CBSE Class 9 Maths Important Questions for 2025 Exam

CBSE Class 9 Maths Important Questions for 2025 Exam:- Scoring well in Class 9 Maths becomes much easier when you have the right set of resources. Our collection of Class 9 Maths important questions is designed to help you score high in your exams and build a solid base for future studies. These questions follow the latest CBSE syllabus and cover all the key topics in a simple and effective way.

By practising these important questions for Class 9 Maths, you can sharpen your problem-solving skills and feel more confident during exams. This collection of Maths Question Bank for Class 9 will also help you stay organised and identify your strong and weak areas. With these resources, understanding maths will become easier, and scoring higher marks will be well within reach. Scroll down for some  class 9th maths important questions here.

Check Out: CBSE Class 9th Books

CBSE Class 9 Maths Important Questions for 2025 Exam

Check out the important questions for class 9 maths for final exam below:- 

Question 1: Find out five rational numbers between 1 and 2.

Answer 1: We have to find five rational numbers between 1 and 2.

So, let us write the numbers with the denominator = 6

Therefore, 6/6 = 1, 12/6 = 2

Now, we can note the five rational numbers between 6/6 and 12/6 as:

7/6, 8/6, 9/6, 10/6, 11/6

Question 2: Express whether the following statement is true or false and give reasons for your answer.

 Every natural number is a whole number.

Answer 2: This statement is true

Natural numbers- Numbers beginning from 1 to infinity (without fractions or decimals)

That is, The required natural numbers= 1,2,3,4…

Whole numbers- Numbers beginning from 0 to infinity (without fractions or decimals)

That is, the required whole numbers= 0,1,2,3…

Or, we can express that whole numbers include all the elements of natural numbers and zero.

Each natural number is a whole number; nevertheless, each whole number is not a natural number.

Question 3: Find five rational numbers between ?/? and ?/?.

Answer 3: Consider 3/5 < 2/3

Now x = 3/5, y = 2/3 and n = 5

d=?−?/?+1 = (2/3−3/5)/5+1 = (10−9/15)/6 = 1/90

Rational numbers in between x as well as y are

x + d, x + 2d, x + 3d, x + 4d and x + 5d

Then we obtain,

= 3/5 + 1/90 , 3/5 + 2(1/90) , 3/5 + 3(1/90) , 3/5 + 4(1/90) as well as 3/5+ 5(1/90)

= 54+1/90 , 3/5 + 1/45 , 3/5 + 1/30 , 3/5 + 2/45 and 3/5+1/18

= 55/90, 27+1/45, 18+1/30 , 27+2/45 and 54+5/90

= 11/18 , 28/45 , 19/30 , 29/45 and 59/90

Question 4:The value of 1.999… in the form p/q, where p and q are integers and q ≠ 0, is

(A) 19/10

(B) 1999/1000

(C) 2

(D) 1/9

Answer 4: (C) 2

Explanation:

(A) 19/10 = 1.9

(B) 1999/1000= 1.999

(C) 2

(D) 1/9= 0.111….

Let x = 1.9999….. — ( 1 )

Multiply equation ( 1 ) with 10

10x = 19.9999….. — ( 2 )

Subtract equation (1) from equation(2) ,

We get,

9x = 18

x = 18 / 9

x = 2

Therefore,

x = 1.9999… = 2

Hence, (C) is the correct option.

Question 5: Find out the required six rational numbers between 3 and 4.

Answer 5:There are an infinite number of rational numbers between 3 and 4.

As we have to see 6 rational numbers between 3 and 4, multiplying both the given numbers 3 and 4, with 6+1 = 7 (or any other number greater than 6)

That is, 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Accordingly, 22/7, 23/7, 24/7, 25/7, 26/7, and 27/7 are the required 6 rational numbers between 3 and 4.

Question 6: If the two  x – 2 and x – ½ are the given factors of px²

 + 5x + r, show that p = r.

