NCERT Solutions for Class 6 Maths Chapter 11 Algebra

NCERT
SOLUTIONS
CLASS - 6TH

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Class : 6th
Subject : Maths
Chapter : 11
Chapter Name : Algebra

Exercise 11.1

Q1 Find the rule which gives the number of matchsticks required to make the following matchstick
patterns. Use a variable to write the rule.
(a) A pattern of letter T as T
(b) A pattern of letter Z as Z
(c) A pattern of letter U as U
(d) A pattern of letter V as V
(e) A pattern of letter E as E
(f) A pattern of letter S as S
(g) A pattern of letter A as A

Answer.

(e)

Page : 226 , Block Name : Exercise 11.1

Q2 We already know the rule for the pattern of letters L. C and F. Some of the letters from some Of
the letters out Of (a) T, (b) Z, (c) U, (d) V, (e) E, (f) S, (g) R
give us the same rule as that given by Which are these? Why does this happen?

Answer. It is known that L requires only two matchsticks. Therefore, the pattern for L is 2n.
Among all the letters given above in question 1, only T and V are the two letters which require two
matchsticks. Hence, (a) and (d)

Page : 227 , Block Name : Exercise 11.1

Q3 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.)

Answer. Let number of rows be n. Number of cadets in one row 5
Total number Of cadets = Number Of cadets in a row x Number Of rows = 5n

Page : 227 , Block Name : Exercise 11.1

Q4 If there are SO 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.)

Answer. Let the number of boxes be b.
Number of mangoes in a box = 50
Total number of mangoes = Number of mangoes in a box x Number of boxes
= 50/b

Page : 227 , Block Name : Exercise 11.1

Q5 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.)

Answer. Let the number of students be s.
Pencils given to each student = 5
Total number of pencils
= Number Of pencils given to each student x Number Of students = 5s

Page : 227 , Block Name : Exercise 11.1

Q6 A bird ies 1 kilometer in one minute. Can you express the distance covered by the bird in
terms of its ying time in minutes? (use t for ying time in minutes.)

Answer. Let the ying time be t minutes.
Distance covered in one minute = 1 km
Distance covered in t minutes = Distance covered in one minute x Flying time
1 x t = t km

Page : 227 , Block Name : Exercise 11.1

Q7 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?

Answer. Number Of dots in 1 row 9
Number of rows = r
Total number of dots in r rows = Number of rows x Number of dots in a row = 9r
Number of dots in 8 rows = 8 x 9 = 72
Number Of dots in 10 rows = 10 x 9 = 90

Page : 227 , Block Name : Exercise 11.1

Q8 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.

Answer. Let Radha's age be x years.
Leela's age = Radha's age - 4
= (x — 4) years

Page : 227 , Block Name : Exercise 11.1

Q9 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?

Answer. Number Of laddus given away = I
Number of laddus remaining = 5
Total number of laddus = Number of laddus given away + Number Of laddus remaining

=l+5

Page : 227 , Block Name : Exercise 11.1

Q10 Oranges are to be transferred from larger boxes into smaller boxes. When a large box is
emptied, the oranges from it ll 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?

Answer. Number of oranges in one small box = x
Number of oranges in two small boxes = 2x
Number Of oranges left = 10
Number Of oranges in the large box = Number Of oranges in two small boxes
+ Number Of oranges left = 2x + 10

Page : 227 , Block Name : Exercise 11.1

Q11 (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 nd 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.)

(b) The given gure gives a matchstick pattern of triangles. Find the general rule that
gives the number of matchsticks in terms of the number of triangles.

Answer. (a) It can be observed that in the given matchstick pattern, the number Of matchsticks are
4, 7, 10, and 13, which is 1 more than thrice of the number of squares in the pattern.
Hence, the pattern is 3n + 1, where n is the number of squares.

(b) It can be observed that in the given matchstick pattern, the number of matchsticks are 3. 5, 7,
and 9, which is 1 more than twice Of the number Of triangles in the pattern.
Hence, the pattern is 2n + i, where n is the number of triangles.

Page : 227 , Block Name : Exercise 11.1

Exercise 11.2

Q1 The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral
triangle using l.

