ncert-solutions-class-7-maths-ch-4-simple-equations

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation

NCERT solutions for class 7 maths chapter 4 simple equations topic 4.3

Q. The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10 y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.

Answer:

(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 class 7 maths chapter 4 simple equations topic 4.7

(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?

Answer:

(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,

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

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

$x\ =\ 9$

Q. There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?

Answer:

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 class 7 maths chapter 4 simple equations exercise 4.1

1. Complete the last column of the table.

Answer:

The table is shown below:-

2. Check whether the value given in the brackets is a solution to the given equation or not:

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

(b) 7n + 5 = 19 (n = – 2)

(e) 4p – 3 = 13 (p = – 4)

(f) 4p – 3 = 13 (p = 0)

Answer:

(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.

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.

3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4

Answer:

(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.

Answer:

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

5. Write the following equations in statement forms:

(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

Answer:

(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).

Answer:

(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 $3y\ =\ 45$

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

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 class 7 maths chapter 4 simple equations exercise 4.2

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

(d) x + 6 = 2

(e) y – 4 = – 7

(f) y – 4 = 4

(g) y + 4 = 4

(h) y + 4 = – 4

Answer:

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

$x\ =\ 1$

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

$x\ =\ -\ 1$

$x\ =\ 6$

(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 :

$y\ =\ -\ 8$

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

(a) 3l = 42

(b) b / 2 = 6

(d) 4x = 25

(e) 8y = 36

(f) z/ 3 = 5 /4

(g) a /5 =7/ 15

(h) 20t = – 10

Answer:

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

$l\ =\ 14$

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

$b\ =\ 12$

$p\ =\ 28$

(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 :

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

3. Give the steps you will use to separate the variable and then solve the equation:

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(d) 3p/10 = 6

Answer:

(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 $m\ =\ 2$

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 $p\ =\ 20$

4. Solve the following equations:

(a) 10p = 100

(b) 10p + 10 = 100

(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

Answer:

(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 : $p\ =\ 9$

$p\ =\ 20$

(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

1. Solve the following equations:

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

(b) 5t + 28 = 10

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

Answer:

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

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 :

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

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

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 :

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

<img alt="m\ =\ \frac{1}{2}" height="37"

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(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 $\frac{3}{2}$

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

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

2. Solve the following equations:

(a) 2(x + 4) = 12

(b) 3(n – 5) = 21

(d) – 4(2 + x) = 8

(e) 4(2 – x) = 8

Answer:

(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 $n\ =\ 12$

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 $x\ =\ 0$

Q3 Solve the following equations:

(a) 4 = 5(p – 2)

(b) – 4 = 5(p – 2)

(d) 4 + 5(p – 1) =34

(e) 0 = 16 + 4(m – 6)

Answer:

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

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 $m\ =\ 2$

4. (a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2

Answer:

(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

1. Set up equations and solve them to find the unknown numbers in the following cases:
(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 add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 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 5/2 of the number, the result is 23.

Answer:

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 $n\ =\ 35$

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

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 $n\ =\ 12$

2. Solve the following:
(a) 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. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is $40\degree$ . What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180\degree$ ).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

Answer:

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

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?

Answer:

(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 $x\ =\ 15$

Then according to question, we have :

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

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

or $z\ =\ 25$

4. Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!

Answer:

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.

Responsive Table

Chapter No.Chapter Name

Chapter 1	NCERT solutions for class 7 maths chapter 1 Integers
Chapter 2	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	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	NCERT solutions for class 7 maths chapter 7 Congruence of Triangles
Chapter 8	NCERT solutions for class 7 maths chapter 8 comparing quantities
Chapter 9	NCERT solutions for class 7 maths chapter 9 Rational Numbers
Chapter 10	NCERT solutions for class 7 maths chapter 10 Practical Geometry
Chapter 11	NCERT solutions for class 7 maths chapter 11 Perimeter and Area
Chapter 12	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	NCERT solutions for class 7 maths chapter 14 Symmetry
Chapter 15	NCERT solutions for class 7 maths chapter 15 visualising-solid-shapes

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NCERT solutions for class 7 maths chapter 4 simple equations topic 4.3

NCERT solutions for class 7 maths chapter 4 simple equations topic 4.7

NCERT solutions for class 7 maths chapter 4 simple equations exercise 4.1

NCERT solutions for class 7 maths chapter 4 simple equations exercise 4.2

NCERT solutions for class 7 maths chapter 4 simple equations exercise 4.3

NCERT solutions for class 7 maths chapter 4 simple equations exercise 4.4

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