NCERT Solutions (Part - 1) - Simple Equations Class 7 Notes | EduRev

Exercise 4.1 
Ques 1: Complete the last column of the table: 

Answer 1: 

Ques 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)
Ans: (a) n + 5 = 19 (n = 1)
Putting n = 1 in L.H.S.,
1 + 5 = 6
∵ L.H.S. ≠ R.H.S.,
∴ n = 1 is not the solution of given equation.
(b) 7n + 5 = 19(n = -2)
Putting n = -2 in L.H.S.,
7(-2) + 5 = -14 + 5 = -9
∵ L.H.S. ≠ R.H.S.,
∴ n = -2 is not the solution of given equation.
(c) 7n + 5 = 19(n = 2)
Putting n = 2 in L.H.S.,
7(2) + 5 = 14 + 5 = 19
∵ L.H.S. = R.H.S.,
∴ n = 2 is the solution of given equation.
(d) 4p - 3 = 13(p = -4)
Putting p = 1 in L.H.S.,
4(1)-3 = 4-3 = 1
∵  L.H.S. ≠ R.H.S.,
∴ p = 1  is not the solution of given equation.
(e) 4p-3 = 13 (p = -4)
Putting p = -4 in L.H.S.,
4(-4) - 3 = -16-3 = -19
∵  L.H.S. ≠ R.H.S.,
∴ p = -4 is not the solution of given equation.
(f) 4p-3 = 13(p = 0)
Putting p = 0 in L.H.S.,
4(0)-3 = 0-3 = -3
∵ L.H.S. ≠ R.H.S.,
∴ p = 0 is not the solution of given equation.

Ques 3: Solve the following equations by trial and error method: 
(i) 5p + 2 = 17
(ii) 3m - 14 = 4
Ans: (i) 5p + 2 = 17
Putting p = -3 in L.H.S. 5(-3) + 2 =-15+2 = -13
∵ -13≠17 Therefore, p = -3 is not the solution.

Putting p = -2 in L.H.S. 5(-2) +2 = -10+2 = -8
∵ -8≠17 Therefore, p = -2 is not the solution.

Putting p = -1 in L.H.S. 5(-1) + 2= -5 + 2 =-3
∵ -3≠17 Therefore, p = -1 is not the solution.

Putting p = 0 in L.H.S. 5(0)+ 2= 0 + 2 = 2
∵ 2≠17 Therefore, p = 0 is not the solution.

Putting p - 1 in L.H.S. 5(1) + 2= 5 + 2 = 7
∵ 7≠17 Therefore, p = 1 is not the solution.

Putting p = 2 in L.H.S. 5(2) + 2= 10+2 = 12
∵ 12≠17 Therefore, p = 2 is not the solution.

Putting p = 3 in L.H.S. 5(3)+ 2= 15 + 2 = 17
∵ 17 = 17 Therefore, p = 3 is the solution,

(ii) 3m-14 = 4
Putting m = -2 inL.H.S. 3(-2) -14 = -6-14 = -20
∵ -20 4 Therefore, m = -2 is not the solution.

Putting m = -1 in L.H.S- 3(-1)-14 = -3-14 = -17
∵ -17≠4 Therefore, m =-1 is not the solution.

Putting m = 0 in L.H.S. 3(0)-14 = 0-14 = -14
∵ -14≠4 Therefore, m = 0 is not the solution.

Putting m = 1 in L.H.S. 3(1)-14 = 3-14 =-11
∵ -11≠4 Therefore, m = 1 is not the solution.

Putting m = 2 in L.H.S. 3(2)- 14 = 6-14 = -8
∵ -8≠4 Therefore, m = 2 is not the solution.

Putting m = 3 in L.H.S. 3(3)- 14 = 9-14 = -5
∵ -5 ≠ 4 Therefore, m = 3 is not the solution.

Putting m = 4 in L.H.S. 3(4)-14 = 12-14 =-2
∵ -2≠4 Therefore, m = 4 is not the solution.

