# NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (Ex 4.1) Exercise 4.1

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## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (EX 4.1) Exercise 4.1

To assist them with their homework and exam preparation, students are required to read through the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 offered by Extramarks experts. Students can use the detailed explanations and step-by-step NCERT Class 7 Maths Chapter 4 Exercise 4.1 provided by Extramarks to help them learn the ideas in Chapter – Simple Equations.

To make sure that students have a thorough comprehension, Extramarks’ academics have carefully curated the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1. Students have access to more than 500 questions to prepare for examinations and ace them. Practise questions are provided for each topic covered in NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 to help students completely understand the ideas.

The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 offered by Extramarks experts cover a wide range of subjects, including the fundamentals of equations and applications of equations. Each topic offers more than 100 practise questions drawn from more than 50 different sources.

Students can also learn the ideas from entertaining and engaging 3D films and explainers on Extramarks. Additionally, they can assess their comprehension by creating tests that suit their preferences.

## Access NCERT Solutions for Class 7 Mathematics Chapter 4- Simple Equations

The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 are available for both online and offline access on the Extramarks website and mobile application.

Below is a brief overview of all the important topics covered in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1:

• An equation is an assertion of equality including one or more variables (literals).
• A linear equation in one variable is one that only uses one literal number (variable) with the highest power one.
• By using the opposite sign, we can move a term from one side of an equation to the other. Transposition is the term for this action.

Students can efficiently complete NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1, based on Simple Equations, and complete the entire curriculum. The team of subject-matter specialists at Extramarks created each of these NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 keeping in mind their complex needs. The step-by-step NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 provided by Extramarks will help the students better understand the ideas.

The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 Simple Equations cover the basics of algebra, starting with the definition of variables, expressions, and equations. Students will be able to comprehend that the equation corresponds to any condition based on the variable with the aid of the offered NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1. Additionally, the values of the two expressions on either side of the equation ought to be identical. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 consists of six questions.

The first question in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 is a little different because it requires writing the equation in statement form, while the final one requires framing equations. The students will find the first four of them simple.

Students should keep in mind that an equation is not deemed to exist if it contains any other signs, such as greater than or less than, between the LHS (Left Hand Side) and RHS (Right Hand Side), in addition to the equal sign.

Students should review the principles of Algebraic Words such as Variables, Expressions, Equations, etc., because the activities in the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 based on Simple Equations include the basics of algebra. The formation of expressions is illustrated throughout the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 with simple real-world examples, that will help students understand the meaning of all the algebraic words.

Students are recommended to carefully read each and every paragraph in the book in order to complete the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 since it properly explains the idea and the reasoning behind each question with examples.  Students can confidently proceed to the activities after practising the NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.1.

## Chapter 4 – Simple Equations

The topics, materials, questions, and solutions in Extramarks’ NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 have been carefully considered. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 are self-explanatory, therefore, students should not worry about them. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 from Extramarks has provided students with a variety of answers to their Mathematical problems. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 are created with an eye towards the exam structure and syllabus. Clicking on the links for Class 7 Maths Chapter 4 will take students to a PDF download of the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1.

Students have the opportunity to learn about the fundamentals of simple equations with the help of NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1. Simple Equations are not always as easy to understand as they may first appear to be. As a result, in addition to teaching students how to create equations, this NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 will also assist them in understanding what exactly they are. Additionally, after finishing these NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1, students will be able to turn assertions into equations. The NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 from Extramarks will be a helpful resource for students to understand the chapter effectively because of this very reason.

Below is an overview of the chapter:

Students are given the correct equation principles in this segment. Extramarks’  NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 assists in offering a variety of examples that enable readers to acquire the clearest understanding possible of how to use equations in the best way. Without any anxiety, the students can easily get all the necessary material that will help them perform well on their tests. Extramarks experts have developed the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 using the most recent CBSE curriculum. Using step-by-step NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 makes it simple to comprehend the equation most practically.

Both sides of an equation should be equal. When LHS and RHS are switched, the same number is added to both sides, the same number is removed from both sides, the same amount is multiplied from both sides of the RHS, and the LHS is halved by the same number,same number. The equation does not change. In Ex. 4.3, students will learn about transferring a number from one side of the equation to another as opposed to adding and subtracting it from both sides. Students will discover how to apply simple equations to the practical questions in Exercise 4.4.

In this section, the students will analyse the universe of straightforward equations and their uses.. In a similar vein, the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 provides real-life examples from the classroom that will aid students in developing a more thorough understanding.

