NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.5) Exercise 9.5

VSAT 2022

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.5) Exercise 9.5

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.5 (EX 9.5) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Vedantu is a platform that provides free NCERT Solution and other study materials for students. Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

Important Concepts Discussed in Exercise 9.5 of Class 8 Maths Chapter 9 

The NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.5 contain eight questions with multiple parts based on four standard algebraic identities. The important concepts that students will learn in this last exercise of Chapter 9 Algebraic Expressions and Identities of Class 8 Maths are:

  • What is an Identity?

  • Standard Identities

  • Applying Identities

Identities are algebraic equations that hold true for all values of the variable. In Exercise 9.5 of Algebraic Expressions and Identities, students are expected to solve the given polynomial expressions using the standard identities. The exercise begins with basic questions where identities are easily recognizable, and levels up to where students are required to rearrange the question to in the form wherein the identity can be applied.

Here are the four standard Algebraic Expressions and Identities you music remember before solving the questions in Exercise 9.5 of Class 8 Maths Chapter 9.

  • (a + b)2 = a2 + 2ab + b2

  • (a – b)2 = a2 – 2ab + b2

  • (a + b) (a – b) = a2 – b2

  • (x + a) (x + b) = x2 + (a + b) x + ab

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Download PDF of NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.5) Exercise 9.5

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Exercise 9.5

1. Use a suitable identity to get each of the following products.

i. $\left( {x + 3} \right)\left( {x + 3} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {x + 3} \right)\left( {x + 3} \right) = {\left( {x + 3} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {x + 3} \right)^2} = {\left( x \right)^2} + 2\left( x \right)\left( 3 \right) + {\left( 3 \right)^2}$

${\left( {x + 3} \right)^2} = {x^2} + 6x + 9$

ii. $\left( {2y + 5} \right)\left( {2y + 5} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {2y + 5} \right)\left( {2y + 5} \right) = {\left( {2y + 5} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {2y + 5} \right)^2} = {\left( {2y} \right)^2} + 2\left( {2y} \right)\left( 5 \right) + {\left( 5 \right)^2}$

${\left( {2y + 5} \right)^2} = 4{y^2} + 20y + 25$

iii. $\left( {2a - 7} \right)\left( {2a - 7} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {2a - 7} \right)\left( {2a - 7} \right) = {\left( {2a - 7} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {2a - 7} \right)^2} = {\left( {2a} \right)^2} - 2\left( {2a} \right)\left( 7 \right) + {\left( 7 \right)^2}$

${\left( {2a - 7} \right)^2} = 4{a^2} - 28a + 49$

iv. $\left( {3a - \dfrac{1}{2}} \right)\left( {3a - \dfrac{1}{2}} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {3a - \dfrac{1}{2}} \right)\left( {3a - \dfrac{1}{2}} \right) = {\left( {3a - \dfrac{1}{2}} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {3a - \dfrac{1}{2}} \right)^2} = {\left( {3a} \right)^2} - 2\left( {3a} \right)\left( {\dfrac{1}{2}} \right) + {\left( {\dfrac{1}{2}} \right)^2}$

${\left( {3a - \dfrac{1}{2}} \right)^2} = 9{a^2} - 3a + \dfrac{1}{4}$

v. $\left( {1.1m - 0.4} \right)\left( {1.1m + 0.4} \right)$

Ans: On writing the given expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get,$\left( {1.1m - 0.4} \right)\left( {1.1m + 0.4} \right) = {\left( {1.1m} \right)^2} - {\left( {0.4} \right)^2}$

On simplifying the above equation further, we get,

$\left( {1.1m - 0.4} \right)\left( {1.1m + 0.4} \right) = 1.21{m^2} - 0.16$

vi. $\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right)$

Ans: Using the commutative rule and writing the given expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get,

$\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right) = \left( {{b^2} + {a^2}} \right)\left( {{b^2} - {a^2}} \right)$

$\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right) = {\left( {{b^2}} \right)^2} - {\left( {{a^2}} \right)^2}$

