# NCERT Solutions Class 7 Maths Chapter 4

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

The NCERT Solutions for Class 7 Mathematics Chapter 4 by Extramarks is a compilation of detailed step-by-step solutions to all exercises included in this chapter.

Students should practise all the exercise questions to get in-depth knowledge about the topics. The solutions are crafted by the subject matter experts, who have framed solutions in a systematic and organised manner  which is easy to understand. Students who refer to these materials will be able to prepare confidently  for their exams and achieve desired  results.

## NCERT Solutions for Class 7 Mathematics Chapter 4 Simple equations

### NCERT Solutions for Class 7 Mathematics Chapter 4

Chapter 4 of Class 7 Mathematics is on simple equations divided into five major sections. It is one of the most important chapters in Class 7 Mathematics, as it brushes-up  basic concepts of algebraic equations.

Students are advised to go through the chapter to get a clear understanding of the concepts in simple equations. The solutions to the questions in this chapter provided by Extramarks will help students to clarify the concepts and they will be able to  solve any questions in the term tests and exams confidently.

Following are the important topics covered under NCERT Class 7 Mathematics Chapter 4.

1. Stepping up of an equation
2. Review of what we know
3. What Equation is?
4. Solving an equation
5. More Equations
6.  From Solution to Equation
7. Application of Simple Equations to practical situations

### NCERT Solutions for Class 7 Mathematics Chapter 4 Exercises

The total number of questions in each of the chapter’s exercises are given in the table below.

 Chapter 4 Simple Equations Exercise 4.1 6   questions & Answers Exercise 4.2 4  Questions & Answers Exercise 4.3 4 Questions & Answers Exercise 4.4 4 Questions & Answers

### Facts

• A variable takes on different numerical values whereas a constant has a fixed value.
• An equation is a statement of a variable in which two expressions of the variable should have equal value.
• An equation remains unchanged if its LHS and RHS are interchanged.
• Transposing a number means moving it to the other side.
• The equations remain unchanged when we:
• Add the same number to both sides.
• Subtract the same number from both sides.
• Multiply  and divide both sides by the same number.
• When we transpose a number from one side of the equation to the other its sign changes

### Variable

A variable does not have a fixed value. The numerical value of the variable changes. These variables are denoted by letters of the alphabet such as l, m, n, p, q, r, s, t, u, v, w, x, y, z, etc. Expressions are formed when we perform operations such as addition, subtraction, multiplication, and division on variables.

• The value of an expression depends upon the chosen value of the variable. If there is only one term in an expression then it is called a monomial expression.
• If there are two terms in an expression then it is called a binomial expression.
• If there are three terms in an expression then it is called a trinomial expression.
• A polynomial expression is an expression that has four terms.

Note: A polynomial expression can have many terms but none of the terms can have a negative exponent for any variable.

### An Equation

An equation is a mathematical statement on a variable where two expressions on either side of the equal sign should have equal value. At least one of the expressions must contain the variable.

Note: An equation does not change when the expression on the left-hand side or the right-hand side is interchanged.

In an equation, there is always an equality sign between two expressions.

Example: Write the following statements in the form of equations.

1. The difference of five times x and 11 is 28.
2. One-fourth of a number minus 8 is 18.

Solution:

1. We have five times x that is 5x

The difference of 5x-11 is 5x-11

5x-11=28

Thus, the required equation is 5x-11=28

1. Let the number be x

One-fourth of x is ¼(x)

Now, one-fourth of x minus 8 is 1/4(x) – 8

Thus, the required equation is ¼(x) – 8=18

Let us see one more example which will help you with Exercise 4.1 of NCERT solutions chapter 4

Example: Write a statement for the equation 2x-5=15

Solution: 2x-5=15

Taking away 5 from twice a number is 15

Solving an Equation

We use this principle when we solve an equation. The equality sign between the LHS and RHS corresponds to the horizontal beam of the balance.

