NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations are provided to aid the students while preparing for their exams, as well as assignments. For students who feel stressed about solving the most comprehensive and detailed NCERT Solutions for Class 7 Maths, we at CoolGyan’S, have prepared step by step solutions with detailed descriptions. We suggest students who aspire to score good marks in Maths, go through these solutions and increase their knowledge.
Chapter 4 – Simple Equations of NCERT Solutions for Class 7 Maths contains 4 exercises. Let’s now look at the important topics covered in this chapter are mentioned below.
- What is Equation
- Solving an Equation
- More Equations
- From Solution to Equation
- Applications of Simple Equations To Practical Situations
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Exercise 4.1 Page: 81
1. Complete the last column of the table.
| S. No. | Equation | Value | Say, whether the equation is satisfied. (Yes/No) |
| (i) | x + 3 = 0 | x = 3 | |
| (ii) | x + 3 = 0 | x = 0 | |
| (iii) | x + 3 = 0 | x = -3 | |
| (iv) | x – 7 = 1 | x = 7 | |
| (v) | x – 7 = 1 | x = 8 | |
| (vi) | 5x = 25 | x = 0 | |
| (vii) | 5x = 25 | x = 5 | |
| (viii) | 5x = 25 | x = -5 | |
| (ix) | (m/3) = 2 | m = – 6 | |
| (x) | (m/3) = 2 | m = 0 | |
| (xi) | (m/3) = 2 | m = 6 |
Solution:-
(i) x + 3 = 0
LHS = x + 3
By substituting the value of x = 3
Then,
LHS = 3 + 3 = 6
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.
(ii) x + 3 = 0
LHS = x + 3
By substituting the value of x = 0
Then,
LHS = 0 + 3 = 3
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.
(iii) x + 3 = 0
LHS = x + 3
By substituting the value of x = – 3
Then,
LHS = – 3 + 3 = 0
By comparing LHS and RHS
LHS = RHS
∴Yes, the equation is satisfied
(iv) x – 7 = 1
LHS = x – 7
By substituting the value of x = 7
Then,
LHS = 7 – 7 = 0
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied
(v) x – 7 = 1
LHS = x – 7
By substituting the value of x = 8
Then,
LHS = 8 – 7 = 1
By comparing LHS and RHS
LHS = RHS
∴Yes, the equation is satisfied.
(vi) 5x = 25
LHS = 5x
By substituting the value of x = 0
Then,
LHS = 5 × 0 = 0
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.
(vii) 5x = 25
LHS = 5x
By substituting the value of x = 5
Then,
LHS = 5 × 5 = 25
By comparing LHS and RHS
LHS = RHS
∴Yes, the equation is satisfied.
(viii) 5x = 25
LHS = 5x
By substituting the value of x = -5
Then,
LHS = 5 × (-5) = – 25
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.
(ix) m/3 = 2
LHS = m/3
By substituting the value of m = – 6
Then,
LHS = -6/3 = – 2
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.
(x) m/3 = 2
LHS = m/3
By substituting the value of m = 0
Then,
LHS = 0/3 = 0
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.
(xi) m/3 = 2
LHS = m/3
By substituting the value of m = 6
Then,
LHS = 6/3 = 2
By comparing LHS and RHS
LHS = RHS
∴Yes, the equation is satisfied.
| S. No. | Equation | Value | Say, whether the equation is satisfied. (Yes/No) |
| (i) | x + 3 = 0 | x = 3 | No |
| (ii) | x + 3 = 0 | x = 0 | No |
| (iii) | x + 3 = 0 | x = -3 | Yes |
| (iv) | x – 7 = 1 | x = 7 | No |
| (v) | x – 7 = 1 | x = 8 | Yes |
| (vi) | 5x = 25 | x = 0 | No |
| (vii) | 5x = 25 | x = 5 | Yes |
| (viii) | 5x = 25 | x = -5 | No |
| (ix) | (m/3) = 2 | m = – 6 | No |
| (x) | (m/3) = 2 | m = 0 | No |
| (xi) | (m/3) = 2 | m = 6 | Yes |
2. Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
Solution:-
LHS = n + 5
By substituting the value of n = 1
Then,
LHS = n + 5
= 1 + 5
= 6
By comparing LHS and RHS
6 ≠ 19
LHS ≠ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b) 7n + 5 = 19 (n = – 2)
Solution:-
LHS = 7n + 5
By substituting the value of n = -2
Then,
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing LHS and RHS
-9 ≠ 19
LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
Solution:-
LHS = 7n + 5
By substituting the value of n = 2
Then,
LHS = 7n + 5
= (7 × (2)) + 5
= 14 + 5
= 19
By comparing LHS and RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.