### Setting up of an Equation

Examples in this subtopic will aid students in formulating equations. This is accomplished by providing students with real-world examples of girls with the names Ameena and Sara. It also includes NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 to the equations that were presented to the students. For easy comprehension, these NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 are provided in a very understandable manner.

Students frequently struggle with understanding how to properly build up an equation. Due to a lack of sufficient solutions, NCERT textbooks fall short of offering an in-depth comprehension of the topic. With the help of NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.1, the professionals at Extramarks can deliver the clearest method possible without causing any uncertainty.Therefore, it is feasible to comprehend Simple Equations where, with the help of the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 it has been made possible to acquire the proper knowledge in an ideal way.

### Review of what we know

This section will teach the specifics of an equation while taking into account what students have learned thus far. It will draw from the knowledge the student gained previously in  NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1. For instance, it has been widely explored here what constitutes a variable. Next, the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 continues to explain more about expressions. Here, examples are presented to help clarify the idea.

### What Equation is?

Simply put, this part of the NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1 will summarise simple equations and wrap up the chapter.It will go over what an equality sign is and how important it is.The learner will also be taught RHS and LHS here. For the sake of the student’s understanding, all of this will be followed by equations and their solutions.

A mathematical statement involving a variable in which the values of two expressions on either side of the equality sign should be equal is called an equation. The links between two expressions on opposite sides of the sign are depicted by the straightforward equation. One variable makes it up.LHS (left-hand side) and RHS (right-hand side) are the names of the expressions on the two sides. A constant has a fixed value, but a variable has a different numeric value. Students will discover what an equation means in Ex 4.1. Any equation has an equality sign that indicates the value of the expression on the left side of the equation is equal to the value on the right side of the equation. Calculation skills are taught in NCERT Solutions For Class 7 Maths Chapter 4 Exercise 4.1.

### NCERT Solutions for Class 7

Students may find Mathematics to be enjoyable with Extramarks’ NCERT Solutions for Class 7 Mathematics. Mathematics practise in Class 7 will help students do better in Class 8, then in Class 9, etc., laying the groundwork for a solid foundation in Mathematics. Experts at Extramarks have created step-by-step solutions with thorough justifications. To assist students in performing well on the Math exam, the solutions are organised for practice.

Students may effortlessly tackle challenging questions with the aid of the NCERT Solutions for Class 7 Mathematics offered by Extramarks. It is advised that students practise the NCERT Solutions, which are offered in both video and PDF versions. Additionally, it makes basic Mathematical ideas simple for students to comprehend. For students taking the CBSE Class 7 Mathematics exam, the NCERT Solutions for Class 7 Mathematics are a dependable study tool. To enhance students’ academic performance and knowledge, Extramarks specialists have answered the questions from the Mathematics NCERT textbooks.

On the Extramarks website and mobile application, students may find the NCERT Solutions for Class 7 Mathematics, which include all the problems organised by chapter for thorough understanding. The knowledgeable team at Extramarks has created the answers in a clear way that aids students in resolving issues quickly. Students can quickly answer a large number of questions from the NCERT textbook with these NCERT Solutions for Class 7 Mathematics.

Students should focus on chapters like comparing amounts, lines and angles, congruence of triangles, and other topics because they can earn high marks in the 15 chapters of the NCERT Solutions for Class 7 Mathematics.

The following chapters are included in Extramarks’ NCERT Solutions for Class 7 Maths:

• Chapter 1 – Integers
• Chapter 2 – Fractions and Decimals
• Chapter 3 – Data Handling
• Chapter 4 – Simple Equations
• Chapter 5 – Lines and Angles
• Chapter 6 – The Triangles and its Properties
• Chapter 7 – Congruence of Triangles
• Chapter 8 – Comparing Quantities
• Chapter 9 – Rational Numbers
• Chapter 10 – Practical Geometry
• Chapter 11 – Perimeter and Area
• Chapter 12 – Algebraic Expressions
• Chapter 13 – Exponents and Powers
• Chapter 14 – Symmetry
• Chapter 15 – Visualising Solid Shapes

Q.1

$\begin{array}{l}\mathrm{Complete}\mathrm{the}\mathrm{last}\mathrm{column}\mathrm{of}\mathrm{the}\mathrm{table}.\end{array}$

 $S.No.$ $Equation$ $Value$ $\begin{array}{l}Say,\text{\hspace{0.17em}}whether\text{}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}equation\\ is\text{\hspace{0.17em}}satisfied.\left(Yes/No\right)\end{array}$ $\left(i\right)$ $x+3=0$ $x=3$ $\left(ii\right)$ $x+3=0$ $x=0$ $\left(iii\right)$ $x+3=0$ $x=–3$ $\left(iv\right)$ $x–7=1$ $x=7$ $\left(v\right)$ $x–7=1$ $x=8$ $\left(vi\right)$ $5x=25$ $x=0$ $\left(vii\right)$ $5x=25$ $x=5$ (viii) $5x=25$ $x=–5$ $\left(ix\right)$ $\frac{m}{3}=2$ $m=–6$ $\left(x\right)$ $\frac{m}{3}=2$ $m=0$ $\left(xi\right)$ $\frac{m}{3}=2$ $m=6$

Ans.