An equation remains undisturbed or unchanged:

1. If LHS and RHS are interchanged.
2. To both the sides, if the same number is added
3. From both sides if the same number is subtracted.
4. When both LHS and RHS are multiplied by the same number
5. When both LHS and RHS are divided by the same number

To understand the concept better, let us try to solve an example. This will help you with exercise 4.2 of NCERT Solutions Chapter 4.

Example: Solve 5x-3=12

Adding 3 to both sides, we get

5x-3+3=12+3

5x=15

Dividing both sides by 5, we get 5x/5=15/5

x=3, which is the required solution.

Note: For checking  the answer, we substitute the value of the variable in the given equation

i.e., L.H.S = (5*3)-3= 15-3= 12= R.H.S

or L.H.S = R.H.S

Example: ½(x) + 5= 65

Subtracting 5 from both sides we have,

½(x) +5-5 = 65-5

½(x) = 60

Multiplying 2 on both sides, we have

½(x) *2 = 60*2

x = 120, is the required solution.

Forming an Equation

We have learned how to solve an equation. Now we shall form or construct the equation when the solution(root) is given. Let us know the following successive steps:

• Multiply both sides by 3

3x = 27

• Subtract 2 from both sides

3x – 2 = 27-2

3x – 2 = 25, which is an equation.

Note: For a given equation, you get one solution; but for a given solution, one can make many equations.

Let us understand this with  more examples so that you can solve exercise 4.3 of NCERT Solutions Chapter 4.

Example: Solve 5(x-3) = 25

(Or) x-3 = 25/5 (Dividing both sides by 5)

(Or) x – 3 = 5

(or) x = 5+3  (Transposing -3 to R.H.S)

x = 8, which is the required solution.

Example: Solve 3(x+1)/2 = 18

Solution: 3(x+1)/2 = 18

(or) (x+1)/2 = 18/2 (Dividing both sides by 2)

(or) (x+1)/2 = 6

(or) x/2 = (6-1)/2 (Transposing 1 to R.H.S)

(or) x = (12-1)/2 = 11/2, which is the required solution.

Application of Simple Equations to Practical Situations

Let us understand this with more examples so that you can solve exercise 4.4 of NCERT Solutions Chapter 4.

Example: The sum of five times a number and 18 is 63. Find the number

Let the required number be x

5 times the number is 5x

According to the condition, we have

5x + 18 = 63

5x = 63 – 18    (Transposing 18 form L.H.S to R.H.S)

5x = 45

(or) dividing both sides by 5, we have

5x/5  = 45/5

x = 9

Thus, the required number is = 9

Related Questions

Question: If 2x-3 = 5, then

• X = 1
• X = -1
• X = 4
• X = -4

Solution: x = 4

Question: If both sides of the equation are divided by the same (non–zero) quantity, the equality -

• Does not change
• Changes
• May or may not change
• None of these

Solutions: Does not change

Q.1 Complete the last column of the table.

 $S.No.$ $Equation$ $Value$ $\begin{array}{l}Say,\text{\hspace{0.17em}}whether\text{​}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}equation\\ is\text{\hspace{0.17em}}satisfied.\left(Yes/No\right)\end{array}$ $\left(i\right)$ $x+3=0$ $x=3$ $\left(ii\right)$ $x+3=0$ $x=0$ $\left(iii\right)$ $x+3=0$ $x=–3$ $\left(iv\right)$ $x–7=1$ $x=7$ $\left(v\right)$ $x–7=1$ $x=8$ $\left(vi\right)$ $5x=25$ $x=0$ $\left(vii\right)$ $5x=25$ $x=5$ (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 Check whether the value given in the brackets is a solutionto the given equation or not:(a) n+5=19(n=1) (b) 7n+5=19(n=2)(c) 7n+5=19(n=2) (d) 4p3=13(p=1)(e) 4p3=13(p=4) (f)4p3=13(p=0)

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 Solve the following equations by trial and error method:(i) 5p+2=17 (ii) 3m14=4

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

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 Set up an equation in the following cases:i Irfan says that he has 7 marbles more than fivetimes the marbles Parmit has. Irfan has 37 marbles.(Take m to be the number of Parmits marbles.(ii) Laxmis father is 49 years old. He is 4 years older thanthree times Laxmis age.(Take Laxmis age to be y years.)iiiThe teacher tells the class that the highest marksobtained by a student in her class is twice the lowestmarks plus 7.The highest score is 87.(Take the lowest score to be l).iv In an isosceles triangle, the vertex angle is twice eitherbase angle.(Let the base angle be b in degrees. Rememberthat the sum of angles of a triangle is 180​ degrees).