(d) 4p – 3 = 13 (p = 1)
Solution:-
LHS = 4p – 3
By substituting the value of p = 1
Then,
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing LHS and RHS
1 ≠ 13
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.
(e) 4p – 3 = 13 (p = – 4)
Solution:-
LHS = 4p – 3
By substituting the value of p = – 4
Then,
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing LHS and RHS
-19 ≠ 13
LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.
(f) 4p – 3 = 13 (p = 0)
Solution:-
LHS = 4p – 3
By substituting the value of p = 0
Then,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
Solution:-
LHS = 5p + 2
By substituting the value of p = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing LHS and RHS
2 ≠ 17
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation.
Let, p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing LHS and RHS
7 ≠ 17
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation.
Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing LHS and RHS
12 ≠ 17
LHS ≠ RHS
Hence, the value of p = 2 is not a solution to the given equation.
Let, p = 3
LHS = 5p + 2
= (5 × 3) + 2
= 15 + 2
= 17
By comparing LHS and RHS
17 = 17
LHS = RHS
Hence, the value of p = 3 is a solution to the given equation.
(ii) 3m – 14 = 4
Solution:-
LHS = 3m – 14
By substituting the value of m = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS and RHS
-5 ≠ 4
LHS ≠ RHS
Hence, the value of m = 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing LHS and RHS
-2 ≠ 4
LHS ≠ RHS
Hence, the value of m = 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS and RHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m = 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is a solution to the given equation.
4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
Solution:-
The above statement can be written in the equation form as,
= x + 4 = 9
(ii) 2 subtracted from y is 8.
Solution:-
The above statement can be written in the equation form as,
= y – 2 = 8
(iii) Ten times a is 70.
Solution:-
The above statement can be written in the equation form as,
= 10a = 70
(iv) The number b divided by 5 gives 6.
Solution:-
The above statement can be written in the equation form as,
= (b/5) = 6
(v) Three-fourth of t is 15.
Solution:-
The above statement can be written in the equation form as,
= ¾t = 15
(vi) Seven times m plus 7 gets you 77.
Solution:-
The above statement can be written in the equation form as,
Seven times m is 7m
= 7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
Solution:-
The above statement can be written in the equation form as,
One-fourth of a number x is x/4
= x/4 – 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
Solution:-
The above statement can be written in the equation form as,
6 times of y is 6y
= 6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30.
Solution:-
The above statement can be written in the equation form as,
One-third of z is z/3
= 3 + z/3 = 30
5. Write the following equations in statement forms:
(i) p + 4 = 15
Solution:-
The sum of numbers p and 4 is 15.
(ii) m – 7 = 3
Solution:-
7 subtracted from m is 3.
(iii) 2m = 7
Solution:-
Twice of number m is 7.
(iv) m/5 = 3
Solution:-
The number m divided by 5 gives 3.
(v) (3m)/5 = 6
Solution:-
Three-fifth of m is 6.
(vi) 3p + 4 = 25
Solution:-
Three times p plus 4 gives you 25.
(vii) 4p – 2 = 18
Solution:-
Four times p minus 2 gives you 18.
(viii) p/2 + 2 = 8
Solution-
If you add half of a number p to 2, you 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.)
Solution:-
From the question it is given that,
Number of Parmit’s marbles = m
Then,
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having
= (5 × m) + 7 = 37
= 5m + 7 = 37
(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.)
