Important Questions Mathematics for Class 9th exams 2025

Sample Paper – Mathematics(Class 9th)

 Section A (1 mark each)

(Short-answer type questions)

 1. Find the value of (256)^0.5

2. Write the coordinates of the origin.

3. Find the HCF of 60 and 72.

4. Find the value of k if one root of the quadratic equation x²+kx−6=0 is 2.

5.If the perimeter of a square is 24 cm, find its area.

6. Write the distance formula to find the distance between two points (x₁,y₁)  and (x₂,y₂) ​).

7. Find the mode of the data: 3, 5, 6, 6, 8, 9, 6, 10, 11.

Section B (2 marks each)

(Medium-length questions)

8. Solve for x: 3x−5=7 .

9. Factorize: x²−5x+6 .

10. Find the sum of first 20 terms of the AP: 2, 4, 6, 8, …

11.The sum of the zeros of a quadratic polynomial is -5 and the product is 6. Write the polynomial.

12.Find the median of the data: 10, 12, 14, 18, 20, 22, 25.

13.Find the probability of getting an even number when a die is rolled.

14.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

                   

Section C (3 marks each)

(Long-answer type questions)

15. Show that √5​ is an irrational number.

16. Find out five rational numbers between 1 and 2.

17. The value of 1.999… in the form p/q, where p and q are integers and q ≠ 0, 

18.Find the coordinates of the centroid of the triangle whose vertices are (2,3), (4,5) and (6,7).

19. Construct a triangle with sides 5 cm, 6 cm, and 7 cm and draw its incircle.

20. The area of a sector of a circle is 154 cm² and its radius is 14 cm. Find the angle of the sector.

21. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

22. In Fig. 10.38, ABC = 69°, ACB = 31°, find BDC.

Section D (4 marks each)

(Lengthy and case study-based questions)

23. Case Study 1: There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively.
All four children roll their ball from centre point O in the direction of  XOY, X’OY, X’OY’ and XOY’. Their balls stopped as shown in the  image.

Answer the following questions:

1. What are the coordinates of the ball of Ashok?

1. (4, 3)
2. (3, 4)
3. (4, 4)
4. (3, 3)

2. What are the coordinates of the ball of Deepa?

1. (2, -3)
2. (3, 2)
3. (2, 3)
4. (2, 2)

3.What the line XOX’ is called?

1. y-axis
2. ordinate
3. x-axis
4. origin

 

4. What the point O (0,0) is called?

1. y-axis
2. ordinate
3. x-axis
4. origin

 

5. What is the ordinate of the ball of Arjun?

1. -3
2. 3
3. 4
4. 2

 

24. Case Study 2: Once four friends Rahul, Arun, Ajay and Vijay went for a picnic at a hill station. Due to peak season, they did not get a proper hotel in the city. The weather was fine so they decided to make a conical tent at a park. They were carrying 300 m² cloth with them. As shown in the figure they made the tent with height 10 m and diameter 14 m. The remaining cloth was used for the floor.

1. How much Cloth was used for the floor?

1. 6 m²
2. 16 m²
3. 10 m²
4. 20 m²

2.What was the volume of the tent?

1. 300 m³
2. 160 m³
3. 3 m³
4. 500 m³

3.What was the area of the floor?

1. 50 m²
2. 100 m²
3. 150 m²
4. 154 m²

4.What was the total surface area of the tent?

1. 400 m²
2. 4 m²
3. 300 m²
4. 400 m²

5.What was the latent height of the tent?

1. 12 m
2. 2 m
3. 15 m
4. 17 m

 

25. Case Study 3: (Mensuration)
    A cylindrical water tank has a radius of 2 meters and a height of 7 meters.
    a) Find the total surface area of the tank.
    b) If the tank is filled with water, find the volume of water it can hold in liters.

26.Case Study 4: (Linear Equations)
A shopkeeper gives discounts on school bags according to the following condition:

• If the price of a bag is less than ₹500, the discount is 10%.
• If the price is between ₹500 and ₹1000, the discount is 15%.
• If the price is more than ₹1000, the discount is 20%.

a) If a bag costs ₹650, what will be the price after discount?

b) If the final price of a bag after discount is ₹800, what was its original price before discount?

 

Here are the answers to your sample paper:

Section A (1 mark each)

1. (256)^0.5=√256=16
2. The coordinates of the origin are (0,0).
3. The HCF of 60 and 72 is 12.
4. Given one root is 2, substitute in the equation: 2²+k(2)−6=0

⇒4+2k−6=0

⇒2k=2

⇒k=1  

1. Perimeter of a square is 4 × side:

24=4s

⇒s=6

⇒Area=6²=36 

6. Distance formula: d= √(x₂−x₁)²+(y₂−y₁)²
7. Mode is the most frequent value: 6 (appears 3 times).

Section B (2 marks each)

8. Solve for x: 3x−5=7

⇒3x=12

⇒x=4

9. Factorize x²−5x+6 :

x²−5x+6

=x²−3x−2x+6

=x(x−3)−2(x−3)

=(x−2)(x−3)

factors are (x−3),(x−2)

 

10. AP: 2, 4, 6, 8, …
Formula for sum: Sn=n/2(2a+(n−1)d) 

 S₂₀=20/2(2(2)+(20−1)2)

=10(4+38)=10×42

=420  

11. Quadratic polynomial with sum = -5 and product = 6: x²+5x+6
12. Median of 10, 12, 14, 18, 20, 22, 25: Middle value=18
13. Probability of rolling an even number on a die: 3/6=1/2  ​
14. Quadrants for given points:
    • (i) (-3, 5) → 2nd quadrant
    • (ii) (-5, -3) → 3rd quadrant
    • (iii) (-5, 3) → 2nd quadrant
    • (iv) (3, 5) → 1st quadrant

Section C (3 marks each)

15. Prove √5 ​ is irrational:

 

Let us prove that √5 is an irrational number.

