.3

1. 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.

(Hint: Three possible expressions are $\mathbf{5}+\left( \mathbf{8}\mathbf{7} \right),\text{ }\mathbf{5}\text{ -}\left( \mathbf{8}-\mathbf{7} \right),\text{ }\left( \mathbf{5}\times \mathbf{8} \right)+\mathbf{7}$make the other expressions)

Ans. So, by using multiplication, addition and subtraction the possible expressions with the given numbers are taken as,

$5\times (7+8)$.

$\left( 8+7 \right)5.$

$5\left( 7+8 \right).$

$5+\left( 87 \right).$

$(8\times 7)-5$.

$\left( 8+5 \right)7.$

$\left( 8\times 5 \right)7$

2. Which out of the following are expressions with numbers only:

(a) $y+3$

(b) $7\times 20-8z$

(c) $5(21-7)+7\times 2$

(d) $5$

(e) $3x$

(f) $5-5n$

(g) $\left( \mathbf{7}\times \mathbf{20} \right)\left( \mathbf{5}\text{ }\times \text{ }\mathbf{10} \right)-\mathbf{45}+p$

Ans. Options (c) and (d) as we can see no variables in these expressions.

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

(a) z + 1, z - 1, y + 17, y -17

z-1 → Subtraction

y -17→ Subtraction

(b) 17y, $\frac{\mathbf{y}}{\mathbf{17}}$, 5z

Ans. 17y → Multiplication

$\frac{y}{17}$→ Division

5z→ Multiplication

(c) 2y + 17, 2y – 17

Ans. 2y + 17→ Multiplication, Addition

2y - 17→ Multiplication, Subtraction

(d) 7m, -7m + 3, -7m – 3

Ans. 7m → Multiplication

-7m - 3→ Multiplication, Subtraction

4. Give expressions for the following cases:

Ans. $p+7$

(b) 7 subtracted from p.

Ans. $p-7$

(c) p multiplied by 7.

Ans. $p\times 7=7p$

(d) p divided by 7.

Ans. $\frac{p}{7}$

(e) 7 subtracted from -m.

Ans. $-m-7$

(f) -p multiplied by 5.

Ans. $-p\times 5=-5p$

(g) -p divided by 5.

Ans. $-\frac{p}{5}$

(h) p multiplied by -5.

Ans. $p\times -5=-5p$

5. Give expression in the following cases:

Ans.$2m+11$

(b) 11 subtracted from 2m.

Ans. $2m-11$

(c) 5 times y to which 3 is added.

Ans. $5y+3$

(d) 5 times y from which 3 is subtracted.

Ans. $5y-3$

(e) y is multiplied by -8.

Ans. $-8y$

(f) y is multiplied by -8 and then 5 is added to the result.

Ans. $-8y+5$

(g) y is multiplied by 5 and result is subtracted from 16.

Ans. $16-5y$

(h) y is multiplied by -5 and the result is added to 16.

Ans. $-5y+16$

6. (a) From expressions using t and 4. Use not more than one number operation. Every expression must have t in it.

Ans. $t+4,\text{ }t4,\text{ }t\times 4,\text{ }\frac{t}{4}$

(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. $2y+7,\text{ }2y7,\text{ }7y+2,7y2$

## Exercise 11.4

(a) Take Sarita’s present age to be y years.

(i) What will be her age 5 years from now?

Ans. Sarita’s age 5 years from now will be $\left( y\text{ }+\text{ }5 \right)$years.

(ii) What was her age 3 years back?

Ans. Sarita’s age 3 years back was $\left( y\text{ }-\text{ }3 \right)$years.

(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?

Ans. Age of Sarita’s grandfather is $6y$ years.

(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?

Ans. Age of Sarita’s grandmother is $\left( 6y-2 \right)$ years.

(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

Ans. Age of Sarita’s father is $\left( 3y+5 \right)$ years.

(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?

Ans. Length of rectangular hall = $\left( 3b-4 \right)$ meters

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

Ans. Height of Box = $h$cm

Length of Box = $5h$ cm

Breadth of Box = $\left( 5h-10 \right)$ cm

(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.

