NCERT Solutions For Class 6 Maths Chapter 11 Algebra (Ex 11.5) Exercise 11.5
NCERT Solutions For Class 6 Maths Chapter 11 Algebra (Ex 11.5) Exercise 11.5
The NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 provide pointbypoint explanations of all questions presented in NCERT textbooks approved by the Central Board of Secondary Education (CBSE). Extramarks provides the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 to help students clarify their questions through a comprehensive understanding of the concepts. The NCERT Class 6 Maths Chapter 11 Exercise 11.5 solutions are accessible in PDF format for students to download and study in offline mode. The book contains a number of different types of questions that students can practise in preparation for different types of competitive exams or their final examinations. These NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 must be used by students, as Extramarks’ experts have organised them in a systematic manner to provide students with the best techniques to solve these questions and to ensure that as much information on these subjects as possible is accessible. It is recommended that students revise the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 prior to the examination in order to achieve the best results. The NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 also helps students establish a solid foundation of the mathematical concepts for the higher classes.
Extramarks also provide students with a variety of resources, such as the NCERT notes and solutions, to help them understand the important topics covered in the subject matter. The NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 provides students with the skills they need to excel in Maths examinations. Understanding new concepts such as numbers and variablebased questions in Maths is very important for higher classes and in examinations. Maths is a very important subject. It is the highestscoring subject among all the subjects. Almost anyone can attain a high score in this subject with the appropriate amount of practice. A little planning can make it possible for students to gain a profound understanding of maths concepts. The NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 can help students get a good command of the concepts introduced in Chapter 11. Students should begin their Maths examination preparation with the NCERT Solutions For Class 6 Maths Chapter 11 Exercise 11.5 and other solutions provided on Extramarks in order to perform well in the final examination.
Students must understand that Maths needs a lot of practice, Extramarks provides proper guidance to the students so that they can understand the techniques needed to solve important problems. Having been in the teaching profession for the better part of a decade, Extramarks’ experts understand the importance of good marks. Students have to get good marks in their Examinations. Extramarks’ NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 help in solving the questions easily so that students can perform well in the examinations. Subject specialists prepare the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 to help students score well in their examinations. All these NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 of NCERT Books have been developed according to the preparatory needs of students. Extramarks provides important instructions for all Class 6 NCERT solutions. Students who are familiar with the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 and practise them regularly are likely to achieve high grades in the Maths exam.
Access Other Exercises of Class 6 Maths Chapter 11
Chapter 11 – Algebra Exercises  
Exercise 11.1 
11 Questions & Solutions

Exercise 11.2 
5 Questions & Solutions

Exercise 11.3 
6 Questions & Solutions

Exercise 11.4 
3 Questions & Solutions

NCERT Solutions For Class 6 Maths Chapter 11 Algebra (Ex 11.5) Exercise 11.5
The CBSE Board prescribes NCERT books for students studying in Classes 1 to 12. The Maths NCERT Book is the course book for students studying under the CBSE Board. The NCERT books are written by subject experts with a profound understanding of the subject matter. The NCERT books are tailored to help students understand and grasp concepts. In terms of obtaining a solid understanding of the basics, NCERT books have consistently proven to be of great assistance. These books provide very elaborate and structured explanations of the theory. It is imperative for students to have a thorough understanding of the theories covered in their syllabus. The NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 make it easier for students to grasp the topics that are taught to them.
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Access NCERT Solutions For Class 6 Maths Chapter 11 Algebra

Chapter 1: Know your numbers
Chapter 1 is an interesting chapter, and it deals with the number system. Numbers play a very important role in everyone’s lives. Without numbers, the value of time cannot be measured. Numbers are important for every subject, whether Science or Commerce. In addition to its importance in the workplace, Maths is also important in everyday life. The use of numbers is ubiquitous wherever people go. Be it simple counting, financial planning or determining time, distance, or weight. Numbers are a key factor in many fields and activities. Students must refer to Extramarks for the solutions of Class 6 Maths Chapter 11 Exercise 11.5.

Chapter 2: Integers
This chapter lays the framework by first introducing students to the concept of integers. It also covers the concept of successors and predecessors. Students are taught addition and subtraction on the number line and the representation of whole numbers on the number line.

Chapter 3: Playing with Numbers
This chapter introduces the concepts of factors and multiples. This chapter teaches students about LCM and HCF. It also deals with prime, composites, and coprime numbers.

