Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises: Exploring Algebraic Identities

Algebraic identities are formulas that remain true for all values of the variables, and they help in expansion, factorisation, products and simplification.
Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises connect these identities with factorisation, rational expressions, products using suitable identities and word problems from Exploring Algebraic Identities.

By the end of Chapter 4, students are expected to use identities not only for expanding expressions but also for factorising, simplifying rational expressions and solving application-based algebra questions. The Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises bring together all these skills from Exploring Algebraic Identities.

These Class 9 Maths Chapter 4 End of Chapter Exercise Solutions cover products using suitable identities, cube identities, factorisation of algebraic expressions, simplification of rational expressions, rectangle and cuboid dimension problems, and proof-based questions. The solutions below follow the textbook order and use clear working steps for revision.

Key Takeaways

Identity-Based Products: Products and powers can be evaluated using suitable identities.
Factorisation: Algebraic expressions can be written as products of simpler factors.
Rational Simplification: Rational expressions are simplified by factorising and cancelling common factors.
Proofs and Applications: Identities help solve geometry, number and proof-based questions.

Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises Structure 2026

Exercise No.	Topic	Question Count
End of Chapter Exercises	Products using suitable identities	9
End of Chapter Exercises	Numerical cubes and products	8
End of Chapter Exercises	Factorisation of algebraic expressions	11
End of Chapter Exercises	Simplifying rational expressions	3
End of Chapter Exercises	Rectangle and cuboid dimensions	2
End of Chapter Exercises	Algebraic word problems	4
End of Chapter Exercises	Proof-based identity questions	3

Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises

The Ganita Manjari Class 9 Chapter 4 End of Chapter Exercises revise all major identities from the chapter. Students use algebraic identities formulas Class 9 such as:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(a + b)(a − b) = a² − b²

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a − b)³ = a³ − 3a²b + 3ab² − b³

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

These identities are used throughout Class 9 Ganita Manjari algebraic identities solutions for calculation, factorisation and simplification.

End of Chapter Exercises Question 1

Use suitable identities to find the following products:

(i) (−3x + 4)²

Solution:

Use:

(a − b)² = a² − 2ab + b²

Here:

a = 4
b = 3x

So,

(−3x + 4)² = (4 − 3x)²

= 4² − 2(4)(3x) + (3x)²

= 16 − 24x + 9x²

Answer: 9x² − 24x + 16

(ii) (2s + 7)(2s − 7)

Solution:

Use:

(a + b)(a − b) = a² − b²

Here:

a = 2s
b = 7

So,

(2s + 7)(2s − 7) = (2s)² − 7²

= 4s² − 49

Answer: 4s² − 49

(iii) (p² + 1/2)(p² − 1/2)

Solution:

Use:

(a + b)(a − b) = a² − b²

Here:

a = p²
b = 1/2

So,

(p² + 1/2)(p² − 1/2)

= (p²)² − (1/2)²

= p⁴ − 1/4

Answer: p⁴ − 1/4

(iv) (2n + 7)(2n − 7)

Solution:

(2n + 7)(2n − 7)

= (2n)² − 7²

= 4n² − 49

Answer: 4n² − 49

(v) (s − 2t)(s² + 2st + 4t²)

Solution:

Use:

a³ − b³ = (a − b)(a² + ab + b²)

Here:

a = s
b = 2t

So,

(s − 2t)(s² + 2st + 4t²)

= s³ − (2t)³

= s³ − 8t³

Answer: s³ − 8t³

(vi) (1/2r − 4r)²

Solution:

Use:

(a − b)² = a² − 2ab + b²

Here:

a = 1/(2r)
b = 4r

So,

(1/2r − 4r)²

= (1/(2r))² − 2(1/(2r))(4r) + (4r)²

= 1/(4r²) − 4 + 16r²

Answer: 16r² − 4 + 1/(4r²)

(vii) (−3m + 4k − l)²

Solution:

Write:

−3m + 4k − l = 4k − 3m − l

Use:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Take:

a = 4k
b = −3m
c = −l

So,

(−3m + 4k − l)²

= (4k)² + (−3m)² + (−l)² + 2(4k)(−3m) + 2(−3m)(−l) + 2(−l)(4k)

= 16k² + 9m² + l² − 24km + 6ml − 8kl

Answer: 16k² + 9m² + l² − 24km + 6ml − 8kl

(viii) (x − y/3)³

Solution:

Use:

(a − b)³ = a³ − 3a²b + 3ab² − b³

Here:

a = x
b = y/3

So,

(x − y/3)³

= x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³

= x³ − x²y + xy²/3 − y³/27

Answer: x³ − x²y + xy²/3 − y³/27

(ix) (7k/2 − 2m/3)³

Solution:

Use:

(a − b)³ = a³ − 3a²b + 3ab² − b³

Here:

a = 7k/2
b = 2m/3

So,

(7k/2 − 2m/3)³

= (7k/2)³ − 3(7k/2)²(2m/3) + 3(7k/2)(2m/3)² − (2m/3)³

= 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27

Answer: 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27

End of Chapter Exercises Question 2

Find the values using suitable identities:

(i) 17 × 21

Solution:

17 × 21 = (19 − 2)(19 + 2)

= 19² − 2²

= 361 − 4

= 357

Answer: 357

(ii) 104 × 96

Solution:

104 × 96 = (100 + 4)(100 − 4)

= 100² − 4²

= 10000 − 16

= 9984

Answer: 9984

(iii) 24 × 16

Solution:

24 × 16 = (20 + 4)(20 − 4)

= 20² − 4²

= 400 − 16

= 384

Answer: 384

(iv) 147³

Solution:

147 = 150 − 3

Use:

(a − b)³ = a³ − 3a²b + 3ab² − b³

147³ = (150 − 3)³

= 150³ − 3(150)²(3) + 3(150)(3)² − 3³

= 3375000 − 202500 + 4050 − 27

= 3176523

Answer: 3176523

(v) 199³

Solution:

199 = 200 − 1

199³ = (200 − 1)³

= 200³ − 3(200)²(1) + 3(200)(1)² − 1³

= 8000000 − 120000 + 600 − 1

= 7880599

Answer: 7880599

(vi) 127³

Solution:

127 = 120 + 7

127³ = (120 + 7)³

= 120³ + 3(120)²(7) + 3(120)(7)² + 7³

= 1728000 + 302400 + 17640 + 343

= 2048383

Answer: 2048383

(vii) (−107)³

Solution:

(−107)³ = −107³

107 = 100 + 7

107³ = (100 + 7)³

= 1000000 + 210000 + 14700 + 343

= 1225043

Therefore:

(−107)³ = −1225043

Answer: −1225043

(viii) (−299)³

Solution:

(−299)³ = −299³

299 = 300 − 1

299³ = (300 − 1)³

= 300³ − 3(300)²(1) + 3(300)(1)² − 1

= 27000000 − 270000 + 900 − 1

= 26730900 − 1

= 26729900? Wait, calculate carefully:

27000000 − 270000 = 26730000
26730000 + 900 = 26730900
26730900 − 1 = 26730899

So,

(−299)³ = −26730899

Answer: −26730899

End of Chapter Exercises Question 3

Factor the following algebraic expressions:

(i) 4y² + 1 + 1/(16y²)

Solution:

4y² = (2y)²

1/(16y²) = (1/(4y))²

1 = 2(2y)(1/(4y))

So,

4y² + 1 + 1/(16y²)

= (2y + 1/(4y))²

Answer: (2y + 1/(4y))²

(ii) 9m² − 1/(25n²)

Solution:

Use:

a² − b² = (a + b)(a − b)

Here:

a = 3m
b = 1/(5n)

So,

9m² − 1/(25n²)

= (3m + 1/(5n))(3m − 1/(5n))

Answer: (3m + 1/(5n))(3m − 1/(5n))

(iii) 27b³ − 1/(64b³)

Solution:

Use:

a³ − b³ = (a − b)(a² + ab + b²)

Here:

a = 3b
b = 1/(4b)

So,

27b³ − 1/(64b³)

= (3b − 1/(4b))[(3b)² + (3b)(1/(4b)) + (1/(4b))²]

= (3b − 1/(4b))(9b² + 3/4 + 1/(16b²))

Answer: (3b − 1/(4b))(9b² + 3/4 + 1/(16b²))

(iv) x² + 5x/6 + 1/6

Solution:

We need two numbers whose sum is 5/6 and product is 1/6.

The numbers are 1/2 and 1/3.

