NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - Download PDF

# NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - Download PDF

Edited By Ramraj Saini | Updated on Feb 07, 2024 05:00 PM IST

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - In this chapter, students will be introduced by a new concept 'Equation'. This topic not only builds knowledge of algebra for maths students of class 7 but also helps to develop analytical thinking. Solutions of NCERT Class 7 Maths chapter 4 Simple Equations have 4 exercises with 18 questions in them. The NCERT solutions for Maths chapter 4 Simple Equations class 7 also discuss some topic-wise questions.

The NCERT Solutions can be extremely helpful for the Class 7 Maths students to understand the basics of this chapter and to clear all their doubts easily. Students can use as worksheets to prepare for their CBSE final exams. Here you will get solutions to all four exercises of this NCERT chapter.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Formulae

Balancing Equation:

LHS = RHS

(LHS ± a) = (RHS ± a)

(LHS × b) = (RHS × b)

(LHS ÷ c) = (RHS ÷ c)

Solution of an Equation: For a variable 'x', if LHS(x) = RHS(x), then 'x' is a solution.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Points

Variable: Something that can vary, Its value is not fixed such as x, y, z, p, q, r, etc.

Equation: It is a mathematical statement that uses an equal sign(=) to show that the value of the expression to the left of the sign (LHS) is equal to the value of the expression to the right of the sign (RHS).

Interchangeability: If the LHS and RHS are interchanged, the equation remains the same.

Balanced equation: the value of LHS remains equal to the value of the RHS. If we,

• Add the same number to both sides

• Subtract the same number from both sides.

• Multiply both sides by the same number.

• Divide both sides by the same number.

Solution of an equation: The value of the variable for which the equation is satisfied ( for that variable, LHS = RHS ).

Transposing: Transposing of a number has the same effect as adding the same number to both sides or subtracting the same number from both sides of the equation.

Free download NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation PDF for CBSE Exam.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation (Intext Questions and Exercise)

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Topic 4.3

(i) Let y = 2 . We have :

$10y\ -\ 20\ =\ 10(2)\ -\ 20\ =\ 0$

(ii) Let y = 3 . We have :

$10y\ -\ 20\ =\ 10(3)\ -\ 20\ =\ 10$

(iii) Let y = 4 . We have :

$10y\ -\ 20\ =\ 10(4)\ -\ 20\ =\ 20$

(iv) Let y = 5 . We have :

$10y\ -\ 20\ =\ 10(5)\ -\ 20\ =\ 30$

(v) Let y = 6 . We have :

$10y\ -\ 20\ =\ 10(6)\ -\ 20\ =\ 40$

Hence 10y - 20 depends upon y.

Now, consider $10y\ -\ 20\ =\ 50$

Transpose - 20 to the RHS :

$10y\ =\ 50\ +\ 20\ =\ 70$

or $y\ =\ 7$

## NCERT solutions for maths chapter 4 simple equation class 7 topic 4.7

(ii) What is that number one third of which added to 5 gives 8?

(i) Let the number be n.

Then according to the question, we have :

$6n\ -\ 5\ =\ 7$

or $6n\ =\ 7\ +\ 5\ =\ 12$

or $n\ =\ 2$

(ii) Let the number be x.

Then according to the question,

$\frac{x}{3}\ +\ 5\ =\ 8$

$\frac{x}{3}\ =\ 8\ -\ 5\ =\ 3$

$x\ =\ 9$

Let the number of mangoes in the smaller box be n.

Then according to the question, we have :

$8n\ +\ 4\ =\ 100$

or $8n\ =\ 100\ -\ 4\ =\ 96$

or $n\ =\ 12$

Hence the number of mangoes in the smaller box is 12.

## NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.1

The table is shown below:-

(a) n + 5 = 19 (n = 1)

(b) 7n + 5 = 19 (n = – 2)

(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)

(e) 4p – 3 = 13 (p = – 4)

(f) 4p – 3 = 13 (p = 0)

(a) Put n = 1 in the equation, we have :

n + 5 = 15

or 1 + 5 = 15

or 6 $\neq$ 15

Thus n = 1 is not a solution.