Solution:-
From the question it is given that,
Let Laxmi’s age to be = y years old
Then,
Lakshmi’s father is 4 years older than three times of her age
= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father
= (3 × y) + 4 = 49
= 3y + 4 = 49
(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.)
Solution:-
From the question it is given that,
Highest score in the class = 87
Let lowest score be l
= 2 × Lowest score + 7 = Highest score in the class
= (2 × l) + 7 = 87
= 2l + 7 = 87
(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).
Solution:-
From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Then,
Vertex angle = 2 × base angle = 2b
= b + b + 2b = 180o
= 4b = 180o
Exercise 4.2 Page: 86
1. Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
Solution:-
We have to add 1 to both the side of given equation,
Then we get,
= x – 1 + 1 = 0 + 1
= x = 1
(b) x + 1 = 0
Solution:-
We have to subtract 1 to both the side of given equation,
Then we get,
= x + 1 – 1 = 0 – 1
= x = – 1
(c) x – 1 = 5
Solution:-
We have to add 1 to both the side of given equation,
Then we get,
= x – 1 + 1 = 5 + 1
= x = 6
(d) x + 6 = 2
Solution:-
We have to subtract 6 to both the side of given equation,
Then we get,
= x + 6 – 6 = 2 – 6
= x = – 4
(e) y – 4 = – 7
Solution:-
We have to add 4 to both the side of given equation,
Then we get,
= y – 4 + 4 = – 7 + 4
= y = – 3
(f) y – 4 = 4
Solution:-
We have to add 4 to both the side of given equation,
Then we get,
= y – 4 + 4 = 4 + 4
= y = 8
(g) y + 4 = 4
Solution:-
We have to subtract 4 to both the side of given equation,
Then we get,
= y + 4 – 4 = 4 – 4
= y = 0
(h) y + 4 = – 4
Solution:-
We have to subtract 4 to both the side of given equation,
Then we get,
= y + 4 – 4 = – 4 – 4
= y = – 8
2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
Solution:-
Now we have to divide both sides of the equation by 3,
Then we get,
= 3l/3 = 42/3
= l = 14
(b) b/2 = 6
Solution:-
Now we have to multiply both sides of the equation by 2,
Then we get,
= b/2 × 2= 6 × 2
= b = 12
(c) p/7 = 4
Solution:-
Now we have to multiply both sides of the equation by 7,
Then we get,
= p/7 × 7= 4 × 7
= p = 28
(d) 4x = 25
Solution:-
Now we have to divide both sides of the equation by 4,
Then we get,
= 4x/4 = 25/4
= x = 25/4
(e) 8y = 36
Solution:-
Now we have to divide both sides of the equation by 8,
Then we get,
= 8y/8 = 36/8
= x = 9/2
(f) (z/3) = (5/4)
Solution:-
Now we have to multiply both sides of the equation by 3,
Then we get,
= (z/3) × 3 = (5/4) × 3
= x = 15/4
(g) (a/5) = (7/15)
Solution:-
Now we have to multiply both sides of the equation by 5,
Then we get,
= (a/5) × 5 = (7/15) × 5
= a = 7/3
(h) 20t = – 10
Solution:-
Now we have to divide both sides of the equation by 20,
Then we get,
= 20t/20 = -10/20
= x = – ½
3. Give the steps you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
Solution:-
First we have to add 2 to the both sides of the equation,
Then, we get,
= 3n – 2 + 2 = 46 + 2
= 3n = 48
Now,