So,

x² + 5x/6 + 1/6

= (x + 1/2)(x + 1/3)

Answer: (x + 1/2)(x + 1/3)

(v) 27u³ − 1/125 − 27u²/5 + 9u/25

Solution:

Rearrange the terms:

27u³ − 27u²/5 + 9u/25 − 1/125

This matches:

(a − b)³ = a³ − 3a²b + 3ab² − b³

Here:

a = 3u
b = 1/5

So,

27u³ − 1/125 − 27u²/5 + 9u/25

= (3u − 1/5)³

Answer: (3u − 1/5)³

(vi) 64y³ + z³/125

Solution:

64y³ = (4y)³

z³/125 = (z/5)³

Use:

a³ + b³ = (a + b)(a² − ab + b²)

So,

64y³ + z³/125

= (4y + z/5)(16y² − 4yz/5 + z²/25)

Answer: (4y + z/5)(16y² − 4yz/5 + z²/25)

(vii) p³ + 27q³ + r³ − 9pqr

Solution:

Use:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Take:

a = p
b = 3q
c = r

So,

p³ + 27q³ + r³ − 9pqr

= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)

Answer: (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)

(viii) 9m² − 12m + 4

Solution:

9m² = (3m)²

4 = 2²

−12m = −2(3m)(2)

So,

9m² − 12m + 4 = (3m − 2)²

Answer: (3m − 2)²

(ix) 9x³ − 8y³/3 + z³/3 + 6xyz

Solution:

Take 1/3 as a common factor.

9x³ − 8y³/3 + z³/3 + 6xyz

= 1/3(27x³ − 8y³ + z³ + 18xyz)

Now use:

a³ + b³ + c³ − 3abc

Take:

a = 3x
b = −2y
c = z

Then:

−3abc = −3(3x)(−2y)(z) = 18xyz

So,

27x³ − 8y³ + z³ + 18xyz

= (3x − 2y + z)(9x² + 4y² + z² + 6xy − 2yz − 3xz)

Therefore:

9x³ − 8y³/3 + z³/3 + 6xyz

= 1/3(3x − 2y + z)(9x² + 4y² + z² + 6xy − 2yz − 3xz)

Answer: 1/3(3x − 2y + z)(9x² + 4y² + z² + 6xy − 2yz − 3xz)

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy

Solution:

The expression does not match a direct perfect-square identity because the cross-terms do not fit the pattern of (2x + 3y + 6z)² or any simple square of three terms.

Answer: This expression is not factorised using the standard identities covered so far.

(xi) 27u³ − 1/216 − 9u²/2 + u/4

Solution:

Rearrange the terms:

27u³ − 9u²/2 + u/4 − 1/216

This matches:

(a − b)³ = a³ − 3a²b + 3ab² − b³

Here:

a = 3u
b = 1/6

So,

27u³ − 1/216 − 9u²/2 + u/4

= (3u − 1/6)³

Answer: (3u − 1/6)³

End of Chapter Exercises Question 4

Simplify the following:

(i) (4x² + 4x + 1) / (4x² − 1)

Solution:

Factor the numerator:

4x² + 4x + 1 = (2x + 1)²

Factor the denominator:

4x² − 1 = (2x + 1)(2x − 1)

So,

(4x² + 4x + 1) / (4x² − 1)

= (2x + 1)² / [(2x + 1)(2x − 1)]

= (2x + 1)/(2x − 1)

Answer: (2x + 1)/(2x − 1)

(ii) 9(3a³ − 24b³) / (9a² − 36b²)

Solution:

Simplify the numerator:

9(3a³ − 24b³) = 27a³ − 216b³

= 27(a³ − 8b³)

= 27(a − 2b)(a² + 2ab + 4b²)

Now factor the denominator:

9a² − 36b² = 9(a² − 4b²)

= 9(a − 2b)(a + 2b)

So,

9(3a³ − 24b³) / (9a² − 36b²)

= 27(a − 2b)(a² + 2ab + 4b²) / [9(a − 2b)(a + 2b)]

= 3(a² + 2ab + 4b²)/(a + 2b)

Answer: 3(a² + 2ab + 4b²)/(a + 2b)

(iii) (s³ + 125t³) / (s² − 2st − 35t²)

Solution:

Factor the numerator:

s³ + 125t³ = s³ + (5t)³

= (s + 5t)(s² − 5st + 25t²)

Factor the denominator:

s² − 2st − 35t²

= (s − 7t)(s + 5t)

So,

(s³ + 125t³) / (s² − 2st − 35t²)

= (s + 5t)(s² − 5st + 25t²) / [(s − 7t)(s + 5t)]

= (s² − 5st + 25t²)/(s − 7t)

Answer: (s² − 5st + 25t²)/(s − 7t)

End of Chapter Exercises Question 5

Find possible expressions for the length and breadth of each rectangle whose area is given.

