essay examples

9th class question paper maths 2022

myCBSEguide

  • Mathematics
  • Class 9 Maths Sample...

Class 9 Maths Sample Paper 2023-24

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

CBSE class 9 Maths sample paper 2023-24 has both subjective and objective type questions. Especially you will find all case studies in subjective format. Unlike class 10 there is only one course in class 9 Mathematics. All students will study the same course content. You can download the model question paper for class 9 maths from myCBSEguide App or our student dashboard for free.

Sample Paper of class 9 Math – in PDF

CBSE sample question papers (solved) class 9 Mathematics is released for session 2023-24.  CBSE, New Delhi has issued the new marking scheme and blueprint for Class 9 Mathematics. We are providing Mathematics guess papers for Class 9 annual examinations 2024. These sample question papers are available for free download in the myCBSEguide app and website in PDF format. CBSE Sample Papers for class 9 Mathematics with solutions will certainly help students to score high in exams.

The Best Model Papers for Class 9 Maths 2023-24

We know that CBSE issues model papers for board classes only. So, the model papers for classes 9th and 11th are prepared by schools only. In this case, the format and difficulty level of the question paper may vary from school to school.

That’s why it is very important that you should have a standard question paper with a moderate difficulty level. Here, the model papers provided by myCBSEguide app and website follow CBSE guidelines word to word. So, these are the best model papers for class 9th students.

Sample Papers of Class 9 Maths 2024 with solution

Download PDF 

CBSE Sample Papers Class 9 Mathematics

As discussed, CBSE does not conduct board exams for class 9 students. It is purely a home exam. But schools have to follow the guidelines issued by CBSE while preparing the question paper. It is necessary to assure uniformity in terms of difficulty level and format of the question paper.

In most cases, the CBSE examination portal provides question papers and schools download them from CBSE official website only.

Class 09 – Mathematics Sample Paper – 01 (2023-24)

Maximum Marks: 80 Time Allowed: : 3 hours

General Instructions:

  • This Question Paper has 5 Sections A-E.
  • Section A has 20 MCQs carrying 1 mark each.
  • Section B has 5 questions carrying 02 marks each.
  • Section C has 6 questions carrying 03 marks each.
  • Section D has 4 questions carrying 05 marks each.
  • Section E has 3 case based integrated units of assessment (04 marks each) with subparts of the values of 1, 1 and 2 marks each respectively.
  • All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2marks questions of Section E.
  • Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

