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# NCERT Solutions for Class 6 Maths Chapter 11: Algebra

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NCERT Solutions for Class 6 Maths Chapter 11 Algebra by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 6 Maths Chapter 11 are designed as per the CBSE Class 6 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 6 Chapter 11.

## NCERT Solutions for Class 6 Maths Chapter 11: Algebra Ex 11.1

#### Question1.Find the rule which gives the number of matchsticks required to make thefollowing matchsticks patterns. Use a variable to write the rule.

Solution:
(a)

From the figure we observe that two matchsticks are required to make a letter T. Hence, the pattern is 2n
(b)

From the figure we observe that three matchsticks are required to make a letter Z. Hence, the pattern is 3n
(c)

From the figure we observe that three matchsticks are required to make a letter U. Hence, the pattern is 3n
(d)

From the figure we observe that two matchsticks are required to make a letter V. Hence, the pattern is 2n
(e)

From the figure we observe that 5 matchsticks are required to make a letter E. Hence, the pattern is 5n
(f)

From the figure we observe that 5 matchsticks are required to make a letter S. Hence, the pattern is 5n
(g)

From the figure we observe that 6 matchsticks are required to make a letter A. Hence, the pattern is 6n

#### Question2:We already know the rule for the pattern of letters L, C and F. Some of theletters from Q.1 (given above) give us the same rule as that given by L.Which are these? Why does this happen?

Solution:
We know that L require only two matchsticks. So, the pattern for letter L is 2n. Among all the letters
given in question 1, only L and V are the letters which require two matchsticks. Hence, (a) and (d).

#### Question3:Cadets are marching in a parade. There are 5 cadets in a row. What is therule which gives the number of cadets, given the number of rows? (Use nfor the number of rows)

Solution:
Let n be the number of rows
Number of cadets in a row = 5
Total number of cadets = number of cadets in a row × number of rows
= 5n

#### Question4:If there are 50 mangoes in a box, how will you write the total number ofmangoes in terms of the number of boxes? (Use b for the number of boxes.)

Solution:
Let b be the number of boxes
Number of mangoes in a box = 50
Total number of mangoes = number of mangoes in a box × number of boxes
= 50b

#### Question5:The teacher distributes 5 pencils per students. Can you tell how manypencils are needed, given the number of students? (Use s for the number ofstudents.)

Solution:
Let s be the number of students
Pencils given to each student = 5
Total number of pencils = number of pencils given to each student × number of students
= 5s

#### Question6:A bird flies 1 kilometer in one minute. Can you express the distancecovered by the birds in terms of its flying time in minutes? (Use t for flyingtime in minutes.)

Solution:
Let t minutes be the flying times
Distance covered in one minute = 1 km
Distance covered in t minutes = Distance covered in one minute × Flying time
= 1 × t
= t km

#### Question7: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 herRangoli have for r rows? How many dots are there if there are 8 rows? Ifthere are 10 rows?

Solution:
Number of dots in a row = 9
Number of rows = r
Total number of dots in r rows = Number of dots in a row × number of rows
= 9r
Number of dots in 8 rows = 8 × 9
= 72
Number of dots in 10 rows = 10 × 9
= 90

#### Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Canyou write Leela’s age in terms of Radha’s age? Take Radha’s age to be xyears.

Solution:
Let Radha’s age be x years
Leela’s age = 4 years younger than Radha
= (x – 4) years

Solution:
Number of laddus mother gave = l
Total number of laddus = number of laddus given away by mother + number of laddus remaining

#### Question10:Oranges are to be transferred from larger boxes into smaller boxes. Whena large box is emptied, the oranges from it fill two smaller boxes and still 10oranges remain outside. If the number of oranges in a small box are takento be x, what is the number of oranges in the larger box?

Solution:
Number of oranges in a small box = x
Number of oranges in two small boxes = 2x
Number of oranges remained = 10
Number of oranges in large box = number of oranges in two small boxes + number of oranges remained = 2x + 10

#### (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 interms of the number of triangles.

Solution:
(a) We may observe that in the given matchstick pattern, the number of matchsticks are 4, 7,10 and 13, which is 1 more than the thrice of the number of squares in the pattern
Therefore the pattern is 3x + 1, where x is the number of squares
(b) We may observe that in the given matchstick pattern, the number of matchsticks are 3, 5, 7 and 9 which is 1 more than the twice of the number of triangles in the pattern.
Therefore the pattern is 2x + 1, where x is the number of triangles.

## NCERT Solutions for Class 6 Maths Chapter 11: Algebra Ex 11.2

#### Question1.The side of an equilateral triangle is shown by l. Express the perimeter ofthe equilateral triangle using l.

