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

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) Pattern of letter = 2n (as two matchstick used in each letter)

(b) Pattern of letter = 3n (as three matchstick used in each letter)

(c) Pattern of letter = 3n (as three matchstick used in each letter)

(d) Pattern of letter = 2n (as two matchstick used in each letter)

(e) Pattern of letter = 5n (as five matchstick used in each letter)

(f) Pattern of letter = 5n (as five matchstick used in each letter)

(g) Pattern of letter = 6n (as six matchstick used in each letter)

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?

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

The letter ‘T’ and ‘V’ that has pattern 2 , n since 2 matchsticks are used in all these letters. 2. Number of rows = n

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)

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Number of rows = n

Cadets in each row = 5

Therefore, total number of cadets = 5n

Q.4 The teacher distributes 5 pencils per students. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Number of students = s

Number of pencils to each student = 5

Therefore, total number of pencils needed are = 5s

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

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

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

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?

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

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

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.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Let Radha’s age be x years

Leela’s age = 4 years younger than Radha

= (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?

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Number of laddus mother gave = l

Remaining laddus = 5

Total number of laddus = number of laddus given away by mother + number of laddus remaining

= (l + 5) laddus

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?

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

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

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

(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.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(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.

Q.2 The side of the regular hexagon (Fig 11.10) is denoted by l. Express the perimeter of the hexagon using l.

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

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Side of a regular hexagon = l

Perimeter = l + l + l + l + l + 1 = 6l

Q.3 A cube is 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.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

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

Q.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 (Fig 11.2) AB is a diameter of a circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Diameter = AB

= AC + CB = r + r = 2r

Q.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 41and 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)

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

For any three whole numbers a, b and c

(a + b) + c = a + (b + c)

Q.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.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(i) 8 + (5 + 7)

(ii) 5 + (8 – 7)

(iii) 8 + (5 x 7)

(iv) 7 – (8 – 5)

(v) 7 x (8 + 5)

(vi) 5 x (8 + 7)

(vii) 8 x (5 + 7)

(viii) 7 + (8 – 5)

(ix) (5 x 7) – 8

(x) 7 + (8 x 5)

Q.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

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

(c) 2y + 17, 2y – 17

(d) 7m, -7m + 3, -7m – 3

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) z + 1 \(\rightarrow \) Addition, z ? 1 \(\rightarrow \) Subtraction

y + 17 \(\rightarrow \) Addition , y ? 17 \(\rightarrow \) Subtraction

(b) 17 y \(\rightarrow \) Multiplication

\(\frac{y}{17} \) \(\rightarrow \) Division

5z \(\rightarrow \) Multiplication

(c) 2y +17 \(\rightarrow \) Multiplication and Addition

2y ? 17 \(\rightarrow \) Multiplication and Subtraction

(d) 7m \(\rightarrow \) Multiplication

?7m + m \(\rightarrow \) Multiplication and Addition

?7m? 3 \(\rightarrow \) Multiplication and Subtraction

Q.4 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

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) p + 7

(b) p – 7

(c) 7p

(d) \(\frac{p}{7} \)

(e) -m – 7

(f) -5p

(g) \(\frac{-p}{5} \)

(h) 5p

Q.5 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) 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 to 16.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) 2m + 11

(b) 2m – 11

(e) 5y + 3

(d) 5y – 3

(e) -8y

(f) -8y+5

(g) 16 – 5y

(h) -5y + 16

Q.6 (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.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) (t + 4), (t – 4), 4t, \(\frac{t}{4} \) , \(\frac{4}{t} \), (4 – t), (4 + t)

(b) 2y + 7, 2y – 7, 7y + 2 , 7y - 2 , \(\frac{7y}{2} \) , \(\frac{2y}{7} \) , \(\frac{2}{7y} \) , \(\frac{7}{2y} \)

Q.1 Answer the following:

(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.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) Sarita’s age is given y years.

(i) After 5 years from now, her age will be (y + 5) years.

(ii) 3 years back from now, she was (y – 3) years of age.

(iii) Age of her grandfather = 6y years.

(iv) Age of her grandmother = (6y – 2) years.

(v) Sarita’s father’s age = (3y + 5) years.

(b) Let T be the length of the rectangular hall

\(\therefore \) length = (3b – 4) metre

Where b represents the breadth.

(c) Height of the rectangular box is ‘h’

\(\therefore \) Length = 5h cm

and Breadth = (5h – 10) cm.

(d) Meena is at step s.

\(\therefore \)Beena is at (s + 8) steps and Leena is at (s – 7) steps.

Total number of steps on to the hill top = (4s – 10)

(e) Distance travelled by Bus in 5 hours = 5v km.

\( Distance from Daspur to Beespur = (5v + 20) km

Q.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 notebook costs Rs p. A book costs Rs 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

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) A book costs 3 times the cost of a notebook.

(b) Tony has 8 times the number of marbles put on the table by him.

(c) The school has 20 times the number of students in a class.

(d) Jaggu’s uncle’s age is 4 times his age and his aunt’s age is 3 years less than the age of his uncle.

