NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations, are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

Question 1.
Complete the last column of the table.

Solution:

Question 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)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = – 4)
(f) 4p – 3 = 13 (p = 0).

Solution:
(a) n + 5 = 19 (n = 1)
L.H.S. = n + 5 = 1 + 5 | when n = 1 = 5
R.H.S. = 19
∵ L.H.S. ≠ R.H.S.
∴ n = 1 is not a solution to the given equation n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)
L.H.S. = 7n + 5 = 7(- 2) + 5 | when n = – 2 = – 14 + 5 = – 9
R.H.S. = 19
∵ L.H.S. ≠ R.H.S.
∴ n = – 2 is not a solution to the given equation 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
L.H.S. = In + 5 = 7(2) + 5 | when n = 2 = 14 + 5 = 19 = R.H.S.
∴ n = 2 is a solution to the given equation 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)
L.H.S. = 4p – 3 = 4(1) – 3 | when p = 1 = 4 – 3 = 1
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = 1 is not a solution to the given equation 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
L.H.S. = 4p – 3 = 4(- 4) – 3 , | when p = – 4 = – 16 – 3 = – 19
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = – 4 is not a solution to the given equation
4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
L.H.S. = 4 (p) – 3 = 4(0) – 3 | when p = 0 = 0 – 3 = – 3
R.H.S. = 13
∵ L.H.S. ≠ R.H.S.
∴ p = 0 is not a solution to the given equation 4p – 3 = 13.

Question 3.
Solve the following equations by trial and error method.

1. 5p + 2 = 17
2. 3m – 14 = 4.

Solution:
(i) 5p + 2 = 17

So, p = 3 is the solution of the given equation 5p + 2 = 17.

(ii) 3m – 14 = 4

So, m = 6 is the solution of the given equation 3m – 14 = 4.

Question 4.
Write equations for the following statements.

1. The sum of numbers x and 4 is 9.
2. 2 subtracted from y is 8.
3. Ten times a is 70.
4. The number b divided by 5 gives 6.
5. Three-fourth oft is 15.
6. Seven times m plus 7 gets you 77.
7. One-fourth of a number x minus 4 gives 4.
8. If you take away 6 from 6 times y, you get 60.
9. If you add 3 to one-third of z, you get 30.

Solution:

1. x + 4 = 9
2. y – 2 = 8
3. 10 α = 70
4. \(\frac { b }{ 5 } \) = 6
5. \(\frac { 3 }{ 4 } \) t = 15
6. 7m + 7 = 77
7. \(\frac { 1 }{ 4 } \) x – 4 = 4
8. 6y – 6 = 60
9. \(\frac { 1 }{ 3 } \) z + 3 = 30

Question 5.
Write the following equations in statement forms:

1. p + 4 = 15
2. m – 7 = 3
3. 2m = 7
4. \(\frac { m }{ 5 } \) = 3
5. \(\frac { 3m }{ 5 } \) = 6
6. 3p + 4 = 25
7. 4p – 2 = 18
8. \(\frac { p }{ 2 } \) + 2 = 8.

Solution:

1. The sum of p and 4 is 15.
2. 7 subtracted from m is 3.
3. Twice a number m is 7.
4. One-fifth of a number m is 3.
5. Three-fifth of a number m is 6.
6. Three times a numberp when added to 4 gives 25.
7. 2 subtracted from four times a number p is 18.
8. Add 2 to half of a number p to get 8.

Question 6.
Set up an equation in the following cases:

1. 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.)
2. 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.)
3. 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.)
4. 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 triangle is 180 degrees).