We have to divide both sides of the equation by 3,
Then, we get,
= 3n/3 = 48/3
= n = 16
(b) 5m + 7 = 17
Solution:-
First we have to subtract 7 to the both sides of the equation,
Then, we get,
= 5m + 7 – 7 = 17 – 7
= 5m = 10
Now,
We have to divide both sides of the equation by 5,
Then, we get,
= 5m/5 = 10/5
= m = 2
(c) 20p/3 = 40
Solution:-
First we have to multiply both sides of the equation by 3,
Then, we get,
= (20p/3) × 3 = 40 × 3
= 20p = 120
Now,
We have to divide both sides of the equation by 20,
Then, we get,
= 20p/20 = 120/20
= p = 6
(d) 3p/10 = 6
Solution:-
First we have to multiply both sides of the equation by 10,
Then, we get,
= (3p/10) × 10 = 6 × 10
= 3p = 60
Now,
We have to divide both sides of the equation by 3,
Then, we get,
= 3p/3 = 60/3
= p = 20
4. Solve the following equations:
(a) 10p = 100
Solution:-
Now,
We have to divide both sides of the equation by 10,
Then, we get,
= 10p/10 = 100/10
= p = 10
(b) 10p + 10 = 100
Solution:-
First we have to subtract 10 to the both sides of the equation,
Then, we get,
= 10p + 10 – 10 = 100 – 10
= 10p = 90
Now,
We have to divide both sides of the equation by 10,
Then, we get,
= 10p/10 = 90/10
= p = 9
(c) p/4 = 5
Solution:-
Now,
We have to multiply both sides of the equation by 4,
Then, we get,
= p/4 × 4 = 5 × 4
= p = 20
(d) – p/3 = 5
Solution:-
Now,
We have to multiply both sides of the equation by – 3,
Then, we get,
= – p/3 × (- 3) = 5 × (- 3)
= p = – 15
(e) 3p/4 = 6
Solution:-
First we have to multiply both sides of the equation by 4,
Then, we get,
= (3p/4) × (4) = 6 × 4
= 3p = 24
Now,
We have to divide both sides of the equation by 3,
Then, we get,
= 3p/3 = 24/3
= p = 8
(f) 3s = – 9
Solution:-
Now,
We have to divide both sides of the equation by 3,
Then, we get,
= 3s/3 = -9/3
= s = -3
(g) 3s + 12 = 0
Solution:-
First we have to subtract 12 to the both sides of the equation,
Then, we get,
= 3s + 12 – 12 = 0 – 12
= 3s = -12
Now,
We have to divide both sides of the equation by 3,
Then, we get,
= 3s/3 = -12/3
= s = – 4
(h) 3s = 0
Solution:-
Now,
We have to divide both sides of the equation by 3,
Then, we get,
= 3s/3 = 0/3
= s = 0
(i) 2q = 6
Solution:-
Now,
We have to divide both sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3
(j) 2q – 6 = 0
Solution:-
First we have to add 6 to the both sides of the equation,
Then, we get,
= 2q – 6 + 6 = 0 + 6
= 2q = 6
Now,
We have to divide both sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3
(k) 2q + 6 = 0
Solution:-
First we have to subtract 6 to the both sides of the equation,
Then, we get,
= 2q + 6 – 6 = 0 – 6
= 2q = – 6
Now,
We have to divide both sides of the equation by 2,
Then, we get,
= 2q/2 = – 6/2
= q = – 3
(l) 2q + 6 = 12
Solution:-
First we have to subtract 6 to the both sides of the equation,
Then, we get,
= 2q + 6 – 6 = 12 – 6
= 2q = 6
Now,
We have to divide both sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3
Exercise 4.3 Page: 89
1. Solve the following equations:
(a) 2y + (5/2) = (37/2)
Solution:-
By transposing (5/2) from LHS to RHS it becomes -5/2
Then,
= 2y = (37/2) – (5/2)
= 2y = (37-5)/2
= 2y = 32/2
Now,
Divide both side by 2,
= 2y/2 = (32/2)/2
= y = (32/2) × (1/2)
= y = 32/4
= y = 8
(b) 5t + 28 = 10
Solution:-
By transposing 28 from LHS to RHS it becomes -28
Then,
= 5t = 10 – 28
= 5t = – 18
Now,
Divide both side by 5,
= 5t/5= -18/5
= t = -18/5
(c) (a/5) + 3 = 2
Solution:-
By transposing 3 from LHS to RHS it becomes -3
Then,
= a/5 = 2 – 3
= a/5 = – 1
Now,
Multiply both side by 5,
= (a/5) × 5= -1 × 5
= a = -5
(d) (q/4) + 7 = 5
Solution:-
By transposing 7 from LHS to RHS it becomes -7
Then,
= q/4 = 5 – 7
= q/4 = – 2
Now,
Multiply both side by 4,
= (q/4) × 4= -2 × 4
= a = -8
(e) (5/2) x = -5
Solution:-
First we have to multiply both the side by 2,
= (5x/2) × 2 = – 5 × 2
= 5x = – 10
Now,
We have to divide both the side by 5,
Then we get,
= 5x/5 = -10/5
= x = -2
(f) (5/2) x = 25/4
Solution:-
First we have to multiply both the side by 2,
= (5x/2) × 2 = (25/4) × 2
= 5x = (25/2)
Now,
We have to divide both the side by 5,
Then we get,
= 5x/5 = (25/2)/5
= x = (25/2) × (1/5)
= x = (5/2)
(g) 7m + (19/2) = 13
Solution:-
By transposing (19/2) from LHS to RHS it becomes -19/2
Then,
= 7m = 13 – (19/2)
= 7m = (26 – 19)/2
= 7m = 7/2
Now,
Divide both side by 7,
= 7m/7 = (7/2)/7
= m = (7/2) × (1/7)
= m = ½
(h) 6z + 10 = – 2
Solution:-
By transposing 10 from LHS to RHS it becomes – 10
Then,
= 6z = -2 – 10
= 6z = – 12
Now,
Divide both side by 6,
= 6z/6 = -12/6
= m = – 2
(i) (3/2) l = 2/3
Solution:-
First we have to multiply both the side by 2,
= (3l/2) × 2 = (2/3) × 2
= 3l = (4/3)
Now,
We have to divide both the side by 3,
Then we get,
= 3l/3 = (4/3)/3
= l = (4/3) × (1/3)
= x = (4/9)
(j) (2b/3) – 5 = 3
Solution:-
By transposing -5 from LHS to RHS it becomes 5
Then,
= 2b/3 = 3 + 5
= 2b/3 = 8
Now,
Multiply both side by 3,
= (2b/3) × 3= 8 × 3
= 2b = 24
And,
Divide both side by 2,
= 2b/2 = 24/2
= b = 12
2. Solve the following equations:
(a) 2(x + 4) = 12
Solution:-
Let us divide both the side by 2,
= (2(x + 4))/2 = 12/2
= x + 4 = 6
By transposing 4 from LHS to RHS it becomes -4
= x = 6 – 4
= x = 2
(b) 3(n – 5) = 21
Solution:-
Let us divide both the side by 3,
= (3(n – 5))/3 = 21/3
= n – 5 = 7
By transposing -5 from LHS to RHS it becomes 5
= n = 7 + 5
= n = 12
(c) 3(n – 5) = – 21
Solution:-
Let us divide both the side by 3,
= (3(n – 5))/3 = – 21/3
= n – 5 = -7
By transposing -5 from LHS to RHS it becomes 5
= n = – 7 + 5
= n = – 2
(d) – 4(2 + x) = 8
Solution:-
Let us divide both the side by -4,
= (-4(2 + x))/ (-4) = 8/ (-4)
= 2 + x = -2
By transposing 2 from LHS to RHS it becomes – 2
= x = -2 – 2
= x = – 4
(e) 4(2 – x) = 8
Solution:-
Let us divide both the side by 4,
= (4(2 – x))/ 4 = 8/ 4
= 2 – x = 2
By transposing 2 from LHS to RHS it becomes – 2
= – x = 2 – 2
= – x = 0
= x = 0
3. Solve the following equations:
(a) 4 = 5(p – 2)
Solution:-
Let us divide both the side by 5,
= 4/5 = (5(p – 2))/5
