SEBA Class 7 Mathematics Chapter 4 Simple Equation

SEBA Class 7 Mathematics Chapter 4 Simple Equation Solutions English Medium, SEBA Class 7 Maths Notes in English Medium, SEBA Class 7 Mathematics Chapter 4 Simple Equation Notes to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 7 Mathematics Chapter 4 Simple Equation Solutions in English and select needs one.

SEBA Class 7 Mathematics Chapter 4 Simple Equation

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 7 General Maths Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 7 Mathematics Chapter 4 Simple Equation Solutions for All Subject, You can practice these here.

Simple Equation

Chapter – 4

1. Express the following statement in the form of an equation:

(i) When 5 is added to 6 times of a number the result is 35.

Ans: Let the number be x

∴ A/Q, 6x + 5 = 35

(ii) One fourth of a number is equal to 9.

Ans: Let the number be x

∴ A/Q, = x/4 = 9

(iii) 5 times of a number is equal to 5 more than 20.

Ans: Let the number be x

∴ A/Q, 5x – 20 = 5

(iv) To get 10 add 3 to 7 times of a number.

Ans: Let the number be x

∴ A/Q, 7x + 3 = 10

(v) When 4 is subtracted from one fifth of a number the result is 2.

Ans: Let the number be x

∴ A/Q, = x/5 – 4 = 2

(vi) 4 times of p is equal to 20.

Ans: Let the number be p 

∴ A/Q, 4p = 20

(vii) When 1 is subtracted from 3 times of a number the result is 2.

Ans: Let the number be x

∴ A/Q, 3x – 1 = 2

(viii) To get 40 divide a number by 10 then subtract 10 from it.

Ans: Let the number be x 

∴ A/Q, x/10 – 10 = 40

2. Write the following equation in statement:

(i) 3x – 4 = 5

Ans: Subtraction of 4 from 3 times of number results 5.

(ii) m/3 + 6 = 11

Ans: Dividing m by 3, add 6 to get 11.

(iii) 7p = 42

Ans: 7 times of a number is equal to 42.

(iv) y/6 = 2

Ans: Dividing a number 6 will be got 2.

(v) 5x + 7 = 2

Ans: Adding 7 with 5 times of a number will be got 2.

(vi) q/2 – 1 = 4

Ans: Substraction of 1 from half of a number results 4.

3. Form equations for each of the following:

(i) Sum of the ages of Anupama, Nirupama and Upoma is 22 years. Anupoma is younger to Nirupoma by 1 year, Upoma is older to Nirupoma by 2 years. Write the equation considering the age of Nirupoma.

Ans: Let the age of Nirupoma be x

Anupama’s age = x – 1 

Upoma’s age = x + 2 

A/Q , (x – 1) + x + (x + 2) = 22

3x + 1 = 22

(ii) Age of Anjan’s grandfather is 72 years. Age of grandfather is 2 years more than seven times of the age of Anjan.

Ans: Let the age of Anjan be x

∴ A/Q, 7x + 2 = 72

(iii) Perimeter of a square is 32 centimeter.

Ans: Let the side of the square be x. 

∴ A/Q, 4x = 32

(iv) Romen’s father bought potato at the rate of Rs. 20 per kg and onion at the rate of Rs. 10 per kg. He paid to the shopkeeper Rs. 50 after buying 1kg less onion than potato.

Ans: Let the weight of potato bought be x. 

Weight of onion bought = x – 1 

Total cost = 20x + 10(x – 1) = 50

(v) Measure of two angles of a triangle is two times and three times of the smallest angle of the triangle. Sum of the three angles of the triangle is 180°.

Ans: Let the smallest angle be x. 

Other angles: 2x, 3x

∴ A/Q, x + 2x + 3x = 180°

4. Tell whether the value of variable inserted within the brackets satisfy the given equation or not:

(i) x + 5 = 0, (x = -5)

Ans: x + 5 = 0

⇒ -5 + 5 = 0

⇒ 0 = 0

∴ For the given value, the equation is satisfied.

(ii) 2x – 8 = 7, (x = 4)

Ans: 2x – 8 = 7

⇒ 2 × 4 – 8 = 7

⇒ 8 – 8 = 7

⇒ 0 ≠ 7

∴ Not satisfied.

(iii) x/3 + 6 = 7, (x = 3)

Ans: x/3 + 6 = 7

⇒ 3/3 + 6 = 7

⇒ 1 + 6 = 7

⇒ 7 = 7

∴ Satisfied.