Section A Class 9 Maths Sample Paper 2023-24

  • The distance of the point (2,3) from the y-axis a) 5 units b) 3 units c) {tex}\sqrt {13} {/tex} units d) 2 units
  • If the area of an equilateral triangle is {tex}36\sqrt 3 \;c{m^2}{/tex} , then the perimeter of the triangle is a) 18 cm b) {tex}12\sqrt 3 \;cm{/tex} c) 36 cm d) 12 cm
  • The value of x – y x-y when x = 2 and y = -2, is a) 14 b) -18 c) 18 d) -14
  • Express ‘x’ in terms of ‘y’ in the equation 2x – 3y – 5 = 0. a) {tex}x = \frac{{3y – 5}}{2}{/tex} b) {tex}x = \frac{{3y + 5}}{2}{/tex} c) {tex}x = \frac{{5 – 3y}}{2}{/tex} d) {tex}x = \frac{{3 + 5y}}{2}{/tex}
  • If  {tex}x+\frac{1}{x}=5{/tex} , then  {tex}x^{2}+\frac{1}{x^{2}}={/tex} a) 23 b) 27 c) 25 d) 10
  • If {tex}\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11} b{/tex} , then the value of b  is a) 3 b) 1 c) -1 d) 2
  • In a {tex}\triangle{/tex} ABC, P, Q and R are the mid-points of the sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ? a)   20 cm b)  80 cm c)  51 cm d)   52 cm
  • The value of x  in 3 + 2 x =  {tex}(64)^{\frac{1}{2}}+(27)^{\frac{1}{3}}{/tex} is a) 14 b) 8 c) 5 d) 3
  • The equation x = 7 in two variables can be written as a) 1.x + 1.y = 7 b) 1.x + 0.y = 7 c) 0.x + 1.y = 7 d) 0.x + 0.y = 7
  • Rationalisation of the denominator of  {tex}\frac{1}{\sqrt{5}+\sqrt{2}}{/tex}  gives a) {tex}\sqrt{5}+\sqrt{2}{/tex} b) {tex}\sqrt{5}-\sqrt{2}{/tex} c) {tex}\frac{1}{\sqrt{10}}{/tex} d) {tex}\frac{\sqrt{5}-\sqrt{2}}{3}{/tex}
  • The perpendicular distance of a point Q(4, 7) from y-axis is a) 4 units b) 3 units c) 7 units d) 11 units
  • If a linear equation has solutions (1, 2), (-1, -16) and (0, -7), then it is of the form a) y = 9x – 7 b) 9x – y + 7 = 0 c) x – 9y = 7 d) x = 9y – 7
  • If x + y + z = 0, then x 3  + y 3  + z 3  is a) 3xyz b) xyz c) 2xyz d) 0
  • Assertion (A): A parallelogram consists of two congruent triangles. Reason (R): Diagonal of a parallelogram divides it into two congruent triangles. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true.
  • Assertion (A): If  {tex}\sqrt{2}{/tex} = 1.414,  {tex}\sqrt{3}{/tex}  = 1.732, then {tex}\sqrt{5}=\sqrt{2}+\sqrt{3}{/tex} . Reason (R): Square root of a positive real number always exists. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true.

Section B Class 9 Maths Sample Paper 2023-24

  • If P, Q, and R are three points on a line and Q is between P and R, then prove that PR – QR = PQ.
  • Name the quadrant in which the point lies :(i) A(1, 1) (ii) (–2, –4) (iii) C(1, –2).

Section C Class 9 Maths Sample Paper 2023-24

  • If  {tex}\sqrt{2}{/tex} =1.4142, find the value of  {tex}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}.{/tex}
  • Write linear equation 3x + 2y =18 in the form of ax + by + c = 0. Also write the values of a, b and c. Are (4, 3) and (1, 2) solution of this equation?

Construct a bar graph to represent the above data so that the bars are drawn horizontally.

Draw a histogram to represent the following grouped frequency distribution:

  • Simplify the following expression: (x + y + z) 2 + (x +  {tex}\frac{y}{2}{/tex}  +  {tex}\frac{z}{3}{/tex} ) 2 – ( {tex}\frac{x}{2}{/tex}  +  {tex}\frac{y}{3}{/tex}  +  {tex}\frac{z}{4}{/tex} ) 2

Section D Class 9 Maths Sample Paper 2023-24

  • The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost of painting one sq. cm of the surface is 7 paise. Find the total cost to paint the vessel all over, (ignore the area of edge).
  • Using factor theorem, factorize the polynomial: x 4 + 10x 3 + 35x 2 + 50x + 24

Section E Class 9 Maths Sample Paper 2023-24

  • What is the second equation formed?
  • What is the present age of Reeta in years?
  • What is the present age of Ranjeet in years?
  • Draw a triangle ABC
  • D and E are found as the mid points of AB and AC
  • DE was joined and DE was extended to F so DE = EF
  • FC was joined.
  • {tex}\triangle{/tex} ADE and {tex}\triangle{/tex} EFC are congruent by which criteria?
  • Show that CF {tex}\parallel{/tex}  AB.
  • Show that CF = BD.
  • Show that the perpendicular drawn from the Centre of a circle to a chord bisects the chord.
  • What is the length of CD?
  • What is the length of AB?

Class 09  Mathematics Sample Paper Solution

Section a solution.

