Assam SCERT Solutions for Class 7 Maths Chapter 4 Simple Equation Exercise 4.3

Free Solutions for The Assam State School Education Board (ASSEB) Class 7 Maths Chapter 4 Exercise 4.3 in English Medium

Assam SCERT Solutions for Class 7 Maths Chapter 4 Simple Equation – Exercise 4.3 (2025-26) | Download Free PDF

Students can now access SCERT Assam Solutions for Class 7 Maths Chapter 4 – Simple Equation (Exercise 4.3, English Medium), designed as per the latest syllabus for the 2025-26 academic session. These solutions, prepared by subject experts at Digital Pipal Academy, help students develop a strong understanding of advanced equation-solving techniques.

Topics Covered in Exercise 4.3:

✔ Solving equations with variables on both sides
✔ Understanding real-world applications of simple equations
✔ Word problems based on equations

Key Features of SCERT Assam Class 7 Maths Solutions:

✔ Step-by-step solutions for better clarity
✔ Practice online to strengthen conceptual understanding
✔ Download free PDF for offline study
✔ Exercise-wise solutions to help students score better marks

SCERT Assam Class 7 Maths Chapter 4 Simple Equation – Exercise 4.3

Click on the link below to access detailed solutions for Simple Equation:

SCERT Assam Class 7 Maths Chapter 4 Solutions

Exercise	Solution Link
Exercise 4.1	Click Here
Exercise 4.2	Click Here
Exercise 4.3	Click Here

 

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1. Form equations from the following statements and solve the equations.

(i) Subtract 7 from 5 times of a number which results 8.

    Solution:

    5x - 7 = 8

   Add 7 to both sides: 5x = 15

   Divide both sides by 5: x = 3

(ii) One-third of a number is 2 more than the number.

    Solution:

    (1/3)x = x + 2

   Subtract x from both sides: (1/3)x - x = 2

   Combine like terms: (-2/3)x = 2

   Multiply both sides by -3/2: x = -3

(iii) To get 10, 4 is added with 3 times of a number.

    Solution: 3x + 4 = 10

   Subtract 4 from both sides: 3x = 6

   Divide both sides by 3: x = 2

(iv) Jehirul added 6 with a number and when the sum is divided by 3 he gets 4.

    Solution:

 (x + 6) / 3 = 4

   Multiply both sides by 3: x + 6 = 12

   Subtract 6 from both sides: x = 6

(v) When 4 is subtracted from two-third of a number, the result is 7.

    Solution:

 (2/3)x - 4 = 7

   Add 4 to both sides: (2/3)x = 11

   Multiply both sides by 3/2: x = 33/2

(vi) 6 times of a number is equal to 24.

    Solution:

 6x = 24

   Divide both sides by 6: x = 4

(vii) When 5 is added to one-fourth of a number, the result is 6.

    Solution:

 (1/4)x + 5 = 6

   Subtract 5 from both sides: (1/4)x = 1

   Multiply both sides by 4: x = 4

(viii) Three-fourths of a number is equal to 12.

    Solution:

 (3/4)x = 12

   Multiply both sides by 4/3: x = 16

2. There are a total of 50 marbles in the pockets of Amal, Ramen, and Anup. There are 11 marbles in the pocket of Anup. In the two pockets of Ramen, there are marbles which are equal to two times the marbles possessed by Amal. Form an equation to find the marbles in the pockets of Amal and Ramen.

Solution:

Let the number of marbles with Amal be x. Since Ramen has marbles equal to twice Amal’s marbles, he has 2x marbles. Anup has 11 marbles.

The total number of marbles is given as 50, so the equation is:

$$x + 2x + 11 = 50$$

Solving,

$$3x = 39$$ $$x = 13$$

Thus, Amal has 13 marbles, and Ramen has 26 marbles.

 

3. The denominator of a fraction is 4 more than the numerator. If 1 is added to both the numerator and denominator, the value of the fraction becomes `\frac{1}{2}`. Form the equation to find the fraction.

Solution:

Let the numerator of the fraction be x. Since the denominator is 4 more than the numerator, it is x + 4.