= 4/5 = p -2
By transposing – 2 from RHS to LHS it becomes 2
= (4/5) + 2 = p
= (4 + 10)/ 5 = p
= p = 14/5
(b) – 4 = 5(p – 2)
Solution:-
Let us divide both the side by 5,
= – 4/5 = (5(p – 2))/5
= – 4/5 = p -2
By transposing – 2 from RHS to LHS it becomes 2
= – (4/5) + 2 = p
= (- 4 + 10)/ 5 = p
= p = 6/5
(c) 16 = 4 + 3(t + 2)
Solution:-
By transposing 4 from RHS to LHS it becomes – 4
= 16 – 4 = 3(t + 2)
= 12 = 3(t + 2)
Let us divide both the side by 3,
= 12/3 = (3(t + 2))/ 3
= 4 = t + 2
By transposing 2 from RHS to LHS it becomes – 2
= 4 – 2 = t
= t = 2
(d) 4 + 5(p – 1) =34
Solution:-
By transposing 4 from LHS to RHS it becomes – 4
= 5(p – 1) = 34 – 4
= 5(p – 1) = 30
Let us divide both the side by 5,
= (5(p – 1))/ 5 = 30/5
= p – 1 = 6
By transposing – 1 from RHS to LHS it becomes 1
= p = 6 + 1
= p = 7
(e) 0 = 16 + 4(m – 6)
Solution:-
By transposing 16 from RHS to LHS it becomes – 16
= 0 – 16 = 4(m – 6)
= – 16 = 4(m – 6)
Let us divide both the side by 4,
= – 16/4 = (4(m – 6))/ 4
= – 4 = m – 6
By transposing – 6 from RHS to LHS it becomes 6
= – 4 + 6 = m
= m = 2
4. (a) Construct 3 equations starting with x = 2
Solution:-
First equation is,
Multiply both side by 6
= 6x = 12 … [equation 1]
Second equation is,
Subtracting 4 from both side,
= 6x – 4 = 12 -4
= 6x – 4 = 8 … [equation 2]
Third equation is,
Divide both side by 6
= (6x/6) – (4/6) = (8/6)
= x – (4/6) = (8/6) … [equation 3]
(b) Construct 3 equations starting with x = – 2
Solution:-
First equation is,
Multiply both side by 5
= 5x = -10 … [equation 1]
Second equation is,
Subtracting 3 from both side,
= 5x – 3 = – 10 – 3
= 5x – 3 = – 13 … [equation 2]
Third equation is,
Dividing both sides by 2
= (5x/2) – (3/2) = (-13/2) … [equation 3]
Exercise 4.4 Page: 91
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.
Solution:-
Let us assume the required number be x
Eight times a number = 8x
The given above statement can be written in the equation form as,
= 8x + 4 = 60
By transposing 4 from LHS to RHS it becomes – 4
= 8x = 60 – 4
= 8x = 56
Divide both side by 8,
Then we get,
= (8x/8) = 56/8
= x = 7
(b) One-fifth of a number minus 4 gives 3.
Solution:-
Let us assume the required number be x
One-fifth of a number = (1/5) x = x/5
The given above statement can be written in the equation form as,
= (x/5) – 4 = 3
By transposing – 4 from LHS to RHS it becomes 4
= x/5 = 3 + 4
= x/5 = 7
Multiply both side by 5,
Then we get,
= (x/5) × 5 = 7 × 5
= x = 35
(c) If I take three-fourths of a number and add 3 to it, I get 21.
Solution:-
Let us assume the required number be x
Three-fourths of a number = (3/4) x
The given above statement can be written in the equation form as,
= (3/4) x + 3 = 21
By transposing 3 from LHS to RHS it becomes – 3
= (3/4) x = 21 – 3
= (3/4) x = 18
Multiply both side by 4,
Then we get,
= (3x/4) × 4 = 18 × 4
= 3x = 72
Then,
Divide both side by 3,
= (3x/3) = 72/3
= x = 24
(d) When I subtracted 11 from twice a number, the result was 15.