Chapter 4: Basic Geometric Ideas
This chapter introduces students of Class 6 to the basic forms of geometry. Line Segments, Different Types of Lines, Triangles, Regular Polygons, and Circles are some of the topics covered in this chapter.

Chapter 5: Understanding Basic Shapes
In this chapter, deepen their understanding of basic shapes. Concepts such as edges, curves (open or closed), and planes are covered. This chapter consists of 9 exercises that help students to understand geometry better.

Chapter 6: Integers
This chapter introduces the basic concepts of integers to Class 6 students. Operations, values, and comparisons of integers are covered in this chapter. It also explains the representation of integers on the number line.

Chapter 7 Fractions
This chapter serves as the foundation for many other concepts. This chapter deals with working with fractions. Students are taught to compare proper and improper fractions. Definitions of various types of fractions are also covered in this chapter.

Chapter 8: Decimal Numbers
The topic of Chapter 8 is Decimal Numbers. Students are taught to represent numbers with whole parts and fractions.

Chapter 9: Data Processing
Under this topic, students learn how to collect data from a variety of sources and present this data in a basic pictorial form. Students are introduced to the concepts of pictograms and bar charts.

Chapter 10: Measurement
Area and perimeter are important concepts that are commonly used in everyday life.This topic helps students build a foundation for many concepts that may be taught to themin higher classes.

Chapter 11: Algebra
This topic introduces basic algebraic concepts such as variables, rules of operation for variables and constants, and expressions for variables. This is the most important chapter in Class 6 Maths. In this chapter, students learn some concepts that prove to be of great value throughout their lives. For the solutions of Class 6 Maths Chapter 11 Exercise 11.5 students must refer to the Extramarks website. It is highly recommended that students of Class 6 study with the help of the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5.

Chapter 12 Ratios and Proportions
This topic begins with the idea of proportions and moves on to explaining proportions in detail. This chapter will teach students how by division, also known as ratio, one expresses two things in comparison, and how two ratios are compared and said to be proportional if they are found to be equal. This chapter describes how to apply these concepts to solve problems.

Chapter 13: Symmetry
This chapter deals with symmetrical figures. The concept of reflection, and lines of symmetry, are important topics covered in this chapter.