(b) Put n = - 2, we have :

7n + 5 = 19

or 7(-2) + 5 = - 14 + 5 = - 9 $\neq$ 19.

So, n = - 2 is not a solution to the given equation.

(c) Put n = 2, we have :

7n + 5 = 19

or 7(2) + 5 = 14 + 5 = 19 = R.H.S

Thus n = 2 is the solution for the given equation.

(d) Put p = 1 , we have :

4p - 3 = 13

or 4(1) - 3 = 1 $\neq$ 13 .

Thus p = 1 is not a solution.

(e) Put p = - 4 , we get :

4p - 3 = 13

or 4(1) - 3 = 1 $\neq$ 13 .

Thus p = 1 is not a solution.

(f) Put p = 0 , we get :

4p - 3 = 13

or 4(0) - 3 = - 3 $\neq$ 13 .

Thus p = 0 is not a solution.

(i) Put p = 1,

We have : $5(1)\ +\ 2\ =\ 7\ \neq\ 17$

Put p = 2,

We have : $5(2)\ +\ 2\ =\ 12\ \neq\ 17$

Put p = 3,

we have : $5(3)\ +\ 2\ =\ 17\ =\ 17$

Thus the solution is p = 3.

(ii) Put m = 4,

we have : $3(4)\ -\ 14\ =\ -2\ \neq\ 4$

Put m = 5,

we have : $3(5)\ -\ 14\ =\ 1\ \neq\ 4$

Now, put m = 6,

we have : $3(6)\ -\ 14\ =\ 4\ =\ 4$

Thus m = 6 is the solution.

(ii) 2 subtracted from y 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 x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.

The equations are given below :

(i) $x\ +\ 4\ =\ 9$

(ii) $y\ -\ 2\ =\ 8$

(iii) $10a\ =\ 70$

(iv) $\frac{b}{5}\ =\ 6$

(v) $\frac{3}{4}t\ =\ 15$

(vi) $7m\ +\ 7\ =\ 77$

(vii) $\frac{x}{4}\ -\ 4\ =\ 4$

(viii) $6y\ -\ 6\ =\ 60$

(ix) $\frac{z}{3}\ +\ 3\ =\ 30$

(i) p + 4 = 15

(ii) m – 7 = 3

(iii) 2m = 7

(iv) m /5 = 3

(v) 3 m/5 = 6

(vi) 3p + 4 = 25

(vii) 4p – 2 = 18

(viii) p /2 + 2 = 8

(i) Add 4 to the number p, we get 15.

(ii) Subtract 7 from m to get 3.

(iii) Twice the number m is 7.

(iv) One-fifth of m is 3.

(v) Three-fifth of m is 6.

(vi) 4 is added to thrice the number p to get 25.

(vii) 2 is subtracted from the product of 4 times p to get 18.

(viii) When 2 is added to half of the number p, we get 8.

(a) Let the Parmit's marbles be m.

Then according to the question we have : $5m\ +\ 7\ =\ 37$

or $5m\ =\ 30$

(b) Let the age of Laxmi be y years.

Then we have : $3y\ +\ 4\ =\ 49$

or $3y\ =\ 45$

(c) Let the lowest marks be l, then :

$2l \ +\ 7\ =\ 87$

or $2l \ =\ 80$

(d) Let the base angle of the triangle be b degree.

Then according to the question we have :

$b\ +\ b\ +\ 2b\ =\ 180^{\circ}$

or $3b\ =\ 180^{\circ}$

## NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.2

(a) x – 1 = 0

(b) x + 1 = 0

(c) x – 1 = 5

(d) x + 6 = 2

(e) y – 4 = – 7

(f) y – 4 = 4

(g) y + 4 = 4

(h) y + 4 = – 4

(a) Add 1 to both the sides, we have :

$x\ =\ 1$

(b) Transpoing 1 to the RHS, we have :

$x\ =\ -\ 1$

(c) Transposing - 1 to the RHS, we have :

