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.
160
(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.
98
(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.
143
(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.
120
NCERT Solutions for Class 7 Maths Chapter 3 Data Handling
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-fourths 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 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 be = y years old
Then,
Lakshmi’s father is 4 years older than three times 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 are 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 the 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 the base angle be b.
Then,
Vertex angle = 2 × base angle = 2b
= b + b + 2b = 180o
= 4b = 180o
SHINING A LIGHT ON SOLAR ENERGY: POWERING A SUSTAINABLE FUTURE
Exercise 4.2
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 sides of the given equation.
Then, we get
= x – 1 + 1 = 0 + 1
= x = 1
(b) x + 1 = 0
Solution:
We have to subtract 1 from both sides of the given equation.
Then, we get
= x + 1 – 1 = 0 – 1
= x = – 1
(c) x – 1 = 5
Solution:
We have to add 1 to both sides of the given equation.
Then, we get
= x – 1 + 1 = 5 + 1
= x = 6
(d) x + 6 = 2
Solution:
We have to subtract 6 from both sides of the given equation.
Then, we get
= x + 6 – 6 = 2 – 6
= x = – 4
(e) y – 4 = – 7
Solution:
We have to add 4 to both sides of the given equation.
Then, we get
= y – 4 + 4 = – 7 + 4
= y = – 3
(f) y – 4 = 4
Solution:
We have to add 4 to both sides of the given equation.
Then, we get
= y – 4 + 4 = 4 + 4
= y = 8
(g) y + 4 = 4
Solution:
We have to subtract 4 from both sides of the given equation.
Then, we get
= y + 4 – 4 = 4 – 4
= y = 0
(h) y + 4 = – 4
Solution:
We have to subtract 4 from both sides of the 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 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 from 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
57
(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 from 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 from 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 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 from 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 from 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
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 sides 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 sides 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 sides by 5.
= (a/5) × 5= -1 × 5
= a = -5
53
(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 sides by 4.
= (q/4) × 4= -2 × 4
= a = -8
(e) (5/2) x = -5
Solution:
First, we have to multiply both sides by 2.
= (5x/2) × 2 = – 5 × 2
= 5x = – 10
Now,
We have to divide both sides by 5.
Then, we get
= 5x/5 = -10/5
= x = -2
(f) (5/2) x = 25/4
Solution:
First, we have to multiply both sides by 2.
= (5x/2) × 2 = (25/4) × 2
= 5x = (25/2)
Now,
We have to divide both sides by 5.
Then, we get
= 5x/5 = (25/2)/5
= x = (25/2) × (1/5)
= x = (5/2)
78
(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 sides 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 sides by 6.
= 6z/6 = -12/6
= m = – 2
(i) (3/2) l = 2/3
Solution:
First, we have to multiply both sides by 2.
= (3l/2) × 2 = (2/3) × 2
= 3l = (4/3)
Now,
We have to divide both sides 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 sides by 3.
= (2b/3) × 3= 8 × 3
= 2b = 24
And,
Divide both sides by 2.
= 2b/2 = 24/2
= b = 12
2. Solve the following equations.
(a) 2(x + 4) = 12
Solution:
Let us divide both sides 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 sides 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 sides 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 sides 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 sides 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 sides 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 sides 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 sides 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 sides 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 sides 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:
The first equation is,
Multiply both sides by 6.
= 6x = 12 … [equation 1]
The second equation is,
Subtracting 4 from both sides,
= 6x – 4 = 12 -4
= 6x – 4 = 8 … [equation 2]
The third equation is,
Divide both sides by 6.
= (6x/6) – (4/6) = (8/6)
= x – (4/6) = (8/6) … [equation 3]
(b) Construct 3 equations starting with x = – 2
Solution:
The first equation is,
Multiply both sides by 5.
= 5x = -10 … [equation 1]
The second equation is,
Subtracting 3 from both sides,
= 5x – 3 = – 10 – 3
= 5x – 3 = – 13 … [equation 2]
The third equation is,
Dividing both sides by 2.
= (5x/2) – (3/2) = (-13/2) … [equation 3]
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.
Solution:
Let us assume the required number is 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 sides 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 is 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 sides 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 is 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 sides by 4.
Then, we get
= (3x/4) × 4 = 18 × 4
= 3x = 72
Then,
Divide both sides 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 is 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 sides by 2.
= (2x/2) = 26/2
= x = 13
(e) Munna subtracts thrice the number of notebooks he has from 50, and he finds the result to be 8.
Solution:
Let us assume the required number is 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 sides 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 is x.
The given above statement can be written in the equation form as,
= (x + 19)/5 = 8
Multiply both sides 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 is 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 sides by 2,
= ((5/2) x) × 2 = 30 × 2
= 5x = 60
Then,
Divide both sides 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 are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
Solution:
Let us assume the lowest score is x.
From the question, it is given that
The highest score is = 87
The highest marks obtained by a student in her class are 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 sides 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 the 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 sides 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 is 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 sides 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 the 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 sides by 5.
= 5m/5 = 30/5
= m = 6
So, the 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 be = y years old
From the question, it is given that
Lakshmi’s father is 4 years older than three times 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 sides 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 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:
Let the number of fruit trees 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 sides by 3.
= 3f/3 = 75/3
= f = 25
So, the number of fruit trees was 25.
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 us assume the number is 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 sides by 7.
= 7x/7 = 210/7
= x = 30
Hence, the number is 30.