According to the given condition:

$$\frac{x + 1}{(x + 4) + 1} = \frac{1}{2}$$ $$\frac{x + 1}{x + 5} = \frac{1}{2}$$

Cross multiplying:

$$2(x + 1) = x + 5$$ $$2x + 2 = x + 5$$ $$2x - x = 5 - 2$$ $$x = 3$$

Thus, the numerator is 3, and the denominator is x + 4 = 3 + 4 = 7.

The required fraction is `\frac{3}{7}`.

 

4. Length of a rectangle is more than its breadth by 5 cm. Perimeter of the rectangle is 26 cm. Find the area of the rectangle.

Solution:  

Let the breadth of the rectangle be x cm. Since the length is 5 cm more than the breadth, it is x + 5 cm.

The perimeter of the rectangle is given as 26 cm, so using the perimeter formula:

`2\times ( \text{Length} + \text{Breadth}) ` 

`\Rightarrow2\times ( \text{x} + \text{x+5})=26`

`\Rightarrow2\times ( \text{2x} + \text{5})=26` 

`\Rightarrow4x + 10=26`

`\Rightarrow4x =26-10` 

`\Rightarrow4x =16`

`\Rightarrow x =\frac{16}{4}` 

`\Rightarrow x =4`

Thus, the breadth is 4 cm, and the length is 4 + 5 = 9 cm.

Now, the area of the rectangle is:

`\text{Length} \times \text{Breadth} = 9 \times 4` 

`=36cm^{2}`

The area of the rectangle is 36 cm².

 

5. Age of Ariful’s mother is 4 times the age of Ariful’s sister Rehena. Rehena is 4 years younger than Ariful. Age of the mother is 32 years. Form the equation to find the age of Ariful and Rehena.

Solution:

Let the age of Ariful be x years. Since Rehena is 4 years younger than Ariful, her age is x - 4 years.

The mother’s age is 4 times the age of Rehena, and it is given as 32 years, so we form the equation:

4(x−4)=32

Solving,

4x−16=32

4x=48

x=12

Thus, Ariful's age is 12 years, and Rehena's age is x - 4 = 12 - 4 = 8 years.

Final Answer: Ariful is 12 years old, and Rehena is 8 years old.

6. Sum of the ages of Rahul, Anupoma, and Jahirul is 22 years. Anupoma is 1 year younger than Rahul, and Jahirul is 2 years older than Rahul. Find their ages.

Solution:

Let Rahul’s age be x years. Then, Anupoma’s age = x - 1, and Jahirul’s age = x + 2.

Given,

(x-1)+x+(x+2)=22

3x+1=22

3x=21

x = 7  

Thus, Rahul = 7 years, Anupoma = 6 years, and Jahirul = 9 years.

 

7. Age of Anjan's Grandfather is 72 years. Age of grandfather is 2 years more than seven times the age of Anjan. Find the age of Anjan.

Solution:

Let Anjan's age be x.

Grandfather's age: 7x + 2

According to Question,

    7x + 2 = 72

 ⇒ 7x = 72 - 2

 ⇒ 7x = 70

 ⇒ x = 70/7

 ⇒ x = 10

∴ The age of Anjan is 10 years

8. Robin, Naren, Shreya, Anubhav, Irfan, and Paruma obtained marks in Mathematics in the following manner: Shreya obtained two times the marks obtained by Naren, Anubhav obtained 5 marks less than marks obtained by Shreya, the sum of the marks of Irfan and Naren is 105, Robin obtained 5 marks less than marks obtained by Paruma, and Paruma obtained 15 marks more than marks obtained by Irfan. Sum of their marks is 435. Find the marks they have obtained.

Solution:

Let Naren's marks be x.