Solution:
Side of equilateral triangle = l
Perimeter = l + l + l = 3l

#### (Hint: A regular hexagon has all its six sides equal in length.)

Solution:
Side of a regular hexagon = l
Perimeter = l + l + l + l + l + 1
= 6l

#### Question3:A cube is three dimensional figure as shown in Fig 11.11. It has six facesand all of them are identical squares. The length of an edge of the cube isgiven by l. Find the formula for the total length of the edges of a cube.

Solution:
Length of an edge of the cube = l
Number of edges = 12
Total length of the edges = Number of edges × length of an edge
=12l

#### Question4:The diameter of a circle is a line which joins two points on the circle andalso passes through the centre of the circle. (In the adjoining figure (Fig11.2) AB is a diameter of a circle; C is its centre.)Express the diameter of the circle (d) in terms of its radius (r).

Solution:
Diameter = AB
= AC + CB
= r + r
= 2r
Hence, the diameter of the circle in terms of its radius is 2r

#### Question5: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 41and then add 13 to it to get thetotal 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)

Solution:
For any three whole numbers a, b and c
(a + b) + c = a + (b + c)

## NCERT Solutions for Class 6 Maths Chapter 11: Algebra Ex 11.3

Exercise 11.3
Question1.
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.

Solution:
Some of the expressions formed by 5, 7 and 8 are as follows:
5 × (8 – 7)
5 × (8 + 7)
(8 + 5) × 7
(8 – 5) × 7
(7 + 5) × 8
(7 – 5) × 8

#### Question2: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

Solution:
(c) and (d) are the expressions with numbers only.

#### Question3:Identify the operations (addition, subtraction, division, multiplication) informing the following expressions and tell how the expressions have beenformed.(a) z + 1, z – 1, y + 17, y – 17(b) 17y, y / 17, 5z(c) 2y + 17, 2y – 17(d) 7m, -7m + 3, -7m – 3

Solution:
(a) z + 1 = 1 is added to z = Addition
z – 1 = 1 is subtracted from z = Subtraction

y – 17 = 17 is subtracted from y = Subtraction

(b) 17y = y is multiplied by 17 = Multiplication
y / 17 = y is divided by 17 = Division
5z = z is multiplied by 5 = Multiplication

(c) 2y + 17 = y is multiplied by 2 and 17 is added to the result = Multiplication and addition
2y – 17 = y is multiplied by 2 and 17 is subtracted from the result = Multiplication and
subtraction

(d) 7m = m is multiplied by 7 = multiplication
-7m + 3 = m is multiplied by -7 and 3 is added to the result = Multiplication and addition
-7m – 3 = m is multiplied by -7 and 3 is subtracted from the result = Multiplication and
subtraction

#### Question4: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

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

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

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

#### Question6:(a) Form expressions using t and 4. Use not more than one numberoperation. 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.

Solution:
(a) (t + 4), (t – 4), 4t, (t / 4), (4 / t), (4 – t), (4 + t) are the expressions using t and 4.
(b) 2y + 7, 2y – 7, 7y + 2, are the expression using y, 2 and 7.

## NCERT Solutions for Class 6 Maths Chapter 11: Algebra Ex 11.4

Exercise 11.4
Question1.
(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 two year 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 three 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 Reena 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.

Solution:
(a)
(i) Sarita’s age aftyer 5 years from now = Sarita’s present age + 5
= (y + 5) years
(ii) Sarita’s age 3 years back = Sarita’s present age – 3
= (y – 3) years
(iii) Grandfather’s age = 6 × Sarita’s present age
= 6y years
(iv) Grandmother’s age = granfather’s present age – 2
= (6y -2) years
(v) Father’s age = 5 + 3 × Sarita’s present age
= (5 + 3y) years

(b) Length = 3 × Breadth – 4
l = (3b – 4) metres
(c) Length = 5 × Breadth
l = 5h cm
Breadth = 5 × length – 10
b = (5h – 10) cm
(d) The step at which Beena is = (step at which Meena is) + 8
= (s + 8)
The step at which Leena is = (step at which Meena is) – 7
= (s – 7)
Total steps = 4 × (step at which Meena is) – 10
= (4s – 10)
(e) Speed = v km / hr
Distance travelled in 5 hours = 5 × v
= 5v km
Total distance travelled between Daspur and Beespur = (5v + 20) km

#### Question2:Change the following statements using expressions into statements inordinary language. (For example, Given Salim scores r runs in a cricketmatch, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15runs more than Salim.)(a) A notebook costs ₹ p. A book costs ₹ 3p(b) Tony put q marbles on the table. He has 8 q marbles in his box.(c) Our class has n students. The school has 20 n 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

Solution:
(a) A book costs 3 times the costs of a notebook.
(b) Tony’s box contains 8 times the number of marbles on the table
(c) Total number of students in the school is 20 times that of our class
(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his
uncle
(e) The total number of dots is 5 times the number of rows.