Ans.

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

Q.7 Give first the step you will use to separate the variableand then solve the equation:(a) x1=0 (b) x+1=0 (c) x1=5 (d) x+6=2(e) y4=7 (f) y4=4 (g) y+4=4 (h) y+4=4

Ans.

$\begin{array}{l}\left(\text{a}\right)x-\text{1}=0\\ \text{Add 1 to both sides to get}\\ \overline{)x=1}\\ \left(\text{b}\right)\text{x}+\text{1}=0\\ \text{Subtract 1 from both sides to get}\\ \overline{)\text{x}=-\text{1}}\\ \left(\text{c}\right)\text{x}-\text{1}=\text{5}\\ \text{Add 1 to both sides to get}\\ \overline{)x=6}\\ \left(\text{d}\right)\text{x}+\text{6}=\text{2}\\ \text{Subtract 6 from both sides to get}\\ \overline{)\text{x}=-\text{4}}\\ \left(\text{e}\right)\text{y}-\text{4}=-\text{7}\\ \text{Add 4 to both sides to get}\\ \overline{)\text{y}=-3}\\ \left(\text{f}\right)\text{y}-\text{4}=\text{4}\\ \text{Add 4 to both sides to get}\\ \overline{)y=0}\\ \left(\text{g}\right)\text{y}+\text{4}=\text{4}\\ \text{Subtract 4 from both sides to get}\\ \overline{)y=0}\\ \left(\text{h}\right)\text{y}+\text{4}=-\text{4}\\ \text{Subtract 4 from both sides to get}\\ \overline{)y=-8}\end{array}$

Q.8 Give first the step you will use to separate the variableand then solve the equation:(a) 3l=42 (b)b2=6 (c)p7=4 (d)4x=25(e) 8y=36 (f)z3=54 (g)a5=715 (h)2t=10

Ans.

$\begin{array}{l}\left(\text{a}\right)3l=42\\ \text{Divide both sides by 3 to get}\\ l=\frac{42}{3}=\overline{)14}\\ \left(\text{b}\right)\frac{b}{2}=6\\ \text{Multiply both sides by 3 to get}\\ \overline{)b=12}\\ \left(\text{c}\right)\frac{p}{7}=\text{4}\\ \text{Multiply both sides by 7 to get}\\ \overline{)p=28}\\ \left(\text{d}\right)\text{4x}=\text{25}\\ \text{Divide both sides by 4 to get}\\ \overline{)x=\frac{25}{4}}\\ \left(\text{e}\right)\text{8y}=36\\ \text{Divide both sides by 8 to get}\\ y=\frac{36}{8}=\overline{)\frac{9}{2}}\\ \left(\text{f}\right)\frac{z}{3}=\frac{5}{4}\\ \text{Multiply both sides by 3 to get}\\ \overline{)z=\frac{15}{4}}\\ \left(\text{g}\right)\frac{a}{5}=\frac{7}{15}\\ \text{Multiply both sides by 5 to get}\\ \text{a=}\frac{7×5}{15}=\frac{7×\overline{)5}}{3×\overline{)5}}=\overline{)\frac{7}{3}}\\ \left(\text{h}\right)2t=-\text{10}\\ \text{Divide both sides by 2 to get}\\ \text{t=}\frac{-10}{2}=\frac{-5×\overline{)2}}{\overline{)2}}=\overline{)-5}\end{array}$

Q.9 Give the steps you will use to separate the variable andthen solve the equation:(a) 3n-2 = 46 (b) 5m+7 = 17(c) 20p3 = 40 (d) 3p10 = 6

Ans.