  • (d) 2 units Explanation: The distance from y-axis is equal to the x-coordinate, so distance = 2 units
  • (c) 36 cm Explanation: {tex}36\sqrt 3 {/tex}  =  {tex}{{\sqrt 3 } \over 4}{/tex} a 2 a 2  =  {tex}{{36\sqrt 3 \times 4} \over {\sqrt 3 }}{/tex} = 144 a = 12 cm Perimeter =  {tex}3 \times 12{/tex}  = 36 cm
  • (d) EC Explanation: By midpoint theorem of a triangle E is the midpoint of AC, hence AE = EC
  • (d) -14 Explanation: x = 2, y = -2 x – y x-y  = 2 – (-2) 2-(-2) = 2 – (-2) 2+2  = 2 – (-2) 4 = 2- (+16) = 2 – 16 = -14
  • (a) 50° Explanation: EC || AB and CD is transverse to it. Now  {tex}\angle{/tex} ECD = {tex}\angle{/tex} AOD = 70° (Corresponding angles) In {tex}\angle{/tex} OBD {tex}\angle{/tex} OBD +  {tex}\angle{/tex} BOD +  {tex}\angle{/tex} ODB = 180° {tex}\angle{/tex} BOD = 180° –  {tex}\angle{/tex} AOD = 180° – 70° = 110° {tex}\angle{/tex} ODB = 20° (Given) So  {tex}\angle{/tex} OBD = 180° –  {tex}\angle{/tex} BOD –  {tex}\angle{/tex} ODB = 180° – 110° – 20° = 50°
  • (b) {tex}x = \frac{{3y + 5}}{2}{/tex} Explanation: {tex}\eqalign{ & 2x – 3y – 5 = 0 \cr & 2x = 3y + 5 \cr & x = {{3y + 5} \over 2} \cr}{/tex}
  • (a) 23 Explanation: Using ,(a + b) 2  = a 2  + b 2  + 2ab {tex}\left(x+\frac{1}{x}\right)^{2}{/tex}  =  {tex}x^{2}+\left(\frac{1}{x^{2}}\right){/tex}  +  {tex}2 x \frac{1}{x}{/tex} {tex}\Rightarrow{/tex}  (5) 2  = x 2  +  {tex}\left(\frac{1}{x^{2}}\right){/tex}  + 2 {tex}\Rightarrow{/tex}   {tex}x^{2}+\frac{1}{x^{2}}{/tex}  = 25 – 2 {tex}x^{2}+\frac{1}{x^{2}}{/tex} = 23
  • (b) 1 Explanation: {tex}\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11} b{/tex} taking LHS, {tex}\Rightarrow \frac{3-\sqrt{5}}{3+2 \sqrt{5}} \times \frac{3-2 \sqrt{5}}{3-2 \sqrt{5}}{/tex} {tex}\Rightarrow \frac{3(3-2 \sqrt{5})-\sqrt{5}(3-2 \sqrt{5})}{9-20}{/tex} {tex}\Rightarrow \frac{9-6 \sqrt{5}-3 \sqrt{5}+10}{-11}{/tex} {tex}\Rightarrow \frac{19-9 \sqrt{5}}{-11}{/tex} {tex}\Rightarrow \frac{-19}{11}+\frac{9 \sqrt{5}}{11}{/tex} equating this with RHS, we get, {tex}\frac{-19}{11} b=-\frac{19}{11}{/tex} {tex}\Rightarrow{/tex}  b = 1