#### .(c) Given n students in the class like football, what may 2n shows? Whatmay n / 2 show?

Solution:
(a) (x – 2) represents the person whose age is (x – 2) years and he is 2 years younger to
Munnu.
(x + 4) represents the person whose age is (x + 4) years and he is 4 years elder than Munnu
(3x + 7) represents the person whose age is (3x + 7) years, elder to Munnu and his age is 7
years more than the three times of the age of Munnu.
(b) In Future
After n years since now, Sara’s age will be (y + n) years
In past, n years ago, Sara’s age was (y – n) years
(y + 7) represents the person whose age is (y + 7) years and is 7 years elder to Sara.
(y – 3) represents the person whose age is (y – 3) years and is 3 years younger to Sara.
y+4

represents the person whose age is y+4

years and is 4

years elder to Sara.
y – 2

represents the person whose age is y – 2

years and is 2

years younger to Sara.
(c) 2n shows the number of students who like either football or some other game like tennis
whereas n / 2 shows the number of students who like tennis out of the total number of
students who like football.

## NCERT Solutions for Class 6 Maths Chapter 11: Algebra Ex 11.5

#### Question1.State which of the following are equations (with a variable). Give reasonfor your answer.Identify the variable from the equations with a variable.(a) 17 = x + 17(b) (t – 7) > 5(c) 4 / 2 = 2(d) (7 × 3) – 19 = 8(e) 5 × 4 – 8 = 2x(f) x – 2 = 0(g) 2m < 30(h) 2n + 1 = 11(i) 7 = (11 × 5) – (12 × 4)(j) 7 = (11 × 2) + p(k) 20 = 5y(l) 3q/ 2 < 5(m) z + 12 > 24(n) 20 – (10 – 5) = 3 × 5(o) 7 – x = 5

Solution:
(a) An equation with variable x
(b) Does not have an equal sign. Not an equation.
(c) No, it’s a numerical equation
(d) No, it’s a numerical equation
(e) An equation with variable x
(f) An equation with variable x
(g) Not an equation.
(h) An equation with variable n
(i) No, it’s a numerical equation
(j) An equation with variable p
(k) An equation with variable y
(l) Not an equation
(m)Not an equation
(n) No, it’s a numerical equation
(o) An equation with variable x

#### Question2:Complete the entries in the third column of the table.

Solution:
(a) 10y = 80
y = 10 is not a solution for this equation because if y = 10,
10y = 10 × 10
= 100 and not 80.
(b) 10y = 80
y = 8 is a solution for this equation because if y = 8,
10y = 10 × 8
= 80
∴ Equation satisfied
(c) 10y = 80
y = 5 is not a solution for this equation because if y = 5,
10y = 10 × 5
= 50 and not 80
(d) 4l = 20
l = 20 is not a solution for this equation because if l = 20,
4l = 4 × 20
= 80 and not 20

(e) 4l = 20
l = 80 is not a solution for this equation because if l = 80,
4l = 4 × 80
= 320 and 20
(f) 4l = 20
l = 5 is a solution for this eqaution because if l = 5,
4l = 4 × 5
= 20
∴ Equation satisfied
(g) b + 5 = 9
b = 5 is not a solution for this equation because if b = 5,
b + 5 = 5 + 5
= 10 and not 9
(h) b + 5 = 9
b = 9 is not a solution for this equation because if b = 9,
b + 5 = 9 + 5
= 14 and not 9
(i) b + 5 = 9
b = 4 is a solution for this equation because if b = 4,
b + 5 = 4 + 5
= 9
∴ Equation satisfied

(j) h – 8 = 5
h = 13 is a solution for this equation because if h = 13,
h – 8 = 13 – 8
= 5
∴ Equation satisfied
(k) h – 8 = 5
h = 8 is not a solution for this equation because if h = 8,
h – 8 = 8 – 8
= 0 and not 5
(l) h – 8 = 5
h = 0 is not a solution for this equation because if h = 0,
h – 8 = 0 – 8
= – 8 and not 5
(m)p + 3 = 1
p = 3 is not a solution for this equation because if p = 3,
p + 3 = 3 + 3
= 6 and not 1