On simplifying the above equation further, we get,

$\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right) = {b^4} - {a^4}\,$

vii. $\left( {6x - 7} \right)\left( {6x + 7} \right)$

Ans: On writing the given expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get,

$\left( {6x - 7} \right)\left( {6x + 7} \right) = {\left( {6x} \right)^2} - {\left( 7 \right)^2}$

On simplifying the above equation further, we get,

$\left( {6x - 7} \right)\left( {6x + 7} \right) = 36{x^2} - 49$

viii. $\left( { - a + c} \right)\left( { - a + c} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( { - a + c} \right)\left( { - a + c} \right) = {\left( { - a + c} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

$\left( { - a + c} \right)\left( { - a + c} \right) = {\left( { - a} \right)^2} + 2\left( { - a} \right)\left( c \right) + {\left( c \right)^2}$

$\left( { - a + c} \right)\left( { - a + c} \right) = {a^2} - 2ac + {c^2}$

ix. $\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = {\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

$\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = {\left( {\dfrac{x}{2}} \right)^2} + 2\left( {\dfrac{x}{2}} \right)\left( {\dfrac{{3y}}{4}} \right) + {\left( {\dfrac{{3y}}{4}} \right)^2}$

$\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = \dfrac{{{x^2}}}{4} + \dfrac{{3xy}}{4} + \dfrac{{9{y^2}}}{{16}}$

x. $\left( {7a - 9b} \right)\left( {7a - 9b} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {7a - 9b} \right)\left( {7a - 9b} \right) = {\left( {7a - 9b} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

$\left( {7a - 9b} \right)\left( {7a - 9b} \right) = {\left( {7a} \right)^2} - 2\left( {7a} \right)\left( {9b} \right) + {\left( {9b} \right)^2}$

$\left( {7a - 9b} \right)\left( {7a - 9b} \right) = 49{a^2} - 126ab + 81{b^2}$


2. Use the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$to find the following products. 

(i). $\left( {x + 3} \right)\left( {x + 7} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {x + 3} \right)\left( {x + 7} \right) = {x^2} + \left( {3 + 7} \right)x + \left( 3 \right)\left( 7 \right)$

On simplifying the above equation further, we get,

$\left( {x + 3} \right)\left( {x + 7} \right) = {x^2} + 10x + 21$

(ii). $\left( {4x + 5} \right)\left( {4x + 1} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {4x + 5} \right)\left( {4x + 1} \right) = {\left( {4x} \right)^2} + \left( {5 + 1} \right)\left( {4x} \right) + \left( 5 \right)\left( 1 \right)$

On simplifying the above equation further, we get,

$\left( {4x + 5} \right)\left( {4x + 1} \right) = 16{x^2} + 6\left( {4x} \right) + 5$

$\left( {4x + 5} \right)\left( {4x + 1} \right) = 16{x^2} + 24x + 5$

(iii). $\left( {4x - 5} \right)\left( {4x - 1} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {4x - 5} \right)\left( {4x - 1} \right) = {\left( {4x} \right)^2} + \left[ {\left( { - 5} \right) + \left( { - 1} \right)} \right]\left( {4x} \right) + \left( { - 5} \right)\left( { - 1} \right)$

On simplifying the above equation further, we get,

\[\left( {4x - 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( { - 5 - 1} \right)\left( {4x} \right) + 5\]

\[\left( {4x - 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( { - 6} \right)\left( {4x} \right) + 5\]

\[\left( {4x - 5} \right)\left( {4x - 1} \right) = 16{x^2} - 24x + 5\]

(iv). $\left( {4x + 5} \right)\left( {4x - 1} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {4x + 5} \right)\left( {4x - 1} \right) = {\left( {4x} \right)^2} + \left[ {\left( 5 \right) + \left( { - 1} \right)} \right]\left( {4x} \right) + \left( 5 \right)\left( { - 1} \right)$

On simplifying the above equation further, we get,

\[\left( {4x + 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( {5 - 1} \right)\left( {4x} \right) - 5\]

\[\left( {4x + 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( 4 \right)\left( {4x} \right) - 5\]

\[\left( {4x + 5} \right)\left( {4x - 1} \right) = 16{x^2} + 16x - 5\]