Answer 6: Given, f(x) = px²+5x+r and factors are x-2, x – ½

g1(x) = 0,

x – 2 = 0

x = 2

Substituting x = 2 in place of the equation, we get

f(x) = px²+5x+r

f(2) = p(2)²+5(2)+r=0

= 4p + 10 + r = 0 … eq.(i)

x – ½ = 0

x = ½

Substituting x = ½ in place of the equation, we get,

f(x) = px²+5x+r

f( ½ ) = p( ½ )² + 5( ½ ) + r =0

= p/4 + 5/2 + r = 0

= p + 10 + 4r = 0 … eq(ii)

On solving eq(i) and eq(ii),

We get,

4p + r = – 10 and p + 4r = – 10

the RHS of both equations are the same,

We get,

4p + r = p + 4r

3p=3r

p = r.

Hence Proved.

Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

(i) – 7 + x

(ii) 6y

(iii) – ? ³

(iv) 1 – y – ? ³

(v) x – ? ³ + ?⁴

(vi) 1 + x + ?²

(vii) -6?²

(viii) -13

(ix) –p

Answer 7: (i) – 7 + x

The degree of – 7 + x is 1.

Hence, it is a linear polynomial.

(ii) 6y

The degree of 6y is 1.

Therefore, it is a linear polynomial.

(iii) – ? ³

We know that the degree of – ? ³ is 3.

Therefore, it is a cubic polynomial.

(iv) 1 – y – ? ³

We know that the degree of 1 – y – ? ³ is 3.

Therefore, it is a cubic polynomial.

(v) x – ? ³ + ?⁴

We know that the degree of x – ? ³ + ?⁴ is 4.

Therefore, it is a quartic polynomial.

(vi) 1 + x + ?²

We know that the degree of 1 + x + ?² is 2.

Therefore, it is a quadratic polynomial.

(vii) -6?²

We know that the degree of -6?² is 2.

Therefore, it is a quadratic polynomial.

(viii) -13

We know that -13 is a constant.

Therefore, it is a constant polynomial.

(ix) – p

We know that the degree of –p is 1.

Therefore, it is a linear polynomial.

Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.

Answer 8: Let the polynomial be f(x) = 5x – 4x² + 3

Now, for x = 2,

f(2) = 5(2) – 4(2)² + 3

=> f(2) = 10 – 16 + 3 = –3

Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is -3.

Similarly, for x = –1,

f(–1) = 5(–1) – 4(–1)² + 3

=> f(–1) = –5 –4 + 3 = -6

The value of the polynomial 5x – 4x² + 3 at x = -1 is -6.

Question 9:Expanding each of the following, using all the suitable identities:

(i) (x+2y+4z)²

(ii) (2x−y+z)²

(iii) (−2x+3y+2z)²

(iv) (3a –7b–c)²

(v) (–2x+5y–3z)²

Answer 9: (i) (x+2y+4z)²

Using identity, (x+y+z)² = x²+²+z²+2xy+2yz+2zx

Here, x = x

y = 2y

z = 4z

(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)

= x²+4y²+16z²+4xy+16yz+8xz

(ii) (2x−y+z)²

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = 2x

y = −y

z = z

(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x²+y²+z²–4xy–2yz+4xz

(iii) (−2x+3y+2z)²

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = −2x

y = 3y

z = 2z

(−2x+3y+2z)² = (−2x)²+(3y)²+(2z)²+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)

= 4x²+9y²+4z²–12xy+12yz–8xz

(iv) (3a –7b–c)²

Using identity (x+y+z)²= x²+y²+z²+2xy+2yz+2zx

Here, x = 3a

y = – 7b

z = – c

(3a –7b– c)² = (3a)²+(– 7b)²+(– c)²+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)

= 9a² + 49b² + c²– 42ab+14bc–6ca

(v) (–2x+5y–3z)²

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = –2x

y = 5y

z = – 3z

(–2x+5y–3z)² = (–2x)² + (5y)² + (–3z)² + (2 × –2x × 5y) + (2 × 5y× – 3z)+(2×–3z ×–2x)

= 4x²+25y² +9z²– 20xy–30yz+12zx

Check out: CBSE Class 9 Sample Papers

Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.