$x\ =\ 6$

(d) Transposing 6 to the RHS, we get :

$x\ =\ -\ 4$

(e) Transposing - 4 to the RHS, we have :

$y\ =\ -\ 3$

(f) Transposing - 4 to the RHS, we get :

$y\ =\ 8$

(g) Transposing 4 to the RHS, we get :

$y\ =\ 0$

(h) Transposing 4 to the RHS, we get :

$y\ =\ -\ 8$

(a) 3l = 42

(b) b / 2 = 6

(c) p /7 = 4

(d) 4x = 25

(e) 8y = 36

(f) z/ 3 = 5 /4

(g) a /5 =7/ 15

(h) 20t = – 10

(a) Divide both sides by 3, we get :

$l\ =\ 14$

(b) Multiply both sides by 2, we get :

$b\ =\ 12$

(c) Multiply both sides by 7, we get :

$p\ =\ 28$

(d) Divide both sides by 4, we get :

$x\ =\ \frac{25}{4}$

(e) Divide both sides by 8, we get :

$y\ =\ \frac{9}{2}$

(f) Multiply both sides by 3, we get :

$z\ =\ \frac{15}{4}$

(g) Multiply both sides by 5, we get :

$a\ =\ \frac{7}{3}$

(h) Divide both sides by 20, we get :

$t =\ -\frac{1}{2}$

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p / 3 = 40

(d) 3p/10 = 6

(a) We have 3n – 2 = 46.

Transposing - 2 to the RHS, we have :

$3n\ =\ 46\ +\ 2\ =\ 48$

or $n\ =\ 16$

(b) We have 5m + 7 = 17

Transposing 7 to the RHS, we have :

$5m\ =\ 17\ -\ 7\ =\ 10$

or $m\ =\ 2$

(c) We have 20p / 3 = 40

Multiply both sides by $\frac{3}{20}$ :

$\frac{20p}{3}\times \frac{3}{20}\ =\ 40\times \frac{3}{20}$

or $p\ =\ 6$

(d) We have 3p/10 = 6

Multiply both sides by $\frac{10}{3}$ :

$\frac{3p}{10}\times \frac{10}{3}\ =\ 6\times \frac{10}{3}$

or $p\ =\ 20$

(a) 10p = 100

(b) 10p + 10 = 100

(c) p /4 = 5

(d) – p/3 = 5

(e) 3 p/4 = 6

(f) 3s = –9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2q – 6 = 0

(k) 2q + 6 = 0

(l) 2q + 6 = 12

(a) Divide both sides by 10, we get :

$p\ =\ 10$

(b) Transposing 10 to the RHS, we get :

$10p\ =\ 100\ -\ 10\ =\ 90$

Now, dividing both sides by 10 gives : $p\ =\ 9$

(c) Multiplying both sides by 4, we have :

$p\ =\ 20$

(d) Multiplying both sides by - 3 , we have :

$p\ =\ 15$

(e) Multiplying both sides by $\frac{4}{3}$ , we have :

$p\ =\ 8$

(f) Dividing both sides by 3, we have :

$s\ =\ -3$

(g) Transposing 12 to the RHS and then dividing both sides by 3, we have :

$s\ =\ -4$

(h) Dividing both sides by 3, we get :

$s\ =\ 0$

(i) Dividing both sides by 2, we get :

$q\ =\ 3$

(j) Transposing - 6 to the RHS and then dividing both sides by 2, we get :

$q\ =\ 3$

(k) Transposing 6 to the RHS and then dividing both sides by 2, we get :

$q\ =\ -\ 3$

(l) Transposing 6 to the RHS and then dividing both sides by 2, we get :

$q\ =\ 3$

## NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.3

(b) 5t + 28 = 10

(c) a /5 + 3 = 2

(d) q/ 4 + 7 = 5

e) 5 x/2 = –5

(f) $\frac{5}{2}x = \frac{25}{4}$

(g) 7m + 19/2 = 13

(h) 6z + 10 = –2

(i) $\frac{3l}{2} = 2/3$

(j) $\frac{2b}{3}- 5 = 3$

$(a) 2y + \frac{5}{2} = \frac{37}{2}$

Transposing $\frac{5}{2}$ to the RHS :