(v). $\left( {2x + 5y} \right)\left( {2x + 3y} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {2x + 5y} \right)\left( {2x + 3y} \right) = {\left( {2x} \right)^2} + \left( {5y + 3y} \right)\left( {2x} \right) + \left( {5y} \right)\left( {3y} \right)$

On simplifying the above equation further, we get,

$\left( {2x + 5y} \right)\left( {2x + 3y} \right) = 4{x^2} + \left( {8y} \right)\left( {2x} \right) + 15{y^2}$

$\left( {2x + 5y} \right)\left( {2x + 3y} \right) = 4{x^2} + 16xy + 15{y^2}$

(vi). $\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right) = {\left( {2{a^2}} \right)^2} + \left( {9 + 5} \right)\left( {2{a^2}} \right) + \left( 9 \right)\left( 5 \right)$

On simplifying the above equation further, we get,

$\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right) = 4{a^4} + \left( {14} \right)\left( {2{a^2}} \right) + 45$

$\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right) = 4{a^4} + 28{a^2} + 45$

(vii). $\left( {xyz - 4} \right)\left( {xyz - 2} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {xyz - 4} \right)\left( {xyz - 2} \right) = {\left( {xyz} \right)^2} + \left[ {\left( { - 4} \right) + \left( { - 2} \right)} \right]\left( {xyz} \right) + \left( { - 4} \right)\left( { - 2} \right)$

On simplifying the above equation further, we get,

\[\left( {xyz - 4} \right)\left( {xyz - 2} \right) = {x^2}{y^2}{z^2} + \left( { - 4 - 2} \right)\left( {xyz} \right) + 8\]

\[\left( {xyz - 4} \right)\left( {xyz - 2} \right) = {x^2}{y^2}{z^2} - 6xyz + 8\]


3. Find the following squares by using the identities.

(i). ${\left( {b - 7} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {b - 7} \right)^2} = {\left( b \right)^2} - 2\left( b \right)\left( 7 \right) + {\left( 7 \right)^2}$

${\left( {b - 7} \right)^2} = {b^2} - 14b + 49$

(ii). ${\left( {xy + 3z} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {xy + 3z} \right)^2} = {\left( {xy} \right)^2} + 2\left( {xy} \right)\left( {3z} \right) + {\left( {3z} \right)^2}$

${\left( {xy + 3z} \right)^2} = {x^2}{y^2} + 6xyz + 9{z^2}$

(iii). ${\left( {6{x^2} - 5y} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {6{x^2} - 5y} \right)^2} = {\left( {6{x^2}} \right)^2} - 2\left( {6{x^2}} \right)\left( {5y} \right) + {\left( {5y} \right)^2}$

${\left( {6{x^2} - 5y} \right)^2} = 36{x^4} - 2\left( {30{x^2}y} \right) + 25{y^2}$

${\left( {6{x^2} - 5y} \right)^2} = 36{x^4} - 60{x^2}y + 25{y^2}$

(iv). ${\left( {\dfrac{2}{3}m + \dfrac{3}{2}n} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {\dfrac{2}{3}m + \dfrac{3}{2}n} \right)^2} = {\left( {\dfrac{2}{3}m} \right)^2} + 2\left( {\dfrac{2}{3}m} \right)\left( {\dfrac{3}{2}n} \right) + {\left( {\dfrac{3}{2}n} \right)^2}$

${\left( {\dfrac{2}{3}m + \dfrac{3}{2}n} \right)^2} = \dfrac{4}{9}{m^2} + 2mn + \dfrac{9}{4}{n^2}$

(v). ${\left( {0.4p - 0.5q} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {0.4p - 0.5q} \right)^2} = {\left( {0.4p} \right)^2} - 2\left( {0.4p} \right)\left( {0.5q} \right) + {\left( {0.5q} \right)^2}$

${\left( {0.4p - 0.5q} \right)^2} = 0.16{p^2} - 2\left( {0.2pq} \right) + 0.25{q^2}$