Ans. Meena’s position = $s$

Beena’s position = $s+8$

Leena’s position = $s-7$

Total number of steps = 4s - 10

(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. Speed of bus = $v$ km/hr

Distance travelled = speed × time = $v\times 5=5v$ km

Distance remaining = 20 km

Total distance from Daspur to Beespur = $\left( 5v+20 \right)$ km

2. 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 note book costs ₹p. A book costs ₹3p.

Ans. A book cost 3 times the notebook.

(b) Tony puts q marbles on the table. He has 8q marbles in his box.

Ans. Number of marbles in box is 8 times the marbles on the table.

(c) Our class has n students. The school has 20n students.

Ans. Total number of students in the school is 20 times that of our class.

(d) Jaggu is z years old. His uncle is 4z years old and his aunt is $\left( \mathbf{4}z\mathbf{3} \right)$ years old.

Ans. Jaggu’s uncle’s age is 4 times the age of Jaggu and Jaggu’s aunt is 3 years younger than his uncle.

(e) In an arrangement of dots there are r rows. Each row contains 5 dots.

Ans. Total number of dots is 5 times the number of rows.

3. (a) Given, Munnu’s age to be x years. Can you guess what $\left( x\mathbf{2} \right)$may show? (Hint: Think of Munnu’s younger brother). Can you guess what $\left( x+\mathbf{4} \right)$ show? What $\left( \mathbf{3}x+\mathbf{7} \right)$ may show?

Ans. $\left( x-2 \right)$years will represent age of Munnu’s younger brother.

($x+4$) years will represent age of Munnu after 4 years.

($3x+7$) years represent age of an elder to Munnu whose age is 7 more than 3 times Munnu’s age.

(b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate? $\mathbf{y}+\mathbf{7},\text{ }\mathbf{y}-\mathbf{3},\text{ }\mathbf{y}+\mathbf{4}\text{ }\frac{1}{2},\text{ }\mathbf{y}-\mathbf{2}\text{ }\frac{1}{2}$

Ans. Age after 7 years = ($y+7$) years

Age 3 year ago = ($y3$) years

Age after 4 and half years = $\left( y+4\frac{1}{2} \right)$ years

Age 2 and half year ago = $\left( y-2\frac{1}{2} \right)$ years

(c) Given, n students in the class like football, what may 2n show? What may $\frac{n}{\mathbf{2}}$ show? (Hint: Think of games other than football).

Ans. Number of students who like swimming is twice the students who like football = 2n

Number of students who like Athletics is half the students who like football =$\frac{n}{2}$

## Exercise 11.5

1.State which of the following are equations (with a variable). Given reason for your answer. Identify the variable from the equations with a variable.

(a)$\mathbf{17}=x+\mathbf{7}$

Ans. It is an equation with variable x.

(b) $\left( t-\mathbf{7} \right)>\mathbf{5}$

Ans. It is not an equation since there is no equal sign.

(c) $\frac{\mathbf{4}}{\mathbf{2}}=\mathbf{2}$

Ans. It is an equation with no variable.

(d) $\left( \mathbf{7}\times \mathbf{3} \right)\mathbf{19}=\mathbf{8}$

Ans. It is an equation with no variable.

(e) $5\text{ }\times \text{ }4\text{ }-\text{ }8\text{ }=\text{ }2\text{ }x\sqrt{{{a}^{2}}+{{b}^{2}}}$

Ans. It is an equation with variable x.

(f) $x-\mathbf{2}=\mathbf{0}$

Ans. It is an equation of variable x.

(g) $\mathbf{2}m<\mathbf{30}$

Ans. It is not an equation since there is no equal sign.

(h) $\mathbf{2}n+\mathbf{1}=\mathbf{11}$

Ans. It is an equation with variable n.

(i) $7\text{ }=\text{ }\left( 11\text{ }\times \text{ }5 \right)-\left( 12\text{ }\times \text{ }4 \right)$

Ans. It is an equation with no variable.

(j) $\mathbf{7}=\left( \mathbf{11}\times \mathbf{2} \right)+p$

Ans. It is an equation with variable p.

(k) $\mathbf{20}=\mathbf{5}y$

Ans. It is an equation with variable y.

(l) $\frac{\mathbf{3}q}{\mathbf{2}}<\mathbf{5}$

Ans. It is not an equation since there is no equal sign.