Chapter 14: Practical Geometry
This chapter is concerned with geometric structures. Students in Class 6 are taught to create shapes such as circles, line segments, perpendiculars to lines, and angles.
Maths is considered one of the most important and highscoring subjects. Students are encouraged to enhance their understanding of mathematical concepts and develop their application skills. Math Class 6 Exercise 11.5 includes the basic concepts of Chapter 11 Algebra. Students can also get solutions to problems based on various concepts such as integers, fractions, decimals, algebra, geometry, and measurements, on the website of Extramarks. It is advisable that students access the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 from the Extramarks website or mobile application for their preparation.
NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercise 11.5
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Q.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.
$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}17=x+7\left(\mathrm{b}\right)(t7)>5\left(c\right)\frac{4}{2}=2\\ \left(d\right)(7\times 3)19=8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5\times 48=2x\left(f\right)x2=0\\ \left(g\right)\text{\hspace{0.17em}}2m<30\left(h\right)\text{\hspace{0.17em}}2n+1=11\left(i\right)\text{\hspace{0.17em}}7=(11\times 5)(12\times 4)\\ \left(j\right)\text{\hspace{0.17em}}7=(11\times 2)+p\left(k\right)\text{\hspace{0.17em}}20=5y\left(l\right)\text{\hspace{0.17em}}\frac{3q}{2}<5\\ \left(m\right)\text{z}+\text{12>24}\left(n\right)20(105)=3\times 5\\ \left(0\right)\text{\hspace{0.17em}}7x=5\end{array}$
Ans.
$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}17=x+7\text{, it is an equation in variable x because equal to sign is there}\text{.}\\ \left(b\right)(t7)>5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is an not equation in variable t because equal}\text{\hspace{0.17em}}\text{to sign is not there}\text{.}\\ \left(c\right)\frac{4}{2}=2,\text{it is numerical equation because there is no variable}\text{.}\\ \left(d\right)(7\times 3)19=8,\text{it is numerical equation because there is}\text{\hspace{0.17em}}\text{no variable}\text{.}\\ \left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}5\times 48=2x,\text{It is an equation in variable x because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(f\right)x2=0,\text{\hspace{0.17em}}\text{It is an equation in variable x because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(g\right)\text{\hspace{0.17em}}2m<30,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(h\right)\text{\hspace{0.17em}}2n+1=11,\text{\hspace{0.17em}}\text{It is an equation in variable n because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(i\right)\text{\hspace{0.17em}}7=(11\times 5)(12\times 4),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is numerical equation because there is no variable}\text{.}\\ \left(j\right)\text{\hspace{0.17em}}7=(11\times 2)+p,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable p because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(k\right)\text{\hspace{0.17em}}20=5y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable y because equal}\text{\hspace{0.17em}}\text{to sign is there}\text{.}\\ \left(l\right)\text{\hspace{0.17em}}\frac{3q}{2}<5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(m\right)\text{z}+\text{12>24,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is not an equation because equal to sign is not there}\text{.}\\ \left(n\right)20(105)=3\times 5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it is numerical equation because there is no variable}\text{.}\\ \left(o\right)\text{\hspace{0.17em}}7x=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is an equation in variable x because equal to sign is there}\text{.}\end{array}$
Q.2 Complete the entries in the third column of the table.
S.No.  Equation  Value of variable  Equation satisfied
Yes/No 
(a)  10y = 80  y = 10  
(b)  10y = 80  y = 8  
(c)  10y = 80  y = 5  
(d)  4l = 20  l = 20  
(e)  4l = 20  l = 80  
(f)  4l = 20  l = 5  
(g)  b + 5 = 9  b = 5  
(h)  b + 5 = 9  b = 9  
(i)  b + 5 = 9  b = 4  
(j)  h – 8 = 5  h = 13  
(k)  h – 8 = 5  h = 8  
(l)  h – 8 = 5  h = 0  
(m)  p + 3 = 1  p = 3  
(n)  p + 3 = 1  p = 1  
(o)  p + 3 = 1  p = 0  
(p)  p + 3 = 1  p = – 1  
(q)  p + 3 = 1  p = – 2 
Ans.
S.No.  Equation  Value of variable  Equation satisfied
Yes/No 
(a)  10y = 80  y = 10  No 
(b)  10y = 80  y = 8  yes 
(c)  10y = 80  y = 5  No 
(d)  4l = 20  l = 20  No 
(e)  4l = 20  l = 80  No 
(f)  4l = 20  l = 5  Yes 
(g)  b + 5 = 9  b = 5  No 
(h)  b + 5 = 9  b = 9  No 
(i)  b + 5 = 9  b = 4  Yes 
(j)  h – 8 = 5  h = 13  Yes 
(k)  h – 8 = 5  h = 8  No 
(l)  h – 8 = 5  h = 0  No 
(m)  p + 3 = 1  p = 3  No 
(n)  p + 3 = 1  p = 1  No 
(o)  p + 3 = 1  p = 0  No 
(p)  p + 3 = 1  p = – 1  No 
(q)  p + 3 = 1  p = – 2  Yes 
Q.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.
$\begin{array}{l}\left(\mathrm{a}\right)\text{\hspace{0.17em}}5\mathrm{m}=60(10,5,12,15)\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\mathrm{n}+12=20(12,8,20,0)\\ \left(\mathrm{c}\right)\mathrm{p}5=5(0,10,5,5)\\ \left(\mathrm{d}\right)\frac{\mathrm{q}}{2}=7(7,2,10,14)\\ \left(\mathrm{e}\right)\mathrm{r}4=0(4,4,8,0)\\ \left(\mathrm{f}\right)\mathrm{x}+4=2(2,0,2,4)\end{array}$
Ans.
