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

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

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.