  • (d) 3 Explanation: 3 + 2 x = {tex}(64)^{\frac{1}{2}}+ (27)^{\frac{1}{3}}{/tex} {tex}\Rightarrow{/tex} 3 + 2 x = {tex}\sqrt {64} + \sqrt[3]{27}{/tex} {tex}\Rightarrow{/tex} 3 + 2 x = 8 + 3 {tex}\Rightarrow{/tex} 2 x = 8 = 2 3 equating both, x = 3
  • (b) 1.x + 0.y = 7 Explanation: The equation x = 7 in two variables can be written as exactly 1.x + 0.y = 7 because it contain two variable x and y and coefficient of y is zero as there is no term containing y in equation x = 7
  • (b) 30°, 60° Explanation: x + y + 90° = 180° (Linear Pair) 2a + a + 90° = 180° (Since, x:y = 2:1) a =  30° x = 2a = {tex} \angle{/tex} COE = 60° (Vertically opposite angles) y = {tex} \angle{/tex} BOD = 30° (Vertically opposite angles)
  • (d) {tex}\frac{\sqrt{5}-\sqrt{2}}{3}{/tex} Explanation: {tex}\frac{1}{\sqrt{5}+\sqrt{2}}{/tex} = {tex}\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}{/tex} =  {tex}\frac{\sqrt{5}-\sqrt{2}}{3}{/tex}
  • (a) 4 units Explanation: Distance of point from y-axis is x -coordinate of given point, So, since, value of x-coordinate is 4 so, distance = 4 units
  • (a) y = 9x – 7 Explanation: Since all the given co- ordinate (1, 2), (-1, -16) and (0, -7) satisfy the given line y = 9x – 7 For point (1, 2) y = 9x – 7 2 = 9(1) – 7 2 = 9 – 7 2 = 2 Hence (2, 1) is a solution. For point (-1, -16) y = 9x – 7 -16 = 9(-1) – 7 -16 = -9 – 7 -16 = -16 Hence (-1, -16) is a solution. For point (0,-7) y = 9x – 7 -7 = 9(0) -7 -7 = -7 Hence (0, -7) is a solution.
  • (a) 3xyz Explanation: x 3  + y 3  + z 3  – 3xyz = (x + y + z) (x 2  + y 2  + z 2  – xy – yz – zx) =  x 3  + y 3  + z 3  – 3xyz = (0) (x 2  + y 2  + z 2  – xy – yz – zx) = x 3  + y 3  + z 3  – 3xyz = 0 = x 3  + y 3  + z 3  – 3xyz If x + y + z = 0, then x 3  + y 3  + z 3  is 3xyz
  • (a) Both A and R are true and R is the correct explanation of A. Explanation: Both A and R are true and R is the correct explanation of A.
  • (d) A is false but R is true. Explanation: {tex}\sqrt{2}+\sqrt{3} \neq 5{/tex} {tex}\sqrt{3} + \sqrt{2}{/tex}  = 1.732 + 1.414 = 3.146  {tex}\ne{/tex}   {tex}\sqrt{5}{/tex}  as  {tex}\sqrt{5}{/tex} = 2.236