$2y\ =\ \frac{37}{2}\ -\ \frac{5}{2}\ =\ 16$

$y\ =\ 8$

(b) 5t + 28 = 10

Transposing 28 to the RHS and then dividing both sides by 5, we get :

$5t\ =\ 10\ -\ 28\ =\ -\ 18$

$t\ =\ -\frac{18}{5}$

(c) a /5 + 3 = 2

Transposing 3 to the RHS and multiplying both sides by 5, we get :

$\frac{a}{5}\ +\ 3\ =\ 2$

$\frac{a}{5}\ =\ -\ 1$

$a\ =\ -\ 5$

(d) q/ 4 + 7 = 5

Transposing 7 to the RHS and multiplying both sides by 4:

$\frac{q}{4}\ +\ 7\ =\ 5$

$\frac{q}{4}\ =\ -\ 2$

$q\ =\ -\ 8$

(e) 5 x/2 = – 5

Multiplying both sides by $\frac{2}{5}$ :

$x\ =\ -5\times \frac{2}{5}\ =\ -\ 2$

(f) $\frac{5}{2}x = \frac{25}{4}$

Multiplying both sides by $\frac{2}{5}$ :

$x\ =\ \frac{25}{4}\times \frac{2}{5}$

$x\ =\ \frac{5}{2}$

(g) 7m + 19/2 = 13

Transposing $\frac{19}{2}$ to the RHS and then dividing both sides by 7 :

$7m\ =\ 13\ -\ \frac{19}{2}\ =\ \frac{7}{2}$

$m\ =\ \frac{1}{2}$

(h) 6z + 10 = –2

Transposing 10 to the RHS and then dividing both sides by 6, we get :

$6z\ =\ -\ 2\ -\ 10\ =\ -\ 12$

$z\ =\ -\ 2$

(i) $\frac{3l}{2} = 2/3$

Multiplying both sides by $\frac{2}{3}$ ,

$l\ =\ \frac{2}{3}\times \frac{2}{3}\ =\ \frac{4}{9}$

(j) $\frac{2b}{3}- 5 = 3$

Transposing 5 to the RHS and then multiplying both sides by $\frac{3}{2}$

$\frac{2b}{3}\ =\ 8$

$b\ =\ 8\times \frac{3}{2}\ =\ 12$

(a) 2(x + 4) = 12

(b) 3(n – 5) = 21

(c) 3(n – 5) = – 21

(d) – 4(2 + x) = 8

(e) 4(2 – x) = 8

(a) We have:

2(x + 4) = 12

Dividing both sides by 2, we have :

$x\ +\ 4\ =\ 6$

Transposing 4 to the RHS, we get :

$x\ =\ 6\ -\ 4\ =\ 2$

Thus $x\ =\ 2$

(b) We have:

3(n – 5) = 21

Dividing both sides by 3, we have :

$n\ -\ 5\ =\ 7$

Transposing - 5 to the RHS, we get :

$n\ =\ 7\ +\ 5\ =\ 12$

Thus $n\ =\ 12$

(c) We have :

3(n – 5) = – 21

Dividing both sides by 3, we have :

$n\ -\ 5\ =\ -\ 7$

Transposing - 5 to the RHS, we get :

$n\ =\ -\ 7\ +\ 5\ =\ -\ 2$

Thus $n\ =\ -\ 2$

(d) We have :

– 4(2 + x) = 8

Dividing both sides by - 4, we have :

$x\ +\ 2\ =\ -\ 2$

Transposing 2 to the RHS, we get :

$x\ =\ -\ 2\ -\ 2\ =\ -\ 4$

Thus $x\ =\ -\ 4$

(e) We have :

4(2 - x) = 8

Dividing both sides by 4, we have :

$2\ -\ x\ =\ 2$

Transposing x to the RHS and 2 to the LHS , we get :