 $S.No.$ $Equation$ $Value$ $\begin{array}{l}Say,\text{\hspace{0.17em}}whether\text{}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}equation\\ is\text{\hspace{0.17em}}satisfied.\left(Yes/No\right)\end{array}$ $\left(i\right)$ $x+3=0$ $x=3$ No $\left(ii\right)$ $x+3=0$ $x=0$ No $\left(iii\right)$ $x+3=0$ $x=–3$ Yes $\left(iv\right)$ $x–7=1$ $x=7$ No $\left(v\right)$ $x–7=1$ $x=8$ Yes $\left(vi\right)$ $5x=25$ $x=0$ No $\left(vii\right)$ $5x=25$ $x=5$ Yes (viii) $5x=25$ $x=–5$ No $\left(ix\right)$ $\frac{m}{3}=2$ $m=–6$ No $\left(x\right)$ $\frac{m}{3}=2$ $m=0$ No $\left(xi\right)$ $\frac{m}{3}=2$ $m=6$ Yes

Q.2

$\begin{array}{l}\mathrm{Check}\mathrm{whether}\mathrm{the}\mathrm{value}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{brackets}\mathrm{is}\mathrm{a}\mathrm{solution}\\ \mathrm{to}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{or}\mathrm{not}:\\ \left(\mathrm{a}\right)\mathrm{n}+5=19\left(\mathrm{n}=1\right)\left(\mathrm{b}\right)7\mathrm{n}+5=19\left(\mathrm{n}=–2\right)\\ \left(\mathrm{c}\right)7\mathrm{n}+5=19\left(\mathrm{n}=2\right)\left(\mathrm{d}\right)4\mathrm{p}–3=13\left(\mathrm{p}=1\right)\\ \left(\mathrm{e}\right)4\mathrm{p}–3=13\left(\mathrm{p}=–4\right)\left(\mathrm{f}\right)4\mathrm{p}–3=13\left(\mathrm{p}=0\right)\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{n}+\text{5}=\text{19}\left(\text{n}=\text{1}\right)\\ \text{Put n}=\text{1 in L}\text{.H}\text{.S to get}\\ \text{n+5}=1+5=6\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{b}\right)\text{7n}+\text{5}=\text{19}\left(\text{n}=-\text{2}\right)\\ \text{Put n}=-\text{2 in L}\text{.H}\text{.S to get}\\ \text{7n+5}=7\left(-2\right)+5=-14+5=-9\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{c}\right)\text{7n}+\text{5}=\text{19}\left(\text{n}=\text{2}\right)\\ \text{Put n}=\text{2 in L}\text{.H}\text{.S to get}\\ \text{7n+5}=7\left(2\right)+5=14+5=19=\text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{d}\right)\text{4p}-\text{3}=\text{13}\left(\text{p}=\text{1}\right)\\ \text{Put p}=1\text{in L}\text{.H}\text{.S to get}\\ \text{4p}-3=4\left(1\right)-3=4-3=1\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{e}\right)\text{4p}-\text{3}=\text{13}\left(\text{p}=-\text{4}\right)\\ \text{Put p}=-\text{4 in L}\text{.H}\text{.S to get}\\ \text{4p}-3=4\left(-4\right)-3=-16-3=-19\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{f}\right)\text{4p}-\text{3}=\text{13}\left(\text{p}=0\right)\\ \text{Put p}=\text{0 in L}\text{.H}\text{.S to get}\\ \text{4p}-3=4\left(0\right)-3=0-3=-3\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\end{array}$