Solution:-
Let us assume the required number be x
Twice a number = 2x
The given above statement can be written in the equation form as,
= 2x –11 = 15
By transposing -11 from LHS to RHS it becomes 11
= 2x = 15 + 11
= 2x = 26
Then,
Divide both side by 2,
= (2x/2) = 26/2
= x = 13
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Solution:-
Let us assume the required number be x
Thrice the number = 3x
The given above statement can be written in the equation form as,
= 50 – 3x = 8
By transposing 50 from LHS to RHS it becomes – 50
= – 3x = 8 – 50
= -3x = – 42
Then,
Divide both side by -3,
= (-3x/-3) = – 42/-3
= x = 14
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Solution:-
Let us assume the required number be x
The given above statement can be written in the equation form as,
= (x + 19)/5 = 8
Multiply both side by 5,
= ((x + 19)/5) × 5 = 8 × 5
= x + 19 = 40
Then,
By transposing 19 from LHS to RHS it becomes – 19
= x = 40 – 19
= x = 21
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
Solution:-
Let us assume the required number be x
5/2 of the number = (5/2) x
The given above statement can be written in the equation form as,
= (5/2) x – 7 = 23
By transposing -7 from LHS to RHS it becomes 7
= (5/2) x = 23 + 7
= (5/2) x = 30
Multiply both side by 2,
= ((5/2) x) × 2 = 30 × 2
= 5x = 60
Then,
Divide both the side by 5
= 5x/5 = 60/5
= x = 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?
Solution:-
Let us assume the lowest score be x
From the question it is given that,
The highest score is = 87
Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7
5/2 of the number = (5/2) x
The given above statement can be written in the equation form as,
Then,
= 2x + 7 = Highest score
= 2x + 7 = 87
By transposing 7 from LHS to RHS it becomes -7
= 2x = 87 – 7
= 2x = 80
Now,
Divide both the side by 2
= 2x/2 = 80/2
= x = 40
Hence, the lowest score is 40
(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°).
Solution:-
From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Then,
= b + b + 40o = 180o
= 2b + 40 = 180o
By transposing 40 from LHS to RHS it becomes -40
= 2b = 180 – 40
= 2b = 140
Now,
Divide both the side by 2
= 2b/2 = 140/2
= b = 70o
Hence, 70o is the base angle of an isosceles triangle.
(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:-
Let us assume Rahul’s score be x
Then,
Sachin scored twice as many runs as Rahul is 2x
Together, their runs fell two short of a double century,
= Rahul’s score + Sachin’s score = 200 – 2
= x + 2x = 198
= 3x = 198
Divide both the side by 3,
= 3x/3 = 198/3
= x = 66
So, Rahul’s score is 66
And Sachin’s score is 2x = 2 × 66 = 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?
Solution:-
Let us assume number of Parmit’s marbles = m
From the question it is given that,
Then,
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having
= (5 × m) + 7 = 37
= 5m + 7 = 37
By transposing 7 from LHS to RHS it becomes -7
= 5m = 37 – 7
= 5m = 30
Divide both the side by 5
= 5m/5 = 30/5
= m = 6
So, Permit has 6 marbles
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age.
What is Laxmi’s age?
Solution:-
Let Laxmi’s age to be = y years old
From the question it is given that,
Lakshmi’s father is 4 years older than three times of her age
= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father
= (3 × y) + 4 = 49
= 3y + 4 = 49
By transposing 4 from LHS to RHS it becomes -4
= 3y = 49 – 4
= 3y = 45
Divide both the side by 3
= 3y/3 = 45/3
= y = 15
So, Lakshmi’s age is 15 years.
(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?
Solution:-
Let the number of fruit tress be f.
From the question it is given that,
3 × number of fruit trees + 2 = number of non-fruit trees
= 3f + 2 = 77
By transposing 2 from LHS to RHS it becomes -2
=3f = 77 – 2
= 3f = 75
Divide both the side by 3
= 3f/3 = 75/3
= f = 25
So, number of fruit tree was 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!
Solution:-
Let us assume the number be x.
Take me seven times over and add a fifty = 7x + 50
To reach a triple century you still need forty = (7x + 50) + 40 = 300
= 7x + 50 + 40 = 300
= 7x + 90 = 300
By transposing 90 from LHS to RHS it becomes -90
= 7x = 300 – 90
= 7x = 210
Divide both side by 7
= 7x/7 = 210/7
= x = 30
Hence the number is 30.