$\begin{array}{l}\left(\text{a}\right)\text{We have},\text{5m}=\text{6}0\dots \left(\text{i}\right)\\ \text{Putting m}=\text{1}0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(\text{1}0\right)\\ \mathrm{}=\text{5}0\ne 60\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{10 is not the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(5\right)\\ =\text{}25\ne 60\\ \Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{5 is not the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\text{\hspace{0.17em}Putting m}=12\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{5m}=\text{5}\left(12\right)\\ =60\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{12 is the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=12\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \mathrm{}\text{5m}=\text{5}\left(12\right)\\ =\text{}60\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{12 is the solution of equation 5m}=\text{6}0.\\ \mathrm{Again},\\ \text{Putting m}=15\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \mathrm{}\text{5m}=\text{5}\left(15\right)\\ =\text{}75\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{m}=\text{15 is not the solution of equation 5m}=\text{6}0.\\ \left(\text{b}\right)\text{We have},\text{n}+\text{12}=\text{2}0\dots \left(\text{i}\right)\\ \text{Putting n}=\text{1}2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{12}+\text{12}\\ =24\ne 20\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{12 is not the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=8\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{8}+\text{12}\\ =20\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{8 is the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=20\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{20}+\text{12}\\ =32\ne 20\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{20 is not the solution of equation n}+\text{12}=\text{2}0.\\ \mathrm{Again},\\ \text{Putting n}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}n}+\text{12}=\text{0}+\text{12}\\ =12\ne 20\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{n}=\text{0 is not the solution of equation n}+\text{12}=\text{2}0.\\ \left(\text{c}\right)\text{We have},\text{p}\text{5}=5\dots \left(\text{i}\right)\\ \text{Putting p}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}\text{5}=\text{0}\text{5}\\ =5\ne 5\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{0 is not the solution of equation p}\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=10\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}\text{5}=1\text{0}\text{5}\\ =5\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=1\text{0 is the solution of equation p}\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}\text{5}=\text{5}\text{5}\\ =0\ne 5\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{5 is not the solution of equation p}\text{5}=5.\\ \mathrm{Again},\\ \text{Putting p}=\text{\hspace{0.17em}}5\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}p}\text{5}=\text{5}\text{5}\\ =10\ne 5\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{p}=\text{\hspace{0.17em}5 is not the solution of equation p}\text{5}=5.\\ \left(\text{d}\right)\text{We have},\text{}\frac{\mathrm{q}}{2}=7\dots \left(\text{i}\right)\\ \text{Putting q}=7\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{7}{2}\\ =3.5\ne 7\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{7 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{2}{2}\\ =1\ne 7\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{2 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=10\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{10}{2}\\ =5\ne 7\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{10 is not the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \mathrm{Again},\\ \text{Putting q}=14\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{q}}{2}=\text{\hspace{0.17em}}\frac{14}{2}\\ =7\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{q}=\text{14 is the solution of equation}\frac{\mathrm{q}}{2}=7.\\ \left(\text{e}\right)\text{We have},\text{}\mathrm{r}4=0\dots \left(\text{i}\right)\\ \text{Putting r}=4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}4=44\\ =0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{4 is the solution of equation}\mathrm{r}4=0.\\ \mathrm{Again},\\ \text{Putting r}=\text{\hspace{0.17em}}4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}4=\text{\hspace{0.17em}}44\\ =\text{\hspace{0.17em}}8\ne 0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{\hspace{0.17em}4 is not the solution of equation}\mathrm{r}4=0.\\ \mathrm{Again},\\ \text{Putting r}=8\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}4=84\\ =4\ne 0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{r}=\text{8 is not the solution of equation}\mathrm{r}4=0.\\ \mathrm{Again},\\ \text{Putting r}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}4=04\\ =\text{\hspace{0.17em}}4\ne 0\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{r}=\text{0 is not the solution of equation}\mathrm{r}4=0.\\ \left(\text{f}\right)\text{We have},\text{}\mathrm{x}+4=2\dots \left(\text{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=\text{\hspace{0.17em}}2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=\text{\hspace{0.17em}}2+4\\ =2\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=\text{\hspace{0.17em}}2\text{is the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=0\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=0+4\\ =4\ne 2\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=0\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=2\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=2+4\\ =6\ne 2\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{x}=2\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\\ \mathrm{Again},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Putting x}=4\text{in L}.\text{H}.\text{S}.\text{of equation}\left(\text{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+4=4+4\\ =8\ne 2\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{x}=4\text{is not the solution of equation x}+\text{4}=\text{\hspace{0.17em}\hspace{0.17em}}2.\end{array}$