Answer 10: Zero of the polynomial,

g1(z) = 0

z-3 = 0

z = 3

Hence, zero of g(z) = – 2a

Let p(z) = az³+4z²+3z-4

Now, substituting the given value of z = 3 in p(z), we get,

p(3) = a (3)³ + 4 (3)² + 3 (3) – 4

⇒p(3) = 27a+36+9-4

⇒p(3) = 27a+41

Let h(z) = z³-4z+a

Now, by substituting the value of z = 3 in h(z), we get,

h(3) = (3)³-4(3)+a

⇒h(3) = 27-12+a

⇒h(3) = 15+a

As per the question,

The two polynomials, p(z) and h(z), leave the same remainder when divided by z-3

So, h(3)=p(3)

⇒15+a = 27a+41

⇒15-41 = 27a – a

⇒-26 = 26a

⇒a = -1

Question 11: Write the answer for the following questions:

(i) What is the name of the lines that are drawn horizontally and vertically to represent the positions of all points in the Cartesian plane?

(ii) What are the names of the various components of the plane that these two lines form?

(iii) Indicate the name of the intersection location of these two lines.

Solution:

(i) The x-axis and y-axis are the names of the horizontal and vertical lines drawn to calculate the position of any point in the Cartesian plane.

(ii) The quadrants are the names of each section of the plane created by the x-axis and y-axis.

(iii)The origin is the location where these two lines intersect.

Question 12: Without plotting any of the points, indicate the quadrant in which they will lie, if

(i) the ordinate is 5 while abscissa is – 3

(ii) the abscissa is – 5 while the ordinate is – 3

(iii) the abscissa is – 5 while ordinate is 3

(iv) the ordinate is 5 while abscissa is 3

Solution:

(i) The Point is (-3,5).

Therefore, the Point lies in the II quadrant.

(ii) The Point is (-5,-3).

Therefore, the Point lies in the III quadrant.

(iii) The Point is (-5,3).

Therefore, the Point lies in the II quadrant.

(iv) The Point is (3,5).

Therefore, the Point lies in the I quadrant.

Question 13: Write the coordinates of any points marked on the axes in the figure given below.

Solution: Part 1

You can see that :

(i) The Point A is at + 4 units distance from the y – axis and at zero distance from the x-axis. Thus, the x – coordinate of A is 4, and the y – coordinate will be 0. Hence, the coordinates of Point A are (4, 0).

(ii) The coordinates of point B are (0, 3).

(iii) The coordinates of point C are (– 5, 0).

(iv) The coordinates of point D are (0, – 4).

(v) The coordinates of E are (23,0).

The y coordinate of any point situated on the x-axis is always zero because every Point on the x-axis is at zero distance from the x-axis. Any point on the x-axis, therefore, has coordinates of the form (x, 0), where x represents the distance of the Point from the y-axis. Similar to the x-axis, any point’s coordinates on the y-axis are of the form (0, y), where y is the Point’s distance from the x-axis.

Part 2: What are the coordinates of the origin O?

Its abscissa and ordinate are both zero since it is at zero distance from both axes. Consequently, the origin’s coordinates are (0, 0).

It is possible that you have noticed the correlation between a point’s coordinate sign and the quadrant in which it  is located.

(i) Since the first quadrant is bounded by the positive x-axis and the positive y-axis, a point in the first quadrant will have the form (+, +).

(ii) Because the second quadrant is bounded by the negative x-axis and the positive y-axis, a point in the second quadrant will have the form (-, +).

(iii)Because the third quadrant is bounded by the negative x-axis and the negative y-axis, a point in the third quadrant will have the form (-, -).

(iv) Given that the fourth quadrant is bounded by the positive x-axis and the negative y-axis, a point in the fourth quadrant will have the form (+, -).

Question 14: Write the answer to the following questions:

(i) What are the names of the lines that are drawn horizontally and vertically to represent the positions of every Point in the Cartesian plane?

(ii) What are the names of the various components of the plane that these two lines form?

(iii) Indicate the name of the intersection location of these two lines.

Solution:

(i)  The x-axis and y-axis are the names of the horizontal and vertical lines drawn to calculate the position of any point in the Cartesian plane.

(ii)  Each section of the plane created by the x- axis and y-axis is referred to as a quadrant.

(iii) The Point where these two lines converge is called the origin.

Question 15: Name each part of the given plane formed by the Vertical and horizontal lines.

Solution:  The vertical line is called the y-axis and the horizontal line is called the x-axis. And these form four quadrants.

