# NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation

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 class 7 maths chapter 4 simple equations also discuss some topic wise questions. NCERT class 7 maths chapter 4 simple equations begins with a fun game 'mind-reading' and introduces 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. 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 NCERT solutions for class 7 maths chapter 4 simple equations as worksheets to prepare for their CBSE final exams. Here you will get solutions to all four exercises of this chapter.

Exercise:4.1

Exercise:4.2

Exercise:4.3

Exercise:4.4

Topics of NCERT Grade 7 Maths Chapter 4 Simple Equations-

4.2  Setting up of an Equation

4.3  Review of what we know

4.4  What Equation is?

4.4.1  Solving an Equation

4.5 More Equations

4.6 From Solution to Equation

4.7 Applications of Simple Equations to Practical Situations

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                                    $y\ =\ 7$

## (i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?

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

(i) Let the number be n.

Then according to question, we have :

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

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

or                                    $n\ =\ 2$

(ii) Let the number be x.

Then according to 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.

## 1. Complete the last column of the table.

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.

## 4. Write equations for the following statements: (i) The sum of numbers x and 4 is 9.

(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.

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

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

## 1. Give first the step you will use to separate the variable and then solve the equation:

(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 class 7 maths chapter 4 simple equations exercise 4.3

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

(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 triangle is $\Theta$.

Then according to 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 runs scored by Sachin is 132.

## 3.  Solve the following: (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have? (ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi's age? (iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the  number of non-fruit trees planted was 77?

(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 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 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.

## NCERT Solutions for Class 7 Maths Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 NCERT solutions for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

## 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.