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NCERT Solutions for class 7 Maths chapter 4 – Simple Equations

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Exercise 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 a = 70
4. 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 number p, 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 Permit has. Irfan has 37 marbles. (Take m to be the number of Permit’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 are 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 the number of marbles with Parmit be m.
Then, 7 added to 5 times mis 5m + 7
It is given that 7 marbles more than five times the marble is 37. Thus, the equation obtained is 5m + 7 = 37.

(ii) Let Laxmi’s age be y years. Then, 4 added to 3 times y is 3y + 4
It is given that the father is 4 years older than 3 times Laxmi’s age. His age is 49years.
Then, we have the following equation : 3y + 4 = 49

(iii) Let the lowest marks be l. Then, twice the lowest marks plus 7 is 2l +7
It is given that, the highest marks 87 obtained by a student is twice the lowest marks plus 7.
So, we have the following equation : 2l + 7 = 87

(iv) Let the base angle be b. Then, the vertex angle = 2b.
Since, sum of the angles of a triangle is 180°
∴ b + b + 2b = 180°
⇒ 4b = 180°
which is the required equation.

Exercise 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:

Exercise 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, 4x = 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) First equation:
Start with x = -2
Multiply both sides by 2, 2x = -4
Subtract 3 from both sides, 2x – 3 = -7

Second equation:
Start with x = – 2
Multiply both sides by – 5, – 5x = 10
Add 10 to both sides, 10 – 5x = 20

Third equation:
Start with x = -2
Divide both sides by 2, $\frac { x }{ 2 }$ = -1
Add 3 to both sides, $\frac { x }{ 2 }$ + 3 = 2

Exercise 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) 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 are 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. The people of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was 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 me 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.