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Here we bring you the CBSE Class 9 Mathematics Solved Practice Paper : Set-I. This practice paper has been designed as per the latest exam structure and coveres the whole syllabus of Class 9 Maths. It will help you get acquainted with the important topics to be prepared for the Class 9th Mathematics Annual Exam 2018.

**CBSE Class 9 Exam Pattern 2017-2018**

**NCERT Solutions for CBSE Class 9 Maths**

With a thorough analysis of the CBSE’s criteria to set papers for finals, subject experts at jagranjosh has designed the CBSE Class 9 Mathematics Solved Practice Paper : Set-I that will make you familiar with important questions that will ultimately help you prepare better for the new assessment structure, i.e., periodic tests as well as the annual exams 2017-2018.

**Some of the sample questions from CBSE Class 9 Mathematics Solved Practice Paper : Set-I are given below:**

**Q.** Find the radius of largest sphere that is carved out of the cube of side 8 cm.

**Sol.**

The largest sphere can be carved out from a cube, if we take diameter of the sphere equal to edge of the cube.

Diameter of the sphere = 8 cm

Thus, radius of the sphere = 8/2 = 4cm

**Q.** An angle is 14^{o} more than its complement. Find its measure.

**Sol.**

Let the measure of the angle be x.

Now, two complementary angles have sum equal to 90^{o}.

So, measure of its complement = 90 − x

From the given condition, we have:

x = 14 + (90 − x)

⟹ 2x = 104

⟹ x = 52^{o}

**Q.** Without actual division, prove that (2*x*^{4} − 6*x*^{3 }+ 3*x*^{2} + 3*x* – 2) is exactly divisible by (*x*^{2 }− 3*x* +2).

**Sol.**

Let *f*(*x*) = 2*x*^{4}– 6*x*^{3} + 3*x*^{2} + 3*x* – 2

And *g*(*x*) = *x*^{2}– 3*x *+ 2

= *x*^{2}– 2*x* −*x *+ 2

= *x* (*x *– 2) −1(*x *– 2) = (*x *– 1) (*x *– 2)

Now if *x*^{2}– 3*x *+ 2 is a factor of *f*(*x*)

Then (*x *– 1) and (*x *– 2) both should be factors of *f*(*x*)

Then, *f *(1) = *f *(2) = 0

So, *f*(1) = 2.1^{4}– 6.1^{3} + 3.1^{2} + 3.1 – 2

= 2 – 6 + 3 + 3 – 2 = 0

*f*(2) = * *2*.*2^{4}– 6.2^{3} + 3.2^{2} + 3.2 – 2

= 32 – 48 + 12 + 6 – 2 = 0

Since *f*(1) = *f*(2) = 0

⟹ *f*(*x*) = 2*x*^{4}– 6*x*^{3} + 3*x*^{2} + 3*x* – 2 is exactly divisible by (*x *– 1) and (*x *– 2).

∴ *x*^{2 }– 3*x *+ 2 is a factor of *f*(*x*).

**Get the complete practice paper by clicking on the following link:**

**CBSE Class 9 Mathematics Solved Practice Paper : Set-I**

Here we bring you the CBSE Class 9 Mathematics Solved Practice Paper : Set-I. This practice paper has been designed as per the latest exam structure and coveres the whole syllabus of Class 9 Maths. It will help you get acquainted with the important topics to be prepared for the Class 9th Mathematics Annual Exam 2018.

**CBSE Class 9 Exam Pattern 2017-2018**

**NCERT Solutions for CBSE Class 9 Maths**

With a thorough analysis of the CBSE’s criteria to set papers for finals, subject experts at jagranjosh has designed the CBSE Class 9 Mathematics Solved Practice Paper : Set-I that will make you familiar with important questions that will ultimately help you prepare better for the new assessment structure, i.e., periodic tests as well as the annual exams 2017-2018.

**Some of the sample questions from CBSE Class 9 Mathematics Solved Practice Paper : Set-I are given below:**

**Q.** Find the radius of largest sphere that is carved out of the cube of side 8 cm.

**Sol.**

The largest sphere can be carved out from a cube, if we take diameter of the sphere equal to edge of the cube.

Diameter of the sphere = 8 cm

Thus, radius of the sphere = 8/2 = 4cm

Q. An angle is 14^{o} more than its complement. Find its measure.

**Sol.**

Let the measure of the angle be x.

Now, two complementary angles have sum equal to 90^{o}.

So, measure of its complement = 90 − x

From the given condition, we have:

x = 14 + (90 − x)

⟹ 2x = 104

⟹ x = 52^{o}

Q. Without actual division, prove that (2*x*^{4} − 6*x*^{3 }+ 3*x*^{2} + 3*x* – 2) is exactly divisible by (*x*^{2 }− 3*x* +2).

**Sol.**

Let *f*(*x*) = 2*x*^{4}– 6*x*^{3} + 3*x*^{2} + 3*x* – 2

And *g*(*x*) = *x*^{2}– 3*x *+ 2

= *x*^{2}– 2*x* −*x *+ 2

= *x* (*x *– 2) −1(*x *– 2) = (*x *– 1) (*x *– 2)

Now if *x*^{2}– 3*x *+ 2 is a factor of *f*(*x*)

Then (*x *– 1) and (*x *– 2) both should be factors of *f*(*x*)

Then, *f *(1) = *f *(2) = 0

So, *f*(1) = 2.1^{4}– 6.1^{3} + 3.1^{2} + 3.1 – 2

= 2 – 6 + 3 + 3 – 2 = 0

*f*(2) = * *2*.*2^{4}– 6.2^{3} + 3.2^{2} + 3.2 – 2

= 32 – 48 + 12 + 6 – 2 = 0

Since *f*(1) = *f*(2) = 0

⟹ *f*(*x*) = 2*x*^{4}– 6*x*^{3} + 3*x*^{2} + 3*x* – 2 is exactly divisible by (*x *– 1) and (*x *– 2).

∴ *x*^{2 }– 3*x *+ 2 is a factor of *f*(*x*).

**Get the complete practice paper by clicking on the following link:**

**CBSE Class 9 Mathematics Solved Practice Paper : Set-I**

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