## NCERT Solutions for class 7 Maths Chapter 4: Simple Equations## Exercise 4.1
1 + 5 = 19 6 = 19 LHS is not equal to RHS Hence, the equation is
7 × (- 2) + 5 = 19 - 14 + 5 = 19 - 9 = 19 LHS is not equal to RHS Hence, the equation is
7 × (2) + 5 = 19 14 + 5 = 19 19 = 19 LHS is equal to RHS Hence, the equation is
4 × (1) - 3 = 13 4 - 3 = 13 1 = 13 LHS is not equal to RHS Hence, the equation is
4 × (- 4) - 3 = 13 - 16 - 3 = 13 - 19 = 13 LHS is not equal to RHS Hence, the equation is
4 × (0) - 3 = 13 0 - 3 = 13 - 3 = 13 LHS is not equal to RHS Hence, the equation is
(i) 5p + 2 = 17
Substituting the value 5 × (1) + 2 = 17 5 + 2 = 17 7 = 17 LHS is not equal to RHS The equation is Substituting the value 5 × (2) + 2 = 17 10 + 2 = 17 12 = 17 LHS is not equal to RHS The equation is Substituting the value 5 × (3) + 2 = 17 15 + 2 = 17 17 = 17 LHS is equal to RHS Hence, the equation is
Substituting the value 3 × (3) - 14 = 4 9 - 14 = 4 -5 = 4 LHS is not equal to RHS The equation is Substituting the value 3 × (4) - 14 = 4 12 - 14 = 4 -2 = 4 LHS is not equal to RHS The equation is Substituting the value 3 × (5) - 14 = 4 15 - 14 = 4 1 = 4 LHS is not equal to RHS The equation is Substituting the value 3 × (6) - 14 = 4 18 - 14 = 4 4 = 4 LHS is equal to RHS Hence, the equation is
Number of marbles with Irfan = 5m + 7
Age of Laxmi's father = 3y + 4, which is equal to 49. Thus, the equation becomes, 3y + 4 = 49
Highest marks scored by a student = 2l + 7 The highest score is equal to 87. Thus, the equation becomes, 2l + 7 = 87
Two base angles of an Let the base angle be b. Vertex angle = 2b Base angle 1 + base angle 2 + vertex angle = 180° b + b + 2b = 180° 4b = 180° ## Exercise 4.2
(a) x - 1 = 0
The variable x has the (-1) value with it. To make it 0, we need to add +1 to it. Adding 1 to both the sides, we get: x - 1 + 1 = 0 + 1
Thus, the equation is solved at the value x = 1.
The variable x has the (+1) value with it. To make it 0, we need to subtract 1 from it. Subtracting 1 from both the sides, we get: x + 1 - 1 = 0 - 1
Thus, the equation is solved at the value x = - 1.
The variable x has the (-1) value with it. To make it 0, we need to add +1 to it. Adding 1 to both the sides, we get: x - 1 + 1 = 5 + 1
Thus, the equation is solved at the value x = 6.
The variable x has the (+6) value with it. To make it 0, we need to subtract 6 from it. Subtracting 6 from both the sides, we get: x + 6 - 6 = 2 - 6
Thus, the equation is solved at the value x = - 4.
The variable y has the (-4) value with it. To make it 0, we need to add +4 to it. Adding 4 to both the sides, we get: y - 4 + 4 = - 7 + 4
Thus, the equation is solved at the value y = - 3.
The variable y has the (-4) value with it. To make it 0, we need to add +4 to it. Adding 4 to both the sides, we get: y - 4 + 4 = 4 + 4
Thus, the equation is solved at the value y = 8.
The variable y has the (+4) value with it. To make it 0, we need to subtract 4 from it. Subtracting 4 from both the sides, we get: y + 4 - 4 = 4 - 4
Thus, the equation is solved at the value y = 0.
The variable y has the (+4) value with it. To make it 0, we need to subtract 4 from it. Subtracting 4 from both the sides, we get: y + 4 - 4 = - 4 - 4
Thus, the equation is solved at the value y = - 8.
The variable l has the value 3 with it. We need to divide the equation by 3 to remove the number with the variable. Dividing 3 on both the sides of the equation we get: 3l/3 = 42/3 l = 14 Thus, the equation is solved at the value l = 14.
The variable b has the value 1/2 with it. We need to multiply the equation by 2 to remove the number with the variable. Multiplying 2 on both the sides of the equation, we get: 2 × (b/2) = 2 × 6 b = 12 Thus, the equation is solved at the value b = 12.
The variable p has the value 1/7 with it. We need to multiply the equation by 7 to remove the number with the variable. Multiplying 7 on both the sides of the equation, we get: 7 × (p/7) = 7 × 4 p = 28 Thus, the equation is solved at the value p = 28.
The variable x has the value 4 with it. We need to divide the equation by 4 to remove the number with the variable. Dividing 4 on both the sides of the equation we get: 4x/4 = 25/4 X = 25/4 Thus, the equation is solved at the value x = 25/4.
The variable y has the value 8 with it. We need to divide the equation by 8 to remove the number with the variable. Dividing 8 on both the sides of the equation we get: 8y/8 = 36/8 y = 36/8 Thus, the equation is solved at the value y = 36/8.
The variable z has the value 1/3 with it. We need to multiply the equation by 3 to remove the number with the variable. Multiplying 3 on both the sides of the equation, we get: 3 × (z/3) = 3 × 5/4 p = 15/4 Thus, the equation is solved at the value p = 15/4.
The variable 'a' has the value 1/5 with it. We need to multiply the equation by 5 to remove the number with the variable. Multiplying 5 on both the sides of the equation, we get: 5 (a/5) = 5 × 7/15 a = (5 × 7)/ 15 a = 7/3 Thus, the equation is solved at the value a = 7/3.
The variable t has the value 20 with it. We need to divide the equation by 20 to remove the number with the variable. Dividing 20 on both the sides of the equation we get: 20t/ 20 = - 10/ 20 t = - 1/2 Thus, the equation is solved at the value t = - 1/2.
Step 2: Divide both sides by 3; n = 16
To solve the equation, we need to find the value of the variable n.
Adding 2 to both the sides, we get: 3n - 2 + 2 = 46 + 2 3n = 48
Dividing the equation by 3, we get: 3n/3 = 48/3 n = 16 Thus, the equation is solved at the value n = 16.
Step 2: Divide both sides by 5; m = 2
To solve the equation, we need to find the value of the variable m.
Subtracting 7 from both the sides, we get: 5m + 7 - 7 = 17 - 7 5m = 10
Dividing the equation by 5, we get: 5m/5 = 10/5 n = 2 Thus, the equation is solved at the value n = 2.
To solve the equation, we need to find the value of the variable p.
Multiplying the equation by 3, we get: 20p/3 = 40 3 × (20p/3) = 3 × 40 20p = 120
Dividing the equation by 20, we get: 20p/20 = 120/20 p = 6 Thus, the equation is solved at the value p = 6.
To solve the equation, we need to find the value of the variable p.
Multiplying the equation by 10, we get: 3p/10 = 6 10 × (3p/10) = 10 × 6 3p = 60
Dividing the equation by 3, we get: 3p/3 = 60/3 p = 20 Thus, the equation is solved at the value p = 20.
10p = 100 Dividing the equation by 10 on both the sides, we get: 10p/10 = 100/10 p = 10
10p + 10 = 100 Subtract 10 from both the sides of the equation, 10p + 10 - 10 = 100 - 10 10p = 90 Dividing the equation by 10 on both the sides, we get: 10p/10 = 90/10 p = 9
p/4 = 5 Multiplying the equation by 4 on both the sides, we get: 4 × (p/4) = 5 × 4 p = 20
-p/3 = 5 Multiplying the equation by 3 on both the sides, we get: 3 × (-p/3) = 3 × 4 p = -15
Multiplying the equation by 4 on both the sides, we get: 4 × (3p/4) = 6 × 4 3p = 24 Dividing the equation by 3 on both the sides, 3p/3 = 24/3 p = 8 (3 × 8 = 24)
Dividing the equation by 3 on both the sides, 3s/3 = -9/3 s = -3 (3 × 3 = 9)
The variable y has the (+12) value with it. To make it 0, we need to subtract 12 from it. Subtracting 12 from both the sides, we get: 3s + 12 - 12 = 0 - 12 3s = - 12 Now, dividing the equation by 3 on both the sides, 3s/3 = - 12/3 s = -4 (3 × 4 = 12)
3s = 0 Dividing the equation by 3 on both the sides, we get: 3s/3 = 0/3 s = 0 (3 × 1 = 3) 0 divided by any number results in 0 only.