$\begin{array}{l}\text{(a)}3n-2=46\\ \text{Add 2 to both sides to get}\\ 3n=48\\ \text{Now, divide both sides by 3 to get}\\ \overline{)n=16}\\ \text{(b)}5m+7=17\\ \text{Subtract 7 from both sides to get}\\ 5m=10\\ \text{Now, divide both sides by 5 to get}\\ \overline{)m=2}\\ \text{(c)}\frac{20p}{3}=40\\ \text{Multiply both sides by 3 to get}\\ 20p=120\\ \text{Now divide both sides by 20, to get}\\ p=\frac{120}{20}=\frac{\overline{)20}×3}{\overline{)20}}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=3}\\ \text{(d)}\text{\hspace{0.17em}}\frac{3p}{10}=6\\ \text{Multiply both sides by 10 to get}\\ 3p=60\\ \text{Now divide both sides by 3, to get}\\ p=\frac{60}{3}=\frac{20×\overline{)3}}{\overline{)3}}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=20}\end{array}$

Q.10 Solve the following equations:(a)10p=100(b)10p+10=100(c)p4=5(d)p3=5(e)3p4=6(f)3s=9(g)3s+12=0(h)3s=0(i)2q=6(j)2q6=0(k)2q+6=0(l)2q+6=12

Ans.

$\begin{array}{l}\left(a\right)\text{}10p=100\\ p=\frac{100}{10}=\overline{)10}\\ \text{(b)}10p+10=100\\ 10p=90\\ p=\frac{90}{10}=\overline{)9}\\ \text{(c)}\frac{p}{4}=5\\ p=5×4=\overline{)20}\\ \text{(d)}\frac{p}{3}=5\\ p=5×3=\overline{)15}\\ \left(e\right)\text{}\frac{3p}{4}=6\\ 3p=6×4=24\\ p=\frac{24}{3}=\overline{)8}\\ \left(f\right)\text{\hspace{0.17em}}3s=-9\\ s=\frac{-9}{3}=\overline{)-3}\\ \left(g\right)\text{\hspace{0.17em}}3s+12=0\\ 3s=-12\\ s=\frac{-12}{3}=\overline{)-4}\\ \left(h\right)\text{\hspace{0.17em}}3s=0\\ s=\frac{0}{3}=\overline{)0}\\ \left(i\right)\text{\hspace{0.17em}}2q=6\\ q=\frac{6}{2}=\overline{)3}\\ \left(j\right)\text{\hspace{0.17em}}2q-6=0\\ 2q=6\\ q=\frac{6}{2}=\overline{)3}\\ \left(k\right)\text{\hspace{0.17em}}2q+6=0\\ 2q=-6\\ q=\frac{-6}{2}=\overline{)-3}\\ \left(l\right)\text{\hspace{0.17em}}2q+6=12\\ 2q=12-6\\ 2q=6\\ q=\frac{6}{3}=\overline{)2}\end{array}$

Q.11 Solve the following equations:(a) 2y+52=372 (b) 5t+28=10 (c)a5+3=2(d) q4+7=5 (e) 52x=10 (f) 52x=254(g) 7m+192=13 (h) 6z+10=2 (i) 3l2=23(j) 2b35=3