(m) $z+\mathbf{12}>\mathbf{24}$

Ans. It is not an equation since there is no equal sign.

(n) $\mathbf{20}\left( \mathbf{10}-\mathbf{5} \right)=\mathbf{3}\times \mathbf{5}$

Ans. It is an equation with no variable.

(o) $\mathbf{7}-x=\mathbf{5}$

Ans. It is an equation with variable x.

2. Complete the entries of the third column of the table:

 Sr. 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.

 Sr. No. Equation Value of Variable Equation Satisfied Yes/No Solution of L.H.S. (a) 10y = 80 y = 10 No 10×10=100 (b) 10y = 80 y = 8 Yes 10×8=80 (c) 10y = 80 y = 5 No 10×5=50 (d) 4l = 20 l = 20 No 4×20=80 (e) 4l = 20 l = 80 No 4×80=320 (f) 4l = 20 l = 5 Yes 4×5=20 (g) b + 5 = 9 b = 5 No 5+5=10 (h) b + 5 = 9 b = 9 Yes 9+5=14 (i) b + 5 = 9 b = 4 Yes 4+5=9 (j) h – 8 = 5 h = 13 Yes 13–8=5 (k) h – 8 = 5 h = 8 No 8–8=0 (l) h – 8 = 5 h = 0 No 0–8=–8 (m) p + 3 = 1 p = 3 No 3+3=6 (n) p + 3 = 1 p = 1 No 1+3=4 (o) p + 3 = 1 p = 0 No 0+3=3 (p) p + 3 = 1 p = -1 No –1+3=2 (q) p + 3 = 1 p = -2 Yes –2+3=1

3. 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.

(a) $5m=\mathbf{60}\left( \mathbf{10},\text{ }\mathbf{5},\text{ }\mathbf{12},\text{ }\mathbf{15} \right)$

Ans. $5m=60$

Putting the given values in L.H.S.,

5×10=50

∴ L.H.S. ≠ R.H.S.

m=10 is not the solution.

5×5=25

∴L.H.S. ≠ R.H.S.

m=5 is not the solution.

5×12=60

∴ L.H.S. = R.H.S.

m=12 is a solution.

5×15=75

∴ L.H.S. ≠ R.H.S.

m=15 is not the solution.

(b) $n+\mathbf{12}=\mathbf{20}\left( \mathbf{12},\text{ }\mathbf{8},\text{ }\mathbf{20},\text{ }\mathbf{0} \right)$

Ans. Putting the given values in L.H.S.,

12+12=24

∴ L.H.S. ≠ R.H.S.

n=12 is not the solution.

8+12=20

∴ L.H.S. = R.H.S.

n=8 is a solution.

20+12=32

∴ L.H.S. ≠ R.H.S.

n=20 is not the solution.

0+12=12

∴ L.H.S. ≠ R.H.S.

n=0 is not the solution.

(c) p–5=5(0, 10, 5, -5)

Ans. Putting the given values in L.H.S.,

0–5=–5

∴ L.H.S. ≠ R.H.S.

p =0 is not the solution.

10–5=5

∴ L.H.S. = R.H.S.

p=10 is a solution.

5–5=0

∴ L.H.S. ≠ R.H.S.

p=5 is not the solution.

–5–5=–10

∴ L.H.S. ≠ R.H.S.

p=5 is not the solution.

(d) $\frac{q}{\mathbf{2}}=\mathbf{7}\left( \mathbf{7},\text{ }\mathbf{2},\text{ }\mathbf{10},\text{ }\mathbf{14} \right)$

Ans. Putting the given values in L.H.S.,

$\frac{7}{2}=\frac{7}{2}$

∴ L.H.S. ≠ R.H.S.

q=7 is not the solution.

$\frac{2}{2}=1$

∴ L.H.S. ≠ R.H.S.

q=2 is not the solution.

$\frac{10}{2}=5$

∴ L.H.S. ≠ R.H.S.

q=10 is not the solution.

$\frac{14}{2}=7$

∴ L.H.S. = R.H.S.

q=14 is a solution.

(e) $r-\mathbf{4}=\mathbf{0}\left( \mathbf{4},\text{ }-\mathbf{4},\text{ }\mathbf{8},\text{ }\mathbf{0} \right)$

Ans. Putting the given values in L.H.S.,

4–4=0

∴ L.H.S. = R.H.S.

r=4 is a solution.

–4–4=–8

∴ L.H.S. ≠ R.H.S.

r=4 is not the solution.

8–4=4

∴ L.H.S. ≠ R.H.S.

r=8 is not the solution.

0–4=–4

∴L.H.S. ≠ R.H.S.

r=0 is not the solution.

(f) $~x+\mathbf{4}=\mathbf{2}\left( -\mathbf{2},\text{ }\mathbf{0},\text{ }\mathbf{2},\text{ }\mathbf{4} \right)$

Ans. Putting the given values in L.H.S.,

–2+4=2

∴ L.H.S. = R.H.S.

x=2 is a solution.

0+4=4

∴ L.H.S. ≠ R.H.S.

x=0 is not the solution.

2+4=6

∴L.H.S. ≠ R.H.S.

x=2 is not the solution.

4+4=8

∴ L.H.S. ≠ R.H.S.

x=4 is not the solution.

4. (a) Complete the table and by inspection of the table find the solution to the equation $m+\mathbf{10}=\mathbf{16}$.

 M 1 2 3 4 5 6 7 8 9 10 … … … m + 10 … … … … … … … … … … … … …

Ans

 M 1 2 3 4 5 6 7 8 9 10 11 12 13 m + 10 11 12 13 14 15 16 17 18 19 20 21 22 23

∵ At $m=6,m+10=16$

is the solution.

(b) Complete the table and by inspection of the table find the solution to the equation $\mathbf{5t}=\mathbf{35}$.

 T 3 4 5 6 7 8 9 10 11 … … … … 5t … … … … … … … … … … … … …

Ans.

 T 3 4 5 6 7 8 9 10 11 12 13 14 15 5t 15 20 25 30 35 40 45 50 55 60 65 70 75

∴ At $t=7$, $5t=35$

$t=7$ is the solution.

(c) Complete the table and by inspection of the table find the solution to the equation $\frac{\mathbf{z}}{\mathbf{3}}$=4

 Z 8 9 10 11 12 13 14 15 16 … … … … $\frac{\mathbf{z}}{\mathbf{3}}$ $2\frac{2}{3}$ 3 3$\frac{\mathbf{1}}{\mathbf{3}}$ … … … … … … … … … …

Ans. 

 Z 8 9 10 11 12 13 14 15 16 17 18 19 20 $\frac{z}{3}$ $\frac{8}{3}$ 3 $\frac{10}{3}$ $\frac{11}{3}$ 4 $\frac{13}{3}$ $\frac{14}{3}$ 5 $\frac{16}{3}$ $\frac{17}{3}$ 6 $\frac{19}{3}$ $\frac{20}{3}$

∴ At $z=12$, $\frac{z}{3}$=4

Hence, $z=12$ is the solution.

(d) Complete the table and by inspection of the table find the solution to the equation $\mathbf{m}-\mathbf{7}=\mathbf{3}$.

 m 5 6 7 8 9 10 11 12 13 … … $\mathbf{m}-\mathbf{7}$ … … … … … … … … … … …

Ans.

 m 5 6 7 8 9 10 11 12 13 14 15 $m-7$ -2 -1 0 1 2 3 4 5 6 7 8

∴ At $m=10$, $m-7=3$

Hence, $m=10$is the solution.

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

### NCERT Solutions For Class 6 Maths Chapter 11 Algebra Free PDF Download

NCERT solutions for Class 6th Maths contain information about algebra, one of the most important parts of mathematics.

The main characteristic of NCERT Class 6 Maths Chapter 11 is that it uses letters to deduce the mathematical analogies. This will give students of Class 6 Maths Ch 11 knowledge about using algebraic values and their applications.

NCERT Maths Book Class 6 Chapter 11 solutions are available in PDF that makes download and accessing the material much more manageable and convenient. The PDF has been designed by the expert team of teachers, who have valuable resources and experience in teaching CBSE students for many years.

The PDF has been developed, keeping in view the latest guidelines issued by CBSE and NCERT. However, students often get confused with the latest structure of syllabus issued by CBSE. In this regard, Vedantu’s NCERT solutions for Class 6 Maths Chapter 11 can resolve this problem.