(i) 25a² − 30ab + 9b²

Solution:

25a² − 30ab + 9b²

= (5a)² − 2(5a)(3b) + (3b)²

= (5a − 3b)²

So, possible length and breadth are:

Length = 5a − 3b
Breadth = 5a − 3b

Answer: 5a − 3b and 5a − 3b

(ii) 36s² − 49t²

Solution:

36s² − 49t²

= (6s)² − (7t)²

= (6s − 7t)(6s + 7t)

So, possible length and breadth are:

Length = 6s + 7t
Breadth = 6s − 7t

Answer: 6s + 7t and 6s − 7t

End of Chapter Exercises Question 6

Find possible expressions for the length, breadth and height of each cuboid whose volume is given.

(i) 6a² − 24b²

Solution:

6a² − 24b²

= 6(a² − 4b²)

= 6(a − 2b)(a + 2b)

So, possible dimensions are:

Length = 6
Breadth = a − 2b
Height = a + 2b

Answer: 6, a − 2b and a + 2b

(ii) 3ps² − 15ps + 12p

Solution:

Take 3p as common factor.

3ps² − 15ps + 12p

= 3p(s² − 5s + 4)

Now factor:

s² − 5s + 4 = (s − 1)(s − 4)

So,

3ps² − 15ps + 12p = 3p(s − 1)(s − 4)

Possible dimensions are:

Length = 3p
Breadth = s − 1
Height = s − 4

Answer: 3p, s − 1 and s − 4

End of Chapter Exercises Question 7

The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.

Solution:

Side of playground = 40 m

Path width around the playground = s m

So, side of outer square:

40 + 2s

Area of outer square:

(40 + 2s)²

Area of playground:

40²

Area of path:

(40 + 2s)² − 40²

Use:

a² − b² = (a + b)(a − b)

Area of path = (40 + 2s)² − 40²

= [(40 + 2s) − 40][(40 + 2s) + 40]

= 2s(80 + 2s)

= 4s(40 + s)

Answer: 4s(40 + s) square metres

End of Chapter Exercises Question 8

If a number plus its reciprocal equals 10/3, find the number.

Solution:

Let the number be x.

Then its reciprocal is 1/x.

Given:

x + 1/x = 10/3

Multiply by 3x:

3x² + 3 = 10x

So,

3x² − 10x + 3 = 0

Factor:

3x² − 10x + 3 = 3x² − 9x − x + 3

= 3x(x − 3) − 1(x − 3)

= (3x − 1)(x − 3)

So,

(3x − 1)(x − 3) = 0

Therefore:

x = 1/3 or x = 3

Answer: The number is 3 or 1/3.

End of Chapter Exercises Question 9

A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length.

Solution:

Area = 2x² + 7x + 3

Width = 2x + 1

Length = Area / Width

Now factor the area:

2x² + 7x + 3 = 2x² + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= (2x + 1)(x + 3)

So,

Length = [(2x + 1)(x + 3)] / (2x + 1)

= x + 3

Answer: The length is x + 3 hastas.

End of Chapter Exercises Question 10

If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.

Solution:

Since x − 2 and x − 1/2 are factors, the roots are:

x = 2 and x = 1/2

So, the quadratic can be written as:

px² + 5x + r = p(x − 2)(x − 1/2)

Now expand:

p(x − 2)(x − 1/2)

= p[x² − (5/2)x + 1]

= px² − (5p/2)x + p

Compare with:

px² + 5x + r

So,

−5p/2 = 5

p = −2

Also,

r = p

Therefore:

p = r

Answer: p = r

End of Chapter Exercises Question 11

If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25.

Solution:

Use the identity:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

First find:

a² + b² + c²

Using:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Given:

a + b + c = 5

ab + bc + ca = 10

So,

25 = a² + b² + c² + 2(10)

25 = a² + b² + c² + 20

a² + b² + c² = 5

Now:

a² + b² + c² − ab − bc − ca = 5 − 10 = −5

Therefore:

a³ + b³ + c³ − 3abc

= (a + b + c)(a² + b² + c² − ab − bc − ca)

= 5(−5)

= −25

Hence proved.

End of Chapter Exercises Question 12

By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.

Solution:

Factor:

n³ − n = n(n² − 1)

= n(n − 1)(n + 1)

So,

n³ − n = (n − 1)n(n + 1)

This is the product of three consecutive natural numbers.

Among any three consecutive natural numbers, one number is divisible by 3.

Also, at least one number is even, so the product is divisible by 2.

Since it is divisible by both 2 and 3, it is divisible by 6.

Answer: n³ − n is always divisible by 6 for all natural numbers n.