(iv) x/7 – 2 = 0, (x = 7)

Ans: x/7 – 2 = 0

⇒ 7/7 – 2 = 0

⇒ 1 – 2 = 0

⇒ -1 ≠ 0

∴ Not satisfied.

(v) 5x = 35, (x = 7)

Ans: 5x = 35 

⇒ 5 × 7 = 35

⇒ 35 = 35

∴ Satisfied.

(vi) 4x + 8 = 4, (x = -1)

Ans: 4x + 8 = 4

⇒ 4x (-1) + 8 = 4

⇒ -4 + 8 = 4

∴ Satisfied.

(vii) 7x + 2 = 9, (x = 2)

Ans: 7x + 2 = 9

⇒ 7 × 2 + 2 = 9

⇒ 14 + 2 = 9

⇒ 16 ≠ 9 

∴ Not Satisfied.

(viii) 2x = 16, (x = 8)

Ans: 2x = 16

⇒ 2 × 8 = 16

⇒ 16 = 16

∴ Satisfied.

(ix) x/5 = 20, (x = 100)

Ans: x/5 = 20

⇒ 100/5 = 20

⇒ 20 = 20

∴ Satisfied.

(x) x/8 + 4 = 9, (x = 1)

Ans: x/8 + 4 = 9

⇒ 1/8 + 4 = 9

⇒ 4 ⅛ ≠ 9

∴ Not Satisfied.

5. Examine whether the value of variable inserted within the brackets is the root of the given equation or not:

(i) 4x + 3 = 7, (x = 1)

Ans: 4x + 3 = 7

⇒ 4 × 1 + 3 = 7

⇒ 4 + 3 = 7

⇒ 7 = 7

∴ x = 1 is the root of the equation.

(ii) 2x/3 + 5 = 7, (x = 3)

Ans: 2x/3 + 5 = 7

⇒ (2 × 3) / 3 + 5 = 7

⇒ 2 + 5 = 7

⇒ 7 = 7 

∴ x = 3 is the root of the given equation.

(iii) x – 4 = 1, (x = 3)

Ans: x – 4 = 1, (x = 3)

⇒ 3 – 4 = 1

⇒ -1 = 1

∴ x = 3 is not the root of the given equation.

(iv) 6x = 18, (x = 2)

Ans: 6x = 18, (x = 2)

⇒ 6 × 2 = 18

⇒ 12 = 18

∴ x = 2 is not the root of the given equation.

(v) 5x – 1 = 7, (x = 2)

Ans: 5x – 1 = 7, (x = 2)

⇒ 5 × 2 – 1 = 7

⇒ 10 – 1 = 7

⇒ 9 = 7

∴ x = 2 is not the root of the given equation.

(vi) x + 9 = 13, (x = 4)

Ans: x + 9 = 13, (x = 4)

⇒ 4 + 9 = 13

⇒ 13 = 13

∴ x = 4 is the root of the given equation.

(vii) 5x – 7 = 8, (x = 3)

Ans: 5x – 7 = 8, (x = 3)

⇒ 5 × 3 – 7 = 8

⇒ 15 – 7 = 8

⇒ 8 = 8

∴ x = 3 is the root of the given equation.

(viii) y/3 + 5 = 8, (y = 9)

Ans: y/3 + 5 = 8, (y = 9)

⇒ 9/3 + 5 = 8

⇒ 3 + 5 = 8

⇒ 8 = 8

∴ y = 9 is the root of the given equation.

(ix) p/5 + 4 = 5, (p = 1)

Ans: p/5 + 4 = 5, (p = 1)

⇒ 1/5 + 4 = 5

⇒ 4 ⅕ = 5

∴ p = 1 is not the root of the given equation.

(x) x/7 = 6, (x = 42)

Ans: x/7 = 6, (x = 42)

⇒ 42/7 = 6

⇒ 6 = 6

∴ x = 42 is the root of the given equation.

6. Try to solve equation by changing the value of the variable (Trial and Error method):

(i) 2x + 5 = 11

Ans: 2x + 5 = 11

x = 0, 1, 2, …….