The students of class 6 are relatively young and just come out of the primary segment of schooling. Therefore it becomes quite evident that the students will need a simplified and detailed approach to get them adjusted to the new pattern of class 6.

The team of teachers kept this in mind while developing the PDF. Parents and guardians can now be at peace, and they can download the PDF from anywhere.

### NCERT Solution For Class 6 Maths Chapter 11

Algebra Class 6 NCERT teaches students how to solve different problems through systematic methods.

NCERT solutions for Class 6 Maths Chapter 11 Algebra contain many new conceptions, problems and solutions, and examples in an uncomplicated and straightforward language.

The chapter further contains illustrated representations, graphs to make Algebra Class 6 student-friendly. This will interest the class 6 students and will help them to develop fundamental knowledge about algebra.

Students of Algebra 6th Class often face a problem of complicated numericals and their solutions. This solution aims to troubleshoot these problems.

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

There are certain benefits of Class 6 Maths PDF which is available online on Vedantu’s website.

The benefits are;

• PDF aims to give uncomplicated solutions and explanations to different numerical using illustrative examples. This will help students prepare promptly before their final exam.

• One of the major problems that students regularly face is memorising formulas.

This solution aims to give an organised pattern and explanations of these formulas to make their task more manageable.

• The PDF is free to download and readily available. This is to make student’s tasks and time management easier before their final exams.

All the numerical examples, questions and solutions are provided in the PDF for students to understand the concepts better and quickly. This will also help them to prepare for higher studies.

### Chapter wise NCERT Solutions for Class 6 Maths

Along with this, students can also view additional study materials provided by Vedantu, for Class 6 Maths Chapter 1 Knowing Our Numbers

### NCERT Solution Class 6 Maths of Chapter 11 All Exercises

Chapter 11 - Algebra Exercises in PDF Format

Exercise 11.1

11 Questions with Solutions

Exercise 11.2

4 Questions with Solutions

Exercise 11.3

6 Questions with Solutions

Exercise 11.4

3 Questions with Solutions

Exercise 11.5

4 Questions with Solutions

1. What is Algebra? What is a Variable?

Variables can be defined as a sign or symbol of which, we are yet to find a valuation. It is generally denoted as X or Y.

An example can be; X+3= 9, where X can be the variable.

When a variable is applied, it is not just one value. It is represented in many numerical forms. Variables can be very productive in graphical illustrations.

In simple terms, variables are changeable. This is the reason why they are called variables. There are three main types of variables, namely; dependent, controlled and independent.

Algebra is a part of mathematics explaining different Greek symbols and regulations for deducing various problems into their solutions. The symbols given constitute specific valuations known as variables.

This branch of mathematics was introduced so that students can learn the art of deducing problems into their solutions.

2. Is Chapter 11 Class 6 Algebra Easy?

Algebra is a branch of Mathematics with its own rules and assumptions wherein you use and manipulate mathematical symbols for solving multiple problems. The rules that govern this manipulation of symbols are also a part of Algebra whose basics the students would learn in their 6th standard along with multiple examples. Chapter 11 is an introduction to Algebra for class 6 students. Since it is an interesting topic, students usually tend to pay more attention making it easier for them. However, if anyone finds it difficult, NCERT Solutions by Vedantu will help you out.

3. How do you solve Chapter 11 Algebra in Class 6?

Solving any algebraic equation would require some basic steps, although the steps would further have their own other smaller steps depending on the question and type of algebraic expression. The basic steps are:

• First understand what the question is asking you to find

• Group the similar terms together and bring them to one side

• Solve both sides simultaneously

• Be careful with the sign changes while shifting the numbers and variables

4. How do you teach Chapter 11 Algebra for 6th Grade?

Teaching Algebra for Class 6 Chapter 11  depends on the teaching techniques implied by the teacher. To each their own. However, some basic steps that could be taught to the students include:

• Proper distribution of the variables

• Teaching them the rules of sign changes

• Teaching them the multiple techniques

• Providing students with questions having a proper structure of difficulty levels

• Adequate practice of different topics.

5. How can I download the Solutions for NCERT Class 6 math Chapter 11??

The solutions are easily available on the Vedantu site. You can follow these steps to avail them:

• The webpage with Vedantu’s solutions for Class 6 Maths Chapter 11 Solutions will open.