(e) Number of dots in a row is 5 times the number of rows.

Q.3 (a) Given Mannu’s age to be x years, Can you guess what (x – 2) may show?

(Hint: Think of Mannu’s younger brother) can you guess what (x + 4) may now? What (3x + 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 + \(4\frac{1}{2} \) , y – \(2\frac{1}{2} \) .

(c) Given n students in the class like football, what may 2n show? What may \(\frac{n}{2} \) show?

(Think of games other than football).

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) Given that Mannu’s age = x years.

\(\therefore \) (x -2) years may be the age of her younger brother or younger sister.

(x + 4) years show the age of her elder brother or elder sister.

(3x + 7) years may be the age of her father, mother or uncle.

(b) y represents the age of Sara in years.

\(\therefore \) y + 7 shows her future age.

y – 3 shows her past age.

y + \(4\frac{1}{2} \) show her future age i.e., the age after four and half years.

y – \(2\frac{1}{2} \) shows her past age i.e., the age before two and half years.

(c) Number of students who like football = n

\(\therefore \) 2n = twice the number of football players may like to play cricket.

and \(\frac{n}{2} \) = half of the number of football 2
players may like to play basket ball.

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

(a) 17 = x + 17

(b) (t – 7) > 5

(c) \(\frac{4}{2} \) = 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) \(\frac{2q}{2} \)< 5

(m) z + 12 > 24

(n) 20 – (10 – 5) = 3 × 5

(o) 7 – x = 5

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) An equation with variable x

(b) An inequality 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) An inequality 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) An inequality equation

(m) An inequality equation

(n) No, it’s a numerical equation

(o) An equation with variable x

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) 10y = 80

y = 10 is not a solution for this equation because if y = 10,

10y = 10 × 10

= 100 \(\neq \)80

(b) 10y = 80

y = 8 is a solution for this equation because if y = 8,

10y = 10 × 8

= 80

\(\therefore \)Equation satisfied

(c) 10y = 80

y = 5 is not a solution for this equation because if y = 5,

10y = 10 × 5

= 50 \(\neq \) 80

(d) 4l = 20

l = 20 is not a solution for this equation because if l = 20,

4l = 4 × 20

= 80 \(\neq \)20

(e) 4l = 20

l = 80 is not a solution for this equation because if l = 80,

4l = 4 × 80

= 320 \(\neq \) 20

(f) 4l = 20

l = 5 is a solution for this eqaution because if l = 5,

4l = 4 × 5

= 20

\(\therefore \) Equation satisfied

(g) b + 5 = 9

b = 5 is not a solution for this equation because if b = 5,

b + 5 = 5 + 5

= 10 \(\neq \) 9

(h) b + 5 = 9

b = 9 is not a solution for this equation because if b = 9,

b + 5 = 9 + 5

= 14 \(\neq \) 9

(i) b + 5 = 9

b = 4 is a solution for this equation because if b = 4,

b + 5 = 4 + 5

= 9

\(\therefore \) Equation satisfied

(j) h – 8 = 5

h = 13 is a solution for this equation because if h = 13,

h – 8 = 13 – 8

= 5

\(\therefore \) Equation satisfied

(k) h – 8 = 5

h = 8 is not a solution for this equation because if h = 8,

h – 8 = 8 – 8

= 0 \(\neq \) 5

(l) h – 8 = 5

h = 0 is not a solution for this equation because if h = 0,

h – 8 = 0 – 8

= – 8 \(\neq \) 5

(m) p + 3 = 1

p = 3 is not a solution for this equation because if p = 3,

p + 3 = 3 + 3

= 6 \(\neq \) 1

(n) p + 3 = 1

p = 1 is not a solution for this equation because if p = 1,

p + 3 = 1 + 3

= 4 \(\neq \) 1

(o) p + 3 = 1

p = 0 is not a solution for this equation because if p = 0,

p + 3 = 0 + 3

= 3 \(\neq \) 1

(p) p + 3 = 1

p = -1 is not a solution for this equation because if p = – 1,

p + 3 = -1 + 3

= 2 \(\neq \) 1

(q) p + 3 = 1

p = -2 is a solution for this equation because if p = -2,

p + 3 = -2 + 3

= 1

\(\therefore \) Equation satisfied

Q.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 = 60 (10, 5, 12, 15)

(b) n + 12 (12, 8, 20, 0)

(c) p – 5 = 5 (0, 10, 5 ,– 5)

(d) \(\frac{q}{2} \) = 7 (7, 2, 10, 14)

(e) r – 4 = 0 (4, -4, 8, 0)

(f) x + 4 = 2 (-2, 0, 2, 4)

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(a) 5m = 60 (10, 5 , 12, 15)
For m = 10, LHS = 5 x 10 = 50, RHS = 60
Here, LHS \(\neq \)RHS