Question 16: Define the following linear equations in the form ax + by + c = 0 and show the values of a, b and c in every individual case:

(i) x – y/5 – 10 = 0

(ii) -2x+3y = 6

(iii) y – 2 = 0

Answer:

(i) The equation x-y/5-10 = 0

(1)x + (-1/5) y + (-10) = 0

Directly compare the above equation with ax + by + c = 0

Therefore, we get;

a = 1

b = -⅕

c = -10

(ii) –2x + 3y = 6

Re-arranging the provided equation, we obtain,

–2x + 3y – 6 = 0

The required equation –2x + 3y – 6 = 0 can be written as,

(–2)x + 3y +(– 6) = 0

Directly comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0

We obtain a = –2

b = 3

c = -6

(iii) y – 2 = 0

y – 2 = 0

The required equation y – 2 = 0 can be written as,

0x + 1y + (–2) = 0

Directly comparing 0x + 1y + (–2) = 0 with ax + by + c = 0

We obtain a = 0

b = 1

c = –2

Question 17: The price of a notebook is twice the cost of a pen. Note a linear equation in two variables to illustrate this statement.

(Taking the price of a notebook to be ₹ x and that of a pen to be ₹ y)

Answer: Let the price of one notebook be = ₹ x

Let the price of one pen be = ₹ y

As per the question,

The price of one notebook is twice the cost of one pen.

i.e., the price of one notebook = 2×price of a pen

x = 2×y

x = 2y

x-2y = 0

x-2y = 0 is the required linear equation in two variables to illustrate the statement, ‘The price of one given notebook is twice the cost of a pen.

Question 18: Give the geometric representations of 2x + 9 = 0 as an equation

(i) in one variable

(ii) in two variables

Answer:  

(i) 2x + 9 = 0

We have, 2x + 9 = 0

2x = – 9

x = -9/2

which is the required linear equation in one variable, that is, x only.

Therefore, x= -9/2 is a unique solution on the number line as shown below:

(ii) 2x +9=0

We can write 2x + 9 = 0 in the two variables as 2x + 0, y + 9 = 0

or x = −9−0.y/2

∴ When y = 1, x =  −9−0.(1)/2 = -9/2

y=2 , x = −9−0.(2)/2 =  -9/2

y = 3, x = −9−0.(3)/2= -9/2

Therefore, we obtain the following table:

Now, plotting the ordered pairs (−9/2,3), (−9/2,3) and (−9/2,3) on graph paper and connecting them, we get a line PQ as the solution of 2x + 9 = 0.

Question 19: Note four solutions individually for the following equations:

(i) 2x + y = 7

Answer: For the four answers of 2x + y = 7, we replace different values for x and y

Let x = 0

Then,

2x + y = 7

(2×0)+y = 7

y = 7

(0,7)

Let x = 1

Now,

2x + y = 7

(2×1)+y = 7

2+y = 7

y = 7 – 2

y = 5

(1,5)

Let y = 1

Now,

2x + y = 7

2x+ 1 = 7

2x = 7 – 1

2x = 6

x = 3

(3,1)

Let x = 2

Now,

2x + y = 7

2(2)+y = 7

4+y = 7

y = 7 – 4

y = 3

(2,3)

The answers are (0, 7), (1,5), (3,1), (2,3)

Question 20: The linear equation 2x -5y = 7 has

(A) A unique solution

(B) Two solutions

(C) Infinitely many solutions

Answer: (C) Infinitely many solutions

Solution:

Linear equation: The equation of two variables which gives a straight line graph is called a linear equation.

Here the linear equation is 2x – 5y = 7

Let y = 0, then the value of x is:

2x – 5(0)=7

2x =7

x = 7/2

Now, let y = 1, then the value of x is:

2x – 5 (1) =7

2x -5 =7

2x = 7 + 5

2x =12

x = 12/2

x = 6

Here for different values of y, we are getting different values of x

Therefore, the equation has infinitely many solutions

Question 21: Represent the following linear equations in the form ax + by + c = 0 and show the required values of a, b and c in every case:

Answer: (i) x –(y/5)–10 = 0

The required equation x –(y/5)-10 = 0 can be written as,

1x+(-1/5)y +(–10) = 0

Comparing the given equation x+(-1/5)y+(–10) = 0 with ax+by+c = 0

We obtain,

a = 1

b = -(1/5)

c = -10

(ii) –2x+3y = 6

–2x+3y = 6

Rearranging the equation, we obtain,

–2x+3y–6 = 0

The required equation –2x+3y–6 = 0 can be written as,

(–2)x+3y+(– 6) = 0

Comparing the given equation (–2)x+3y+(–6) = 0 with ax+by+c = 0

We obtain a = –2

b = 3

c =-6

(iii) x = 3y

x = 3y

Rearranging the equation, we obtain,

x-3y = 0

The required equation x-3y=0 can be written as,

1x+(-3)y+(0)c = 0

Comparing the given equation 1x+(-3)y+(0)c = 0 with ax+by+c = 0

We obtain a = 1

b = -3

c =0

(iv) 2x = –5y

2x = –5y

Rearranging the equation, we obtain,

2x+5y = 0

The required equation 2x+5y = 0 can be written as,

2x+5y+0 = 0

Comparing the given equation 2x+5y+0= 0 with ax+by+c = 0

We obtain a = 2

b = 5

c = 0

(v) 3x+2 = 0

3x+2 = 0

The required equation 3x+2 = 0 can be written as,

3x+0y+2 = 0

Comparing the given equation 3x+0+2= 0 with ax+by+c = 0

We obtain a = 3

b = 0

c = 2

(vi) y–2 = 0

y–2 = 0

The required equation y–2 = 0 can be written as,

0x+1y+(–2) = 0

Comparing the given equation 0x+1y+(–2) = 0 with ax+by+c = 0

We obtain a = 0

b = 1

c = –2

(vii) 5 = 2x

5 = 2x

Rearranging the equation, we obtain,

2x = 5

i.e., 2x–5 = 0

The required equation 2x–5 = 0 can be written as,

2x+0y–5 = 0

Comparing the given equation 2x+0y–5 = 0 with ax+by+c = 0

We obtain a = 2

b = 0

c = -5

Question 22: Note four solutions individually for the following equations:

 πx + y = 9

Answer: For the four answers of πx + y = 9, we replace other values for x and y

Let x = 0

Now,

πx + y = 9

(π × 0)+y = 9

y = 9

(0,9)

Let x = 1

Now,

πx + y = 9

(π×1)+y = 9

π+y = 9

y = 9-π

(1,9-π)

Let y = 0

Now,

πx + y = 9

πx +0 = 9

πx = 9

x =9/π

(9/π,0)

Let x = -1

Now,

Put x=2, we have

πx + y = 9

π(2) + y = 9

y = 9 – 2π

The answers are (0,9), (1,9-π),(9/π,0),(2,9 – 2π)

Question 23:  Find out the value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k.

Answer: The provided equation is

2x + 3y = k

As per the given question, x = 2 and y = 1.

Then, Replacing the values of x and y in the equation 2x + 3y = k,

We get,

⇒(2 x 2)+ (3 × 1) = k

⇒4+3 = k

⇒7 = k

⇒k = 7

The required value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k, is 7.

Q.24: What are the five postulates of Euclid’s Geometry?

Answer: Euclid’s postulates were:

1. A straight line may be drawn from one point to any other point.

2. A terminated line can be produced indefinitely.

3. A circle can be drawn with any centre and any radius.

4. All right angles are equal to one another.

5. If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

Q.25: If a point C lies between two points A and B such that AC = BC, then prove that AC =1/2 AB. Explain by drawing the figure.

Solution:

Given, AC = BC

Now, add AC on both sides.

L.H.S + AC = R.H.S + AC

AC + AC = BC + AC

2AC = BC + AC

Since, we know,

BC +AC = AB (as it coincides with line segment AB, from figure)

∴ 2 AC = AB (If equals are added to equals, the wholes are equal.)

⇒ AC = 1/2 AB.

Q.26: If in Q.2, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.

Solution:

Let, AB be the line segment

Assume that points P and Q are the two different midpoints of AB.

Therefore,

AP=PB ………(1)

and AQ = QB …..(2)

Also,

PB + AP = AB (as it coincides with line segment AB)

Similarly, QB + AQ = AB.

Now,

Adding AP to the L.H.S and R.H.S of the equation (1)

We get, AP + AP = PB + AP (If equals are added to equals, the wholes are equal.)