Answer. Side of equilateral triangle = l
Perimeter = l + l + l = 3l

Page : 230 , Block Name : Exercise 11.2

Q2 The Side of a regular hexagon (see the given gure) is denoted by l. Express the perimeter of
the hexagon using l. (Hint: A regular hexagon has all its six sides equal in length.)

Answer. Side of regular hexagon = I
perimeter = 6l

Page : 230 , Block Name : Exercise 11.2

Q3 A cube is a three-dimensional gure as shown in the given gure. 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.

Answer. Length of edge = L
Number of edges = 12
Total length of the edges = Number of edges x Length of one edge = 12L

Page : 230 , Block Name : Exercise 11.2

Q4 The diameter Of a circle is a line which joins two points on the circle and also passed through
the centre of the circle. (In the adjoining gure AB is a diameter of the circle; C is its centre.)
Express the diameter of the circle (d) in terms of its radius(r).

Answer. Diameter = AB AC + CB = r+ r = 2r
d = 2r

Page : 231 , Block Name : Exercise 11.2

Q5 To nd sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may rst 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.

Answer. For any three whole numbers a, b, and c,
( a + b ) + c = a + ( b + c)

Page : 231 , Block Name : Exercise 11.2

Exercise 11.3

Q1 Make up as many expressions with numbers (no variables) as you can from three numbers S, 7
and 8. Every number should be used not more than once. Use only addition, subtraction and
multiplication.
(Hint: possible expressions are S + (8 — 7), S _ (8 _ 7), (5 x 8) + 7; make the other expressions.)

Answer. Many expressions can be formed by using the three numbers S, 7, and 8.
Some Of these are as follows.
5x(8-7)
5x(8+7)
(8 + 5) x 7
(8 - 5) x 7
(7+5)x8
(7-5)x8

Page : 233 , Block Name : Exercise 11.3

Q2 Which out Of the following are expressions with numbers only?
(a) y + 3
(b) (7 x 20) -8z
(c) 5 (21 - 7) + 7 x 2
(d) 5
(e) (7 x 20) - (5 x 10) - 45 + p

Answer. It can be observed that the expressions in alternatives (c) and (d) are formed by using
numbers only.

Page : 233 , Block Name : Exercise 11.3

Q3 Identify the operations (addition, subtraction, division, multiplication) in forming the
following expressions and tell how the expressions have been formed.

y
(a) z + 1, z − 1, y + 17, y − 17 (b) , 5z
17

(c) 2y + 17, 2y − 17 (d) 7m, −7m + 3, −7m − 3

Answer. (a) Addition as 1 is added to z.
Subtraction as 1 is subtracted from z.
Addition as 17 is added to y.
Subtraction as 17 is subtracted from y.

(b) Multiplication as y is multiplied with 17.
Division as y is divided by 17.
Multiplication as y is multiplied with 5.

(c) Multiplication and addition
y is multiplied with 2, and 17 is added to the result.
Multiplication and subtraction
y is multiplied with 2, and 17 is subtracted from the result.

(d) Multiplication as m is multiplied with 7.
Multiplication and addition as m is multiplied with - 7, and 3 is added to the result.
Multiplication and subtraction as m is multiplied by —7, and 3 is
subtracted from the result.

Page : 233 , Block Name : Exercise 11.3

Q4 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

Answer. (a) p + 7
(b) p - 7
(c) 7p
(d)
p

7

(e) - m - 7
(f) - 5p
−p
(g) 5

(h) - 5p

Page : 233 , Block Name : Exercise 11.3

Q5 Give expressions in the following cases.

(a) 11 added to 2m
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added
(d) S 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 to 16

Answer.(a) 2m + 11
(b) 2m - 11
(c) 5y + 3
(d) 5y - 3
(e) - 8y
(f) - 8y + 5
(g) 16 - 5y
(h) - 5y + 16

Page : 234 , Block Name : Exercise 11.3

Q6 (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.

t 4
(a) t + 4, t − 4, 4t, , , 4 − t, 4 + t
Answer. 4 t

(b) 2y + 7, 2y − 7, 7y + 2, …

Page : 234 , Block Name : Exercise 11.3

Exercise 11.4

Q1 Answer the following:
(a) Take Santa'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) Santa's grandfather 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 S 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 Mena? 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 S
hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using
v.

Answer. (a) (i) Sarita's age after S years from now — Sarita's present age + 5 = y + 5
(ii) 3 years ago, Sarita's age = Sarita’s present age — 3 = y-3
(iii) Grandfather's age 6 x Santa's present age = 6y
(iv) Grandmother's age Grandfather's present age — 2 = 6y — 2
(v) Fathers age 5 + 3 x Savita's present age 5 + 3y

(b) Length = 3 Breadth - 4
L = (3b — 4) metres

(c) Length = 5 x Height
l = 5h cm
Breadth = 5 x Height — 10
b = (5h — 10) cm

(d) Step at which Beena is = (Step at which Meena is) + 8
=s+8
Step at which leena is = (Step at which Meena is) — 7
=s-7
Total steps 4 x (Step at which Meena is) - 10 = 4s - 10

(e) Speed = v km/hr
Distance travelled in 5 hrs = 5 x v = 5v km
Total distance between Daspur and Beespur = (5v + 20) km

Page : 235 , Block Name : Exercise 11.4

Q2 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 Rs p. A book costs Rs 3 p.
(b) Tony puts 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 4 z 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.

Answer. (a) A book costs three times the cost of a notebook.
(b) Tony's box contains 8 times the 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 S times the number of rows.

Page : 236 , Block Name : Exercise 11.4

Q3 (a) Given Munnu’s age to be x years, can you guess what (x – 2) may show?
(Hint : Think of Munnu’s younger brother.)
Can you guess what (x + 4) may show? What (3 x + 7) may show?
(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?y + 7, y − 3, y + 4 , y − 2
1

2
1

2

(c) Given n students in the class like football, what may 2n show? What may n/2 show?
(Hint : Think of games other than football).

Answer. (a) (x — 2) represents that the person, whose age is (x — 2) years, is 2 years younger to
Munnu.
(x + 4) represents that the person, whose age is (x + 4) years. is 4 years elder to Munnu.
(3x + 7) represents that the person, whose age is (3x + 7) years, is elder to Munnu and his age is 7
years more than three times of the age of Munnu.

(b) In future
After n years from now, Sara's age will be (y + n) years.
In past
n years ago, Sara's age was (y — n) years.
(y + 7) represents that the person, whose age is (y + 7) years, is 7 years elder to Sara.
(y — 3) represents that the person, whose age is (y — 3) years, is 3 years younger to Sara.
( y + 4 ) represents that the person, whose age is ( y + 4 ) years, is 4 years elder to Sara.
1

2
1

2
1

2

( y - 2 ) represents that the person, whose age is ( y - 2 ) years, is 2
1

2
1

2
1

2
years elder to Sara.

(c) 2n may represent the number of students who like either football or some other game such as
cricket whereas n/2 represents the number Of students who like cricket, Out of the total number of
students who like football.

Page : 236 , Block Name : Exercise 11.4

Exercise 11.5

Q1 State which of the following are equations (with a variable). Give reason for your
answer. Identify the variable from equations with a variable.
(a) 17 = x + 7
(b) (t — 7) > 5
(c) 41

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
3q
(k) 20 = 5y (1) < 5
2

(m) (m)z + 12 > 24 (n)20 − (10 − 5) = 3 × 5

(o) 7-x = 5

Answer. (a) An equation with variable x

(b) An inequality
(c) No, it is a numerical equation.
(d) No, it is a numerical equation.
(e) An equation With variable x
(f) An equation with variable x
(g) An inequality
(h) An equation with variable n
(i) No, it is a numerical equation.
(j) An equation with variable p
(k) An equation With variable y
(l) An inequality
(m) An inequality
(n) No, it is a numerical equation.
(o) An equation with variable x

Page : 240 , Block Name : Exercise 11.5

Q2 Complete the entries in the third column of the table.

Answer.
(a) 10 y = 80
y = 10 is not a solution to the given equation because for y = 10,
10y = 10 x 100, and not 80

(b) 10y = 80
y = 8 is a solution to the given equation because for y = 8,
10y = 10 x 8 = 80 and hence, the equation is satis ed.

(c) 10y = 80
y = S is not a solution to the given equation because for y = 5,

10y = 10 x 5 =50, and not 80

(d) 4l = 20
l = 20 is not a solution to the given equation because for l =20,
4l = 4 x 20 = 80, and not 20

(e)4l = 20
I = 80 is not a solution to the given equation because for / = 80,
4l = 4 x 80 = 320, and not 20

(f) 4l = 20
l = 5 is a solution to the given equation because for I = 5
4l = 4 x 5 = 20 and hence, the equation is satis ed.