${\left( {0.4p - 0.5q} \right)^2} = 0.16{p^2} - 0.4pq + 0.25{q^2}$

(vi). ${\left( {2xy + 5y} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {2xy + 5y} \right)^2} = {\left( {2xy} \right)^2} + 2\left( {2xy} \right)\left( {5y} \right) + {\left( {5y} \right)^2}$

${\left( {2xy + 5y} \right)^2} = 4{x^2}{y^2} + 2\left( {10x{y^2}} \right) + 25{y^2}$

${\left( {2xy + 5y} \right)^2} = 4{x^2}{y^2} + 20x{y^2} + 25{y^2}$


4. Simplify the following expressions. 

(i). ${\left( {{a^2} - {b^2}} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {{a^2} - {b^2}} \right)^2} = {\left( {{a^2}} \right)^2} - 2\left( {{a^2}} \right)\left( {{b^2}} \right) + {\left( {{b^2}} \right)^2}$

${\left( {{a^2} - {b^2}} \right)^2} = {a^4} - 2{a^2}{b^2} + {b^4}$

(ii). ${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2}$

Ans: On expanding the term ${\left( {2x + 5} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and expanding the term ${\left( {2x - 5} \right)^2}$ using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = {\left( {2x} \right)^2} + 2\left( {2x} \right)\left( 5 \right) + {\left( 5 \right)^2} - \left[ {{{\left( {2x} \right)}^2} - 2\left( {2x} \right)\left( 5 \right) + {{\left( 5 \right)}^2}} \right]$

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = 4{x^2} + 20x + 25 - \left[ {4{x^2} - 20x + 25} \right]$

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = 4{x^2} + 20x + 25 - 4{x^2} + 20x - 25$

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = 40x$

(iii). ${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2}$

Ans: On expanding the term ${\left( {7m - 8n} \right)^2}$ using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and expanding the term ${\left( {7m + 8n} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = {\left( {7m} \right)^2} - 2\left( {7m} \right)\left( {8n} \right) + {\left( {8n} \right)^2} + \left[ {{{\left( {7m} \right)}^2} + 2\left( {7m} \right)\left( {8n} \right) + {{\left( {8n} \right)}^2}} \right]$

${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = 49{m^2} - 112mn + 64{n^2} + \left[ {49{m^2} + 112mn + 64{n^2}} \right]$${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = 49{m^2} - 112mn + 64{n^2} + 49{m^2} + 112mn + 64{n^2}$

${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = 98{m^2} + 128{n^2}$

(iv). ${\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2}$

Ans: On expanding the terms ${\left( {4m + 5n} \right)^2}$ and ${\left( {5m + 4n} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

\[{\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = {\left( {4m} \right)^2} + 2\left( {4m} \right)\left( {5n} \right) + {\left( {5n} \right)^2} + \left[ {{{\left( {5m} \right)}^2} + 2\left( {5m} \right)\left( {4n} \right) + {{\left( {4n} \right)}^2}} \right]\]

\[{\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = 16{m^2} + 40mn + 25{n^2} + \left[ {25{m^2} + 40mn + 16{n^2}} \right]\]

\[{\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = 16{m^2} + 40mn + 25{n^2} + 25{m^2} + 40mn + 16{n^2}\]\[{\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = 41{m^2} + 80mn + 41{n^2}\]

(v). ${\left( {2.5p - 1.5q} \right)^2} -{\left( {1.5p - 2.5q} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

Solving ${\left( {2.5p - 1.5q} \right)^2}$ we get

$=\, 2.5p^2+1.5q^2-2(2.5p)(1.5q)$

$={\dfrac{25}{10}}^2\times p^2+{\dfrac{15}{10}}^2\times q^2 -(2\times\dfrac{25}{10}\times\dfrac{15}{10})\times(p \times q)$