The variable has number 2 with it. We need to divide the equation by 2 to find the value of the variable. Dividing the equation by 2 on both the sides, we get: 2q/2 = 6/2 q = 3
The variable y has the (-6) value with it. To make it 0, we need to add 6 to it. Adding 6 to both the sides, we get: 2q + 6 - 6 = 0 + 6 2q = 6 Dividing the equation by 2 on both the sides, we get: 2q/2 = 6/2 q = 3
The variable y has the (+6) value with it. To make it 0, we need to subtract 6 from it. Subtracting 6 from both the sides, we get: 2q + 6 - 6 = 0 - 6 2q = - 6 Dividing the equation by 2 on both the sides, we get: 2q/2 = -6/2 q = -3
The variable y has the (+6) value with it. To make it 0, we need to subtract 6 from it. Subtracting 6 from both the sides, we get: 2q + 6 - 6 = 12 - 6 2q = 6 Dividing the equation by 2 on both the sides, we get: 2q/2 = 6/2 q = 3 ## Exercise 4.3
4y + 5 = 37 Moving 5 to RHS, 4y = 37 - 5 4y = 32 y = 32/4 y = 8
Substitute the value of y in LHS, 2y + 5/2 = 2 × 8 + 5/2 = 16 + 5/2 = (32 + 5)/2 = 37/2 LHS = RHS Hence, verified.
Moving 28 to the other side, 5t = 10 - 28 When the number is moved to the other side, we change its sign. 5t = -18 t = -18/5
Substitute the value of t in LHS, 5t + 28 = 5 × (-18)/5 + 28 = -18 + 28 = 10 LHS = RHS Hence, verified.
Moving 3 to the other side, a/5 = 2 - 3 a/5 = -1 When the number is moved to the other side, we change its sign. a = (-1)(5) a = -5
Substitute the value of a in LHS, a/5 + 3 (-5)/5 + 3 = -1 + 3 = 3 - 1 = 2 LHS = RHS Hence, verified.
q/4 = 5 - 7 q/4 = -2 Multiplying the equation by 4 on both the sides, we get: q = -8
Substitute the value of q in LHS, q/4 + 7 (-8)/4 + 7 -2 + 7 = 7 - 2 = 5 LHS = RHS Hence, verified.
Multiplying the equation by 2 on both the sides, we get: 5x = -10 x = -10/5 x = -2
Substitute the value of x in LHS, 5x/2 5 (-2)/2 = -5 LHS = RHS Hence, verified.
Multiplying the equation by 4 on both the sides, we get: 10x = 25 Dividing the equation by 10 on both the sides, we get: 10x/10 = 25/10 x = 25/10 x = 5/2
Substitute the value of x in LHS, 5x/2 = 5(5/2) /2 = 25/4 LHS = RHS Hence, verified.
Multiplying the equation by 2 on both the sides, we get: 14m + 19 = 26 Taking 19 to the other side, 14m = 26 - 19 14m = 7 m = 7/14 m = 1/2 (7 × 2 = 14)
Substitute the value of m in LHS, 7m + 19/2 7(1/2) + 19/2 7/2 + 19/2 = (7 + 19)/2 = 26/2 = 13 LHS = RHS Hence, verified.
Taking 10 to the other side, we get: 6z When the number is moved to the other side, we change its sign. 6z = -12 z = -12/6 z = -2
Substitute the value of z in LHS, 6z + 10 6(-2) + 10 -12 + 10 = -2 LHS = RHS Hence, verified.
Multiplying the equation by 6 on both the sides, we get: (3l/2) × 6 = (2/3) × 6 (3l) × 3 = (2) × 2 9l = 4 l = 4/9
Substitute the value of l in LHS, 3l/2 = 3/2 × 4/9 = (3 × 4)/ (2 × 9) = 2/3 LHS = RHS Hence, verified.
Multiplying the equation by 3 on both the sides, we get: 2b - 15 = 9 Taking 15 to the other side, we get: 2b = 9 + 15 When the number is moved to the other side, we change its sign. 2b = 24 b = 24/2 b = 12
To verify, substitute the value of 12 in LHS, 2b/3 - 5 (2 × 12)/3 - 5 = 24/3 - 5 = 8 - 5 = 3 LHS = RHS Hence, verified.
Dividing both the sides by 2 to remove the brackets, we get: x + 4 = 6 Moving 4 to the other side, x = 6 - 4 x = 2
Substituting the value of x in LHS, 2(x + 4) = 2 (2 + 4) = 2 (6) = 12 LHS = RHS Hence, verified.
Dividing both the sides by 3 to remove the brackets, we get: n - 5 = 7 Moving 5 to the other side, n = 7 + 5 n = 12
Substituting the value of n in LHS, 3(n - 5) = 3 (12 - 5) = 3 (7) = 21 LHS = RHS Hence, verified.
Dividing both the sides by 3 to remove the brackets, we get: n - 5 = - 7 Moving 5 to the other side, n = -7 + 5 n = -2
Substituting the value of n in LHS, 3(n - 5) = 3 (-2 - 5) = 3 (-7) = - 21 LHS = RHS Hence, verified.