$x\ =\ 2\ -\ 2\ =\ 0$

Thus $x\ =\ 0$

(a) 4 = 5(p – 2)

(b) – 4 = 5(p – 2)

(c) 16 = 4 + 3(t + 2)

(d) 4 + 5(p – 1) =34

(e) 0 = 16 + 4(m – 6)

(a) 4 = 5(p – 2)

Dividing both sides by 5, we get :

$p\ -\ 2\ =\ \frac{4}{5}$

or $p\ =\ \frac{4}{5}\ +\ 2\ =\ \frac{14}{5}$

(b) – 4 = 5(p – 2)

Dividing both sides by 5, we get :

$p\ -\ 2\ =\ \frac{-4}{5}$

or $p\ =\ \frac{-4}{5}\ +\ 2\ =\ \frac{6}{5}$

(c) 16 = 4 + 3(t + 2)

Transposing 4 to the LHS and then dividing both sides by 3, we get :

$16\ -\ 4\ =\ 3(t\ +\ 2)$

or $t\ +\ 2\ =\ 4$

or $t\ =\ 2$

(d) 4 + 5(p – 1) =34

Transposing 4 to the RHS and then dividing both sides by 5, we get :

$5(p\ -\ 1)\ =\ 30$

or $p\ -\ 1\ =\ 6$

or $p\ =\ 7$

(e) 0 = 16 + 4(m – 6)

Transposing 16 to the LHS, we get :

$-\ 16\ =\ 4(m\ -\ 6)$

or $m\ -\ 6\ =\ -4$

or $m\ =\ 2$

(a) The 3 required equations can be :

$7x\ =\ 14$

$7x\ +\ 2 =\ 16$

$3x\ =\ 6$

(b) The required equations are :

$7x\ =\ -\ 14$

$7x\ +\ 2 =\ -\ 12$

$-\ 3x\ =\ 6$

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4

Let the number in each case be n.

(a) According to the question: $8n\ +\ 4\ =\ 60$

or $8n\ =\ 60\ -\ 4\ =\ 56$

or $n\ =\ 7$

(b) We have :

$\frac{n}{5}\ -\ 4\ =\ 3$

or $\frac{n}{5}\ =\ 7$

or $n\ =\ 35$

(c) The equation is :

$\frac{3n}{4}\ +\ 3\ =\ 21$

or $\frac{3n}{4}\ =\ 18$

or $n\ =\ 24$

(d) We have :

$2n\ -\ 11\ =\ 15$

or $2n\ =\ 26$

or $n\ =\ 13$

(e) The equation is :

$50\ -\ 3n\ =\ 8$

or $3n\ =\ 42$

or $n\ =\ 14$

(f) We have :

$\frac{n+19}{5}\ =\ 8$

or $n\ +\ 19\ =\ 40$

or $n\ =\ 21$

(g) We have :

$\frac{5n}{2}\ -\ 7\ =\ 23$

or $\frac{5n}{2}\ =\ 30$

or $n\ =\ 12$

(a) Let the lowest score be l.

Then according to the question, we have :

$2l\ +\ 7\ =\ 87$

or $2l\ =\ 80$

or $l\ =\ 40$

Thus the lowest marks are 40.

(b) Let the base angle of a triangle is $\Theta$ .

Then according to the question, we get :

$\Theta \ +\ \Theta\ +\ 40^{\circ}\ =\ 180^{\circ}$

or $2\Theta\ +\ 40^{\circ}\ =\ 180^{\circ}$

or $2\Theta\ =\ 140^{\circ}$

or $\Theta \ =\ 70^{\circ}$

(c) Let the runs scored by Rahul is x. Then runs by Sachin is 2x.

Further, it is given that their runs fell two short of a double century.

Thus we have : $x\ +\ 2x\ =\ 198$

or $3x\ =\ 198$

or $x\ =\ 66$ .

Hence runs by Rahul is 66 and the runs scored by Sachin is 132.

(a) Let the number of Parmit's marble be n.

