Karnataka Class 9 Maths Model Question Papers 2023 Set 3

Karnataka Class 9 Maths Model Question Papers Set 3

Choose the correct alternative and Write the complete answer along with its alphabet in the sheet provided:

Question 1. Which of the following is an irrational number

1. \(\sqrt{9}\)
2. \(\sqrt{4}\)
3. \(\sqrt[3]{27}\)
4. \(\sqrt{10}\)

Solution: 4. \(\sqrt{10}\)

Question 2. The zero of the polynomial p(x) = x3 +1 is

1. \(\sqrt{7}\)
2. – 1
3. 2
4. 1

Solution:  2. – 1

Question 3. The expansion form of (x+ a)(x+ 6) is

1. x²-ab-a
2. x²  -(a-b)x-ab
3. x²+(a+b)x+ab
4. x²+2ab

Solution: 3. x²+(a+b)x+ab

Question 4.y = 3x+ 5 has

1. A unique solution
2. Only two solutions
3. No solutions
4. Infinitely many solutions

Solution: 4. Infinitely many solutions

Karnataka 9th Standard Maths Model Question Paper 2023 Set 3 Free PDF

Question 5. A sum of the angles of a quadrilateral is

1. 180°
2. 270°
3. 360°
4. 400°

Solution: 3.360°

Question 6. In a triangle ABC, E is the midpoint of median AD. Then ar (ABED) =

1. 1/4 ar(ΔABC)
2. 1/3 ar(ΔABC)
3. ar(ΔABC)
4. 1/6 ar(ΔABC)

Solution: 1. 1/4 ar(ΔABC)

Question 7. The total surface area of a hemisphere is

1. 4πr2
2. 2πr2
3. 3πr2
4. πr2

Solution: 3. 3πr2

Question 8. Heron’s formula –

1. \(\sqrt{s(s-a)(s-b)}\)
2. \(\sqrt{s(s-a)(s-c)}\)
3. \(\sqrt{s(s-a)}\)
4. \(\sqrt{s(s-a)(s-b)(s-c)}\)

Solution: 4. \(\sqrt{s(s-a)(s-b)(s-c)}\)

Class 9 Karnataka Maths Model Question Paper 2023 Set 3 with Answer Key

Question 9. The mode of the following score is 4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4

1. 5
2. 7
3. 6
4. 4

Solution: 2. 7

Question 10. In which quadrant does (-2, 4) lie?

1. I
2. II
3. III
4. IV

Solution: 2. II

Answer the following: 

Question 1. Define a rational number

Solution: A number that can be expressed in the form of p/q, where ‘p’ and ‘q’ are integers and (q ≠0)

Example: 1/2,-7/5 ……..

Question 2. Is zero a rational number? can you write it  in the form of p/q,  where p and q are integers and q ≠ 0

Solution: yes, zero is a rational number as it can be represented as (0/1) or (0/2)…………….

Question 3. Write 36/100  in decimal form.

Solution = 36/100 = 0.36, terminating………..

Question 4. Simplify \((\sqrt{5}+\sqrt{2})^2\)

Solution:

\(\begin{aligned}
& (\sqrt{5}+\sqrt{2})^2 \\
& (a+b)^2=a^2+2 a b+b^2 \\
& (\sqrt{5}+\sqrt{2})^2=(\not \sqrt{5})^2+2(\sqrt{5})(\sqrt{2})+(\sqrt{2})^2 \\
& (\sqrt{5}+\sqrt{2})^2=7+2 \sqrt{10}
\end{aligned}\)

Question 5. Write the general form of linear equation in two variable

Solution: ax + by + c = 0, where a, b and c are real numbers.

Question 6. Find the value of ‘x’ in the figure,

Solution: x = 1/2 x 80°

Question 7. The diameter of the base of a cone is 10cm and its slant height is 10cm. Find the curved surface area

Solution:

CSA of cone =πrl

= 22/7 x 5 x 5

= 7805cm²

Question 8. Find the value of (125)1/3

Solution: 

\((125)^{\frac{1}{3}}=\left(125^3\right)^{\frac{1}{3}}=5^{\not p \times \frac{1}{3}}=5\)

Question 9. Write any two polynomials whose degree is 2.

Solution: 2X² +X= 0  ,  3 y²+l = 0

Question 10. Name the quadrants in which the following points lie:

1. P(4,4)
2. Q (-4, 4)
3. R(-4,4)
4. S(-4,8)

Solution:

1. P(4,4) = I
2. Q(-4,4) = II
3. R(-4,-4) = III
4. S(-4,8) =II

Karnataka Board 9th Class Maths Model Paper 2023 Set 3 Latest Pattern

Answer the following :

Question 1. Express locate \(\sqrt{2}\) on the number line.
Solution:

Question 2. Express \(0 \ overlines {99}\) in the form of p/q, where p and q are integers and 0.

Solution:

\(0 \ overlines {99}\)= 0.99999……..

Let x = 0.99999…………

100x= 100 x 0.9999…………..

100x=99.9999…………..

100x=99+0.9999……………

100x = 99 + x 1

100x-x = 99

99x=9

x = 99/99 = 1

Thus 0.99999 = 1 = 1/1

1 Here P = 1, q = 1

Question 3. Simplify: \((\sqrt{5}+\sqrt{3})(2+\sqrt{3})\)

Solution:

\(\begin{aligned}
& (5+\sqrt{3})(2+\sqrt{3}) \\
& =5(2+\sqrt{3})+\sqrt{3}(2+\sqrt{3}) \\
& =10+5 \sqrt{3}+2 \sqrt{3}+(\sqrt{3})^2 \\
& =10+5 \sqrt{3}+2 \sqrt{3}+(\sqrt{3})^2 \\
& =10+5 \sqrt{3}+2 \sqrt{3}+3 \\
& =13+5 \sqrt{3}+2 \sqrt{3} \\
& =13+7 \sqrt{3}
\end{aligned}\)

Question 4. Find the value of k, if (x-l) is a factor of
\(p(x)=k x^2-\sqrt{2} x+1\)

Solution: 

\(\begin{aligned}
& p(x)=k x^2-\sqrt{2} x+1 \\
& x=1, k(1)^2-\sqrt{2}(1)+1 \\
& p(1)=k-\sqrt{2}+1 \\
& 0=k-\sqrt{2}+1 \\
& k=\sqrt{2}-1
\end{aligned}\)

Question 5.Factorise : 6X² + 5x – 6

Solution:

6X² +5x-6

= 6X² +9x—4x — 6

= 3x(2x + 3) — 2(2x + 3)

= (2x + 3)(3x-2)

Question 6. Expand \(\left(x+\frac{2}{3} y\right)^3 \)using a suitable identity.

Solution:

\(\left(x+\frac{2}{3} y\right)^3\)

(a + b)³=a³+b³ + 3ab(a + b)

\(\begin{gathered}
\left(x+\frac{2}{3} y\right)^3=(x)^3\left(\frac{2}{3} y\right)^3+\not \beta(x)\left(\frac{2}{\not p} y\right)\left(x+\frac{2}{3} y\right) \\
\left(x+\frac{2}{3} y\right)^3=x^3+\frac{8}{27} y^3+2 x y\left(x+\frac{2}{3} y\right)
\end{gathered}\)

Question 7.Which of the following options is true and why?
y = 3xr + 5

1. a unique solution
2. only two solutions
3. infinitely many solutions.

Solution: 3. infinitely many solutions.

As for every value of x, there will be a value for satisfying the above equation and vice-versa.

Question 8.Find the value of k, if x = 2,y = 1 is a solution of the equation 2x + 3y=k

Solution:

2x + 3y = k

2(2) + 3(1) = k

4 + 3 = k

∴ k = l

9th Standard Karnataka Maths Sample Paper 2023 Set 3 with Solutions PDF

Question 9. In AABC 5 AD is the perpendicular bisector of BC. S.T. ΔABCis an isosceles triangle in which AB = AC.