(n) p + 3 = 1
p = 1 is not a solution for this equation because if p = 1,
p + 3 = 1 + 3
= 4 and not 1
(o) p + 3 = 1
p = 0 is not a solution for this equation because if p = 0,
p + 3 = 0 + 3
= 3 and not 1
(p) p + 3 = 1
p = -1 is not a solution for this equation because if p = – 1,
p + 3 = -1 + 3
= 2 and not 1
(q) p + 3 = 1
p = -2 is a solution for this equation because if p = -2,
p + 3 = -2 + 3
= 1
∴ Equation satisfied

#### Question3:Pick out the solution from the values given in the bracket next to eachequation. Show that the other values do not satisfy the equation.(a) 5m = 60 (10, 5, 12, 15)(b) n + 12 (12, 8, 20, 0)(c) p – 5 = 5 (0, 10, 5 – 5)(d) q / 2 = 7 (7, 2, 10, 14)(e) r – 4 = 0 (4, -4, 8, 0)(f) x + 4 = 2 (-2, 0, 2, 4)

Solution:
(a) 5m = 60
m = 12 is a solution for this equation because for m = 12,
5m = 5 × 12
= 60
∴ Equation satisfied
m = 10 is not a solution for this equation because for m = 10,
5m = 5 × 10
= 50 and not 60
m = 5 is not a solution for this equation because for m = 5,
5m = 5 × 5
= 25 and not 60
m = 15 is not a solution for this equation because for m = 15

5m = 5 × 15
= 75 and not 60
(b) n + 12 = 20
n = 8 is a solution for this equation because for n = 8,
n + 12 = 8 + 12
= 20
∴ Equation satisfied
n = 12 is not a solution for this equation because for n = 12,
n + 12 = 12 + 12
= 24 and not 20
n = 20 is not a solution for this equation because for n = 20,
n + 12 = 20 + 12
= 32 and not 20
n = 0 is not a solution for this equation because for n = 0,
n + 12 = 0 + 12
= 12 and not 20

(c) p – 5 = 5
p = 10 is a solution for this equation because for p = 10,
p – 5 = 10 – 5
= 5
∴ Equation satisfied
p = 0 is not a solution for this equation because for p = 0,
p – 5 = 0 – 5
= -5 and not 5
p = 5 is not a solution for this equation because for p = 5,
p – 5 = 5 – 5
= 0 and not 5
p = -5 is not a solution for this equation because for p = -5,
p – 5 = -5 – 5
= – 10 and not 5
(d) q / 2 = 7
q = 14 is a solution for this equation because for q = 14,
q / 2 = 14 / 2
= 7
∴ Equation satisfied
q = 7 is not a solution for this equation because for q = 7,
q / 2 = 7 / 2 and not 7
q = 2 is not a solution for this equation because for q = 2,
q / 2 = 2 / 2
= 1 and not 7
q = 10 is not a solution for this equation because for q = 10,
q / 2 = 10 / 2
= 5 and not 7

(e) r – 4 = 0
r = 4 is a solution for this equation because for r = 4,
r – 4 = 4 – 4
= 0
∴ Equation satisfied
r = -4 is not a solution for this equation because for r = – 4,
r – 4 = – 4 – 4
= -8 and not 0
r = 8 is not a solution for this equation because for r = 8,
r – 4 = 8 – 4
= 4 and not 0
r = 0 is not a solution for this equation because for r = 0,
r – 4 = 0 – 4
= – 4 and not 0

(f) x + 4 = 2
x = -2 is a solution for this equation because for x = -2,
x + 4 = – 2 + 4
= 2
∴ Equation satisfied
x = 0 is not solution for this equation because for x = 0,
x + 4 = 0 + 4
= 4 and not 2
x = 2 is not a solution for this equation because for x = 2,
x + 4 = 2 + 4
= 6 and not 2
x = 4 is not a solution for this equation because for x = 4,
x + 4 = 4 + 4
= 8 and not 2

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

Solution:
(a) For m + 10, the table is represented as below

Now, by inspection we may conclude that m = 6 is the solution of the above equation since,
for m = 6,
m + 10 = 6 + 10 = 16

(b) For 5t, the table is represented as below

Now, by inspection we may conclude that t = 7 is the solution of the above equation since,
for t = 7,
5t = 5 × 7 = 35

(c) For z / 3, the table is represented as below

Now, by inspection we may conclude that z = 12 is the solution of the above equation since
for z = 12,
z / 3 = 4

(d) For m – 7, the table is represented as below

Now, by inspection we may conclude that m = 10 is the solution of the above equation since,
for m = 10,
m – 7 = 10 – 7 = 3

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