Section B Sample Paper Solution

  • We have AX = CY [Given] Now, by Euclid’s axiom 6, we have things which are double of the same thing are equal to one another, so 2AX = 2CY Hence, AC = BC. [ {tex}\because{/tex} X and Y are the mid- points of AC and BC]
  • (i) (+, +) are the signs of the co-ordinates of points in the I quadrant. ∴ A(1, 1) lies in the I quadrant. (ii) (–, –) are the signs of the co-ordinates of points in the III quadrant. ∴ B(–2, –4) lies in the III quadrant. (iii) (+, –) are the signs of the co-ordinates of points in the IV quadrant. ∴ C(1, –2) lies in the IV quadrant.

Section C Sample Paper Solution

  • Given, {tex}\sqrt{2}{/tex}   = 1.4142 Now,  {tex}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)} \times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}}{/tex}  [by rationalising] {tex}=\sqrt{\frac{(\sqrt{2}-1)^{2}}{2-1}}=\frac{\sqrt{(\sqrt{2}-1)^{2}}}{1}{/tex}  [ {tex}\because{/tex} (a + b)(a – b) = a 2  – b 2 ] {tex}=\sqrt{2}{/tex}  – 1 = 1.4142 – 1 [ {tex}\because \sqrt{2}{/tex} = 1.4142] = 0.4142
  • In order to prove that BE is the median, it is sufficient to show that E is the mid-point of AC. Now, AD is the median in {tex}\triangle{/tex} ABC {tex}\Rightarrow{/tex}  D is the mid-point of BC. Since DE is a line drawn through the mid-point of side BC of  {tex}\triangle{/tex} ABC and is parallel to AB (given). Therefore, E is the mid-point of AC. Hence, BE is the median of {tex}\triangle{/tex} ABC.
  • We have the equation as 3x + 2y = 18 In standard form 3x + 2y – 18 = 0 Or 3x + 2y + (-18) =0 But standard linear equation is ax + by + c = 0 On comparison we get, a = 3, b = 2, c = -18 If (4, 3) lie on the line, i.e., solution of the equation LHS = RHS {tex}\therefore{/tex}  3(4) + 2(3) = 18 12 + 6 = 18 18 = 18 As LHS = RHS, Hence (4, 3) is the solution of given equation. Again for (1,2) 3x + 2y = 18 {tex}\therefore{/tex}  3(1)+2(2)=18 3 + 4 = 18 7 = 18 LHS  {tex}\neq{/tex}  RHS Hence (1, 2) is not the solution of given equation. Therefore (4,3) is the point where the equation of the line 3x + 2y = 18 passes through where as the line for the equation 3x + 2y =18 does not pass through the point (1,2).

The given table is in inclusive form. So, we first convert it into an exclusive form, as given below.

  • We have, (x + y + z) 2 + (x +  {tex}\frac{y}{2}{/tex}  +  {tex}\frac{z}{3}{/tex} ) 2 – ( {tex}\frac{x}{2}{/tex}  +  {tex}\frac{y}{3}{/tex}  +  {tex}\frac{z}{4}{/tex} ) 2 = [x 2 + y 2 + z 2 + 2(xy + yz + zx)] + [x 2 +  {tex}\frac{y^2}{4}{/tex}  +  {tex}\frac{z^2}{9}{/tex}  + 2( {tex}\frac{xy}{2}{/tex}  +  {tex}\frac{yz}{6}{/tex}  +  {tex}\frac{zx}{3}{/tex} )] – [ {tex}\frac{x^2}{4}{/tex}  +  {tex}\frac{y^2}{9}{/tex}  +  {tex}\frac{z^2}{16}{/tex}  + 2( {tex}\frac{xy}{6}{/tex}  +  {tex}\frac{yz}{12}{/tex}  +  {tex}\frac{zx}{8}{/tex} )] = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx + x 2 +  {tex}\frac{y^{2}}{4}{/tex}  +  {tex}\frac{z^{2}}{9}{/tex}  +  {tex}\frac{2 x y}{2}{/tex}  +  {tex}\frac{2 y z}{6}{/tex}  +  {tex}\frac{2 z x}{3}{/tex}  –  {tex}\frac{x^{2}}{4}{/tex}  –  {tex}\frac{y^2}{9}{/tex} –  {tex}\frac{z^{2}}{16}{/tex}  –  {tex}\frac{2 x y}{6}{/tex}  –  {tex}\frac{2 y z}{12}{/tex}  –  {tex}\frac{2 z x}{8}{/tex} = 2x 2 –  {tex}\frac{x^{2}}{4}{/tex}  + y 2 +  {tex}\frac{y^{2}}{4}-\frac{y^{2}}{9}{/tex}  + z 2 +  {tex}\frac{z^{2}}{9}-\frac{z^{2}}{16}{/tex}  + 2xy + xy –  {tex}\frac{x y}{3}{/tex}  + 2yz +  {tex}\frac{y z}{3}-\frac{y z}{6}{/tex}  + 2zx +  {tex}\frac{2 z x}{3}-\frac{z x}{4}{/tex} =  {tex}\frac{8 x^{2}-x^{2}}{4}{/tex}  +  {tex}\frac{36 y^{2}+9 y^{2}-4 y^{2}}{36}{/tex}  +  {tex}\frac{144 z^{2}+16 z^{2}-9 z^{2}}{144}{/tex}  +  {tex}\frac{6 x y+3 x y-x y}{3}{/tex}  +  {tex}\frac{12 y z+2 y z-y z}{6}{/tex}  +  {tex}\frac{24 z x+8 z x-3 z x}{12}{/tex} =  {tex}\frac{7 x^{2}}{4}{/tex}  +  {tex}\frac{41 y^{2}}{36}{/tex}  +  {tex}\frac{151 z^{2}}{144}{/tex}  +  {tex}\frac{8 x y}{3}{/tex}  +  {tex}\frac{13 y z}{6}{/tex}  +  {tex}\frac{29 z x}{12}{/tex}

Section D Sample Paper Solution

  • Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel. Then, R = 12.5 cm, r = 12 cm. Now, External surface area of the vessel = 2 {tex}\pi R ^ { 2 }{/tex}  = 2  {tex}\times \frac { 22 } { 7 } \times{/tex}  (12.5) 2  cm 2 Internal surface area of the vessel = 2 {tex}\pi r ^ { 2 }{/tex}  = 2  {tex}\times \frac { 22 } { 7 } \times{/tex}  (12) 2  cm 2 {tex}\therefore{/tex}  Total area to be painted = 2  {tex}\times \frac { 22 } { 7 } \times{/tex}  (12.5) 2  + 2  {tex}\times \frac { 22 } { 7 } \times{/tex}  12 2  cm 2 {tex}\Rightarrow{/tex}  Total area to be painted = 2  {tex}\times \frac { 22 } { 7 } \times{/tex}  {( {tex}\frac { 25 } { 2 }{/tex} ) 2  + 12 2 } cm 2 {tex}\Rightarrow{/tex}  Total area to be painted = 2  {tex}\times \frac { 22 } { 7 } \times{/tex}  ( {tex}\frac { 625 } { 4 }{/tex}  + 144) cm 2  =  {tex}\frac { 13211 } { 7 }{/tex}  cm 2 Cost of painting at the rate of 7 paise per sq. cm = Rs.  {tex}\frac { 13211 } { 7 } \times \frac { 7 } { 100 }{/tex} = Rs. 132.11
  • Given, f(x) = x 4  + 10x 3  + 35x 2  + 50x + 24 The constant term in f(x) is equal to 24 The factors of 24 are {tex}\pm{/tex} 1, {tex}\pm{/tex} 2, {tex}\pm{/tex} 3, {tex}\pm{/tex} 4, {tex}\pm{/tex} 6, {tex}\pm{/tex} 8, {tex}\pm{/tex} 12, {tex}\pm{/tex} 24 Let, x + 1 = 0 {tex}\Rightarrow{/tex}  x = -1 Substitute the value of x in f(x) f(-1) = (-1) 4  + 10(-1) 3  + 35(-1) 2  + 50(-1) + 24 = 1-10 + 35 – 50 + 24 = 0 {tex}\Rightarrow{/tex}  (x + 1) is the factor of f(x) Similarly, (x + 2), (x + 3), (x + 4) are also the factors of f(x) Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors. {tex}\Rightarrow{/tex}  f(x) = k(x + 1)(x + 2)(x + 3)(x + 4) {tex}\Rightarrow{/tex}  x 4  + 10x 3  + 35x 2  + 50x + 24 = k(x + 1)(x + 2)(x + 3)(x + 4) Substitute x = 0 on both sides {tex}\Rightarrow{/tex}  0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4) {tex}\Rightarrow{/tex}  24 = k(24) {tex}\Rightarrow{/tex}  k = 1 Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4) f(x) = (1)(x + 1)(x + 2)(x + 3)(x + 4) f(x) = (x + 1)(x + 2)(x + 3)(x + 4) hence, x 4  + 10x 3  + 35x 2  + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4) This is the required factorisation of f(x).

Section E Sample Paper Solution

  • x – 2y = 10
  • x + y = 55 …(i) and x – 2y = 10 …(ii) Subtracting (ii) from (i) x + y – x + 2y = 55 – 10 {tex}\Rightarrow{/tex}  3y = 45 {tex}\Rightarrow{/tex}  y = 15 So present age of Reeta is 15 years.
  • x + y = 55 …(i) and x – 2y = 10 …(ii) Subtracting (ii) from (i) x + y – x + 2y = 55 – 10 {tex}\Rightarrow{/tex}  3y = 45 {tex}\Rightarrow{/tex}  y = 15 Put y = 15 in equation (i) x + y = 55 {tex}\Rightarrow{/tex}  x + 15 = 55 {tex}\Rightarrow{/tex}  x = 55 − 15 = 40​​ ​​​​​So Ranjeet’s present age is 40 years.
  • {tex}\triangle{/tex} ADE and {tex}\triangle{/tex} CFE DE = EF (By construction) {tex}\angle{/tex} AED = {tex}\angle{/tex} CEF (Vertically opposite angles) AE = EC(By construction) By SAS criteria {tex}\triangle{/tex} ADE {tex}\cong{/tex} {tex}\triangle{/tex} CFE
  • {tex}\triangle{/tex} ADE {tex}\cong{/tex} {tex}\triangle{/tex} CFE Corresponding part of congruent triangle are equal {tex}\angle{/tex} EFC = {tex}\angle{/tex} EDA alternate interior angles are equal {tex}\Rightarrow{/tex} AD ∥ FC {tex}\Rightarrow{/tex}  CF ∥ AB
  • {tex}\triangle{/tex} ADE {tex}\cong{/tex} {tex}\triangle{/tex} CFE Corresponding part of congruent triangle are equal. CF = AD We know that D is mid point AB {tex}\Rightarrow{/tex} AD = BD {tex}\Rightarrow{/tex}  CF = BD
  • In  {tex}\Delta{/tex} AOP and  {tex}\Delta{/tex} BOP {tex}\angle{/tex} APO = {tex}\angle{/tex} BPO (Given) OP = OP (Common) AO = OB (radius of circle) {tex}\Delta{/tex} AOP  {tex}\cong{/tex}   {tex}\Delta{/tex} BOP AP = BP (CPCT)
  • ​​​In right  {tex}\Delta{/tex} COQ CO 2 = OQ 2 + CQ 2 {tex}\Rightarrow{/tex}  10 2 = 8 2 + CQ 2 {tex}\Rightarrow{/tex}  CQ 2 = 100 – 64 = 36 {tex}\Rightarrow{/tex}  CQ = 6 CD = 2CQ {tex}\Rightarrow{/tex}  CD = 12 cm​​
  • In right {tex}\Delta{/tex} AOB AO 2 = OP 2 + AP 2 {tex}\Rightarrow{/tex}  10 2 = 6 2 + AP 2 {tex}\Rightarrow{/tex}  AP 2 = 100 – 36 = 64 {tex}\Rightarrow{/tex}  AP = 8 AB = 2AP {tex}\Rightarrow{/tex}  AB = 16 cm

Sample Papers for Class 9 all subjects 2023-24

  • Social Science
  • English Language and Literature

To download sample papers for class 9 Science, Social Science, Mathematics, English Communicative, English language and literature, Hindi Course A, and Hindi Course B; do check myCBSEguide app or website. myCBSEguide provides sample papers with solutions, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample papers all are made available through the best app for CBSE students and the myCBSEguide website.

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

  • Class 9 English Language Sample Paper 2024
  • Class 9 Social Science Sample Papers 2024
  • Class 9 Science Sample Papers 2024
  • Class 9 Hindi B Sample Papers 2024
  • Class 9 Hindi A Sample Papers 2024
  • Class 12 Maths sample papers 2024
  • CBSE Sample Paper Class 12 Physics 2024-25
  • Class 10 Science Sample Papers 2024

4 thoughts on “Class 9 Maths Sample Paper 2023-24”

please give the solution of this sample paper

PLEASE !!!!!

This paper was kind of challenging

the answer for q. no. 36 (second choice )is wrong for the value of y . which is given 40 but it is really 20

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

COMMENTS

  1. Class 9 Maths Sample Paper 2023-24

    myCBSEguide App. Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes. Install Now. CBSE class 9 Maths sample paper 2023-24 has both subjective and objective type questions. Especially you will find all case studies in subjective format.