Dividing both the sides by (-4) to remove the brackets, we get: 2 + x = -2 x = -2 - 2 x = -4
Substituting the value of x in LHS, - 4 (2 + x) = - 4 (2 - 4) = - 4 (- 2) = 8 LHS = RHS Hence, verified.
Dividing both the sides by 4 to remove the brackets, we get: 2 - x = 2 x = 2 - 2 x = 0
Substituting the value of x in LHS, 4(2 - x) = 4 (2 - 0) = 4 (2) = 8 LHS = RHS Hence, verified.
Dividing both the sides by 5 to remove the brackets, we get: 4/5 = p - 2 (We can also reverse the sides as per our convenience. The answer in both the cases would be the same.) p - 2 = 4/5 Moving 2 to the other side, we get: p = 4/5 + 2 p = (4 + 10)/5 p = 14/5
To verify, we need to substitute the value of the variable present on the either side. Here, the variable is present on the right side. So, we will substitute the value of p in RHS. 5(p - 2) = 5 (14/5 - 2) = 5 ((14 - 10)/5) = 5 (4/5) = 4 LHS = RHS Hence, verified.
Dividing both the sides by 5 to remove the brackets, we get: - 4/5 = p - 2 (We can also reverse the sides as per our convenience. The answer in both the cases would be the same.) p - 2 = - 4/5 Moving 2 to the other side, p = - 4/5 + 2 p = (10 - 4)/5 p = 6/5
To verify, we need to substitute the value of the variable present on the either side. Here, the variable is present on the right side. So, we will substitute the value of p in RHS. 5(p - 2) = 5 (6/5 - 2) = 5 ((6 - 10)/5) = 5 (-4/5) = - 4 LHS = RHS Hence, verified.
(We can also reverse the sides as per our convenience. The answer in both the cases would be the same.) 4 + 3(t + 2) = 16 Moving 4 to the other side, 3(t + 2) = 16 - 4 3(t + 2) = 12 Dividing both the sides by 3 to remove the brackets, we get: t + 2 = 12/3 t + 2 = 4 t = 4 - 2 t = 2
To verify, we need substitute the value of p in RHS. 4 + 3(t + 2) = 4 + 3 (2 + 2) = 4 + 3 (4) = 4 + 12 = 16 LHS = RHS Hence, verified.
Moving 4 to the other side, 5 (p - 1) = 34 - 4 5 (p - 1) = 30 Dividing both the sides by 5 to remove the brackets, we get: p - 1 = 6 p = 6 + 1 p = 7
To verify, we need substitute the value of p in LHS. 4 + 5(p - 1) = 4 + 5(7 - 1) = 4 + 5 (6) = 4 + 30 = 34 LHS = RHS Hence, verified.
Moving 16 to the other side, -16 = 4 (m - 6) Dividing both the sides by 4 to remove the brackets, we get: -4 = m - 6 Moving 6 to the other side, -4 + 6 = m 2 = m
To verify, we need substitute the value of p in RHS. 16 + 4(m - 6) = 16 + 4 (2 - 6) = 16 + 4(-4) = 16 + (-16) = 16 - 16 = 0 LHS = RHS Hence, verified.
10x + 2 = 22 We can substitute any values to make the LHS and RHS equal. At x = 2, LHS will be: 10x + 2 = 10 (2) + 2 = 20 + 2 = 22 LHS = RHS Other examples of equation are: 5x + 5 = 15 6x + 2 = 14
x/5 = 2/5 Dividing the equation (x = 2) by 5 on both the sides, we get: x/5 = 2/5 We can divide the equation with any number of our choice. For example, x/3 = 2/3 x/10 = 2/10
5x - 3 = 7 We can substitute any values to make the LHS and RHS equal. At x = 2, LHS will be: 5 (2) - 3 = 10 - 3 = 7 LHS = RHS Other examples of equation are: 4x - 1 = 7 8x - 5 = 11
3x = - 6 We can substitute any values to make the LHS and RHS equal. Multiply the equation by 3 on both the sides, 3x = - 6 LHS: 3 (- 2) = - 6 LHS = RHS Other examples of equation are: 2x = - 4 7x = - 14
3x + 7 = 1 We can substitute any values to make the LHS and RHS equal. LHS: 3 (- 2) + 7 = - 6 + 7 = 1 LHS = RHS Other examples of equation are: 4x + 2 = - 6 6x + 1 = - 11
3x + 10 = 4 We can substitute any values to make the LHS and RHS equal. LHS: 3 (- 2) + 10 = - 6 + 10 = 4 LHS = RHS Other examples of equation are: 5x + 13 = 3 4x + 15 = 7 ## Exercise 4.4