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{2}y+\frac{5}{2}=\frac{37}{2}\\ 2y=\frac{37}{2}-\frac{5}{2}=\frac{37-5}{2}=\frac{32}{2}=16\\ 2y=16\\ y=\frac{16}{2}=8\\ Thus,\text{\hspace{0.17em}}\overline{)y=8}\\ \left(\text{b}\right)\text{5t}+\text{28}=\text{1}0\\ 5t=10-28\\ 5t=-18\\ \overline{)t=\frac{-18}{5}}\\ \left(\text{c}\right)\frac{\text{a}}{5}+3=2\\ \frac{a}{5}=2-3\\ \frac{a}{5}=-1\\ \overline{)a=-5}\\ \left(\text{d}\right)\frac{q}{4}+7=5\\ \frac{q}{4}=5-7\\ \frac{q}{4}=-2\\ \overline{)q=-8}\\ \left(\text{e}\right)\frac{5}{2}x=-10\\ 5x=-10×2\\ x=\frac{-10×2}{5}=\frac{-\overline{)5}×2×2}{\overline{)5}}=-4\\ Thus,\text{\hspace{0.17em}}\overline{)x=-4}\\ \left(\text{f}\right)\frac{5}{2}x=\frac{25}{4}\\ 5x=\frac{25×2}{4}\\ x=\frac{\overline{)5}×5×\overline{)2}}{\overline{)2}×2×\overline{)5}}\\ Thus,\text{\hspace{0.17em}}\overline{)x=\frac{5}{2}}\\ \left(\text{g}\right)7m+\frac{19}{2}=13\\ 7m=13-\frac{19}{2}=\frac{26-19}{2}=\frac{7}{2}\\ 7m=\frac{7}{2}\\ m=\frac{\overline{)7}}{2×\overline{)7}}\\ Thus,\text{\hspace{0.17em}}\overline{)m=\frac{1}{2}}\\ \left(\text{h}\right)\text{6z}+\text{1}0=-\text{2}\\ 6z=-2-10\\ 6z=-12\\ z=\frac{-12}{6}=\frac{-2×\overline{)6}}{\overline{)6}}0\\ Thus,\text{\hspace{0.17em}}\overline{)z=-2}\\ \left(\text{i}\right)\frac{3l}{2}=\frac{2}{3}\\ 3l=\frac{4}{3}\\ \overline{)l=\frac{4}{9}}\\ \left(\text{j}\right)\frac{2b}{3}-5=3\\ \frac{2b}{3}=8\\ 2b=24\\ b=\frac{24}{2}=\frac{\overline{)2}×12}{\overline{)2}}\\ Thus,\text{\hspace{0.17em}}\overline{)b=12}\end{array}$

Q.12 Solve the following equations:(a) 2(x+4)=12 (b) 3(n5)=21 (c) 3(n5)=21(d) 32(2y)=7 (e)4(2x)=9 (f) 4(2x)=9(g) 4+5(p1)=34 (h)345(p1)=4

Ans.

$\begin{array}{l}\text{(a}\right)\text{}2\left(x+\text{}4\right)=12\\ \text{Divide both sides by 2 to get}\\ x+4=6\\ \text{Thus,}\text{\hspace{0.17em}}\overline{)x=2}\\ \left(\text{b}\right)\text{}3\left(n-5\right)=-21\\ \text{Divide both sides by 3 to get}\\ n-5=-\text{7}\\ \text{Thus,}\overline{)\text{}n=-2}\\ \left(\text{c}\right)\text{}3\left(n-5\right)=-\text{21}\\ \text{Divide both sides by 2 to get}\\ n-5=-7\\ Thus,\text{\hspace{0.17em}}\overline{)n=-12}\\ \left(\text{d}\right)\text{3}-\text{2}\left(\text{2}-y\right)=\text{7}\\ -2\left(2-y\right)=4\\ \text{Divide both sides by}-\text{2 to get}\\ 2-y=-2\\ \text{Multiply both sides by}-\text{1 to get}\\ y-2=2\\ \text{Thus,}\overline{)y=4}\\ \left(\text{e}\right)\text{}-4\left(2-x\right)=\text{9}\\ \text{Divide both sides by}-\text{4 to get}\\ \text{2}-x=\frac{-9}{4}\\ \text{Multiply both sides by}-\text{1 to get}\\ x-2=\frac{9}{4}\\ x=\frac{9}{4}+2=\frac{9+8}{4}\\ Thus,\text{\hspace{0.17em}}\overline{)x=\frac{17}{4}}\\ \left(\text{f}\right)\text{}4\left(2-x\right)=\text{9}\\ \text{Divide both sides by 4 to get}\\ \text{2}-x=\frac{9}{4}\\ \text{Multiply both sides by -1 to get}\\ \text{x}-2=\frac{-9}{4}\\ x=\frac{-9}{4}+2=\frac{-9+8}{4}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)x=\frac{-1}{4}}\\ \left(\text{g}\right)\text{}4+5\text{}\left(p-1\right)=34\\ 5\left(p-1\right)=30\\ \text{Divide both sides by 5 to get}\\ \text{p}-1=6\\ p=6+1=7\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=7}\\ \left(\text{h}\right)\text{}34-5\left(p-1\right)=4\\ -5\left(p-1\right)=-30\\ \text{Divide both sides by}-\text{5 to get}\\ p-1=6\\ p=6+1=7\\ \text{Thus,}\text{\hspace{0.17em}}\overline{)p=7}\end{array}$

