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In this article you will get a collection of most important 2 marks questions for CBSE class 9 Mathematics. All these questions have been provided with detailed solution to help students make an easy and quick preparation for the Class 9 Maths annual exam 2018.

In CBSE Class 9 Mathematics Exam 2018, section - B will comprise 6 questions of 2 marks each.

In order to track their preparedness for the final exam, students are advised to practice the important questions given here. This will help them to fine-tune their preparation. Moreover, the solution provided here give students an idea to write proper solution to each question in the exam so that they may score full marks.

**Given below are some sample questions for 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 5*x*^{ }.

Then,

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

⟹ 6*x*^{ }= 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

**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°

**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 = 11*x*

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

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

= 11x – (10*x* – 20)

But *x* + 20*x* + 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:**

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