Q.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 
(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 
(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  –  –  – 
z/3  $2\frac{2}{3}$  3  $3\frac{1}{3}$ 
(d) Complete the table and find the solution to the equation m – 7 = 3.
m  5  6  7  8  9  10  11  12  13  –  –  – 
m7 
Ans.
(a) Table having solution of m+10 = 16, is given below
m  m+10 
1  1+10 = 11 
2  2 + 10 = 12 
3  3 + 10 = 13 
4  4 + 10 = 14 
5  5 + 10 = 15 
6  6 + 10 = 16 
7  7 + 10 = 17 
8  8 + 10 = 18 
9  9 + 10 = 19 
10  10 + 10 = 20 
By inspection of the table, the solution to the equation m + 10 = 16 is m = 6.
(b) Table having solution of 5t = 35, is given below:
t  5t 
3  5(3) = 15 
4  5(4) = 20 
5  5(5) = 25 
6  5(6) = 30 
7  5(7) = 35 
8  5(8) = 40 
9  5(9) = 45 
10  5(10) = 50 
11  5(11) = 55 
12  5(12) = 60 
By inspection of the table, the solution to the equation 5t = 35 is t = 7.
(c) Table having solution of z/3 = 4, is given below:
z  z/3 
8  \frac{\text{8}}{\text{3}}=2\frac{2}{3} 
9  9/3 = 3 
10  \frac{\text{10}}{\text{3}}=3\frac{1}{3} 
11  \frac{\text{11}}{\text{3}}=3\frac{2}{3} 
12  12/3 = 4 
13  \frac{\text{13}}{\text{3}}=4\frac{1}{3} 
14  \frac{\text{14}}{\text{3}}=4\frac{2}{3} 
15  15/3 = 5 
16  \frac{\text{16}}{\text{3}}=5\frac{1}{3} 
17  \frac{\text{17}}{\text{3}}=5\frac{2}{3} 
By inspection of the table, the solution to the equation z/3 = 4 is z = 12.
(d) Table having solution of m – 7 = 3, is given below:
m  m7 
5  5 – 7 = – 2 
6  6 – 7 = – 1 
7  7 – 7 = 0 
8  8 – 7 = 1 
9  9 – 7 = 2 
10  10 – 7 = 3 
11  11 – 7 = 4 
12  12 – 7 = 5 
13  13 – 7 = 6 
By inspection of the table, the solution to the equation m – 7 = 3 is m = 10.
Q.5 Solve the following riddles, you may yourself construct such riddles. Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!
(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!
(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!
(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
Ans.
(i) Let value of me be x.
Number of corners in a square = 4
Thrice of 4 = 12
Then, x + 12 = 24
x = 24 – 12
= 22
Therefore, the value of me is 22.
(ii) Let value of me be x.
Upcount on Sunday from me = x + 1
Upcount on Monday from me = x + 2
Upcount on Tuesday from me = x + 3
Upcount on Wednesday from me = x + 4
Upcount on Thursday from me = x + 5
Upcount on Friday from me = x + 6
Upcount on Saturday from me = x + 7
In this way, x + 7 = 23
x = 16
Therefore, the value of me is 16.
(iii) Let the special number be x
Number of player in cricket team
= 11
Subtracted number = 6
Then, according to condition,
x – 6 = 11
(iv) Let the value of me be x.
Then, according to condition,
22 – x = x
22 = 2x
x = 11
Therefore, the special number is 11.
FAQs (Frequently Asked Questions)
1. Are the Extramarks NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 easily accessible for practice?
Students can access exercisebased NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 from the Extramarks mobile application for their preparation. The same can be accessed by going to Extramarks’s website and then going to the solutions page, where students can find links for specific exercises on the website.
2. How helpful are the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5?
Most students need to know how to deal with important problems so that they can perform well in important subjects such as Maths. But most of the students find it difficult to understand such concepts, and without mastering certain skills, it may be difficult to score high marks. It’s one of the core subjects that are taught to students from an early age. The mathematical concepts should be explained to students in a way they can understand easily. Students must refer to the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 and follow NCERT textbook and the latest solutions to thoroughly understand Math Exercise 11.5 Class 6.
3. How many questions are covered in the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5?
The NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5 consists of a total of 5 questions from the NCERT textbook of Maths.
4. How can students use the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5?
It is advisable that students should ensure that they have a thorough understanding of theoretical concepts before practising the NCERT Solutions Class 6 Maths Chapter 11 Exercise 11.5. It is recommended that students move on to solving new questions with different concepts after they have gained a thorough understanding of a particular topic.