\(\therefore \)m = 10 is not the solution of the equation

For m = 5, LHS = 5×5 = 25, RHS = 60

Here, LHS \(\neq \) RHS

\(\therefore \) m = 5 is not the solution of the equation

For m = 12, LHS = 5 x 12 = 60, RHS = 60

Here, LHS = RHS

\(\therefore \) m = 12 is the solution of the equation

For m = 15 LHS = 5 x 15 = 75, RHS = 60

Here, LHS \(\neq \) RHS

\(\therefore \) m = 15 is not the solution of the equation

(b) n + 12 = 20 (12, 8, 20, 0)

For n = 12, LHS = 12 + 12 = 24, RHS = 20

Here, LHS \(\neq \) RHS

\(\therefore \) n = 12 is not the solution of the equation

For n = 8, LHS = 8 + 12 = 20, RHS = 20

Here, LHS = RHS

\(\therefore \) n = 8 is the solution of the equation

For n = 20, LHS = 20 + 12 = 32, RHS = 20

Here, LHS \(\neq \) RHS

\(\therefore \) n = 20 is not the solution of the equation

For n = 0, LHS = 0 + 12 – 12, RHS = 20

Here, LHS \(\neq \) RHS

\(\therefore \) n= 0 is not the solution of the equation

(c) p – 5 = 5 (0, 10, 5, -5)

For p = 0, LHS = 0 – 5 = -5, RHS = 5

Here, LHS \(\neq \) RHS

\(\therefore \) p = 0 is not the solution of the equation

For p = 10, LHS = 10 – 5 = 5, RHS = 5

Here, LHS = RHS

\(\therefore \) p = 10 is the solution of the equation

For p = 5, LHS = 5-5-0, RHS = 5

Here LHS \(\neq \) RHS

\(\therefore \) p = 5 is not the solution of the equation

For p = 5, LHS = 5 – 5 = 0, RHS = 5

Here, LHS \(\neq \) RHS

\(\therefore \) p = -5 is not the solution of the equation

(d) q/2 = 7 (7, 2, 10, 14)

For q = 7, LHS = 7/2 , RHS = 7

Here LHS \(\neq \) RHS

\(\therefore \) q = 7 is not the solution of the equation

For q = 2, LHS = 2/2 = 1, RHS = 7

Here, LHS \(\neq \) RHS

\(\therefore \) q = 2 is not the solution of the equation

For q = 10, LHS = 10/2 = 5, RHS = 7

Here, LHS \(\neq \) RHS

For q = 14, LHS = 14/2 = 7, RHS = 7

Here, LHS = RHS

\(\therefore \) q = 14 is the solution of the equation

(e) r – 4 = 0 (4, -4, 8, 0)

For r = 4, LHS = 4 – 4 = 0, RHS = 0

Here, LHS = RHS

\(\therefore \) r = 4 is the solution of the equation

For r = -4, LHS = -4 – 4 = -8, RHS = 0

Here, LHS \(\neq \) RHS

\(\therefore \) r = -4 is not the solution of the equation

For r = 8, LHS = 8 – 4 = 4, RHS = 0

Here, LHS \(\neq \) RHS

For r = 8 is not the solution of the equation

For r = 0, LHS = 0 – 4 = – 4, RHS = 0

Here, LHS \(\neq \) RHS

\(\therefore \) r = 0 is not the solution of the equation

(f) x + 4 = 2 (-2, 0, 2, 4)

For x = -2, LHS = -2 + 4 = 2, RHS = 2

Here, LHS – RHS

\(\therefore \) x = -2 is the solution of the equation

For x = 0, LHS = 0 + 4 – 4, RHS = 2

Here, LHS \(\neq \) RHS

\(\therefore \) x = 0 is not the solution of the equation

For x = – 2, LHS = -2 + 4 – 6, RHS = 2

Here, LHS \(\neq \) RHS

\(\therefore \) x = 2 is not the solution of the equation

For r = 4, LHS = 4 + 4 = 8, RHS = 2

Here, LHS \(\neq \) RHS

\(\therefore \)x = 4 is not the solution of the equation

Q.4 (a)Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.

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

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

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

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

Q.5 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!

Add the count to me

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!

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

Answer :

(i) According to the condition,

I + 12 = 34 or x + 12 = 34

\(\therefore \) By inspection, we have

22 + 12 = 34

So, I am 22.

(ii) Let I am ‘x’.

We know that there are 7 days in a week.

\(\therefore \) upcounting from x for 7, the sum = 23

By inspections, we have

16 + 7 = 23

\(\therefore \) x = 16

Thus I am 16.

(iii) Let the special number be x and there are 11 players in cricket team.

\(\therefore \)Special Number -6 = 11

\(\therefore \) x – 6 = 11

By inspection, we get

17 – 6 = 11

\(\therefore \)x = 17

Thus I am 17.

(iv) Suppose I am ‘x’.

\(\therefore \) 22 – I = I

or 22 – x = x

By inspection, we have

22 – 11 = 11

\(\therefore \)x = 11

Thus I am 11.

There are total 29 questions present in ncert solutions for class 6 maths chapter 11 algebra

There are total 2 long question/answers in ncert solutions for class 6 maths chapter 11 algebra

There are total 5 exercise present in ncert solutions for class 6 maths chapter 11 algebra