Class 7 Maths Chapter 4 HOTS Questions - Simple Equations

Q1: Solve the following equations:
(a) 10p = 100
Ans:
Now,
We have to divide both of these sides of the equation by 10,
Then, we get,
= 10p/10 = 100/10
= p = 10

(b) 10p + 10 = 100
Ans:
Firstly we have to subtract 10 from both sides of the following equation,
Then, we get,
= 10p + 10 – 10 = 100 – 10
= 10p = 90
Now,
We have to divide both of these sides of the equation by 10,
Then, we get,
= 10p/10 = 90/10
= p = 9

(c) p/4 = 5
Ans:
Now,
We have to multiply both of these sides of the equation by 4,
Then, we get,
= p/4 × 4 = 5 × 4
= p = 20

(d) – p/3 = 5
Ans:
Now,
We have to multiply both of these sides of the equation by – 3,
Then, we receive,
= – p/3 × (- 3) = 5 × (- 3)
= p = – 15

(e) 3p/4 = 6
Ans:
First, we have to multiply both of these sides of the equation by 4,
Then, we get,
= (3p/4) × (4) = 6 × 4
= 3p = 24
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3p/3 = 24/3
= p = 8

(f) 3s = – 9
Ans:
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = -9/3
= s = -3

(g) 3s + 12 = 0
Ans:
First, we have to subtract the number 12 from the both sides of the equation,
Then, we get,
= 3s + 12 – 12 = 0 – 12
= 3s = -12
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = -12/3
= s = – 4

(h) 3s = 0
Ans:
Now,
We have to divide both of these sides of the equation by 3,
Then, we get,
= 3s/3 = 0/3
= s = 0

(i) 2q = 6
Ans:
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3

(j) 2q – 6 = 0
Ans:
Firstly we have to add 6 to both the sides of the following equation,
Then, we get,
= 2q – 6 + 6 = 0 + 6
= 2q = 6
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3

(k) 2q + 6 = 0
Answer :
First, we have to subtract the number 6 from the both sides of the following equation,
Then, we get,
= 2q + 6 – 6 = 0 – 6
= 2q = – 6
Now,
We have to divide both of these sides of the equation by the number 2,
Then, we get,
= 2q/2 = – 6/2
= q = – 3

(l) 2q + 6 = 12
Ans:
First, we have to subtract the number 6 to the both sides of the following equation,
Then, we get,
= 2q + 6 – 6 = 12 – 6
= 2q = 6
Now,
We have to divide both of these sides of the equation by 2,
Then, we get,
= 2q/2 = 6/2
= q = 3

Q2: Solve the following equations:
(a) 2(x + 4) = 12
Ans:
Let us divide both these sides by 2,
= (2(x + 4))/2 = 12/2
= x + 4 = 6
By transposing four from the LHS to RHS, it becomes -4
= x = 6 – 4
= x = 2

(b) 3(n – 5) = 21
Ans:
Let us divide both these sides by 3,
= (3(n – 5))/3 = 21/3
= n – 5 = 7
By transposing -5 from the LHS to the RHS, it becomes 5
= n = 7 + 5
= n = 12

(c) 3(n – 5) = – 21
Ans:
Let us divide both these sides by 3,
= (3(n – 5))/3 = – 21/3
= n – 5 = -7
By transposing -5 from the LHS to the RHS, it becomes 5
= n = – 7 + 5
= n = – 2

(d) – 4(2 + x) = 8
Ans:
Let us divide both these sides by -4,
= (-4(2 + x))/ (-4) = 8/ (-4)
= 2 + x = -2
By transposing two from the LHS to the RHS, it becomes – 2
= x = -2 – 2
= x = – 4

(e) 4(2 – x) = 8
Ans:
Let us divide both these sides by 4,
= (4(2 – x))/ 4 = 8/ 4
= 2 – x = 2
By transposing the number 2 from the LHS to the RHS, it becomes – 2
= – x = 2 – 2
= – x = 0
= x = 0

Q3: (a) Construct 3 equations starting with x = 2
Ans:
First equation is,
Multiply both these side by 6
= 6x = 12 … [equation 1]
Second equation is,
Subtracting the number 4 from both side,
= 6x – 4 = 12 -4
= 6x – 4 = 8 … [equation 2]
Third equation is,
Divide both these sides by 6
= (6x/6) – (4/6) = (8/6)
= x – (4/6) = (8/6) … [equation 3]