Solution:

ΔABC, AD ⊥ BC,

In ΔADB and ΔADC

∠APB = ∠ADC = 90°

DB = DC

∴ AD is the perpendicular bisector of BC

AD =AD (common)

∴ AADB ≅ AADC (SAS)

∴ AB = AC (CPCT)

Question 10. Find the volume of a sphere whose 7cm radius is

Solution:

The volume of the sphere = 4/3πr³

\(=\frac{4}{3} \times \frac{22}{7} \times(7)^{\frac{2}{z}}\)

\(\begin{aligned}
& =\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \\
& =\frac{4312}{3} \\
& =1437.33 \mathrm{~cm}^3
\end{aligned}\)

Question 11. Define a line segment.

Solution: A straight line joining any two distinct points is called a line segment.

Question 12. In the fig, sides QP and RQ of APQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ

Solution: ∠SPR + ∠QPR = 180° (linear pair)

135° + ∠QPR = 180°

∠QPR = 45°

∠PQT + ∠PQR = 180° (linear pair)

110″+ ∠PQR =180°

∠PQR = 70°

In APQR, ∠PQR + ∠QPR + ∠PRQ = 180°

70° + 45° + ∠PRQ = 180°

∴∠PRQ = 65°

Question 13. In fig, if x + y = w + z, then prove that AOB is a line.

Solution:

Let x + y — w + z = k

∴ x + y + w + z = 360°

k + k = 360°

2k = 360°

k = 360°/2

k= 180°

∴ x + y = w + z = k

x + y = w + z = 180°

Question 14.ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal  S.T.

1. ΔABE ≅ΔACF
2. AB = AC

Solution:

In ΔABE and ΔACF

∠AEB = ∠AFC = 90° (given)

altitude BE = altitude CF (given)

∠A is common

∴ΔABE = ΔACF

AB = AC (CPCT)

∴AB is an isosceles triangle

Question 15.ABCD is a quadrilateral in which P, Q, R, and S are midpoints of the sides AB, BC, CD, and DA. AC diagonal S.T. SR ||  AC and SR= 1/2 AC
Solution:

In A ADC, S, and R are the midpoints of the sides AD and CD respectively.

∴By the midpoint theorem, the line joining the midpoints of any two sides of a triangle is parallel to the third side and half of the third side.

∴SR || AC

and SR 1/2 AC

Question 16. The following table gives the lifetimes of 400 neon lamps.
Solution:

Represent the given information with the help of a histogram.

Scale : x-axis ; 1cm- limit y-axis; lcm = limit

Question 17. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Solution:

Total number of balls played = 30

Number of boundaries hit = 6

Number of balls in which she has not hit the boundary =30-6 = 24 ∴Probability=

(No. of times she does not hit the boundary)/

(Total number of balls played)

p= 24/30 = 4/5

Question 18. Use deductive reasoning to answer the following.

1. Martians have red tongues. Gulag is a Martian. Based on these two statements, what can you conclude about Gulag?
2. If it rains for more than four hours on a particular day, the gutters will have to be cleaned the next day. It has rained for 6 hours today. What can we conclude about the condition of the gutters tomorrow?

Solution:

1. Gulag his red tongue.
2. we conclude that the gutters will have to be cleaned tomorrow.

Answer the following :

Question 1.Check whether 7 + 3x is a factor of 3x³+ 7x

Solution:

consider p (x) = 3x³ + 7x

\(\begin{gathered}
\mathrm{P}\left(-\frac{7}{3}\right)=3\left(-\frac{7}{3}\right)^3+7\left(-\frac{7}{3}\right) \\
={ }^1 \not \beta\left(-\frac{343}{27}\right)^3+7\left(-\frac{7}{3}\right) \\
=\frac{-343}{9}-\frac{49}{3} \\
=\frac{-343-147}{9} \\
\mathrm{P}\left(-\frac{7}{3}\right)=\frac{-490}{9}
\end{gathered}\)

Karnataka Class 9 Maths Previous Year Model Paper 2023 Set 3 Solved PDF

Question 2. Visualize \(4. \overline{26}\) on the number line up to 4 decimal places.