$=\dfrac{625}{100}p^2+\dfrac{225}{100} q^2 - \dfrac{750}{100} pq$ 

Solving ${\left( {1.5p - 2.5q} \right)^2}$ we get

$=\, 1.5p^2\,+\,2.5q^2\,-\,2(1.5p)(2.5q)$

$={\dfrac{15}{10}}^2\times p^2+{\dfrac{25}{10}}^2\times q^2 -(2\times\dfrac{15}{10}\times\dfrac{25}{10})\times(p \times q)$

$=\dfrac{225}{100}p^2+\dfrac{625}{100} q^2 - \dfrac{750}{100} pq$

$\therefore$  solving the given equation we get,

${\left( {2.5p - 1.5q} \right)^2} -{\left( {1.5p - 2.5q} \right)^2}$ = $\dfrac{625}{100}p^2+\dfrac{225}{100} q^2 - \dfrac{750}{100} pq -$  ($\dfrac{225}{100}p^2+\dfrac{625}{100} q^2 - \dfrac{750}{100} pq$)

$=\dfrac{400}{100} p^2 - \dfrac{400}{100} q^2$

$=4p^2-4q^2$

(vi). ${\left( {ab + bc} \right)^2} - 2a{b^2}c$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {ab + bc} \right)^2} - 2a{b^2}c = {\left( {ab} \right)^2} + 2\left( {ab} \right)\left( {bc} \right) + {\left( {bc} \right)^2} - 2a{b^2}c$

${\left( {ab + bc} \right)^2} - 2a{b^2}c = {a^2}{b^2} + 2a{b^2}c + {b^2}{c^2} - 2a{b^2}c$

${\left( {ab + bc} \right)^2} - 2a{b^2}c = {a^2}{b^2} + {b^2}{c^2}$

(vii). ${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2} = {\left( {{m^2}} \right)^2} - 2\left( {{m^2}} \right)\left( {{n^2}m} \right) + {\left( {{n^2}m} \right)^2} + 2{m^3}{n^2}$

${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2} = {m^4} - 2{m^3}{n^2} + {n^4}{m^2} + 2{m^3}{n^2}$

${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2} = {m^4} + {n^4}{m^2}$


5. Show that the following equations are true.

(i). ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {3x + 7} \right)^2} - 84x$

On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {3x + 7} \right)^2} - 84x$

${\text{LHS}} = {\left( {3x} \right)^2} + 2\left( {3x} \right)\left( 7 \right) + {\left( 7 \right)^2} - 84x$

${\text{LHS}} = 9{x^2} + 42x + 49 - 84x$

${\text{LHS}} = 9{x^2} - 42x + 49$

On considering the ${\text{RHS}}$ of the expression, we get,

${\text{RHS}} = {\left( {3x - 7} \right)^2}$

On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{RHS}} = {\left( {3x - 7} \right)^2}$

${\text{RHS}} = {\left( {3x} \right)^2} - 2\left( {3x} \right)\left( 7 \right) + {\left( 7 \right)^2}$

${\text{RHS}} = 9{x^2} - 42x + 49$

As ${\text{LHS = RHS}}$, so the given expression is proved. 

(ii). ${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {9p - 5q} \right)^2} + 180pq$

On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {9p - 5q} \right)^2} + 180pq$

${\text{LHS}} = {\left( {9p} \right)^2} - 2\left( {9p} \right)\left( {5q} \right) + {\left( {5q} \right)^2} + 180pq$

${\text{LHS}} = 81{p^2} - 90pq + 25{q^2} + 180pq$

${\text{LHS}} = 81{p^2} + 90pq + 25{q^2}$

On considering the ${\text{RHS}}$ of the expression, we get,

${\text{RHS}} = {\left( {9p + 5q} \right)^2}$

On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\text{RHS}} = {\left( {9p + 5q} \right)^2}$

${\text{RHS}} = {\left( {9p} \right)^2} + 2\left( {9p} \right)\left( {5q} \right) + {\left( {5q} \right)^2}$

${\text{RHS}} = 81{p^2} + 90pq + 25{q^2}$

As ${\text{LHS = RHS}}$, so the given expression is proved. 