Q.13 Solve the following equations.(a) 4=5 (p2) (b)4=5 (p2) (c)16=5(2p)(d) 10=4+3 (t+2) (e) 28=4+3 (t+5)(f) 0=16+4 (m6)

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{4}=\text{5}\left(p-\text{2}\right)\\ \text{Divide both sides by 5 to get}\\ \frac{4}{5}=p-2\\ p=\frac{4}{5}+2=\frac{4+10}{5}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=\frac{14}{5}}\\ \left(\text{b}\right)-\text{4}=\text{5}\left(p-2\right)\\ \text{Divide both sides by 5 to get}\\ \frac{-4}{5}=p-2\\ p=\frac{-4}{5}+2=\frac{-4+10}{5}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=\frac{6}{5}}\\ \left(\text{c}\right)-16=-5\text{}\left(2-p\right)\\ \text{Divide both sides by}-\text{5 to get}\\ \frac{16}{5}=2-p\\ \text{Multiply both sides by}-\text{1 to get}\\ \frac{-\text{16}}{5}=p-2\\ p=\frac{-16}{5}+2=\frac{-16+10}{5}\\ \text{Thus},\text{\hspace{0.17em}}\overline{)p=\frac{-6}{5}}\\ \left(\text{d}\right)\text{1}0=\text{4}+\text{3}\left(t+2\right)\\ 6=3\left(t+2\right)\\ \text{Divide both sides by 3 to get}\\ 2=t+2\\ \text{Thus,}\overline{)t=0}\\ \left(\text{e}\right)\text{28}=\text{4}+\text{3}\left(t+\text{5}\right)\\ 24=3\left(t+5\right)\\ \text{Divide both sides by 3 to get}\\ 8=t+5\\ \text{Thus,}\overline{)t=3}\\ \left(\text{f}\right)0=\text{16}+\text{4}\left(m-\text{6}\right)\\ -16=4\left(m-6\right)\\ \text{Divide both sides by 4 to get}\\ -4=m-6\\ \text{Thus,}\overline{)m=2}\end{array}$

Q.14 (a) Construct 3 equations starting with x=2(b) Construct 3 equations starting with x=2

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=2\\ \left(\mathrm{b}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=-2\end{array}$

Q.15 Set up equations and solve them to find the unknownnumbers in the following case:a Add 4 to eight times a number; you get 60.b One fifth of a number minus 4 gives 3.c If I take three fourths of a number and count up 3 more, I get 21.(d) When I subtracted 11 from twice a number, the resultwas 15.(e) Munna subtracts thrice the number of notebooks he has from 50,he finds the result to be 8.f Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.g Anwar thinks of a number. If he takes away 7 from 52 of the number,the result is 112.

Ans.

