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# NCERT Solutions for Class 6 Maths Chapter 11: Algebra - Exercise 11.5

Last updated date: 12th Aug 2024
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## NCERT Solutions for Class 6 Maths Chapter 11 (Ex 11.5)

Free PDF download of NCERT Solutions for Class 6 Maths Chapter 11 Exercise 11.5 (Ex 11.5) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 6 Maths Chapter 11 Algebra Exercise 11.5 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 6 Subject: Class 6 Maths Chapter Name: Chapter 11 - Algebra Exercise: Exercise - 11.5 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2023-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

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## Access NCERT Solutions for Class 6 Maths Chapter 11- Algebra

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.

 S. No. Equation Value of variable Equation satisfiedYes/No $10y = 80$ $y = 10$ $10y = 80$ $y = 8$ $10y = 80$ $y = 5$ $4l = 20$ $l = 20$ $4l = 20$ $l = 80$ $4l = 20$ $l = 5$ $b + 5 = 9$ $b = 5$ $b + 5 = 9$ $b = 9$ $b + 5 = 9$ $b = 4$ $h - 8 = 5$ $h = 13$ $h - 8 = 5$ $h = 8$ $h - 8 = 5$ $h = 0$ $p + 3 = 1$ $p = 3$ $p + 3 = 1$ $p = 1$ $p + 3 = 1$ $p = 0$ $p + 3 = 1$ $p = - 1$ $p + 3 = 1$ $p = - 2$

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.

 S. No. Equation Value of variable Equation satisfiedYes/No Solution of L.H.S. $10y = 80$ $y = 10$ No $10 \times 10 = 100$ $10y = 80$ $y = 8$ Yes $10 \times 8 = 80$ $10y = 80$ $y = 5$ No $10 \times 5 = 50$ $4l = 20$ $l = 20$ No $4 \times 20 = 80$ $4l = 20$ $l = 80$ No $4 \times 80 = 320$ $4l = 20$ $l = 5$ Yes $4 \times 5 = 20$ $b + 5 = 9$ $b = 5$ No $5 + 5 = 10$ $b + 5 = 9$ $b = 9$ No $9 + 5 = 14$ $b + 5 = 9$ $b = 4$ Yes $4 + 5 = 9$ $h - 8 = 5$ $h = 13$ Yes $13 - 8 = 5$ $h - 8 = 5$ $h = 8$ No $8 - 8 = 0$ $h - 8 = 5$ $h = 0$ No $0 - 8 = - 8$ $p + 3 = 1$ $p = 3$ No $3 + 3 = 6$ $p + 3 = 1$ $p = 1$ No $1 + 3 = 4$ $p + 3 = 1$ $p = 0$ No $0 + 3 = 3$ $p + 3 = 1$ $p = - 1$ No $- 1 + 3 = 2$ $p + 3 = 1$ $p = - 2$ Yes $- 2 + 3 = 1$

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 $m + 10 = 16$.

 $m$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ _ _ _ $m + 10$ _ _ _ _ _ _ _ _ _ _ _ _ _

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.

 $m$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ $m + 10$ $11$ $12$ $13$ $14$ $15$ $16$ $17$ $18$ $19$ $20$ $21$ $22$ $23$

$\because$   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 $5t = 35$.

 $t$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ _ _ _ _ _ $5t$ _ _ _ _ _ _ _ _ _ _ _ _ _ _

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.

 $t$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $15$ $16$ $5t$ $15$ $20$ $25$ $30$ $35$ $40$ $45$ $50$ $55$ $60$ $65$ $70$ $75$ $80$

$\because$   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.

 $z$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $15$ $16$ _ _ _ _ $\dfrac{z}{3}$ _ _ _ _ _ _ _ _ _ _ _ _ _

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.

 $z$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $15$ $16$ $17$ $18$ $19$ $20$ $\dfrac{z}{3}$ $2\dfrac{2}{3}$ $3$ $3\dfrac{1}{3}$ $3\dfrac{2}{3}$ $4$ $4\dfrac{1}{3}$ $4\dfrac{2}{3}$ $5$ $5\dfrac{1}{3}$ $5\dfrac{2}{3}$ $6$ $6\dfrac{1}{3}$ $6\dfrac{2}{3}$

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

So, $z = 12$ is the solution.

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

 $m$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ _ _ $m - 7$ _ _ _ _ _ _ _ _ _ _ _

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.

 $m$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $13$ $14$ $15$ $m - 7$ $- 2$ $- 1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$\because$   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.

Besides these NCERT solutions for Class 6 Maths Chapter 11 Exercise 11.5, 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 6 Maths Chapter 11 Exercise 11.5 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.

## FAQs on NCERT Solutions for Class 6 Maths Chapter 11: Algebra - Exercise 11.5

1. How many questions are present in NCERT Solutions for Class 6 Maths Chapter 11 Algebra?

Class 6 Maths Chapter 11 has a total of ten exercises -

• Ex. 11.1 - Introduction

• Ex. 11.2 - Matchstick Patterns

• Ex. 11.3 - Idea of a variable

• Ex. 11.4 - Some more matchstick patterns

• Ex. 11.5 - Some more Variables

• Ex. 11.6 - Variables and standard rules

• Ex. 11.7 - Expressions with variables

• Ex. 11.8 - Use of Expressions

• Ex. 11.9 - Equations

• Ex. 11.10 - Solution to Equations.

2. What is an equation?

An equation is a mathematical expression in which both sides are equal to each other. In other words, the left-hand side of the expression is equal to the right-hand side.

An equation always has an equal sign. Without it, it can't be called an equation.

For example:

x + y + 10 = 35

5x - 15 = 10

Besides Maths, equations can be used in other subjects like Physics and Chemistry.

3. Which online website should I refer to solve NCERT questions?

It is important for students to know the right manner to solve a math problem. It is known that each step of a solution contains respective marks in exams. Therefore, you can refer to Vedantu for NCERT Solutions of Class 6 Maths, which gives solutions to all NCERT questions in a proper step-wise manner. You can improve your quality of answers by practicing the given solutions of NCERT questions.

4. How can I practice important topics covered in Class 6 Maths Chapter 11?

If you are done with imbibing all the important concepts, now is the time for the practice session. We know that practice makes a man perfect. First off, get started with NCERT Solutions. Practice all the questions provided in the NCERT book because they are really important from an exam point of view. Then, you can move onto questions from other guidebooks. Doing this will eventually improve your problem-solving.