This question can be proved with the help of the contradiction method. Let’s assume that √5 is a rational number. If √5 is rational, that means it can be written in the form of a/b, where a and b integers that have no common factor other than 1 and b ≠ 0. i.e., a and b are coprime number.

√5/1 = a/b

√5b = a

Squaring both sides,

5b² = a² … (1)

This means 5 divides a².

From this, 5 also divides a.

Then a = 5c, for some integer ‘c’.

On squaring, we get

a² = 25c²

Put the value of a² in equation (1).

5b² = 25c²

b² = 5c²

This means b² is divisible by 5 and so b is also divisible by 5. Therefore, a and b have 5 as common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √5 is a rational number. So, we conclude that √5 is irrational.

 

16. Five rational numbers between 1 and 2:

To find five rational numbers between 1 and 2, we multiply & divide both the numbers by 6, as shown:

1 × 6 /6 = 6/ 6  And,  2 × 6/ 6 = 12 /6   Therefore, five rational numbers between 1 and 2 are: 7/ 6 , 8/ 6 , 9 /6 , 10 /6 , 11/ 6

17. Convert 999… into p/q :

x=1.999…

10x=19.999…

10x=18+1.999…=18+x
Therefore, 10x−x=18, i.e., 9x=18
i.e., x=18/9=2/1=2

18. Centroid formula: (x₁+x₂+x₃ / 3, y₁+y₂+y₃ /3)   

 (2+4+6/3, 3+5+7/3)=(4,5)  

19. Triangle construction (Use compass to construct a triangle with given sides, bisect angles to draw incircle).

Steps of Construction:
1. Construct △ABC having sides BC=7cm,AC=5cm and AB=6cm.
2. Draw angle bisector of any two angles say ∠B and ∠C of △ABC and let these intersect at a point say O.
3. Draw perpendicular from O on any side of triangle, say OP on BC.
4. Taking O as centre and OP as radius, draw a circle.
5. The circle touches the other two sides of triangle. This will be the required in circle of the triangle.
Radius, OP of incircle is 1.6 cm.

20. Find the angle of the sector: A= θ/360×πr²   

Given: 154=θ/360×22/7×14² 

Solving, θ=90∘  

21. Chord equal to radius:

Here, the chord AB is equal to the radius of the circle. In the above diagram, OA and OB are the two radii of the circle.

Now, consider the ΔOAB. Here,

AB = OA = OB = radius of the circle.

So, it can be said that ΔOAB has all equal sides and thus, it is an equilateral triangle.

∴ AOC = 60°

And, ACB = ½ AOB

So, ACB = ½ × 60° = 30°

Now, since ACBD is a cyclic quadrilateral,

ADB +ACB = 180° (Since they are the opposite angles of a cyclic quadrilateral)

So, ADB = 180°-30° = 150°

So, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.

22. We know that angles in the segment of the circle are equal so,

BAC = BDC

Now in the in ΔABC, sum of all the interior angles will be 180°

So, ABC+BAC+ACB = 180°

Now, by putting the values,

BAC = 180°-69°-31°

So, BAC = 80°

Section D (4 marks each)

Case Study 1 (Square Park)

23.  

• Ashok’s ball: (4,3)
• Deepa’s ball: (2,-3)
• Line XOX’ is x-axis
• Point O(0,0)  is origin
• Ordinate of Arjun’s ball: -3

Case Study 2 (Conical Tent)

24.  

• Cloth for floor: 16 m²
• Volume: V=1/3πr²h

=13×22/7×7²×10=513.33≈500 m³  

 

• Area of floor: πr²=154 m²

• Total surface area: πrl+πr²=400 m²  
• Slant height: 15 m (by Pythagoras).

Case Study 3 (Cylindrical Water Tank)

25.  

• Total surface area: 2πr(r+h)=2×22/7×2×(2+7)=114.28 m²  
• Volume in liters: V=πr²h=22/7×2²×7=88 m³=88000 L 

Case Study 4 (Linear Equations – Discounts)

26. (a) If a bag costs ₹650, what will be the price after discount?

• The price of the bag is ₹650, which falls in the range ₹500 – ₹1000.
• The discount given is 15%.
• Discount amount = 15% of 650:

15/100×650=97.5    

• Price after discount:

650−97.5=552.5     

Final price after discount = ₹552.50

(b) If the final price of a bag after discount is ₹800, what was its original price before discount?

Let the original price be ₹x.

• Based on the conditions:
  • If x < 500, discount is 10%.
  • If 500 ≤ x ≤ 1000, discount is 15%.
  • If x > 1000, discount is 20%.

Since the final price is ₹800, let’s assume the bag had a 15% discount, meaning it was in the ₹500 – ₹1000 range.

• The formula for final price after discount:

Final Price=x−15% of x 

800=x−15/100 x  

800=x(1−0.15)

 800=x×0.85 

 x=800 / 0.85=941.18