$\begin{array}{l}\text{(a) Let the number be x}\text{.}\\ \text{8 times of this number = 8x}\\ \text{So, we get}\\ 8x+4=6\text{0}\\ 8x=\text{56}\\ \text{x=}\frac{56}{8}\\ Thus,\overline{)x=7}\\ \text{(b) Let the number be}x.\\ \text{One-fifth of this number=}\frac{x}{5}\\ \text{So, we get}\\ \frac{x}{5}-4=3\\ \frac{x}{5}=7\\ \overline{)x=35}\\ \text{(c) Let the number be}x.\\ \text{Three-fourth of this number}=\frac{3x}{4}\\ \text{So,}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{get}\\ \frac{3x}{4}+3=21\\ \frac{3x}{4}=18\\ 3x=72\\ x=\frac{72}{3}\\ Thus,\text{\hspace{0.17em}}\overline{)x=24}\\ \left(d\right)\text{\hspace{0.17em}}\text{Let the number be}x.\\ \text{So, we have}\\ \text{2x-11=15}\\ \text{2x=26}\\ \text{x=}\frac{26}{13}=13\\ Thus,\overline{)x=13}\\ \left(e\right)\text{\hspace{0.17em}}\text{Let the number be}x\\ \text{Thrice the number of books}=\text{3}x\\ \text{So, we get}\\ \text{50}-3x=\text{8}\\ -3x=8-50=-42\\ \text{Divide both sides by -3 to get}\\ \text{x=}\frac{-42}{-3}=14\\ Thus,\overline{)x=14}\\ \left(f\right)\text{\hspace{0.17em}}\text{Let the number be}x.\\ \text{We, have}\\ \frac{x+19}{5}=8\\ x+19=8×5=40\\ x=40-19=21\\ \text{Thus,}\text{\hspace{0.17em}}\overline{)x=21}\\ \left(g\right)\text{\hspace{0.17em}}\text{Let the number be}x.\\ \overline{)So,\text{\hspace{0.17em}}we\text{\hspace{0.17em}}have}\\ \frac{5x}{2}-7=23\\ \frac{5x}{2}=23+7=30\\ 5x=30×2=60\\ x=\frac{60}{5}=12\\ \text{Thus},\text{\hspace{0.17em}}\overline{)x=12}\end{array}$

Q.16 Solve the following:a The teacher tells the class that the highest marksobtained by a student in her class is twice the lowestmarks plus 7.The highest score is 87. What is thelowest score?b In an isosceles triangle, the base angles are equal.The vertex angle is 40°. What are the base angles of thetriangle? (Remember, the sum of three angles of a triangleis 180°)

Ans.

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{a}\right)\mathrm{The}\mathrm{teacher}\mathrm{tells}\mathrm{the}\mathrm{class}\mathrm{that}\mathrm{the}\mathrm{highest}\mathrm{marks}\\ \mathrm{obtained}\mathrm{by}\mathrm{a}\mathrm{student}\mathrm{in}\mathrm{her}\mathrm{class}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{lowest}\\ \mathrm{marks}\mathrm{plus}7.\mathrm{The}\mathrm{highest}\mathrm{score}\mathrm{is}87.\mathrm{What}\mathrm{is}\mathrm{the}\\ \mathrm{lowest}\mathrm{score}?\\ \left(\mathrm{b}\right)\mathrm{In}\mathrm{an}\mathrm{isosceles}\mathrm{triangle},\mathrm{the}\mathrm{base}\mathrm{angles}\mathrm{are}\mathrm{equal}.\\ \mathrm{The}\mathrm{vertex}\mathrm{angle}\mathrm{is}40°.\mathrm{What}\mathrm{are}\mathrm{the}\mathrm{base}\mathrm{angles}\mathrm{of}\mathrm{the}\\ \mathrm{triangle}? \left(\mathrm{Remember},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\\ \mathrm{is}180°\right)\end{array}$

Q.17 Solve the following:(i)Irfan says that he has 7 marbles more than five timesthe marbles Parmithas. Irfan has 37 marbles. How manymarbles does Parmit have?(ii)Laxmis father is 49 year sold.He is 4 years older thanthree times Laxmis age.What is Laxmi sage?(iii) Maya, Madhura and Mohsina are friends studying in thesame class.In a class testin geography, Maya got 16out of 25. Madhura got 20. Their average score was 19.How much did Mohsina score?

$\begin{array}{l}\left(\mathrm{iv}\right)\mathrm{People}\mathrm{of}\mathrm{Sundargram}\mathrm{planted}\mathrm{a}\mathrm{total}\mathrm{of}102\mathrm{trees}\mathrm{in}\mathrm{the}\\ \mathrm{village}\mathrm{garden}.\mathrm{Some}\mathrm{of}\mathrm{the}\mathrm{trees}\mathrm{were}\mathrm{fruit}\mathrm{trees}.\mathrm{The}\\ \mathrm{number}\mathrm{of}\mathrm{non}-\mathrm{fruit}\mathrm{trees}\mathrm{were}\mathrm{two}\mathrm{more}\mathrm{than}\mathrm{three}\mathrm{times}\\ \mathrm{the}\mathrm{number}\mathrm{of}\mathrm{fruit}\mathrm{trees}.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{fruit}\\ \mathrm{trees}\mathrm{planted}?\end{array}$

Ans.