Q.3

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}\mathrm{by}\mathrm{trial}\mathrm{and}\mathrm{error}\mathrm{method}:\\ \left(\mathrm{i}\right)5\mathrm{p}+2=17\left(\mathrm{ii}\right)3\mathrm{m}–14=4\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{5p}+\text{2}=\text{17}\\ \text{Put p}=\text{1 to L}\text{.H}\text{.S to get}\\ \text{5}\left(1\right)\text{+2}=\text{5+2}=\text{7}\ne \text{R}\text{.H}\text{.S}\\ \text{Put p}=\text{2 to L}\text{.H}\text{.S to get}\\ \text{5}\left(2\right)\text{+2}=\text{10+2}=\text{12}\ne \text{R}\text{.H}\text{.S}\\ \text{Put p}=\text{3 to L}\text{.H}\text{.S to get}\\ \text{5}\left(3\right)\text{+2}=\text{15+2}=\text{17}=\text{R}\text{.H}\text{.S}\\ \text{Thus, p}=\text{3 is a solution to the given equation}\\ \left(\text{ii}\right)\text{3m}-\text{14}=\text{4}\\ \text{Put m}=\text{4 in L}\text{.H}\text{.S to get}\\ \text{3}\left(4\right)-14=12-14=-2\ne \text{R}\text{.H}\text{.S}\\ \text{Put m}=\text{5 in L}\text{.H}\text{.S to get}\\ \text{3}\left(5\right)-14=15-14=-1\ne \text{R}\text{.H}\text{.S}\\ \text{Put m}=\text{6 in L}\text{.H}\text{.S to get}\\ \text{3}\left(4\right)-14=18-14=4=\text{R}\text{.H}\text{.S}\\ \text{Thus, m}=\text{6 is a solution to the given equation}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Write}\mathrm{equations}\mathrm{for}\mathrm{the}\mathrm{following}\mathrm{statements}:\\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{numbers}\mathrm{x}\mathrm{and}4\mathrm{is}9.\\ \left(\mathrm{ii}\right)\mathrm{The}\mathrm{difference}\mathrm{between}\mathrm{y}\mathrm{and}2\mathrm{is}8.\\ \left(\mathrm{iii}\right)\mathrm{Ten}\mathrm{times}\mathrm{a}\mathrm{is}70.\\ \left(\mathrm{iv}\right)\mathrm{The}\mathrm{number}\mathrm{b}\mathrm{divided}\mathrm{by}5\mathrm{gives}6.\\ \left(\mathrm{v}\right)\mathrm{Three}\mathrm{fourth}\mathrm{of}\mathrm{t}\mathrm{is}15.\\ \left(\mathrm{vi}\right)\mathrm{Seven}\mathrm{times}\mathrm{m}\mathrm{plus}7\mathrm{gets}\mathrm{you}77.\\ \left(\mathrm{vii}\right)\mathrm{One}\mathrm{fourth}\mathrm{of}\mathrm{a}\mathrm{number}\mathrm{minus}4\mathrm{gives}4.\\ \left(\mathrm{viii}\right)\mathrm{If}\mathrm{you}\mathrm{take}\mathrm{away}6\mathrm{from}6\mathrm{times}\mathrm{y},\mathrm{you}\mathrm{get}60.\\ \left(\mathrm{ix}\right)\mathrm{If}\mathrm{you}\mathrm{add}3\mathrm{toone}\mathrm{third}\mathrm{of}\mathrm{z},\mathrm{you}\mathrm{get}30.\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.17em}}x+4=9\\ \left(\text{ii)}\text{\hspace{0.17em}}y-2=8\\ \text{(iii)}\text{\hspace{0.17em}}10a=70\\ \text{(iv)}\text{\hspace{0.17em}}\frac{b}{5}=6\\ \text{(v)}\text{\hspace{0.17em}}\frac{3}{4}t=15\\ \left(\text{vi}\right)\text{\hspace{0.17em}}7m+7=77\\ \left(\text{vii}\right)\text{\hspace{0.17em}}\frac{x}{4}-4=4\\ \left(\text{viii}\right)\text{\hspace{0.17em}}6y-6=60\\ \left(\text{ix}\right)\text{\hspace{0.17em}}\frac{z}{3}+3=30\end{array}$