Solution:
(i) Let Parmit has m marbles. Then, five times the marbles Parmit has = 5m
Irfan has 7 marbles more than five times the marbles Parmit has = 5m + 7,
i.e., Irfan has (5m + 7) marbles
But it is given that Irfan has 37 marbles. Therefore, 5m + 7 = 37

(ii) Let the age of Laxmi be y years.
Then, three times Laxmi’s age = 3y years Laxmi’s father is 4 years older than three times Laxmi’s age,
i.e., Age of Laxmi’s father = (3y + 4) years But it is given that Laxmi’s father is 49 years old Therefore, 3y + 4 = 49

(iii) Let the lowest marks be l.
Then, twice the lowest marks = 21
Highest score obtained by a student in her class is twice the lowest marks plus 7, i.e.,
Highest score = 2l + 7
But this is given to be 87
Therefore, 2l + 7 = 87

(iv) Let the base angle be b in degrees. Then the vertex angle is 2b in degrees.
∵ Sum of the angles of a triangle is 180 degrees.
∴ 2 b + b + b = 180° or 4 b = 180°

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Question 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
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4

Solution:
(a) The given equation is x – 1 = 0
Add 1 to both sides,
x – 1 + 1 = 0 + 1 ⇒ x = 1
It is the required solution.
Check. Put the solution x = 1 back into the equation.
L.H.S. = x – 1 = 1 = 1 – 0 = R.H.S.
The solution is thus checked for its correctness.

(b) The given equation is x + 1 = 0
Subtract 1 from both sides, x + 1 – 1 = 0 – 1 ⇒ x = – 1
It is the required solution.
Check. Put the solution x = – 1 back into the equation.
L.H.S. = x + 1 = (-1)+1
= 0 = R.H.S.
The solution is thus checked for its correctness.

(c) The given equation is
x – 1 = 5
Add 1 to both sides,
x + 1 – 1 = 5 + 1 ⇒ x = 6
It is the required solution
Check. Put the solution x = 6 back into the equation.
L.H.S. = x – 1 = 6 – 1 = 5 = R.H.S.
The solution is thus checked for its correctness.

(d) The given equation is x + 6 = 2
Subtract 6 from both sides, x + 6 – 6 = 2 – 6 ⇒ x = – 4
It is the required solution.
Check. Put the solution x = – 4 back into the equation.
L.H.S. = x + 6 = – 4 + 6 = 2 = R.H.S.
The solution is thus checked for its correctness.

(e) The given equation is y – 4 = – 7
Add 4 to both sides, y – 4 + 4 = – 7 + 4 ⇒ y = – 3
It is the required solution.
Check. Put the solution
L.H.S. = y – 4 = – 3 – 4 = – 7 = R.H.S.
The solution is thus checked for its correctness.

(f) The given equation is y – 4 = 4
Add 4 to both sides,
y – 4 + 4 = 4 + 4 ⇒ y = 8
It is the required solution.
Check. Put the solution y = 8 back into the equation.
L.H.S. = y – 4 = 8 – 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(g) The given equation is y + 4 = 4
Subtract 4 from both sides, y + 4 – 4 = 4 – 4 ⇒ y = 0
It is the required solution.
Check. Put the solution y = 0 back into the equation.
L.H.S. =y + 4 = 0 + 4 = 4 = R.H.S.
The solution is thus checked for its correctness.

(h) The given equation is y + 4 = – 4
Subtract 4 from both sides, y + 4 – 4 = – 4 – 4 ⇒ y = -8
It is the required solution.
Check. Put the solution y = – 8 back into the equation.
L.H.S. = y + 4 = -8 + 4 = – 4 = R.H.S.
The solution is thus checked for its correctness.

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

(a) 3l = 42
(b) \(\frac { b }{ 2 } \) = 6
(c) \(\frac { p }{ 7 } \) = 4
(d) 4x = 25
(e) 8y = 36
(f) \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)
(g) \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)
(h) 20t = – 10

Solution:

Question 3.
Give the steps you will use to separate the variable and then solve the equation :
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) \(\frac { 20p }{ 3 } \) = 40
(d) \(\frac { 3p }{ 10 } \) = 6
Solution:

Question 4.
Solve the following equations:

Solution:

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

Question 1.
Solve the following equations :

Solution:

Question 2.
Solve the following equations :

(a) 2 (x + 4) = 12
(6) 3 (n – 5) = 21
(c) 3 (n – 5) = – 21
(d) -4 (2 + x) = 8
(e) 4 (2 – x) = 8

Solution:

Question 3.
Solve the /bilowing equations :

(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).