Shreya's marks: 2x

Anubhav's marks: 2x - 5

Irfan's marks: 105 - x

Paruma's marks: 105 - x + 15

                                    = 120 - x

Robin's marks: 120 - x - 5 = 115 - x

  According to Question,

Total marks: x + 2x + (2x - 5) + (105 - x) + (120 - x) + (115 - x) = 435

 ⇒ x + 2x + 2x - 5 + 105 - x + 120 - x + 115 - x = 435

⇒ (x + 2x + 2x - x - x - x) + (-5 + 105 + 120 + 115) = 435

 ⇒ 2x + 335 = 435

⇒ 2x = 435 - 335

⇒ 2x = 100

⇒ x = `\frac{100}{2}`

⇒ x = 50

 ∴ Naren's marks (x)= 25

 Shreya's marks: 2 × 50 = 100

Anubhav's marks: 100 - 5 = 95

Irfan's marks: 105 - 50 = 55

 Paruma's marks: 120 - 50 = 70

Robin's marks: 115 - 50 = 65

9. The sum of three consecutive odd numbers is 75. Find the numbers.

Solution:

Let the first odd number be x.

Second odd number: x + 2

Third odd number: x + 4

According to question, 

x + (x + 2) + (x + 4) = 75

⇒ x + x + 2 + x + 4 = 75

⇒ 3x + 6 = 75

⇒ 3x = 75 - 6

 ⇒ 3x = 69

⇒ x = `\frac{69}{3}`

⇒ x = 23

The three consecutive odd numbers are : 23, 25, 27

10. The sum of two consecutive even numbers is 38. Find the numbers.

Solution:

Let the first even number be x.

Second even number: x + 2

According to question, 

x + (x + 2) = 38

⇒ x + x + 2 = 38

⇒ 2x + 2 = 38

⇒ 2x = 38 - 2

⇒ 2x = 36

⇒ x = `\frac{36}{2}`

⇒ x = 18

The two consecutive even numbers are 18, 20

 

11. In a two-digit number, the digit in the tens place is thrice the digit in the ones place. The sum of the original number and the new number obtained by interchanging the digits is equal to 88. Find the original number.

Solution:

Let the digit in the ones place be x.

Digit in the tens place: 3x

Original number: 10(3x) + x = 31x

New number: 10x + 3x = 13x

According to question,

31x + 13x = 88

⇒ 44x = 88

⇒ x = `\frac{88}{44}`

⇒ x = 2   

So, the Original number is: 31 × 2 = 62

12. The sum of three consecutive numbers is 48. Find the numbers.

Solution:

Let the first number be x.

Second number: x + 1

Third number: x + 2

According to question,

x + (x + 1) + (x + 2) = 48

⇒ x + x + 1 + x + 2 = 48

⇒ 3x + 3 = 48

⇒ 3x = 48 - 3

⇒ 3x = 45

⇒ x = `\frac{45}{3}`

⇒ x = 15

So,The three consecutive numbers are : 15, 16, 17

13. The sum of two consecutive numbers is 40. One number is 10 more than other number. Find the numbers.

Solution:

Let the smaller number be x.

Larger number: x + 10

 x + (x + 10) = 40

 ⇒  2x+10 = 40

 ⇒  2x = 40-10

 ⇒  2x= 30

 ⇒ x = `\frac{30}{2}`

 ⇒ x = 15

∴ The two numbers are 15, 25

14. Ratio of two numbers is 8:3. Difference of the numbers is 60. Find the numbers.

Solution:

Let , the two numbers are:  8x and  3x

According to question,

⇒ 8x - 3x = 60

⇒ 5x = 60

⇒ x = `\frac{60}{5}`

⇒ x = 12

∴ Larger number: 8x = 8 × 12 = 96

Smaller number: 3x = 3 × 12 = 36

15. Length of a rectangle is two times its breadth. Perimeter of the rectangle is 72 unit. Find the length and breadth of the rectangle.

Solution:

Let the breadth be x.
Then, the length is 2x.

Perimeter: 2(length + breadth)

⇒ 2(2x + x) = 72

⇒ 2× 3x  = 72

⇒ 6x = 72

⇒ x = `\frac{72}{6}`

⇒ x = 12

∴ breadth(x) = 12 (breadth)

   Length (2x) =  2 × 12 = 24

The breadth is 12 units and the length is 24 units.

