In this article, we are providing a collection of most important 2 marks questions for CBSE Class 9 Maths Exam 2020. All these questions have been provided with detailed solutions for easy understanding of the concepts used which will make your exam preparation easy and effective.

In CBSE Class 9 Mathematics Exam 2020, section - B will comprise 6 questions of 2 marks each.

In order to track your preparedness for the final exam, practice the important questions given here. This will help you fine-tune your preparation. Moreover, the detailed answers provided here will give you an idea to write perfect solution to each question in the exam so as to grab maximum marks.

**Given below are some sample questions from the set of CBSE Class 9 Mathematics: Important 2 Marks Questions:**

**Q.** Determine the number of sides of polygon whose exterior and interior angles are in the ratio 1:5.

**Sol.**

Let exterior and interior angles of polygon be x^{ }and 5x .

Then,

x^{ }+ 5x = 180^{o} [As exterior and interior angles form a linear pair]

⟹ 6x = 180^{o}

⟹ x = 180/6

⟹ x = 30^{o}

Also,since sum of all exterior angles is always 360^{}

So, for a polygon having n sides and x^{o} as its exterior angle, we have:

x.n = 360^{o}

⟹ 30^{o}.n = 360^{o}

⟹ n = 12 sides

**CBSE Class 9 Maths Chapter-wise Important Topics and Questions**

**Q.** If two opposite angles of a parallelogram are (63 − 3x)° and (4x − 7)°. Find all the angles of the parallelogram.

**Sol.**

In a parallelogram, the opposite angles are equal.

∴ (63 -3x)° = (4x -7)°

⟹ 4x + 3x = 63 +7

⟹ 7x = 70

⟹ x = 10

(63 -3x)° = 33°

(4x -7)° = 33°

Now, sum of all interior angles of a parallelogram = 360°

∴ Sum of the other two opposite angles = 360° - (33° + 33°) = 360° - 66° = 294°

∴ Each of the other two opposite angles = 294/2 = 147°

Hence the four angles of a parallelogram are 33°, 147°, 33°, 147°

**CBSE Class 9 Mathematics Practice Papers**

**Q.** A batsman in his 11 th innings makes a score of 68 runs and there by increases his average score by 2. What is his average score after the 11 th innings.

**Sol.**

Let the average score of 11 innings be x.

Then the average score of 10 innings = x – 2

Total score of 11 innings = 11x

Total score of 10 innings = 10(x -2) = 10x – 20

Score of the 11th innings = Total score of 11 innings – Total score of 10 innings

= 11x – (10x – 20)

But x + 20x + 20 = 68 ( Given )

⟹ x = 48

Hence, the average score after the 11 th innings is 48.

**To get the complete set of questions, click on the following link:**

**Students may also check the following links to explore more stuff, important for CBSE Class 9 Annual exam preparations:**