Eight times a number = 8x Adding 4 to eight times a number = 8x + 4 So, 8x + 4 = 60 Let's solve the above equation. 8x + 4 = 60 Moving 4 to the other side, we get: 8x = 60 - 4 When a number is moved to the other side, we change its sign. 8x = 56 x = 56/8 x = 7
One-fifth of a number = x/5 One-fifth of a number minus 4 = x/5 - 4 So, x/5 - 4 = 3 Let's solve the above equation. x/5 - 4 = 3 Moving 4 to the other side, we get: x/5 = 3 + 4 x/5 = 7 When a number is moved to the other side, we change its sign. x = 7 × 5 x = 35
We can take any variable, such as, x, y, z, m, and n. Three-fourths of a number = 3z/4 Adding 3 to three-fourths of a number = 3z/4 + 3 So, 3/4z + 3 = 21 Let's solve the above equation. 3/4z + 3 = 21 Moving 4 to the other side, we get: 3z/4 = 21 - 3 3z/4 = 18 When a number is moved to the other side, we change its sign. z = (18 × 4)/3 z = (6 × 4) z = 24
Twice a number = 2y Subtracting 11 from twice a number = 2y - 11 So, 2y - 11 = 15 Let's solve the above equation. 2y - 11 = 15 Moving 11 to the other side, we get: 2y = 15 + 11 2y = 26 When a number is moved to the other side, we change its sign. y = 26/2 = 13
Thrice the number = 3y Subtracting 3y from 50 = 50 - 3y So, 50 - 3y = 8 Let's solve the above equation. 50 - 3y = 8 Moving 50 to the other side, we get: 50 = 8 + 3y Moving again 8 to the other side, 3y = 50 - 8 3y = 42 When a number is moved to the other side, we change its sign. y = 42/3 y = 14
Adding 19 to the number = x + 19 Dividing the sum by 5 = (x + 19)/5 So, (x + 19)/5 = 8 Let's solve the above equation. Multiplying the equation by 5, we get: x + 19 = 8 × 5 x + 19 = 40 Moving 19 to the other side, x = 40 - 19 x = 21 When a number is moved to the other side, we change its sign.
5/2 of the number = 5n/2 Taking away 7 from 5/2 of the number = 5n/2 - 7 So, 5n/2 - 7 = 23 Let's solve the above equation. 5n/2 - 7 = 23 Moving 7 to the other side, we get: 5n/2 = 23 + 7 5n/2 = 30 When a number is moved to the other side, we change its sign. 5n = 30 × 2 5n = 60 n = 60/5 n = 12
Let the lowest marks be Highest marks = 2x + 7 Highest score = 87 Equating both the values, we get: 2x + 7 = 87 2x = 87 - 7 2x = 80 x = 80/2 x = 40 Thus, the lowest score is obtained by a student in her class is 40.
Two base angles of an Let the base angle be x. Base angle 1 + base angle 2 + vertex angle = 180° x + x + 40° = 180° 2x + 40° = 180° 2x = 180° - 40° 2x = 140° x = 140°/2 x = 70° Thus, the value of each base angle is 70°.
Score of Sachin = 2x (Two short of a double century = 200 - 2 = 198) (1 century = 100 scores) Score of Rahul + Score of Sachin = Total score Total score = 198 x + 2x = 198 3x = 198 x = 198/3 x = 66 Thus, Score of Rahul = 66 runs 2x = 2 (66) = 132 Thus, Score of Sachin = 132 runs
(We can take any variable of our choice, such as x, y, z, m, n, l, k, and p.) Number of marbles with Irfan = 5z + 7 5z + 7 = 37 5z = 37 - 7 5z = 30 z = 30/5 z = 6 Thus, Parmit has 6 marbles.
Laxmi's father age = 3x + 4 (4 years older than three times Laxmi's age = 3x + 4) 3x + 4 = 49 Let's solve the above equation. 3x + 4 = 49 3x = 49 - 4 3x = 45 x = 45/3 x = 15 Thus, the age of Laxmi is 15 years.
Number of non-fruit trees = 3y + 2 3y + 2 = 77 3y = 77 - 2 3y = 75 y = 75/3 y = 25 Thus, the number of fruit trees planted was 25.
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!
Let the number be y.
Seven time over = 7 times the number = 7y
Adding 50 = 7y + 50
1 century = 100 Triple century = 300 = 7y + 50 = 300
= 7y + 50 + 40 = 300 7y + 90 = 300 Taking 90 to the other side, we get: 7y = 300 - 90 7y = 210 y = 210/7 y = 30 Next TopicClass 7 Maths chapter 5 |