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## 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 the**

**following 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 the**

**letters 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 the**

**rule which gives the number of cadets, given the number of rows? (Use n**

**for 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 of**

**mangoes 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 many**

**pencils are needed, given the number of students? (Use s for the number of**

**students.**)

**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 distance**

**covered by the birds in terms of its flying time in minutes? (Use t for flying**

**time 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 her**

**Rangoli have for r rows? How many dots are there if there are 8 rows? If**

**there 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

**Question8:**

**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.**

**Solution**:

Let Radha’s age be x years

Leela’s age = 4 years younger than Radha

= (x – 4) years

**Question9:**

**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?**

**Solution**:

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

**Question10:**

**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?**

**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

**Question11:**

**(a) Look at the following matchstick pattern of squares (Fig 11.6). The**

**squares are not separate. Two neighboring squares have a common**

**matchstick. Observe the patterns and find the rule that gives the**

**number of matchsticks (a) a square (b) an equilateral triangle (c) a**

**regular hexagon?**

**(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.**

**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 of**

**the equilateral triangle using l.**

**Solution**:

Side of equilateral triangle = l

Perimeter = l + l + l = 3l

**Question2:**

**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.)**

**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 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.**

**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 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).**

**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 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)**

**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) in**

**forming the following expressions and tell how the expressions have been**

**formed.**

**(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 added to y = Addition

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)

**Question5:**

**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.**

**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 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.**

**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.**

**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.**

**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 in**

**ordinary language. (For example, Given Salim scores r runs in a cricket**

**match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15**

**runs more than Salim.)**

**(a) A notebook costs ₹ p. A book costs ₹ 3p**

**(b) Tony 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.

**Question3:**

**(a) Given Munnu’s age to be x years, can you guess what (x – 2) may show?**

**Can you guess what (x + 4) may show? What (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**

**,**

**y – 2**

**.**

**(c) Given n students in the class like football, what may 2n shows? What**

**may 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 reason**

**for 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 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) 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

**Question4:**

**(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.**

**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