(b) Construct three equations that start with x = – 2
Ans:
First equation is,
Multiply both these side by 5
= 5x = -10 … [equation 1]
Second equation is,
Subtracting the number 3 from both side,
= 5x – 3 = – 10 – 3
= 5x – 3 = – 13 … [equation 2]
Third equation is,
Dividing both these side by 2
= (5x/2) – (3/2) = (-13/2) … [equation 3]

Q4: Solve the following equations:
(a) 2y + (5/2) = (37/2)
Ans:
By transposing the fraction (5/2) from LHS to RHS, so it becomes -5/2
Then,
= 2y = (37/2) – (5/2)
= 2y = (37-5)/2
= 2y = 32/2
Now,
Divide both the side by the number 2,
= 2y/2 = (32/2)/2
= y = (32/2) × (1/2)
= y = 32/4
= y = 8

(b) 5t + 28 = 10
Answer :
By transposing 28 from the LHS to RHS, it becomes -28
Then,
= 5t = 10 – 28
= 5t = – 18
Now,
Dividing both these sides by 5,
= 5t/5= -18/5
= t = -18/5

(c) (a/5) + 3 = 2
Ans:
By transposing three from the LHS to RHS, it becomes -3
Then,
= a/5 = 2 – 3
= a/5 = – 1
Now,
Multiply both these sides by 5,
= (a/5) × 5= -1 × 5
= a = -5

(d) (q/4) + 7 = 5
Ans:
By transposing the number 7 from the LHS to RHS, it becomes -7
Then,
= q/4 = 5 – 7
= q/4 = – 2
Now,
Multiply both these sides by 4,
= (q/4) × 4= -2 × 4
= a = -8

(e) (5/2) x = -5
Ans:
First, we have to multiply both these sides by 2,
= (5x/2) × 2 = – 5 × 2
= 5x = – 10
Now,
We have to divide both these sides by 5,
Then we get,
= 5x/5 = -10/5
= x = -2

(f) (5/2) x = 25/4
Ans:
First, we have to multiply both these sides by 2,
= (5x/2) × 2 = (25/4) × 2
= 5x = (25/2)
Now,
We have to divide both these sides by 5,
Then we get,
= 5x/5 = (25/2)/5
= x = (25/2) × (1/5)
= x = (5/2)

Q5: Solve the following equations:
(a) 4 = 5(p – 2)
Ans:
Let us divide both these sides by 5,
= 4/5 = (5(p – 2))/5
= 4/5 = p -2
By transposing – 2 from the RHS to the LHS, it becomes 2
= (4/5) + 2 = p
= (4 + 10)/ 5 = p
= p = 14/5

(b) – 4 = 5(p – 2)
Ans:
Let us divide both these sides by 5,
= – 4/5 = (5(p – 2))/5
= – 4/5 = p -2
Now by transposing – 2 from the RHS to the LHS, it becomes 2
= – (4/5) + 2 = p
= (- 4 + 10)/ 5 = p
= p = 6/5

(c) 16 = 4 + 3(t + 2)
Ans:
By transposing four from the RHS to the LHS, it becomes – 4
= 16 – 4 = 3(t + 2)
= 12 = 3(t + 2)
Let us divide both these side by 3,
= 12/3 = (3(t + 2))/ 3
= 4 = t + 2
By transposing 2 from the RHS to the LHS it becomes – 2
= 4 – 2 = t
= t = 2

(d) 4 + 5(p – 1) =34
Ans:
By transposing 4 from the LHS to the RHS it becomes – 4
= 5(p – 1) = 34 – 4
= 5(p – 1) = 30
Let us divide both these side by 5,
= (5(p – 1))/ 5 = 30/5
= p – 1 = 6
By transposing – 1 from the RHS to the LHS it becomes 1
= p = 6 + 1
= p = 7

(e) 0 = 16 + 4(m – 6)
Ans:
By transposing 16 from the RHS to the LHS it becomes – 16
= 0 – 16 = 4(m – 6)
= – 16 = 4(m – 6)
Let us divide both these side by the number 4
= – 16/4 = (4(m – 6))/ 4
= – 4 = m – 6
Now by transposing – 6 from the RHS to the LHS it becomes 6
= – 4 + 6 = m
= m = 2