∴ 2 × 0 + 5 = 11

⇒ 0 + 5 = 11

⇒ 5 ≠ 11

2 × 1 + 5 = 11

⇒ 2 + 5 = 11

⇒ 7 ≠ 11

2 × 3 + 5 = 11

⇒ 6 + 5 = 11

⇒ 11 = 11

∴ x = 3

(ii) x/5 + 5 = 7

Ans: x/5 + 5 = 7

x = 0, 5, 10,……

∴ 0/5 + 5 = 7

⇒ 0 + 5 = 7

⇒ 5 ≠ 7

5/5 + 5 = 7

⇒ 1 + 5 = 7

⇒ 6 ≠ 7

∴ 10/5 + 5 = 7

⇒ 2 + 5 = 7

⇒ 7 = 7

∴ x = 10

(iii) 7x – 4 = 24

Ans: 7x – 4 = 24

x = 0, 2, 4,…..

∴ 7 × 0 – 4 = 24

⇒ 0 – 4 = 24

⇒ -4 ≠ 24

7 × 2 – 4 = 24

⇒ 14 – 4 = 24

⇒ 10 ≠ 24

7 × 4 – 4 = 24

⇒ 28 – 4 = 24

⇒ 24 = 24

∴ x = 4

1. Solve the equation and write down the step while separating the variable.

(i) x + 5 = 12

Ans: x + 5 = 12

= x = 12 – 5 

= 7

(ii) x – 7 = 0

Ans: x – 7 = 0

= x = 7

= 7

(iii) y – 3 = 6

Ans: y – 3 = 6

= y  = 6 + 3

= y = 9

(iv) z + 6 =- 5

Ans: z + 6 = – 5

= z = – 5 – 6

= z = -11

(v) 3x = 42

Ans: 3x = 42

= x = 42/3

= x = 14

(vi) x/5 = 6

Ans: x/5 = 6

= x = 6 × 5

= x = 30

(vii) 12x = -36

Ans: 12x = -36

= x = -36/12

= x = -3

(viii) x/4 = 3/2 

Ans: x/4 = 3/2

= x = 3/2 × 4

= x = 6

(ix) 7x = 35

Ans: 7x = 35

= x = 35/7

= x = 5

(x) p/4 = 3

Ans: p/4 = 3

= p = 3 × 4

= p = 12

2. Solve the equation and write down the step while separating the variable.

(i) 4x + 5 = 45

Ans: 4x + 5 = 45

⇒ 4x = 45 – 5

⇒ 4x = 40

⇒ x= 40/4

⇒ x = 10

(ii) 3x – 7 = 11

Ans: 3x – 7 = 11

⇒ 3x = 11 + 7

⇒ 3x = 18

⇒ x= 18/3

⇒ x = 6

(iii) 2x/3 + 5 = 7

Ans: 2x/3 + 5 = 7

⇒ 2x/3  = 7 – 5

⇒ 2x/3 = 2

⇒ 2x  = 2 × 3

 ⇒ 2x = 6

⇒ x = 6/2

⇒ x = 3

(iv) 4y/3 – 7 = 5

Ans: 4y/3 – 7 = 5

⇒ 4y/5  = 5 + 7

⇒ 4y/5 = 12

⇒ 4y = 12 × 5

⇒ 4y = 60

⇒ y = 15

3. Solve the following equations:

(i) 4x = 64

Ans: 4x = 64

⇒ x = 64/4

⇒ x = 16

(ii) 4x + 7 = 15

Ans: 4x + 7 = 15

⇒ 4x = 15 – 7

⇒ 4x = 8

⇒ x = 8/4

∴ x = 2

(iii) y/4 = 6

Ans: y/4 = 6

⇒ y = 6 × 4 

⇒ y = 24

(iv) 3y = 60

Ans: 3y = 60

⇒ y = 60/3

⇒ y = 20

(v) 6p + 7 = 37

Ans: 6p + 7 = 37

⇒ 6p = 37 – 7

⇒ 6p = 30

⇒ p = 30/6

⇒ p = 5

(vi) 7p – 9 = 5

Ans: 7p – 9 = 5

⇒ 7p  = 5 + 9

⇒ 7p = 14

⇒ p = 14/7

⇒ p = 2

(vii) 5x – 7 = 8

Ans: 5x – 7 = 8

⇒ 5x  = 8 + 7

⇒ 5x = 15

⇒ x= 15/5

⇒ x = 3

(viii) x/5 + 2 = 3

Ans: x/5 + 2 = 3

⇒ x/5  = 3 – 2

⇒ x/5 = 1

⇒ x = 1 x 5

x = 5 

(ix) q/3 – 1 = 2

Ans: q/3 – 1 = 2

⇒ q/3  = 2 + 1 

⇒ q/3 = 3 

⇒ q = 3 × 3 

⇒ q = 9

(x) 3x + 11 = 50

Ans: 3x + 11 = 50

⇒ 3x  = 50 – 11 

⇒ 3x = 39 

⇒ x = 39/3 

⇒ x = 13

(xi) 4x + 10 = 26

Ans: 4x + 10 = 26

⇒ 4x  = 26 – 10

⇒ 4x = 16

⇒ x= 16/4

⇒ x = 4

(xii) x/3 + 4 = 6

Ans: x/3 + 4 = 6

⇒ x/3  = 6 – 4 

⇒ x/3 = 2

⇒ x  = 2 × 3

