NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercise 11.5 - 2025-26

Exercise 11.5

1. 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) \[17 = x + 7\] 

Ans: It is an equation with variable $x$.

(b) \[\left( {t - 7} \right) > 5\]

Ans: It is not an equation since there is no equal sign.

(c) \[\dfrac{4}{2} = 2\]

Ans: It is an equation with no variable.

(d) \[\left( {7 \times 3} \right) - 19 = 8\]

Ans: It is an equation with no variable.

(e) \[5 \times 4 - 8 = 2x\]

Ans: It is an equation with variable $x$.

(f) \[x - 2 = 0\]

Ans: It is an equation with variable $x$.

(g) \[2m < 30\]

Ans: It is not an equation since there is no equal sign.

(h) \[2n + 1 = 11\]

Ans: It is an equation with variable $n$.

(i) \[7 = \left( {11 \times 5} \right) - \left( {12 \times 4} \right)\]

Ans: It is an equation with no variable.

(j) \[7 = \left( {11 \times 2} \right) + p\]

Ans: It is an equation with variable $p$.

(k) \[20 = 5y\]

Ans: It is an equation with variable $y$.

(l) \[\dfrac{{3q}}{2} < 5\]

Ans: It is not an equation since there is no equal sign.

(m) \[z + 12 > 24\]

Ans: It is not an equation since there is no equal sign.

(n) \[20 - \left( {10 - 5} \right) = 3 \times 5\]

Ans: It is an equation with no variable.

(o) \[7 - x = 5\]

Ans: It is an equation with variable $x$.

2. Complete the entries in the third column of the table.

Ans: Create the table with one additional column as “Solution of L.H.S.” and substitute the value of column 3 in the second column and fill Yes/No in the fourth column.

3. 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) $5m = 60$             $\left( {10,5,12,15} \right)$ 

Ans: Substitute the given values in L.H.S. of equation,

$5 \times 10 = 50$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $m = 10$ is not the solution.

$5 \times 5 = 25$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $m = 5$ is not the solution.

$5 \times 12 = 60$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $m = 12$ is a solution.

$5 \times 15 = 75$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $m = 15$ is not the solution.  

The value $m = 12$ is a solution.

(b) $n + 12 = 20$          $\left( {12,8,20,0} \right)$ 

Ans: Substitute the given values in L.H.S. of equation,

$12 + 12 = 24$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $n = 12$ is not the solution.

$12 + 8 = 20$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $n = 8$ is a solution.

$12 + 20 = 32$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $n = 20$ is not the solution.

$0 + 12 = 12$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $n = 0$ is not the solution.  

The value $n = 8$ is a solution.

(c) $p - 5 = 5$              $\left( {0,10,5, - 5} \right)$   

Ans: Substitute the given values in L.H.S. of equation,

$0 - 5 =  - 5$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $p = 0$ is not the solution.

$10 - 5 = 5$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $p = 10$ is a solution.

$5 - 5 = 0$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $p = 5$ is not the solution.

$ - 5 - 5 =  - 10$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $p =  - 5$ is not the solution.  

The value $p = 10$ is a solution.

(d) $\dfrac{q}{2} = 7$                    $\left( {7,2,10,14} \right)$

Ans: Substitute the given values in L.H.S. of equation,

$\dfrac{7}{2}$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $q = 7$ is not the solution.

$\dfrac{2}{2} = 1$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $q = 2$ is not the solution.

$\dfrac{{10}}{2} = 5$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $q = 10$ is not the solution.

$\dfrac{{14}}{2} = 7$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $q = 14$ is a solution.  

The value $q = 14$ is a solution.

(e) $r - 4 = 0$                $\left( {4, - 4,8,0} \right)$

Ans: Substitute the given values in L.H.S. of equation,

$4 - 4 = 0$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $r = 4$ is a solution.

$ - 4 - 4 =  - 8$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $r =  - 4$ is not the solution.

$8 - 4 = 4$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $r = 8$ is not the solution.

$0 - 4 =  - 4$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $r = 0$ is not the solution.  

The value $r = 4$ is a solution.

(f) $x + 4 = 2$                 $\left( { - 2,0,2,4} \right)$

Ans: Substitute the given values in L.H.S. of equation,

$ - 2 + 4 = 2$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $x =  -2$ is a solution.

$0 + 4 = 4$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $x = 0$ is not the solution.

$2 + 4 = 6$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $x = 2$ is not the solution.

$4 + 4 = 8$

$\because {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$

So, $x = 4$ is not the solution.  

The value $x =  - 2$ is a solution.

4. (a) Complete the table and by inspection of the table find the solution to the

equation $m + 10 = 16$.

Ans: Substitute the values of first row in the equation and write the sum in second row then see the value which satisfies the equation.

$\because $   At $m = 6$, $m + 10 = 16$.

So, $m = 6$ is the solution.

(b) Complete the table and by inspection of the table, find the solution to the

equation $5t = 35$.

Ans: Substitute the values of first row in the equation and write the product in second row then see the value which satisfies the equation.

$\because $   At $t = 7$, $5t = 35$.

So, $t = 7$ is the solution.

(c) Complete the table and find the solution of the equation $z/3 = 4$ using the

table.

Ans: Substitute the values of first row in the equation and write the result in second row then see the value which satisfies the equation.

$\because $   At $z = 12$, $\dfrac{z}{3} = 4$.

So, $z = 12$ is the solution.

(d) Complete the table and find the solution to the equation $m - 7 = 3$.

Ans: Substitute the values of first row in the equation and write the difference in second row then see the value which satisfies the equation.

$\because $   At $m = 10$, $m - 7 = 3$.

So, $m = 10$ is the solution.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.5

Opting for the NCERT solutions for Ex 11.5 Class 6 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 11.5 Class 6 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.

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 6 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 6 Maths Chapter 11 Exercise 11.5 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.

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