Q.5

$\begin{array}{l}\mathrm{Write}\mathrm{the}\mathrm{following}\mathrm{equations}\mathrm{in}\mathrm{statement}\mathrm{forms}:\\ \left(\mathrm{i}\right)\mathrm{ }\mathrm{p}+4=15\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{m}–7=3\\ \left(\mathrm{iii}\right)\mathrm{ }2\mathrm{m}=7\\ \left(\mathrm{iv}\right)\frac{\mathrm{m}}{5}=3\\ \left(\mathrm{v}\right) \frac{3}{5}\mathrm{m}=6\\ \left(\mathrm{vi}\right)3\mathrm{p}+4=25\\ \left(\mathrm{vii}\right)4\mathrm{p}–2=18\\ \left(\mathrm{viii}\right)\frac{\mathrm{p}}{2}+2=8\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{The sum of p and 4 is 15}\text{.}\\ \text{(ii) 7 subtracted from m is 3}\text{.}\\ \text{(iii) Twice of a number m is 7}\text{.}\\ \text{(iv) One-fifth of m is 3}\text{.}\\ \text{(v) Three-fifth of m is 6}\text{.}\\ \text{(vi) Three times of a number p, when add to 4 gives 25}\text{.}\\ \text{(vii) When 2 is subtracted from four times of a number p,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{it gives 18}\text{.}\\ \text{(viii) When 2 is added to half of a number p, it gives 8}\text{.}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Set}\mathrm{up}\mathrm{an}\mathrm{equation}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{cases}:\\ \left(\mathrm{i}\right)\mathrm{Irfan}\mathrm{says}\mathrm{that}\mathrm{he}\mathrm{has}7\mathrm{marbles}\mathrm{more}\mathrm{than}\mathrm{five}\\ \mathrm{times}\mathrm{the}\mathrm{marbles}\mathrm{Parmit}\mathrm{has}.\mathrm{Irfan}\mathrm{has}37\mathrm{marbles}.\\ \left(\mathrm{Take}\mathrm{m}\mathrm{to}\mathrm{be}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{Parmit}’\mathrm{s}\mathrm{marbles}.\\ \left(\mathrm{ii}\right)\mathrm{Laxmi}’\mathrm{s}\mathrm{father}\mathrm{is}49\mathrm{years}\mathrm{old}.\mathrm{He}\mathrm{is}4\mathrm{years}\mathrm{older}\mathrm{than}\\ \mathrm{three}\mathrm{times}\mathrm{Laxmi}’\mathrm{s}\mathrm{age}.\left(\mathrm{Take}\mathrm{Laxmi}’\mathrm{s}\mathrm{age}\mathrm{to}\mathrm{be}\mathrm{y}\mathrm{years}.\right)\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{teacher}\mathrm{tells}\mathrm{the}\mathrm{class}\mathrm{that}\mathrm{the}\mathrm{highest}\mathrm{marks}\\ \mathrm{obtained}\mathrm{by}\mathrm{a}\mathrm{student}\mathrm{in}\mathrm{her}\mathrm{class}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{lowest}\\ \mathrm{marks}\mathrm{plus}7.\mathrm{The}\mathrm{highest}\mathrm{score}\mathrm{is}87.\\ \left(\mathrm{Take}\mathrm{the}\mathrm{lowest}\mathrm{score}\mathrm{to}\mathrm{be}\mathrm{l}\right).\\ \left(\mathrm{iv}\right)\mathrm{In}\mathrm{an}\mathrm{isosceles}\mathrm{triangle},\mathrm{the}\mathrm{vertex}\mathrm{angle}\mathrm{is}\mathrm{twice}\mathrm{either}\\ \mathrm{base}\mathrm{angle}.\left(\mathrm{Let}\mathrm{the}\mathrm{base}\mathrm{angle}\mathrm{be}\mathrm{b}\mathrm{in}\mathrm{degrees}.\mathrm{Remember}\\ \mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\mathrm{degrees}\right).\end{array}$

Ans.

$\begin{array}{l}\text{(i) Let Parmit has}m\text{marbles}\text{.}\\ \text{Then, according to the question, we have}\\ \text{5}×\text{Number of marbles Parmit has +7}=\text{Number of marbles}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{Irfan has}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}×m+\text{7}=\text{37}\\ \text{So, we get}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\overline{)\text{5}m+\text{7}=\text{37}}\\ \text{(ii) Let Laxmi be}y\text{years old}\\ \text{Then, according to the question, we have}\\ \text{3}×\text{Laxmi’s age}+\text{4}=\text{Laxmi’s father age}\\ \text{3}×y\text{+4}=\text{49}\\ \overline{)\text{3}y+\text{4}=\text{49}}\\ \text{(iii) Let the lowest marks be}l\text{.}\\ \text{Then, according to the question, we have}\\ 2×\text{lowest marks}+\text{7}=\text{Highest marks}\\ \text{2}×l+\text{7}=\text{87}\\ \overline{)2l+7=87}\\ \text{(iv) An isoceles triangle has two angles equal}\text{.}\\ \text{Let the base angle be}b\text{.}\\ \text{Then, according to the question, we have}\\ b+b+2b=\text{180}°\\ \overline{)4b=180°}\end{array}$