⇒ x = 6

(xiii) p/3 + 5 = 12

Ans: p/3 + 5 = 12

⇒ p/3  = 12 – 5 

⇒ p/3 = 7 

⇒ p = 7 × 3 

⇒ p = 21

(xiv) q/2 + 4 = 7

Ans: q/2 + 4 = 7

⇒ q/2 =  7 – 4 

⇒ q/2 = 3 

⇒ q = 3 × 2

⇒ q = 6

(xv) 2(x + 3) = x + 7

Ans: 2(x + 3) = x + 7

⇒ 2x + 6 = x + 7

⇒ 2x  – x  =   + 7 – 6 

⇒ x = 1

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

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

Ans: Let the number be x

According to question. 5x – 7 = 8

⇒ 5x -= 8 + 7 

⇒ 5x = 15

⇒ x = 15/5

⇒ x = 3 

∴ x = 3

(ii) One third of a number is 2 more than the number 5.

Ans: Let the number be x

∴ A/Q, x/3 – 5 = 2 

⇒ x/3 – 2 + 5 = 7 

⇒ x = 7 × 3

⇒ x = 21

∴ The required no = 21.

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

Ans: Let the number be x

∴ A/Q, 3x + 4 = 10

⇒ 3x = 10 – 4

⇒ 3x = 6

⇒ x = 6/3

⇒ x = 2

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

Ans: Let the number be x

∴ A/Q, x + 6/3 = 4

⇒ x + 6 = 3 × 4

⇒ x = 12 – 6

⇒ x = 6

∴ The required number = 6

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

Ans: Let the number be x

∴ A/Q, 2/3x – 4 = 7

⇒ 2x/3 = 7 + 4

⇒ 2x = 11 × 3 

⇒ 2x = 33 

⇒ x = 33/2

∴ The required number = 33/2

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

Ans: Let the number be x

∴ A/Q, 6x = 24

⇒ x = 24/6 

⇒ x = 4

∴ The required number = 4

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

Ans: Let the number be x

∴ A/Q, x × 1/4 + 5 = 6

⇒ x/4  = 6 – 5

⇒ x/4 = 1

⇒  x = 4 × 1

⇒ x = 4

∴ The required number = 4

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

Ans: Let the number be x

∴ A/Q,  x × 3/4 = 12

⇒ 3x/4 = 12 

⇒ 3x = 4 × 12

⇒ 3x = 48 

⇒ x = 48/3 

⇒ x = 16

∴ The required number = 16

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

Ans: Let the breadth of a rectangle = x cm.

∴ Length = (x + 5) cm.

∴ A/Q, 2(L + B) = Perimeter

⇒ 2{(x +5) + x} = 26

⇒ 2(2x + 5) = 26

⇒ 2x + 5 = 26/2 = 13

⇒ 2x = 13 – 5

⇒ 2x = 8

⇒ x = 8/2

⇒ x = 4

∴ Length = (4 + 5) cm = 9 cm and breadth = 4 cm.

∴ Area of the rectangle = L × B

= 9 cm × 4 cm

= 36 sq. cm.

3. Sum of the ages of Anupam, Rahul, and Jahirul is 32 years. Anupam is 1 year younger to Rahul, Jahirul is 2 years older than Rahul. Find their ages.

Ans: Let the age of Rahul = x years,

age of Anupam = (x – 1) years

and age of Jahirul = (x + 2) years.

A/Q, (x – 1) + x (x + 2) = 22 

⇒ 3x + 1 = 22

⇒ 3x = 22 – 1

⇒ 3x = 21

⇒ x = 21/3

⇒ x = 7

∴ Age of Rahul = 7 years, 

age of Anupam = (7 – 1) years = 6 years.