$\begin{array}{l}\text{(i) Let Parmit has}m\text{marbles}\text{.}\\ \text{Then, according to the question, we have}\\ \text{5}×\text{Number of marbles Parmit has +7}=\text{Number of marbles}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{Irfan has}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}×m+\text{7}=\text{37}\\ \text{So, we get}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}m+\text{7}=\text{37}\\ 5m=37-7=30\\ 5m=30\\ m=\frac{30}{5}=6\\ \text{Therefore,}\text{​}\text{Parmit}\text{\hspace{0.17em}}\text{has}\text{\hspace{0.17em}}\overline{)\text{6}\text{\hspace{0.17em}}\text{marbles}}\text{.}\\ \text{(ii) Let Laxmi be}y\text{years old}\\ \text{Then, according to the question, we have}\\ \text{3}×\text{Laxmi’s age}+\text{4}=\text{Laxmi’s father age}\\ \text{3}×y+4=\text{49}\\ \text{3}y+\text{4}=\text{49}\\ 3y=49-4=45\\ 3y=45\\ y=\frac{45}{3}=15\\ \text{Therefore,}\text{\hspace{0.17em}}\text{laxmi’s}\text{\hspace{0.17em}}\text{age}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\overline{)\text{15}\text{\hspace{0.17em}}\text{years}}\text{.}\\ \text{(iii) Let the number of fruit trees be}x.\\ \text{So, we have}\\ \text{3}×\text{Number of fruit trees}+2=\text{Number of non-fruit trees}\\ 3x+2=77\\ 3x=77–2=75\\ 3x=75\\ x=\frac{75}{3}=25\\ \text{Therefore,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{number}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{fruit}\text{\hspace{0.17em}}\text{trees}\text{\hspace{0.17em}}\text{was}\text{\hspace{0.17em}}\overline{)25}.\end{array}$

Q.18 Solve the following riddle.I am a number,Tell my identity!Take me seven times overAnd add a fifty!To reach a triple centuryYou still need forty

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{be}\mathrm{x}\mathrm{.}\\ \mathrm{Then}\mathrm{we}\mathrm{have}\\ \left(7\mathrm{x}+50\right)+40=300\\ 7\mathrm{x}+50+40=300\\ 7\mathrm{x}+90+300\\ 7\mathrm{x}=300-90\\ 7\mathrm{x}=210\\ \mathrm{x}=\frac{210}{7}=30.\\ \mathrm{Therefore},\mathrm{the}\mathrm{number}\mathrm{is}\overline{)\mathrm{30}}\mathrm{.}\end{array}$

1. What are the important topics covered in NCERT Solutions Class 7 Mathematics Chapter 4?

A few of the important topics covered under Chapter 7 are given below:

• Constants and variables
• L.H.S and R.H.S
• Equations

2. How do you avail the study materials for NCERT Solutions for Class 7 Mathematics?

NCERT Solutions for Class 7 Mathematics are available on Extramarks. Subject matter experts have crafted the solutions in a step-by-step method that is easy to understand. Students can revise and solve the questions to master this chapter.

3. Are there any theoretical questions in chapter Simple Equations of Class 7 Mathematics?

There aren’t any theoretical questions in this chapter. The questions in this chapter are mostly practical. Even if a theoretical question is asked, the response will be one line or one word long. As a result, your primary focus should be on problem solving rather than learning theory.

4. How many questions are there in NCERT Solutions for class 11 chapter 4?

There are a variety of questions found in NCERT solutions for Class 11 Chapter 4. The chapter is divided into four exercises. Exercise 4.1 has a total of 6 questions, exercise 4.2 has 4 questions, exercise 4.3 has four questions and exercise 4.4 has four questions.