Putting m = 5 in L.H.S. 3(5)-14 = 15-14 = 1
∵ 1≠4 Therefore, m = 5 is not the solution.

Putting m = 6 in L.H.S. 3(6)-14 = 18-14 = 4
∵ 4 = 4 Therefore, m = 6 is the solution.

Ques 4: Write equations for the following statements: 
(i) The sum of numbers x and 4 is 9. 
(ii) 2 subtracted from y is 8. 
(iii) Ten times a is 70. 
(iv) The number b divided by 5 gives 6. 
(v) Three-fourth of t is 15. 
(vi) Seven times m plus 7 gets you 77. 
(vii) One-fourth of a number x minus 4 gives 4. 
(viii) If you take away 6 from 6 times y, you get 60. 
(ix) If you add 3 to one-third of z, you get 30. 
Ans: (i) x + 4 = 9
(ii) y - 2 = 8
(iii) 10a = 70
(iv) b/5 = 6
(v) 
(vi) 7m +7 = 77
(vii) 
(viii) 6y - 6 = 60
(ix) 

Ques 5: Write the following equations in statement form: 
(i) p + 4 = 15  
(ii)  m - 7 = 3
(iii) 2m = 7  
(iv) m/5 = 3
(v) 3m/5 = 6  
(vi) 3p + 4 = 25
(vii) 4p - 2 = 18
(viii) 
Ans: (i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Two times m is 7.
(iv) The number m is divided by 5 gives 3.
(v) Three-fifth of the number m is 6.
(vi) Three times p plus 4 gets 25.
(vii) If you take away 2 from 4 times p, you get 18.
(viii) If you added 2 to half is p, you get 8.

Ques 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.) 
(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.) 
(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) 
(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^o) 
Ans: (i) Let m be the number of Permit's marbles.
∴ 5m + 7 = 37
(ii) Let the age of Laxmi be y years.
∴  3y + 4 = 49
(ii) Let the lowest score be l.
∴  2l + 7 = 87
(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b.
∴  2b+b+b= 180°
⇒ 46 = 180°  [Angle sum property of a Δ]

Exercise 4.2 
Ques 1: Give first the step you will use to separate the variable and then solve the equations: 
(a) x - 1 = 0  
(b) x + 1 = 0
(c) x + 1 =5  
(d) x + 6 = 2
(e) y - 4 = -7  
(f) y - 4 = 4
(g) y + 4 = 4  
(h) y + 4 = -4
Ans: (a) x - 1 =0
⇒ x - 1 + 1 = 0 + 1  [Adding 1 both sides]
⇒  x = 1
(b) x + 1 = 0
⇒  x + 1 - 1 = 0 - 1   [Subtracting 1 both sides]
⇒  x = -1
(c)  x - 1 = 5
⇒  x - 1 + 1 = 5 + 1  [Adding 1 both sides]
⇒ x = 6
(d)  x + 6 = 2
⇒  x + 6 - 6 = 2 - 6  [Subtracting 6 both sides]
⇒ x = - 4
(e)  y - 4 = - 1
⇒ y - 4 + 4 = -7 +  4  [Adding4 both sides]
⇒  y = -3
(f)  y-4 = 4
⇒  y-4 + 4 = 4 + 4  [Adding 4 both sides]
⇒  y = 8
(g) y + 4 = 4
⇒  y + 4 - 4 = 4 - 4  [Subtracting 4 both sides]
⇒  y = 0
(h) y + 4 = -4
⇒  y + 4 - 4 = - 4 - 4  [Subtracting 4 both sides]
⇒  y = -8

Ques 2: Give first the step you will use to separate the variable and then solve the equations 
(a) 3l = 42
(b) b/2 = 6
(c) p/7 = 4
(d) 4x = 25
(e) 8y = 36
(f) z/3 = 5/4
(g) a/5 = 7/15 
(h) 20t = -10
Ans: (a) 3l = 42
⇒  [Dividing both sides by 3]
⇒ l = 14
(b) b/2 = 6  
⇒  [Dividing both sides by 3]
⇒ b = 12