16. Ajoy is 5 years younger to Bijoy. After 4 years, age of Bijoy will be 2 times the age of Ajoy. What are their present ages?

Solution:     

Let,

Present age of Bijoy = x

Then  ,  Present age of Ajoy = x-5

According to question,

x + 4 = 2(x - 5 + 4)

⇒ x + 4 = 2(x - 1)

⇒ x + 4 = 2x - 2

⇒ 4 + 2 = 2x - x

⇒ 6 = x

⇒ x = 6

∴   Present age of Bijoy (x)= 6 years

And  Present age of Ajoy ( x-5 ) = 6-1=1 years

17. Age of Ramen's father is 4 times the age of Ramen. After 5 years the age of Ra- men's father will be 3 times the age of Ramen. What are their present ages?

Solution:

Let Ramen’s present age be x.
Then, Ramen’s father’s present age = 4x.

After 5 years, their ages will be:

Ramen's age = x + 5

Father's age = 4x + 5

According to question,

   (4x + 5) = 3(x + 5)

⇒ 4x + 5 = 3x + 15

⇒ 4x - 3x = 15 - 5

⇒ x = 10

∴ So, Ramen’s present age = 10 years.
Ramen’s father’s present age = 4 × 10 = 40 years.

 

18. Total cost of 2 tables and 3 chairs is Rs. 705. Cost price of a

table is Rs. 40, more than the cost price of a chair, What is the cost price of a table and a chair?

Solution: 

Let the cost price of a chair be x.
Then, the cost price of a table = x + 40

A/q,

 2(x + 40) + 3x = 705

 ⇒ 2x + 80 + 3x = 705

 ⇒ 5x + 80 = 705

 ⇒ 5x = 705 - 80

 ⇒ 5x = 625

 ⇒ x = `\frac{625}{5}`

 ⇒ x = 125

So, the cost price of a chair = Rs. 125.
The cost price of a table = x + 40 = 125 + 40 = Rs. 165.

 

19. Difference of measures of two complementary angles is 12°. What are the measures of two angles?

Solution:

Let the measure of the smaller angle be x.
Then, the larger angle = x + 12.

Since the angles are complementary, their sum is 90°:

According to question,

x+(x+12) =90

⇒ 2x+12 = 90

⇒ 2x = 90-12

⇒ 2x = 78

⇒ x =`\frac{78}{2}`

⇒ x = 39

So, the smaller angle = 39°, and the larger angle = 39° + 12° = 51°.

 

New Questions (Similar to Exercise 4.3)

 

1. Solve the  Solution: 7x - 15 = 12

2. Solve the  Solution: 4y + 9 = 25

3. Solve the  Solution: 5z - 18 = 17

4. Solve the  Solution: 6p + 13 = 37

5. Solve the  Solution: 8q - 21 = 33

6. Solve the  Solution: 9r + 17 = 50

7. Solve the  Solution: 10s - 23 = 47

8. Solve the  Solution: 11t + 25 = 61

9. Solve the  Solution: 12u - 32 = 40

10. Solve the  Solution: 13v + 37 = 71

11. The sum of three consecutive numbers is 63. Find the numbers.

12. The difference between two numbers is 12. Their sum is 34. Find the numbers.

13. One-fifth of a number is 7 more than the number itself. Find the number.

14. The length of a rectangle is 4 cm more than its breadth. If the perimeter is 32 cm, find the length and breadth.

15. The age of a father is 5 times the age of his son. After 3 years, the father will be 4 times the son's age. Find their present ages.

16. The cost of 7 pens is Rs. 40 more than the cost of 5 pencils. If the cost of 4 pens and 3 pencils is Rs. 51, find the cost of a pen and a pencil.

17. The numerator of a fraction is 4 less than the denominator. If 2 is added to both the numerator and denominator, the fraction becomes 3/4. Find the original fraction.

18. The sum of the digits of a two-digit number is 11. If the digits are interchanged, the new number is 9 less than the original number. Find the original number.

19. The sum of three consecutive even numbers is 54. Find the numbers.

20. The difference between two complementary angles is 18°. Find the measures of the angles.

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