Jagranjosh presents here the CBSE Class 9 Mathematics Solved Practice Paper to prepare for the upcoming exams.
Mathematics is one such subject in which students can score full marks by following the right preparatrion technique and intensive practice. You cannot learn Maths by just memorizing a set of formulae given in the book. So, it’s very important to solve practice papers, guess papers and sample papers.
CBSE Class 9 Mathematics Solved Practice Paper: Set-III, has been specially prepared by the subject experts after the brief analysis of previous year question papers and the latest examination format. Practicing this paper will definitely help to fine tune your preparations for the board exam 2018.
Some significant features of this paper are:
Practicing this paper will help the students in understanding the depth with which a topic should be studied in order to prepare in a more effective way to score a meritorious position.
Structure of this paper is as below:
Some sample questions and their solutions from the CBSE Class 9 Mathematics Solved Practice Paper are given below:
Q. If xa/b =1, then find the value of ‘a’.
⟹ xa/b =x [As x = 1]
⟹ a/b =0
⟹ a =0
Q. If (1, −2) is a solution of the equation 2x – y = p, then find the value of p.
2x − y = p
Putting x = 1, y = −2, in above equation, we get:
2(1) – (−2) = p
⟹ p = 4
Q. If two opposite angles of a parallelogram are (63 − 3x)° and (4x − 7)°. Find all the angles of the parallelogram.
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°.
Get the complete practice paper by clicking on the following link: