# CBSE Class 9 Mathematics Exam 2018: Important 2 Marks Questions

Feb 2, 2018 17:40 IST
CBSE Class 9 Mathematics: Important 2 Marks Questions

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 5x .

Then,

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

⟹       6x = 180o

⟹         x = 180/6

⟹         x = 30o

Also,since sum of all exterior angles is always 360

So, for a polygon having n sides and xo as its exterior angle, we have:

x.n = 360o

⟹       30o.n = 360o

⟹       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.

&there4;          (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°

&there4; Sum of the other two opposite angles = 360° - (33° + 33°) = 360° - 66° = 294°

&there4; 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 = 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:

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