And age of Jahirul = (7 + 2 ) = 9 years.

4. Sum of the ages of Anupam, Rahul, and Jahirul is 32 years. Anupam is 1 year younger to Rahul, Jahirul is 2 years older than Rahul. Find their ages.

Ans: Let Anjan’s age be x.

Grandfather’s age: 7x + 2

A/Q, 7x + 2 = 72

⇒ 7x = 72 – 2 

⇒ 7x = 70 

⇒ x = 70/7

⇒ x = 10 

∴ The age of Anjan = 10 years.

5. 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.

Ans: 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 

Robin marks = 120 – x

Robin’s marks: 120 – x – 5 = 115 – x

∴ A/Q, 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 = 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

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

Ans: Let the first odd number be x.

Second odd number: x + 2

Third odd number: x + 4

∴ A/Q, x + (x + 2) + (x + 4) = 75

⇒ x + x + 2 + x + 4 = 75

⇒ 3x + 6 = 75

⇒ 3x = 75 – 6

⇒ 3x = 69

⇒ x = 69/3 

⇒ x = 23 

So the Numbers are: 23, 25, 27

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

Ans: Let the first even number be x.

Second even number: x + 2

∴ A/Q,  x + (x + 2) = 38

⇒ x + x + 2 = 38

⇒ 2x + 2 = 38

⇒ 2x = 38 – 2

⇒ 2x = 36

⇒ x = 36/2

⇒ x = 18

8. 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.

Ans: 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

∴ A/Q, 3x + 10x + x + 10(3x) = 88 

⇒ 44x = 88

⇒ x = 88/44 

⇒ x = 2    

So, the Original number: 31 × 2 = 62

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

Ans: Let the three consecutive numbers be x, x + 1 and x + 2

∴ A/Q, x + (x + 1) + (x + 2) = 48

⇒ 3x + 3 = 48

⇒ 3x = 48 – 3

⇒ 3x = 45

⇒ x = 45/3 = 15

∴ The reqd three consecutive numbers are 15, 16 and 17.

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

Ans: Let the one number be x and other number: ( 40 -x ) 

 A/Q x + (x + 10) = 40

⇒ 2x+10 = 40

⇒ 2x = 40 -10

⇒ 2x= 30

⇒ x = 30/2

⇒ x = 15

Other number = (40 – 15) = 25

∴ Numbers: 15, 25

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

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

A/Q,

⇒ 8x – 3x = 60

⇒ 5x = 60

⇒ x = 60/5

⇒ x = 12

∴ Larger number: 8x = 8 × 12 = 96

Smaller number: 3x = 3 × 12 = 36

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

Ans: Let the breadth = x and length = 2x

A/Q, 2(2x + x) = 72

⇒ 2 × 3x = 72

⇒ 6x = 72

⇒ x = 72/6 = 12

∴ Length = (2 × 12) Unit = 24 and breadth = 12 Unit.

13. 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?

Ans: Let the present age of Bijoy = x years and the present age of Ajoy = (x – 5) years.

∴ A/Q,(x+4) = 2 {(x – 5 ) + 4}

⇒ x + 4 = 2(x – 1) 

⇒  x + 4 = 2x – 2

⇒ -x = -4 -2 

⇒ -x = -6 

⇒ x = 6

∴ Present age of Bijoy = 6 years and the present age of Ajoy = (6 – 5 ) = 1 years.  

14. 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?

Ans: Let, 

Ramen’s age = x

Father’s age = 4x

After 5 years: 

Ramen’s age = x + 5

Father’s age = 4x + 5

A/Q, (4x + 5) = 3(x + 5)

⇒ 4x + 5 = 3x + 15

⇒ 4x – 3x = 15 – 5 

⇒ x = 10

∴ Ramen’s age is 10 years old

And his father’s age = 4 × 10 = 40 years old.

15. 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?

Ans: Let, 

Cost of a chair = x

Cost 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 = 625/5

⇒ x = 125 

∴ The cost of a chair is Rs. 125 

And the cost of a table is Rs. 125 + 40 = Rs. 165.

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

Ans: Let,

one angle be x.

Other angle: 90 – x

A/Q, (90 – x) – x = 12

⇒ 90 – x – x = 12

⇒ 90 – 2x = 12

⇒ -2x = 12 – 90

⇒ -2x = -78

⇒ x = -78/-2

⇒ x = 39 

So, the Angles are: 39°, 51

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