Solution:  \(4. \overline{26}\)= 4.2626

Question 3. In fig, if PQ ⊥PS. PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y

Solution:

PQ || SR and QR is a transversal line

∴ ∠PQR = ∠QRT (Alternate interior angles)

x + 28° = 65°

x = 65°- 28°

x = 37°

By using the angle sum property in ΔSPQ

∠SPQ + x + y = 180°

90°+37° +y = 180°

y = 53°

∴ x = 37°

y = 53°

Karnataka 9th Maths Sample Paper 2023 Set 3 for Exam Preparation

Question 4.In fig. AC =AE, AB = AD, and ∠BAD = ∠EAC S.T. BC = DE

Solution:

∠BAD = ∠EAC (given)

Adding ∠DAC on both sides,

∠BAD + ∠DAC = ∠EAC + ∠DAC

∴∠BAC-∠DAE

In ΔBAC and ΔDAE

AB =AD (given)

∠BAC = ∠DAE

AC = AE (given)

∴ ΔBAC ≅ ΔDAE (SAS)

∴ BC=DE (CPCT)

Question 4. S.T. the diagonals of a parallelogram divide it into four triangles of equal area.

Solution:

We know that the diagonals A and B of a parallelogram bisect, each other.

∴‘O’ is the midpoint of AC and BD.

BO is the median in ΔABC.

∴It will divide it into two triangles of equal areas.

∴Area (ΔAOB) =Area (ΔBOC)…………………..(1)

In ΔBCD, CO is the median

∴Area (ΔBOC) =Area (ΔCOD)……………………. (2)

similarly, Area(ΔCOD)=Area (ΔAOD) …(3)

From (1) (2) & (3)

Area (ΔAOB) = Area (ΔBOC)

= Area (ΔCOD) = Area (ΔAOD)

The diagonals of a parallelogram divide it into four triangles of equal area.

Question 5. Construct a triangle ABC in which Be = 8cm,
∠B = 45° and AB – AC = 3.5cm
Solution:

Answer the following:

Question 1. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution:

The volume of the moon = 4/3 πr³
   
\(\begin{aligned}
& =\frac{4}{3} \pi r^3\left(\frac{d}{8}\right)^3 \\
& =\frac{4}{3} \times \frac{\pi d^3}{512}
\end{aligned}\)

The volume of the earth = 4/3 πr³

\(\begin{aligned}
& =\frac{4}{3} \pi\left(\frac{d}{2}\right)^3 \\
& =\frac{4}{3} \times \frac{\pi d^3}{8}
\end{aligned}\) \(\frac{\text { volume of the moon }}{\text { volume of the earth }}=\frac{\frac{4}{3} \times \frac{\pi d^3}{512}}{\frac{4}{3} \times \frac{\pi d^3}{8}}\) \(\frac{\text { volume of the moon }}{\text { volume of the earth }}=\frac{1}{64}\)

The volume of the moon=1/64 volume of the earth.

Question 2. Find the area of a triangle with two sides which are 18cm and 10cm and the perimeter is 42cm.

Solution:

Let the third side of the triangle 42cm

perimeter = 42cm

18 + 10 + x = 42

28 + x = 42

x = 42 – 28

x = 14

x = perimeter/2 =42/2 = 21cm

By Heron’s formula

Area of a triangle= \(\sqrt{s(s-a)(s-b)(s-c)}\)

\(\begin{aligned}
& =\sqrt{21(21-18)(21-10)(21-14)} \\
& =\sqrt{21 \times 3 \times 11 \times 7} \\
& =21 \sqrt{11} \mathrm{~cm}^2
\end{aligned}\)

Question 3. Prove that equal chords in a circle are equidistant from the center.

Solution:

Data: A circle with center‘ O’, chords AB and CD such that AB = CD. Draw OP ⊥ AB and OQ ⊥ CD.

To prove: OP = OQ

Construction: Join  OB and OD

Proof: AP = PB

CQ = QD

PB = 1/2 AB and QD = 1/2 CD.

In ΔOPB and ΔOQD

∠OPB= ∠OQD (both equal to 90°)

OB = OD (radii of the circle)

PB = QD

∴ OPB≅ QQD

OB= OD (radii of the circle)

PB = QD

∴OPB≅ OQD (RHS theorem)

OP = OQ (CPCT)

9th Class Karnataka Maths Model Paper 2023 Set 3 Important Questions PDF

Question 4. Draw the graph of 4x-y = 5
Solution:

Scale : x-axis ; 1cm = 1unit y-axis ; 1cm = 1unit