Solution:

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2.
Solution:
(a)
1. Start with x = 2
Multiply both sides by 3, 3x = 6
Subtract 2 from both sides, 3x – 2 = 4 …(1)

2. Start with x = 2
Multiply both sides by 4, 4 x = 8
Add 5 to both sides, 4x + 5 = 13 …(2)

3. Start with x = 2 Multiply both sides by 5 5x = 10
Subtract 1 from both sides, 5x – 1 = 9 …(3)

(b)
1. Start with x = – 2
Multiply both sides by 3, 3x = – 6
Subtract 2 from both sides, 3x – 2 = – 8 …(1)

2. Start with x = – 2
Multiply both sides by 5, 5x = – 10
Subtract 1 from both sides, 5x – 1 = – 11 …(2)

3. Start with x = – 2
Multiply both sides by 4, 4x = – 8
Add 5 to both sides 4x + 5 = – 3 …(3)

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 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) What 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 \(\frac { 5 }{ 2 } \) of the number, the result is 23.

Solution:

Question 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 mark 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°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(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?

Solution:

Question 3.
Solve the following :

1. 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?
2. Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
3. 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?

Solution:

Question 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 !
Solution:
Let ‘x’ be the number,
Then, according to the question, we get (x × 7) + 50 = 300 – 40
7x + 50 = 260
7x = 210
x = \(\frac { 210 }{ 7 } \) = 30
So, the number is 30.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations MCQ

Important Multiple Choice Questions

1. Write the following statement in the form of an equation:
The sum of three times x and 10 is 13.

(a) 3x + 10 = 13
(b) 3x – 10 = 13
(c) 3x + 13 = 10
(d) none of these.

2. Write the following statement in the form of an equation:
If you subtract 3 from 6 times a number, you get 9

(a) 3x – 6 = 9
(b) 6x – 3 = 9
(c) 6x + 3 = 9
(d) 3x + 6 = 9.

3. Write the following statement in the form of an equation:
One fourth of n is 3 more than 2

(a) \(\frac { n }{ 4 } \) – 2 = 3
(b) \(\frac { n }{ 4 } \) + 2 = 3
(c) \(\frac { n }{ 2 } \) – 4 = 3
(d) \(\frac { n }{ 2 } \) + 4 = 3

4. Write the following statement in the form of an equation:
One third of a number plus 2 is 3

(a) \(\frac { m }{ 3 } \) – 2 = 3
(b) \(\frac { m }{ 3 } \) + 2 = 3
(c) \(\frac { m }{ 2 } \) – 3 = 3
(d) \(\frac { m }{ 2 } \) + 3 = 3

5. Write the following statement in the form of an equation:
Taking away 5 from x gives 10

(a) x – 5 = 10
(b) x + 5 = 10
(c) x – 10 – 5
(d) none of these.

6. Write the following statement in the form of an equation:
Four times a number p is 8.

(a) 4P = 8
(b) P + 4 = 8
(c) p – 4 = 8
(d) p ÷ 4 = 8.

7. Write the following statement in the form of an equation:
Add 1 to three times n to get 7

(a) 3n + 1 = 7
(b) 3n – 1 = 7
(c) 3n + 7 = 1
(d) none of these.

8. Write the following statement in the form of an equation:
The number b divided by 6 gives 5.

(a) \(\frac { b }{ 6 } \) = 5
(b) b – 5 = 6
(c) 5b = 6
(d) b + 5 = 6.

9. The solution of the equation x + 3 = 0 is

(a) 3
(b) – 3
(c) 0
(d) 1

10. The solution of the equation x – 6 = 1 is

(a) 1
(b) 6
(c) – 7
(b) 7.

11. The solution of the equation 5x = 10 is

(a) 1
(b) 2
(c) 5
(d) 10.

12. The solution of the equation \(\frac { m }{ 2 } \) = 3 is

(a) 2
(b) 3
(c) 12
(d) 6.

13. The solution of the equation 7n + 5 = 12 is

(a) 0
(b) – 1
(c) 1
(d) 5.

