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CBSE Class 7  >  Class 7 Notes  >  Mathematics (Maths) (Old NCERT)  >  HOTS Question: Simple Equations

HOTS Question: 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

The document HOTS Question: Simple Equations is a part of the Class 7 Course Mathematics (Maths) Class 7 (Old NCERT).
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FAQs on HOTS Question: Simple Equations

1. What are simple equations?
Ans. Simple equations are mathematical expressions that involve one variable and can be solved to find the value of that variable. They typically consist of an equal sign (=) and contain basic arithmetic operations such as addition, subtraction, multiplication, and division.
2. How do you solve simple equations?
Ans. To solve a simple equation, you need to isolate the variable on one side of the equal sign. You can do this by performing the same operation on both sides of the equation, such as adding or subtracting a number, multiplying or dividing by a number, or applying inverse operations.
3. Can simple equations have more than one solution?
Ans. Yes, simple equations can have more than one solution. This occurs when different values of the variable satisfy the equation. For example, in the equation 2x = 10, both x = 5 and x = -5 are valid solutions.
4. What is the role of balance in solving simple equations?
Ans. Balance is crucial in solving simple equations because the goal is to keep both sides of the equation equal. By applying the same operation to both sides, you maintain the balance and ensure that the equation remains true. This allows you to manipulate the equation and solve for the variable.
5. How are simple equations useful in real-life situations?
Ans. Simple equations are used in various real-life situations, such as calculating expenses, determining time or distance, and solving basic problems involving quantities. They provide a way to represent and solve problems using mathematical expressions, allowing for efficient problem-solving and decision-making.
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