(c) p/7 = 4
⇒  [Dividing both sides by 2]
⇒ p = 28
(d) 4x = 25
⇒  [Dividing both sides by 7]
⇒ x = 25/4
(e) 8y = 36
⇒  [Dividing both sides by 4]
⇒ y = 9/2

⇒  [Dividing both sides by 8]
⇒ z = 15/4

⇒  [Dividing both sides by 3]
⇒ a = 7/3
(h) 20t = -10
⇒  [Dividing both sides by 5]
⇒ t = -1/2

Ques 3: Give first the step you will use to separate the variable and then solve the equations 
(a) 3n - 2 = 46
(b) 5m + 7 = 17
(c) 20p/3 =40
(d) 3p/10 = 6
Ans: (a) 3n - 2 = 46
Step l: 3n-2+2 = 46 + 2
⇒ 3n = 48   [Adding 2 both sides]
Step II: 3n/3 = 48/3
⇒ n = 16    [Dividing both sides by 3]

(b) 5m+7 = 17
Step l: 5m + 7-7 = 17-7
⇒ 5m = 10     [Subtracting 7 both sides]
Step II:  5m/5 = 10/5
⇒ m = 2    [ Dividing both sides by 5]

(c) 20p/3 = 40
Step I: [Multiplying both sides by 3]
⇒ 20p = 120
Step II:    [Dividing both sides by 20]
⇒ p = 6

(d) 3p/10 = 6
Step I:   [Multiplying both sides by 10]
⇒ 3p = 60
Step II:  [Dividing both sides by 3]
⇒ p = 20

Ques 4: Solve the following equation: 
(a) 10p = 100
(b) 10p + 10 = 100
(c) p/4 = 5
(d) -p/3 = 5
(e) 3p/4 = 6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q - 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Ans:
(a) 10p = 100
⇒    [Dividing both sides by 10]
⇒ p = 10

(b) 10p+10 = 100
⇒  10p + 10-10 = 100-10    [Subtracting both sides 10]
⇒ 10 p = 90
 [Dividing both sides by 10]
⇒ p = 9

(c) p/4 = 5
⇒  [Multiplying both sides by 4]
⇒ p = 20

(d) -p/3 = 5
⇒  [Multiplying both sides by - 3]
⇒ p = -15

(e) 3p/4 = 6
⇒   [Multiplying both sides by 4]
⇒ 3p = 24
⇒  [Dividing both sides by 3]
⇒ p = 8

(f) 3s = -9
⇒    [Dividing both sides by 3]
⇒ s = -3

(g) 3s + 12 = 0
⇒ 3s + 12 - 12 = 0 - 12  [Subtracting both sides 10]
⇒ 3s = -12
⇒  [Dividing both sides by 3]
⇒ s = -4

(h) 3s = 0
⇒ 3s/3 = 0/3
⇒ 3s/3 = 0/3 [Dividing both sides by 3]
⇒ s = 0

(i) 2q = 6
⇒ 2q/2 = 6/2 [Dividing both sides by 2]
⇒ q = 3

(j) 2q - 6 = 0
⇒ 2q - 6 + = 0+6 [Adding both sides 6]
⇒ 2q = 6
⇒ 2q/2 = 6/2  [Dividing both sides by 2]
⇒ q = 3

(k) 2q + 6 = 0
⇒ 2q + 6 - 6 = 0-6 [Subtracting both sides 6]
⇒ 2q = -6
⇒ 2q/2 = -6/2 [Dividing both sides by 2]
⇒ q = -3

(l) 2q + 6 = 12
⇒ 2q + - 6 = 12 - 6  [Subtracting both sides 6]
⇒ 2q = 6
⇒ 2q/2 = 6/2  [Dividing both sides by 2]
⇒ q = 3