End of Chapter Exercises Question 13

Find the value of:

(i) x³ + y³ − 12xy + 64, when x + y = −4

Solution:

Use:

x³ + y³ = (x + y)³ − 3xy(x + y)

Given:

x + y = −4

So,

x³ + y³ = (−4)³ − 3xy(−4)

= −64 + 12xy

Now substitute in the expression:

x³ + y³ − 12xy + 64

= (−64 + 12xy) − 12xy + 64

= 0

Answer: 0

(ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6

Solution:

Given:

x = 2y + 6

So,

x − 2y = 6

Let:

a = x
b = 2y

Then:

a − b = 6

The expression is:

x³ − 8y³ − 36xy − 216

= x³ − (2y)³ − 36xy − 216

Now:

36xy = 3(x)(2y)(6)

Since x − 2y = 6,

36xy = 3ab(a − b)

So,

x³ − (2y)³ − 36xy

= a³ − b³ − 3ab(a − b)

= (a − b)³

= 6³

= 216

Therefore:

x³ − 8y³ − 36xy − 216

= 216 − 216

= 0

Answer: 0

Final Answers for Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises

Question	Final Answer
1(i)	9x² − 24x + 16
1(ii)	4s² − 49
1(iii)	p⁴ − 1/4
1(iv)	4n² − 49
1(v)	s³ − 8t³
1(vi)	16r² − 4 + 1/(4r²)
1(vii)	16k² + 9m² + l² − 24km + 6ml − 8kl
1(viii)	x³ − x²y + xy²/3 − y³/27
1(ix)	343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27
2(i)	357
2(ii)	9984
2(iii)	384
2(iv)	3176523
2(v)	7880599
2(vi)	2048383
2(vii)	−1225043
2(viii)	−26730899
3(i)	(2y + 1/(4y))²
3(ii)	(3m + 1/(5n))(3m − 1/(5n))
3(iii)	(3b − 1/(4b))(9b² + 3/4 + 1/(16b²))
3(iv)	(x + 1/2)(x + 1/3)
3(v)	(3u − 1/5)³
3(vi)	(4y + z/5)(16y² − 4yz/5 + z²/25)
3(vii)	(p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)
3(viii)	(3m − 2)²
3(ix)	1/3(3x − 2y + z)(9x² + 4y² + z² + 6xy − 2yz − 3xz)
3(x)	Not factorised using the standard identities covered so far
3(xi)	(3u − 1/6)³
4(i)	(2x + 1)/(2x − 1)
4(ii)	3(a² + 2ab + 4b²)/(a + 2b)
4(iii)	(s² − 5st + 25t²)/(s − 7t)
5(i)	5a − 3b and 5a − 3b
5(ii)	6s + 7t and 6s − 7t
6(i)	6, a − 2b, a + 2b
6(ii)	3p, s − 1, s − 4
7	4s(40 + s) square metres
8	3 or 1/3
9	x + 3 hastas
10	p = r
11	a³ + b³ + c³ − 3abc = −25
12	n³ − n is divisible by 6
13(i)	0
13(ii)	0

Concept Used in Exploring Algebraic Identities End of Chapter Exercises

Exploring Algebraic Identities End of Chapter Exercises combine all the identities studied in Chapter 4. Students use identities to expand products, factor expressions, simplify rational expressions and solve proof-based questions.

The main concepts are:

• algebraic identities Class 9,
• products using suitable identities Class 9,
• factorisation using identities Class 9,
• factorisation of algebraic expressions Class 9,
• simplifying rational expressions Class 9,
• square and cube identities,
• difference of squares,
• sum and difference of cubes,
• rational expression cancellation,
• application of identities in geometry and word problems.

These concepts are central to Class 9 Maths Ganita Manjari Chapter 4 Solutions because they prepare students for higher algebra, polynomial factorisation and simplification.

About Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises

Ganita Manjari Class 9 Chapter 4 End of Chapter Exercises appear after students study visual identities, algebra tiles, factorisation without algebra tiles, new identities and rational algebraic expressions.

These Class 9 Maths algebraic identities answers help students revise the whole chapter in one place. The exercises are useful for understanding how one identity can be used in different ways: expansion, factorisation, quick calculation, simplification and proof.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4

Section	NCERT Solutions
Class 9 Maths Ganita Manjari 2026	NCERT Class 9 Maths Ganita Manjari 2026
Chapter 4	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4
Exercise 4.1	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.1
Exercise 4.2	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.2
Exercise 4.3	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.3
Exercise 4.4	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.4
Exercise 4.5	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exercise 4.5
End of Chapter Exercises	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 End of Chapter Exercises