14. The solution of the equation 4p – 3 = 9 is

(a) 1
(b) 2
(c) 3
(d) 4

15. The solution of the equation 5p + 2 = 7 is

(a) 0
(b) 1
(c) – 1
(d) 2

16. The solution of the equation 3p – 2 = 4 is

(a) 0
(b) 1
(c) 2
(d) 3

17. The solution of the equation p + 4 = 4 is

(a) 0
(b) 4
(c) – 4
(d) 8

18. The solution of the equation m – 1 = 2 is

(a) 1
(b) 2
(c) 3
(d) 6.

19. The solution of the equation 2m = 4 is

(a) 1
(b) 2
(c) -1
(d) -2.

20. The solution of the equation \(\frac { m }{ 3 } \) = 3 is

(a) 3
(b) 6
(c) 9
(d) 12.

21. The solution of the equation = 6 is

(a) 6
(b) 7
(c) 14
(d) 3

22. The solution of the equation 3p + 5 = 8 is

(a) -1
(b) 1
(c) 3
(d) 5

23. The solution of the equation 4p – 2 = 10 is

(a) 1
(b) 2
(c) 3
(d) 4

24. The solution of the equation \(\frac { p }{ 2 } \) + 1 = 3 is

(a) 1
(b) 2
(c) 3
(d) 4

25. The solution of the equation 3 m + 7=16 is

(a) 1
(b) 2
(c) 3
(d) 4

26. The solution of the equation 2p – 1 = 3 is

(a) 1
(b) 2
(c) 3
(d) 4

27. The solution of the equation 4x + 5 = 9 is

(a) 1
(b) 2
(c) 3
(d) 4

28. The solution of the equation 10 y – 20 = 30 is

(a) 1
(b) 2
(c) 3
(d) 5

29. The solution of the equation y – 4 = – 1 is

(a) 1
(b) 2
(c) 3
(d) 4

30. The solution of the equation y + 2 = – 2 is

(a) 2
(b) -2
(c) 4
(d) -4

31. The solution of the equation 10t = – 20 is

(a) 1
(b) -1
(c) 2
(d) -2

32. The solution of the equation \(\frac { z }{ 2 } \) = \(\frac { 3 }{ 4 } \) is

(a) \(\frac { 1 }{ 2 } \)
(b) \(\frac { 3 }{ 2 } \)
(c) \(\frac { 1 }{ 4 } \)
(d) \(\frac { 3 }{ 4 } \)

33. The solution of the equation 2s = 0 is

(a) 2
(b) -2
(c) 0
(d) \(\frac { 1 }{ 2 } \)

34. The solution of the equation – \(\frac { p }{ 7 } \) = 3 is

(a) -3
(b) -7
(c) -21
(d) 21

35. The solution of the equation 3s + 6 = 0 is

(a) 1
(b) -1
(c) 2
(d) -2

36. The solution of the equation 10p = 10 is

(a) 1
(b) – 1
(c) 10
(d) – 10.

37. The solution of the equation 10p + 10 = 110 is

(a) 10
(b) -10
(c) 100
(d) 110

38. The solution of the equation 12p – 11 = 13 is

(a) 1
(b) 2
(c) 3
(d) 4

39. The solution of the equation 2 (m + 3) = 8 is

(a) 1
(b) 2
(c) 3
(d) 4

40. The solution of the equation – 2(x + 3) = 4 is

(a) -2
(b) -3
(c) -4
(d) -5

41. The solution of the equation \(\frac { 5 }{ 2 } \) x = 15 is

(a) 2
(b) 4
(c) 6
(d) 10

42. The solution of the equation 4(2 – x) = 4 is

(a) 1
(b) 2
(c) 3
(d) 4

43. The solution of the equation – 4(2 + x) = 4 is

(a) – 1
(b) – 2
(c) – 3
(d) – 4.

44. The solution of the equation – 4 = 2 (p – 2) is

(a) 0
(b) 1
(c) 2
(d) 4.

45. The solution of the equation 0 = 4 + 4(m + 1) is

(a) 1
(b) – 1
(c) 2
(d) – 2.

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