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NCERT Solutions for Class 7 Maths Chapter 10 - Algebraic Expressions Exercise 10.2

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NCERT Solutions for Maths Class 7 Chapter 10 Exercise 10.2 - FREE PDF Download

NCERT Solution for Maths Class 7 Chapter 10 Exercise 10.2 focuses on deepening your understanding of algebraic terms and their operations. This exercise helps you practice identifying and working with different components of algebraic expressions, such as terms, coefficients, and constants.

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Table of Content
1. NCERT Solutions for Maths Class 7 Chapter 10 Exercise 10.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 10 Exercise 10.2 Class Class 7 | Vedantu
3. Access NCERT Solutions for Maths Class 7 Chapter 10 - Algebraic Expressions
4. Class Class 7 Maths Chapter 10: Exercises Breakdown
5. CBSE Class Class 7 Maths Chapter 10 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs


Maths Exercise 10.2 Class 7 is designed to enhance your skills in simplifying expressions and performing basic algebraic operations like addition and subtraction of algebraic terms. By solving these problems, you will develop a solid foundation in handling algebraic expressions, which is essential for progressing to more advanced topics in mathematics. Access Class 7 Maths Syllabus here.


Glance on NCERT Solutions Maths Chapter 10 Exercise 10.2 Class Class 7 | Vedantu

  • In NCERT Class 7 Chapter 10 solutions cover topics such as Evaluating Expressions, Simplifying Expressions, Order of Operations, Evaluating expressions, and simplifying expression.

  • Evaluating Expressions involves finding the numerical value of an expression by substituting numerical values for the variables.

  • Simplifying Expressions means combining terms in the expression that are alike (have the same variable raised to the same power).

  • Understanding Like Terms are considered "like" if they have the same variable raised to the same power. are not (because the exponents are different).

  • Order of Operations (PEMDAS/BODMAS)  refers to the correct sequence for evaluating an expression that includes multiple operations. Evaluating expressions with given values for the variables.

  • Simplifying expressions by combining like terms and Recognizing like terms based on their variable and exponent.

  • Applying PEMDAS/BODMAS for correct evaluation when an expression has multiple operations in Maths Exercise 10.2 Class 7.

  • The Class 7 Maths Exercise 10.2 has 10 Questions & Solutions

Access NCERT Solutions for Maths Class 7 Chapter 10 - Algebraic Expressions

Exercise 10.2

1. If $\mathbf{m = 2}$, find the value of:

(i)$\mathbf{m - 2}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow m - 2 = 2 - 2 \\ $

$ \Rightarrow m - 2 = 0  \\ $

The value is $0$.


(ii) $\mathbf{3m - 5}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 3m - 5 = 3\left( 2 \right) - 5 \\$

$ \Rightarrow 3m - 5 = 6 - 5 \\ $

$ \Rightarrow 3m - 5 = 1 \\ $

The value is $1$.


(iii) $\mathbf{9 - 5m}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 9 - 5m = 9 - 5\left( 2 \right) \\ $

$ \Rightarrow 9 - 5m = 9 - 10  \\ $

$ \Rightarrow 9 - 5m =  - 1 \\ $

The value is $ - 1$.


(iv) $\mathbf{3{m^2} - 2m - 7}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow 3{m^2} - 2m - 7 = 3{\left( 2 \right)^2} - 2\left( 2 \right) - 7 \\ $

$ \Rightarrow 3{m^2} - 2m - 7 = 3\left( 4 \right) - 4 - 7  \\ $

 $ \Rightarrow 3{m^2} - 2m - 7 = 12 - 11  \\ $

  $ \Rightarrow 3{m^2} - 2m - 7 = 1 \\  $

The value is $1$.


(v) $\mathbf{\dfrac{{5m}}{2} - 4}$

Ans: Substitute $2$ for $m$ and solve.

$ \Rightarrow \dfrac{{5m}}{2} - 4 = \dfrac{{5 \times 2}}{2} - 4 \\ $

 $ \Rightarrow \dfrac{{5m}}{2} - 4 = 5 - 4 \\ $

 $ \Rightarrow \dfrac{{5m}}{2} - 4 = 1 \\ $

The value is $1$.


2. If $\mathbf{p =  - 2}$, find the value of:

(i) $\mathbf{4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$ \Rightarrow 4p + 7 = 4\left( { - 2} \right) + 7 \\ $

$ \Rightarrow 4p + 7 =  - 8 + 7 \\ $

 $ \Rightarrow 4p + 7 =  - 1 \\ $

 The value is $ - 1$.


(ii) $\mathbf{ - 3{p^2} + 4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$\Rightarrow  - 3{p^2} + 4p + 7 =  - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 3\left( 4 \right) - 8 + 7 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 12 - 1 \\ $

  $ \Rightarrow  - 3{p^2} + 4p + 7 =  - 13 \\ $

The value is $ - 13$.


(iii) $\mathbf{ - 2{p^3} - 3{p^2} + 4p + 7}$

Ans: Substitute $ - 2$ for $p$ and solve.

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 =  - 2{\left( { - 2} \right)^3} - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 7 \\ $

 $ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 =  - 2 \times \left( { - 8} \right) - 3\left( 4 \right) - 8 + 7 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 16 - 12 - 1 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 4 - 1 \\ $

$ \Rightarrow  - 2{p^3} - 3{p^2} + 4p + 7 = 3  \\ $

The value is $3$.


3. Find the value of the following expressions, when $\mathbf{x =  - 1}$:

(i) $\mathbf{2x - 7}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow 2x - 7 = 2\left( { - 1} \right) - 7 \\ $

$ \Rightarrow 2x - 7 =  - 2 - 7 \\ $

$ \Rightarrow 2x - 7 =  - 9 \\ $

The value of expression $2x - 7$ is $ - 9$.


(ii) $ \mathbf{- x + 2}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow  - x + 2 =  - \left( { - 1} \right) + 2  \\ $

$ \Rightarrow  - x + 2 = 1 + 2 \\ $

 $ \Rightarrow  - x + 2 = 3  \\ $ 

The value of expression $ - x + 2$ is $3$.


(iii) $\mathbf{{x^2} + 2x + 1}$

Ans: Substitute $ - 1$ for $x$ and solve.

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

$ \Rightarrow {x^2} + 2x + 1 = 1 - 2 + 1  \\ $

$ \Rightarrow {x^2} + 2x + 1 =  - 1 + 1  \\ $

$ \Rightarrow {x^2} + 2x + 1 = 0 \\ $

The value of expression ${x^2} + 2x + 1$ is $0$.


(iv) $\mathbf{2{x^2} - x - 2}$

Ans: Substitute $ - 1$ for $x$ and solve.

$ \Rightarrow 2{x^2} - x - 2 = 2{\left( { - 1} \right)^2} - \left( { - 1} \right) - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 2 \times 1 + 1 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 2 + 1 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 3 - 2 \\ $

$ \Rightarrow 2{x^2} - x - 2 = 1 \\ $

The value of expression $2{x^2} - x - 2$ is $1$.


4. If $\mathbf{a = 2}$, ${b =  - 2}$, find the value of:

(i) $\mathbf{{a^2} + {b^2}}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

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

$ \Rightarrow {a^2} + {b^2} = 4 + 4  \\ $

$ \Rightarrow {a^2} + {b^2} = 8 \\ $

The value is $8$.


(ii) ${a^2} + ab + {b^2}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

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

$ \Rightarrow {a^2} + ab + {b^2} = 4 - 4 + 4 \\ $

$ \Rightarrow {a^2} + ab + {b^2} = 4 \\ $

The value is $4$.


(iii) ${a^2} - {b^2}$

Ans: Substitute $2$ for $a$ and $ - 2$ for $b$ then solve.

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

$ \Rightarrow {a^2} - {b^2} = 4 - 4 \\ $

$ \Rightarrow {a^2} - {b^2} = 0  \\ $

The value is $0$.


5. When $\mathbf{a = 0}$, ${b =  - 1}$, find the value of the given expressions:

(i) $\mathbf{2a + 2b}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2a + 2b = 2\left( 0 \right) + 2\left( { - 1} \right)  \\ $

$ \Rightarrow 2a + 2b = 0 - 2  \\ $

$ \Rightarrow 2a + 2b =  - 2  \\ $

The value of expression $2a + 2b$ is $ - 2$.


(ii) $\mathbf{2{a^2} + {b^2} + 1}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2{\left( 0 \right)^2} + {\left( { - 1} \right)^2} + 1  \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2 \times 0 + 1 + 1 \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 0 + 2 \\ $

$ \Rightarrow 2{a^2} + {b^2} + 1 = 2 \\ $

The value of expression $2{a^2} + {b^2} + 1$ is $2$.


(iii) $\mathbf{2{a^2}b + 2a{b^2} + ab}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 2{\left( 0 \right)^2}\left( { - 1} \right) + 2\left( 0 \right){\left( { - 1} \right)^2} + \left( 0 \right)\left( { - 1} \right) \\ $

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 + 0 + 0 \\ $

$ \Rightarrow 2{a^2}b + 2a{b^2} + ab = 0 \\ $

The value of expression $2{a^2}b + 2a{b^2} + ab$ is $0$.


(iv)$\mathbf{{a^2} + ab + 2}$

Ans: Substitute $0$ for $a$ and $ - 1$ for $b$ then solve.

$ \Rightarrow {a^2} + ab + 2 = {\left( 0 \right)^2} + \left( 0 \right)\left( { - 1} \right) + 2 \\$

$ \Rightarrow {a^2} + ab + 2 = 0 + 0 + 2 \\ $

$ \Rightarrow {a^2} + ab + 2 = 2 \\ $

The value of expression ${a^2} + ab + 2$ is $2$.


6. Simplify the expressions and find the value of $\mathbf{x}$ is equal to $\mathbf{2}$:

(i) $\mathbf{x + 7 + 4\left( {x - 5} \right)}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 7 + 4x - 20  \\ $

 $ \Rightarrow x + 7 + 4\left( {x - 5} \right) = x + 4x + 7 - 20 \\ $

 $ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 5x - 13 \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 5\left( 2 \right) - 13  \\ $

$ \Rightarrow x + 7 + 4\left( {x - 5} \right) = 10 - 13 \\ $

$  \Rightarrow x + 7 + 4\left( {x - 5} \right) =  - 3 \\  $

 The determined value is $ - 3$.


(ii) $\mathbf{3\left( {x + 2} \right) + 5x - 7}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 6 + 5x - 7 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 3x + 5x + 6 - 7 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8x - 1 \\  $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 8\left( 2 \right) - 1 \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 16 - 1  \\ $

$ \Rightarrow 3\left( {x + 2} \right) + 5x - 7 = 15 \\ $

The determined value is $15$.


(iii) $\mathbf{6x + 5\left( {x - 2} \right)}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 6x + 5x - 10 \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 11x - 10  \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 11 \times 2 - 10  \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 22 - 10   \\ $

$ \Rightarrow 6x + 5\left( {x - 2} \right) = 12  \\  $

The determined value is $12$.


(iv) $\mathbf{4\left( {2x - 1} \right) + 3x + 11}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x - 4 + 3x + 11  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 8x + 3x - 4 + 11  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11x + 7   \\ $

Substitute $2$ for $x$ and then solve.

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 11 \times 2 + 7 \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 22 + 7  \\ $

$ \Rightarrow 4\left( {2x - 1} \right) + 3x + 11 = 29  \\ $

The determined value is $29$.


7. Simplify these expressions and find their values if $\mathbf{x = 3}$, ${a =  - 1}$, ${b =  - 2}$:

(i) $\mathbf{3x - 5 - x + 9}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3x - 5 - x + 9 = 3x - x - 5 + 9 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 2x + 4 \\ $

Substitute $3$ for $x$ and then solve.

$ \Rightarrow 3x - 5 - x + 9 = 2 \times 3 + 4 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 6 + 4 \\ $

$ \Rightarrow 3x - 5 - x + 9 = 10 \\ $

The determined value is $10$.


(ii) $\mathbf{2 - 8x + 4x + 4}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 2 - 8x + 4x + 4 =  - 8x + 4x + 2 + 4 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 4x + 6  \\ $

Substitute $3$ for $x$ and then solve.

$ \Rightarrow 2 - 8x + 4x + 4 =  - 4 \times 3 + 6 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 12 + 6 \\ $

$ \Rightarrow 2 - 8x + 4x + 4 =  - 6 \\ $

The determined value is $ - 6$.


(iii) $\mathbf{3a + 5 - 8a + 1}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 3a + 5 - 8a + 1 = 3a - 8a + 5 + 1 \\ $

 $ \Rightarrow 3a + 5 - 8a + 1 =  - 5a + 6 \\ $

Substitute $ - 1$ for $a$ and then solve.

$ \Rightarrow 3a + 5 - 8a + 1 =  - 5\left( { - 1} \right) + 6 \\ $

$ \Rightarrow 3a + 5 - 8a + 1 = 5 + 6 \\ $

$ \Rightarrow 3a + 5 - 8a + 1 = 11 \\ $

The determined value is $11$.


(iv) $\mathbf{10 - 3b - 4 - 5b}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 10 - 3b - 4 - 5b =  - 3b - 5b + 10 - 4 \\ $

$ \Rightarrow 10 - 3b - 4 - 5b =  - 8b + 6 \\  $

Substitute $ - 2$ for $b$ and then solve.

$ \Rightarrow 10 - 3b - 4 - 5b =  - 8\left( { - 2} \right) + 6  \\ $

$ \Rightarrow 10 - 3b - 4 - 5b = 16 + 6 \\ $

$ \Rightarrow 10 - 3b - 4 - 5b = 22 \\ $

The determined value is $22$.


(v) $\mathbf{2a - 2b - 4 - 5 + a}$

Ans: Simplify the brackets and combine the like terms.

$ \Rightarrow 2a - 2b - 4 - 5 + a = 2a + a - 2b - 4 - 5  \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a = 3a - 2b - 9 \\  $

 Substitute $ - 2$ for $b$ and $ - 1$ for $a$ and then solve.

$ \Rightarrow 2a - 2b - 4 - 5 + a = 3\left( { - 1} \right) - 2\left( { - 2} \right) - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a =  - 3 + 4 - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a = 1 - 9 \\ $

$ \Rightarrow 2a - 2b - 4 - 5 + a =  - 8 \\  $

The determined value is $ - 8$.


8. (i) If $\mathbf{z = 10}$, find the value of $\mathbf{{z^3} - 3\left( {z - 10} \right)}$.

Ans: Substitute the value of $z$ as $10$ then find a cube of $10$ to solve.

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = {\left( {10} \right)^3} - 3\left( {10 - 10} \right) \\ $

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 - 3 \times 0 \\ $

$ \Rightarrow {z^3} - 3\left( {z - 10} \right) = 1000 \\ $

The value of ${z^3} - 3\left( {z - 10} \right)$ is $1000$.

(ii) If $\mathbf{p =  - 10}$, find the value of $\mathbf{{p^2} - 2p - 100}$

Ans: Substitute the value of $p$ as $ - 10$ then find a square of $10$ to solve.

$  \Rightarrow {p^2} - 2p - 100 = {\left( {10} \right)^2} - 2\left( { - 10} \right) - 100  \\ $

$ \Rightarrow {p^2} - 2p - 100 = 100 + 20 - 100 \\ $

$ \Rightarrow {p^2} - 2p - 100 = 20  \\  $

The value of ${p^2} - 2p - 100$ is $20$.


9. What should be the value of $a$ if the value of $\mathbf{2{x^2} + x - a}$ equals to $5$, when $\mathbf{x = 0}$?

Ans: Substitute $0$ for $x$ in the expression $2{x^2} + x - a$ and equate it to $5$ in order to solve for $a$.