Then according to question, we have :

$5n\ +\ 7\ =\ 37$

or $5n\ =\ 37\ -\ 7$

or $5n\ =\ 30$

or $n\ =\ 6$

(b) Let the age of Laxmi be x.

Then according to the question, we have :

$3x\ +\ 4\ =\ 49$

or $3x\ =\ 49\ -\ 4\ =\ 45$

or $x\ =\ 15$

(c) Let the number of fruit trees planted be z.

Then according to the question, we have :

$3z\ +\ 2\ =\ 77$

or $3z\ =\ 77\ -\ 2\ =\ 75$

or $z\ =\ 25$

Let the number be x.

According to the question the equation is :

$7x\ +\ 50\ +\ 40\ =\ 300$

or $7x\ +\ 90\ =\ 300$

or $7x\ =\ 300\ -\ 90\ =\ 210$

or $x\ =\ 30$

Hence the number is 30.

## More About Simple Equations Class 6 Maths NCERT Chapter 4

NCERT Class 7 Maths book chapter 4 Simple Equations begin with a fun game 'mind-reading' and introduce the concept of the equation. So what is an equation? It is a condition on the variable such that two expressions in the variable should have equal value. The value of the variable for which an equation is satisfied is called the solution of the equation. In the equation, there is always an equality(=) sign. The equality sign shows that the value of the expression to the left-hand side (LHS) of the equality sign is equal to the value of the expression to the right-hand side (RHS) of the equality sign. For example, the equation 3x+ 7 =2x - 35 has the expression (3x + 7) on the left of the equality sign and (2x - 35) on the right of the equality sign. But it is not an equation if there is a sign other than the equality sign. For example, 3x + 5 > 75 is not an equation.

## Simple Equations Class 6 Maths Chapter 4-Topic

• Setting Up Of An Equation
• Review Of What We Know
• What Equation Is?
• More Equations
• From Solution To Equation
• Application Of Simple Equations To Practical Situations

### NCERT Solutions for Class 7 Maths Chapter Wise

 Chapter No. Chapter Name Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangle 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

### NCERT Solutions for Class 7 Subject Wise

 NCERT Solutions for Class 7 Maths NCERT Solutions for Class 7 Science

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

As far as the subject Mathematics is concerned, the practising problem is important to score well in exams. It is important to know how to apply the concepts studied in an application-level problem. This is achieved through practice and the CBSE NCERT solutions for Class 7 Maths chapter 4 Simple Equations help for the same. The solutions of NCERT Class 7 Maths chapter 4 Simple Equations are helpful in solving homework problems.

Also Check NCERT Books and NCERT Syllabus here:

1. What is a simple equation, and how do I solve one?

According to chapter 4 maths class 7, A simple equation is a mathematical statement that shows that two expressions are equal. It has an equal sign (=) between the two sides.

To solve a simple equation, aim to isolate the variable (usually represented by a letter) on one side of the equation.

Perform the same operations on both sides of the equation to keep it balanced. This maintains the equality of the expressions.

The goal is to find the value of the variable that satisfies the equation. This value will make both sides of the equation equal.

Also Practice chapter 4 maths class 7 solutions to command the concepts.

2. What are extraneous solutions discussed n class 7th chapter 4 maths, and how do I avoid them?

According to class 7th chapter 4 maths, an extraneous solution is a solution that appears to work in the process of solving an equation but doesn't satisfy the original equation.

They often arise from operations that introduce restrictions, like division by zero or square roots of negative numbers.

To avoid extraneous solutions, always check the solutions you find by substituting them back into the original equation. If the substituted value doesn't satisfy the equation, it's not a valid solution.

Practice class 7th chapter 4 maths solutions to get in-depth understanding of the concepts.

3. What are the topics covered in NCERT solution Maths Class 7 Chapter 4?

Here is the  topics are covered in NCERT solution Maths Class 7 chapter 4

• Setting Up Of An Equation
• Review Of What We Know
• What Equation Is?
• More Equations
• From Solution To Equation
• Application Of Simple Equations To Practical Situations

Also To get command on the concepts, practice class 7 chapter 4 maths solutions.

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