(iii). ${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn$

On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn$

${\text{LHS}} = {\left( {\dfrac{4}{3}m} \right)^2} - 2\left( {\dfrac{4}{3}m} \right)\left( {\dfrac{3}{4}n} \right) + {\left( {\dfrac{3}{4}n} \right)^2} + 2mn$

${\text{LHS}} = \dfrac{{16}}{9}{m^2} - 2mn + \dfrac{9}{{16}}{n^2} + 2mn$

${\text{LHS}} = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$

${\text{LHS}} = {\text{RHS}}$

As ${\text{LHS = RHS}}$, so the given expression is proved. 

(iv). ${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2}$

On expanding the term ${\left( {4pq + 3q} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and expanding the term ${\left( {4pq - 3q} \right)^2}$ using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2}$

${\text{LHS}} = {\left( {4pq} \right)^2} + 2\left( {4pq} \right)\left( {3q} \right) + {\left( {3q} \right)^2} - \left[ {{{\left( {4pq} \right)}^2} - 2\left( {4pq} \right)\left( {3q} \right) + {{\left( {3q} \right)}^2}} \right]$

${\text{LHS}} = 16{p^2}{q^2} + 24p{q^2} + 9{q^2} - \left[ {16{p^2}{q^2} - 24p{q^2} + 9{q^2}} \right]$

${\text{LHS}} = 16{p^2}{q^2} + 24p{q^2} + 9{q^2} - 16{p^2}{q^2} + 24p{q^2} - 9{q^2}$

\[{\text{LHS}} = 48p{q^2}\]

\[{\text{LHS}} = {\text{RHS}}\]

As ${\text{LHS = RHS}}$, so the given expression is proved. 

(v). $\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = \left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right)$

On simplifying the terms using the identity $\left( {x - y} \right)\left( {x + y} \right) = {x^2} - {y^2}$ and solving it further, we get,

\[{\text{LHS}} = \left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right)\]

\[{\text{LHS}} = \left( {{a^2} - {b^2}} \right) + \left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right)\]

\[{\text{LHS}} = \left( {{a^2} - {a^2}} \right) + \left( {{b^2} - {b^2}} \right) + \left( {{c^2} - {c^2}} \right)\]

\[{\text{LHS}} = 0\]

\[{\text{LHS}} = {\text{RHS}}\]

As ${\text{LHS = RHS}}$, so the given expression is proved.


6. Using the different algebraic identities, evaluate the following.

(i). ${71^2}$

Ans: On expressing ${71^2}$ as the square of the sum of two numbers, we get,

${71^2} = {\left( {70 + 1} \right)^2}$

On expanding the above expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {70 + 1} \right)^2} = {\left( {70} \right)^2} + 2\left( {70} \right)\left( 1 \right) + {\left( 1 \right)^2}$

${\left( {70 + 1} \right)^2} = 4900 + 140 + 1$

${\left( {70 + 1} \right)^2} = 5041$

(ii). ${99^2}$

Ans: On expressing ${99^2}$ as the square of the difference of two numbers, we get,

${99^2} = {\left( {100 - 1} \right)^2}$

On expanding the above expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {100 - 1} \right)^2} = {\left( {100} \right)^2} - 2\left( {100} \right)\left( 1 \right) + {\left( 1 \right)^2}$

${\left( {100 - 1} \right)^2} = 10000 - 200 + 1$

${\left( {100 - 1} \right)^2} = 9801$

(iii). ${102^2}$

Ans: On expressing ${102^2}$ as the square of the sum of two numbers, we get,

${102^2} = {\left( {100 + 2} \right)^2}$

On expanding the above expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {100 + 2} \right)^2} = {\left( {100} \right)^2} + 2\left( {100} \right)\left( 2 \right) + {\left( 2 \right)^2}$

${\left( {100 + 2} \right)^2} = 10000 + 400 + 4$

${\left( {100 + 2} \right)^2} = 10404$

(iv). ${998^2}$

Ans: On expressing ${998^2}$ as the square of the difference of two numbers, we get,

${998^2} = {\left( {1000 - 2} \right)^2}$

On expanding the above expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {1000 - 2} \right)^2} = {\left( {1000} \right)^2} - 2\left( {1000} \right)\left( 2 \right) + {\left( 2 \right)^2}$