(g) b +5 = 9
b = 5 iS not a solution to the given equation because for b = 5,
b + 5 = 5 + 5 = 10, and not 9

(h) b + 5 = 9
b= 9 is not a solution to the given equation because for b = 9,
b + 5 = 5 + 5 = 10, and not 9

(i) b + 5 = 9
B = 4 is a solution to the given equation because for b = 4,
b + 5 = 4 + 5 = 9 and hence, the equation is satis ed.

(j) h - 8 = 5
h = 13 is a solution to the given equation because for h = 13,
H - 8 = 13 - 8 = 5 and hence, the equation is satis ed.

(k) h - 8 = 5
h = 8 is not a solution to the given equation because for h = 8,
H - 8 = 8 - 8 = 0, and not 5

(l) h - 8 =5
h = 0 is not a solution to the given equation because for h = 0,
H - 8 = 0 - 8 = -8, and not 5

(m) p + 3 = 1
p = 3 is not a solution to the given equation because for p = 3,
p + 3 = 3 + 3 = 6, and not 1

(n) p + 3 = 1
p = 1 is not a solution to the given equation because for p = 3,
p + 3 = 1 + 3 = 4, and not 1

(o) p + 3 = 1

p = 0 is not a solution to the given equation because for p = 0,
p + 3 = 0 + 3 = 4, and not 1

(p) p + 3 = 1
p = -1 is not a solution to the given equation because for p = -1,
p + 3 = -1 + 3 = 4, and not 1

(q) p + 3 = 1
p = -2 is a solution to the given equation because for p = -2,
p + 3 = -2 + 3 = 1 and hence, the equation is satis ed.

Page : 241 , Block Name : Exercise 11.5

Q3 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 = 60 (10, 5, 12, 15)
(b) n + 12 = 20 (12, 8, 20, 0)
(c) p - 5 = 5(0, 10, 5 — 5)
(d) = 7 (7, 2, 10, 14)
q

2

(e) r - 4 = 0 (4, -4 , 8, 0)
(f) x + 4 = 2 (-2, 0, 2, 4)

Answer. (a) 5m = 60
m = 12 is a solution to the given equation because for m = 12,
5m - 5 x 12 = 60 and hence, the equation is satis ed.
M = not a solution to the given equation because for m 10,
5m = 5 x 10 = 50, and not 60
m = 5 is not a solution to the given equation because for m = 5,
5m = 5 x 5 = 25, and not 60
m 15 is not a solution to the given equation because for m = 15
5m = 5 x 15 = 75, and not 60

(b) n + 12 = 20
n = 8 is a solution to the given equation because for n = 8,
n + 12 = 8 + 12 = 20 and hence, the equation is satis ed.
n = 12 is not a solution to the given equation because for n = 12,
n + 12 = 12 + 12 = 24, and not 20
n = 20 is not a solution to the given equation because for n = 20,
n + 12 = 20 + 12 = 32, and not 20
n = 0 is not a solution to the given equation because for n = 0,
n + 12 = 0 + 12 = 12, and not 20

(c) p - 5 = 5
p = 10 is a solution to the given equation because for p = 10,
p - 5 = 10 - 5 = 5 and hence, the equation is satis ed.
p = 0 is not a solution to the given equation because for p = 0,
p - 5 = 0 - 5 = -5, and not 5

p = 5 is not a solution to the given equation because for p = 5,
P - 5 = 5 - 5 = 0, and not 5
p = - 5 is not a solution to the given equation because for p = - 5,
p - 5 = - 5 - 5 = -10, and not 5

(d) q/2 = 7
q = 14 is a solution to the given equation because for q = 14,
q/2 = 14/2 = 7 and hence, the equation is satis ed.
q = 7 is not a solution to the given equation because for q = 7,
q/2 = 7/2, and not 7
q = 2 is not a solution to the given equation because for q = 2,
q/2 = 2/2 =1, and not 7
q = 10 is not a solution to the given equation because for q = 10,
q/2 = 10/2 = 5, and not 7

Page : 236 , Block Name : Exercise 11.5