Important Questions: Simple Equations - Grade 7 MCQ

Begin with the equation: m⁄5 + 20 = 45

Subtract 20 from both sides to get: m⁄5 = 25

Multiply both sides by 5 to find: m = 125

The value of m is 125, which corresponds to option 1.

0.2 ( 2x - 1 ) - 0.5 ( 3x - 1 ) = 0.4
0.4x - 0.2 - 1.5x + 0.5 = 0.4
( 0.4 - 1.5 ) x - 0.2 + 0.5 = 0.4
-1.1x + 0.3 = 0.4
-1.1x = 0.4 - 0.3
-1.1x = 0.1

Multiply Both Sides by 10 .
-1.1x × 10 = 0.1 × 10
-11x = 1
x = - 1/11 

the equation m/5 = 4 means m divided by  5 gives 4. 

If we write this , it'll be one fifth of m is 4 or the fifth part of m is 4.

To solve the equation m/3 = 2, follow these steps:

• Multiply both sides by 3 to isolate m.
• This gives: m = 2 × 3
• Calculating this results in m = 6.

Therefore, the correct answer is 6, which corresponds to option A.

The correct answer is option B
Given  x = 2:

3x - 4= 2

Putting x = 2;

⇒ (3 x 2)  - 4 = 2

Let one number  = x
other number      = 2x-3 
(x+2x−3)+7=37
3x = 33
x = 11
∴ The two number are 11 and 19

Let's denote the number of marbles Riya has as x.

According to the problem, Aman has 9 marbles more than five times Riya's marbles:

• Equation: 5x + 9 = 54

Now, we will solve for x:

• Subtract 9 from both sides:
• 5x = 45
• Divide both sides by 5:
• x = 9

Therefore, Riya has 9 marbles.

Let Rohit's age be x.

According to the problem, the father's age is 5 years younger than three times Rohit's age. This can be written as:

3x - 5 = 55

Solving for x:

• Add 5 to both sides:

3x = 60

• Divide both sides by 3:

x = 20

Therefore, Rohit's age is 20 years.

Let's denote the number of candies Riya has as x. According to the problem, Arjun has 8 more than twice the number of candies Riya has. This can be written algebraically as:

2x + 8 = 30

To find x, we'll solve the equation step by step:

1. Subtract 8 from both sides:

   2x = 30 - 8

2. This simplifies to:

   2x = 22

3. Divide both sides by 2:

   x = 22/2

4. This results in:

   x = 11

Therefore, Riya has 11 candies.

Let Sneha's age be x 

4x + 8 = 52

4x = 52 - 8

4x = 44

x = 11

Let number be x 
Five times a number = 5x 
ATQ equation becomes
5x = 50-6 

The statement "The sum of 3x and 11 is 32" means adding 3x and 11 gives 32.
This is written as: 3x+11=32

- Option A: Incorrect. This states "One-fourth of 10 is m," which is not accurate.
- Option B: Incorrect. It says "One-fourth of m is 3 more than 3," which is not relevant to the equation.
- Option C: Correct. This option correctly states "One-fourth of m is 10," aligning with the equation.
- Option D: Incorrect. It mentions "Four times m is 10," which is a different equation.
- Option E: Irrelevant.
Therefore, the correct answer is Option C: "One-fourth of m is 10."

- Vinay's father's age is given as 44 years.
- He is described as being 5 years older than thrice Vinay's age.
- Let Vinay's age be ( x ).
- Thrice Vinay's age is ( 3x ).
- Five years older than thrice Vinay's age is ( 3x + 5 ).
- Therefore, the equation is ( 3x + 5 = 44 ).
- Simplifying ( 3x + 5 = 44 ) to find ( x ) correctly describes the relationship.
- Thus, option A, ( 3x + 5 = 44 ), is correct.

Let's check each statement carefully step by step to see which one is false.

a) The solution of 4x = 60 is 12

• Solve: 4x = 60 → x = 60 ÷ 4 = 15
  So this statement is false, because the solution is x = 15, not 12.

b) y = 7 satisfies the equation y 0 = 7

• The equation seems to have a typo: "y 0 = 7". If it means y × 0 = 7, then 7 × 0 = 0, not 7.

• That would make this statement false too, but assuming it’s a misprint, we focus on the first clearly false one.

c) p = 5/2 is the solution of 12p − 5 = 25

• Substitute p = 5/2: 12 × 5/2 − 5 = 30 − 5 = 25  True

d) m = 3/2 is the solution of 4(m + 3) = 18

• Substitute m = 3/2: 4 × (3/2 + 3) = 4 × (3/2 + 6/2) = 4 × 9/2 = 18  True

Answer: a) is false.