${\left( {1000 - 2} \right)^2} = 1000000 - 4000 + 4$

${\left( {1000 - 2} \right)^2} = 996004$

(v). ${\left( {5.2} \right)^2}$

Ans: On expressing ${\left( {5.2} \right)^2}$ as the square of the sum of two numbers, we get,

${\left( {5.2} \right)^2} = {\left( {5.0 + 0.2} \right)^2}$

On expanding the above expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {5.0 + 0.2} \right)^2} = {\left( {5.0} \right)^2} + 2\left( {5.0} \right)\left( {0.2} \right) + {\left( {0.2} \right)^2}$

${\left( {5.0 + 0.2} \right)^2} = 25 + 2 + 0.04$

${\left( {5.0 + 0.2} \right)^2} = 27.04$

(vi). $297 \times 303$

Ans: On expressing 297 and 303 as the difference and sum of the same two numbers respectively, we get,

$297 \times 303 = \left( {300 - 3} \right) \times \left( {300 + 3} \right)$

On writing the above expression using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ and solving it further, we get,

$\left( {300 - 3} \right) \times \left( {300 + 3} \right) = {\left( {300} \right)^2} - {\left( 3 \right)^2}$

$\left( {300 - 3} \right) \times \left( {300 + 3} \right) = 90000 - 9$

$\left( {300 - 3} \right) \times \left( {300 + 3} \right) = 89991$

(vii). $78 \times 82$

Ans: On expressing 78 and 82 as the difference and sum of the same two numbers respectively, we get,

$78 \times 82 = \left( {80 - 2} \right) \times \left( {80 + 2} \right)$

On writing the above expression using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ and solving it further, we get,

$\left( {80 - 2} \right) \times \left( {80 + 2} \right) = {\left( {80} \right)^2} - {\left( 2 \right)^2}$

$\left( {80 - 2} \right) \times \left( {80 + 2} \right) = 6400 - 4$

$\left( {80 - 2} \right) \times \left( {80 + 2} \right) = 6396$

(viii). ${\left( {8.9} \right)^2}$

Ans: On expressing ${\left( {8.9} \right)^2}$ as the square of the difference of two numbers, we get,

$\,{\left( {8.9} \right)^2} = {\left( {9.0 - 0.1} \right)^2}$

On expanding the above expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {9.0 - 0.1} \right)^2} = {\left( {9.0} \right)^2} - 2\left( {9.0} \right)\left( {0.1} \right) + {\left( {0.1} \right)^2}$

${\left( {9.0 - 0.1} \right)^2} = 81 - 1.8 + 0.01$

${\left( {9.0 - 0.1} \right)^2} = 79.21$

(ix). ${\text{1}}{\text{.05}} \times {\text{9}}{\text{.5}}$

Ans: On expressing ${\text{9}}{\text{.5}}$ as the product of two numbers, we get,

$1.05 \times 9.5 = 1.05 \times 0.95 \times 10$

On expressing $1.05$ and $0.95$ as the sum and difference of the same two numbers respectively, we get,

$1.05 \times 0.95 \times 10 = \left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10$

On writing the above expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ and solving it further, we get,

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = \left[ {{{\left( 1 \right)}^2} - {{\left( {0.05} \right)}^2}} \right] \times 10$

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = \left[ {1 - 0.0025} \right] \times 10$

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = 0.9975 \times 10$

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = 9.975$


7. Use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$to evaluate the following expressions. 

(i). ${51^2} - {49^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${51^2} - {49^2} = \left( {51 + 49} \right)\left( {51 - 49} \right)$

On simplifying the above equation further, we get,

${51^2} - {49^2} = \left( {100} \right)\left( 2 \right)$

${51^2} - {49^2} = 200$

(ii). ${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2} = \left( {1.02 + 0.98} \right)\left( {1.02 - 0.98} \right)$

On simplifying the above equation further, we get,

${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2} = \left( 2 \right)\left( {0.04} \right)$

${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2} = 0.08$

(iii). ${153^2} - {147^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${153^2} - {147^2} = \left( {153 + 147} \right)\left( {153 - 147} \right)$

On simplifying the above equation further, we get,

${153^2} - {147^2} = \left( {300} \right)\left( 6 \right)$

${153^2} - {147^2} = 1800$

(iv). ${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2} = \left( {12.1 + 7.9} \right)\left( {12.1 - 7.9} \right)$

On simplifying the above equation further, we get,

${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2} = \left( {20.0} \right)\left( {4.2} \right)$

${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2} = 84$


8. Use the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$to find the following products. 

(i). $103 \times 104$

Ans: On expressing 103 and 104 as the sum of two different numbers, we get,

$103 \times 104 = \left( {100 + 3} \right)\left( {100 + 4} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {100 + 3} \right)\left( {100 + 4} \right) = {\left( {100} \right)^2} + \left( {3 + 4} \right)\left( {100} \right) + \left( 3 \right)\left( 4 \right)$

On simplifying the above equation further, we get,

$\left( {100 + 3} \right)\left( {100 + 4} \right) = 10000 + 7\left( {100} \right) + 12$

$\left( {100 + 3} \right)\left( {100 + 4} \right) = 10000 + 700 + 12$

$\left( {100 + 3} \right)\left( {100 + 4} \right) = 10712$

(ii). $5.1 \times 5.2$

Ans: On expressing $5.1$ and $5.2$ as the sum of two different numbers, we get,

$5.1 \times 5.2 = \left( {5 + 0.1} \right)\left( {5 + 0.2} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = {\left( 5 \right)^2} + \left( {0.1 + 0.2} \right)\left( 5 \right) + \left( {0.1} \right)\left( {0.2} \right)$

On simplifying the above equation further, we get,

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = 25 + 0.3\left( 5 \right) + 0.02$

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = 25 + 1.5 + 0.02$

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = 26.52$

(iii). $103 \times 98$

Ans: On expressing 103 and 98 as the sum and difference of two different numbers respectively, we get,

$103 \times 98 = \left( {100 + 3} \right)\left( {100 - 2} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {100 + 3} \right)\left( {100 - 2} \right) = {\left( {100} \right)^2} + \left[ {3 + \left( { - 2} \right)} \right]\left( {100} \right) + \left( 3 \right)\left( { - 2} \right)$

On simplifying the above equation further, we get,

$\left( {100 + 3} \right)\left( {100 - 2} \right) = 10000 + \left( {3 - 2} \right)\left( {100} \right) - 6$

$\left( {100 + 3} \right)\left( {100 - 2} \right) = 10000 + 100 - 6$

$\left( {100 + 3} \right)\left( {100 - 2} \right) = 10094$

(iv). $9.7 \times 9.8$

Ans: On expressing $9.7$ and $9.8$ as the difference of two different numbers, we get,

$9.7 \times 9.8 = \left( {10 - 0.3} \right)\left( {10 - 0.2} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = {\left( {10} \right)^2} + \left[ {\left( { - 0.3} \right) + \left( { - 0.2} \right)} \right]\left( {10} \right) + \left( { - 0.3} \right)\left( { - 0.2} \right)$

On simplifying the above equation further, we get,

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 100 + \left[ { - 0.3 - 0.2} \right]\left( {10} \right) + 0.06$

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 100 + \left( { - 0.5} \right)\left( {10} \right) + 0.06$

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 100 - 5 + 0.06$

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 95.06$


NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Opting for the NCERT solutions for Ex 9.5 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.5 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.


Chapter wise NCERT Solutions for Class 8 Maths


Class 8 Maths Chapter 9 Includes:

Chapter 9 Algebraic Expressions and Identities All Exercises in PDF Format

Exercise 9.1

4 Questions & Solutions

Exercise 9.2

5 Questions & Solutions

Exercise 9.3

5 Questions & Solutions

Exercise 9.4

3 Questions & Solutions

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 9 Exercise 9.5 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 9 Exercise 9.5, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 9 Exercise 9.5 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

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