$ \Rightarrow 2{x^2} + x - a = 5 \\ $

$ \Rightarrow 2{\left( 0 \right)^2} + 0 - a = 5 \\ $

$ \Rightarrow 0 + 0 - a = 5 \\ $

$ \Rightarrow  - a = 5 \\ $

$ \Rightarrow a =  - 5 \\ $

So, the value of $a$ is $ - 5$.


10. Simplify the expression and find its value when $\mathbf{a = 5}$ and $\mathbf{b =  - 3}$:

$\mathbf{2\left( {{a^2} + ab} \right) + 3 - ab}$

Ans: Solve the brackets and combine the like terms with the variable $ab$. 

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab + 3 - ab  \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + 2ab - ab + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{a^2} + ab + 3 \\  $

Now substitute $5$ for $a$ and $ - 3$ for $b$ to solve.

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2{\left( 5 \right)^2} + \left( 5 \right)\left( { - 3} \right) + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 2 \times 25 - 15 + 3 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 50 - 12 \\ $

$ \Rightarrow 2\left( {{a^2} + ab} \right) + 3 - ab = 38  \\ $

The value of expression $2\left( {{a^2} + ab} \right) + 3 - ab$ is $38$.


Conclusion

Class 7 Maths Chapter 10 Exercise 10.2 Solutions, By working through the problems in this exercise, you have gained valuable practice in identifying, simplifying, and performing operations on algebraic expressions. This foundational knowledge is essential for tackling more complex algebraic concepts in future studies. Class 7 Maths Exercise 10.2 Answer provides detailed explanations, helping you to comprehend and apply these concepts with confidence, thereby preparing you effectively for your exams and everyday mathematical problems.


Class Class 7 Maths Chapter 10: Exercises Breakdown

Exercise

Number of Questions

Exercise 10.1

7 Questions & Solutions


CBSE Class Class 7 Maths Chapter 10 Other Study Materials


Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions for Class 7 Maths Chapter 10 - Algebraic Expressions Exercise 10.2

1. What is the main focus of Class 7 Maths Chapter 10 Exercise 10.2 Solutions?

Exercise 10.2 deals with evaluating and simplifying algebraic expressions.

2. What are algebraic expressions according to class 7 maths chapter algebraic expressions exercise 10.2?

Algebraic expressions are combinations of variables, numbers, and mathematical operations (+, -, *, /).

3. What does it mean to evaluate an expression in Maths Class 7 Chapter 10 Exercise 10.2?

Evaluating an expression means finding its numerical value by substituting specific numbers for the variables.

4. How do I know which terms to combine when simplifying expressions?

As we studied in class 7 maths chapter algebraic expressions exercise 10.2, You can combine terms only if they are "like terms." Like terms have the same variable raised to the same power.

6. What are some examples of like terms in Class 7 Maths Exercise 10.2 Answer?

  • 3x^2, 5x^2 (both have x raised to the power of 2)

  • -2y, 7y (both have y raised to the power of 1, which is understood)

7. What are some examples of, unlike terms we studied in Class 7 Maths Exercise 10.2 Answer?

  • 3x^2, 5x (different exponents)

  • 2a, 3b (different variables)

8. What's the importance of PEMDAS/BODMAS?

PEMDAS/BODMAS tells you the correct order of operations to use when evaluating expressions with multiple operations (+, -, *, /).

9. How does PEMDAS/BODMAS work?

PEMDAS/BODMAS stands for Parentheses, Exponents, Multiplication and Division (left to right), Addition and Subtraction (left to right).


You perform operations within parentheses first, then exponents, then multiplication and division (whichever comes first from left to right), and finally addition and subtraction (from left to right).

10. What are some tips for solving problems in class 7 maths chapter algebraic expressions exercise 10.2?

  • Read the question carefully to understand what's being asked.

  • Identify the variables and their values (if provided).

  • Use PEMDAS/BODMAS to evaluate expressions with multiple operations.

  • Combine like terms whenever possible to simplify the expression.

11. Where can I find solutions to the problems in Maths Class 7 Chapter 10 Exercise 10.2?

You can refer to your NCERT textbook, solution manuals, or online resources for detailed solutions to the problems in Exercise 10.2 available at Vedantu.