⇒ 2 AP = AB — (3)

Similarly,

2 AQ = AB — (4)

From (3) and (4),

2 AP = 2 AQ

⇒ AP = AQ

Thus, we conclude that P and Q are the same points.

This contradicts our assumption that P and Q are two different midpoints of AB.

Thus, it is proved that every line segment has one and only one mid-point.

Hence Proved.

Q.27: In the given figure, if AC = BD, then prove that AB = CD.

Solution:

It is given, AC = BD

From the given figure, we get,

AC = AB + BC

BD = BC + CD

⇒ AB + BC = BC + CD [Given: AC=BD]

We know that, according to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.

Subtracting BC from the L.H.S and R.H.S of the equation AB + BC = BC + CD, we get,

AB + BC – BC = BC + CD – BC

AB = CD

Hence proved.

Q.28: Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.

Solution:

Yes, Euclid’s fifth postulate does imply the existence of the parallel lines.

If the sum of the interior angles is equal to the sum of the right angles, then the two lines will not meet each other at any given point, hence making them parallel to each other.

So,

∠1+∠2=180°

Or ∠3+∠4=180°

Q.29: It is known that x + y = 10 and that x = z. Show that z + y = 10.

Solution:

According to the question,

We have,

x+y=10 …(i)

And, x=z …(ii)

Applying Euclid’s axiom,

“if equals are added to equals, the wholes are equal”

We get,

From Eq. (i) and (ii)

x+y=z+y ….(iii)

From Eqs. (i) and (iii)

z+y=10

Q30. Two salesmen make equal sales during the month of August. In September, each salesmen doubles his sale of the month of August. Compare their sales in September.

Solution: Let the equal sales of two salesmen in August be y. In September, each salesman doubles his sale of August.

Thus, sale of first salesman is 2y and sale of second salesman is 2y.

According to Euclid’s axioms, things which are double of the same things are equal to one another.

So, in September their sales are again equal.

Check out: CBSE Class 9th Revision Books

CBSE Class 9 Maths Preparation Tips

1. Understand the Syllabus

The Class 9 Maths syllabus is vast and divided into 15 chapters. It is essential to focus on each chapter’s concepts, formulae, and theorems. Below is the syllabus with a list of chapters and class 9 maths important questions topics for each:

2. Revise Formulae and Theorems

Keep a dedicated notebook to revise:

• Algebraic identities.

• Geometry theorems like Pythagoras theorem, congruence rules, and properties of parallelograms.

• Formulae for surface areas and volumes.

3. Practice Mock Tests

Take mock tests to evaluate your understanding of the syllabus. Focus on weak areas and improve accuracy.

4. Stay Consistent

Maths requires regular practice. Solve problems daily to build confidence.

5. NCERT Exercises and Exemplar Problems

NCERT exercises cover the entire syllabus and are highly exam-relevant. Ensure to:

• Solve all examples before starting exercises.

• Practice miscellaneous problems for better conceptual understanding.

• Attempt NCERT Exemplar Problems for advanced preparation.

Read More: CBSE Class 9 Important Questions for All Subjects

CBSE Class 9 Maths Important Questions FAQs

1. What is the most important chapter in Class 9 Maths?

All chapters are important, but focus more on Polynomials, Linear Equations in Two Variables, Triangles, Surface Areas and Volumes, and Statistics, as these form the base for Class 10 Maths.

2. How can I improve my problem-solving speed in Maths?

Practise NCERT exercises and exemplar problems daily. Time yourself while solving sample papers or mock tests. Memorise important formulae and theorems to reduce calculation time.

3.  How can I prepare for the graph-related questions?

Understand plotting points in the Cartesian plane (Chapter 3). Practise drawing and interpreting graphs for linear equations (Chapter 4). Use graph paper to ensure neatness and accuracy.

4. Are NCERT exercises enough for CBSE Class 9 Maths exams?

Yes, NCERT exercises cover most of the syllabus and exam-relevant topics. However, solving additional reference books and sample papers can help you prepare for higher-order thinking skills (HOTS) questions.

5. How can I memorise formulas effectively?

Write formulas in a separate notebook. Revise them daily. Use flashcards for quick recall. Solve formula-based questions repeatedly to understand their application.