CBSE Class 9

CBSE Sample Papers for Class 9 Maths Paper 5 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 5

CBSE Sample Papers for Class 9 Maths Paper 5

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Write the value of \(\frac { \sqrt { 32 } +\sqrt { 48 } }{ \sqrt { 8 } +\sqrt { 12 } } \)
OR
Find \(\sqrt [ 4 ]{ \sqrt [ 3 ]{ { 2 }^{ 2 } } } \)

Question 2.
For what value of a, x – 3 is a factor of x³ + x² – 17x + a?

Question 3.
In figure below, ∆PQR is an isosceles right triangle right angled at Q. Find angle P (∠P).

Question 4.
In a parallelogram ABCD, E and F are any two points on the sides AB and BC respectively. If ar (∆DCE) is 12 cm², then find ar (∆ADF).

Question 5.
What is the volume of the hollow right circular cylinder?

Question 6.
In a class consisting of 18 boys and 22 girls, one student is absent. Find the probability that the absent student is boy.

SECTION-B

Question 7.
If \(\frac { { x }^{ 2 }+1 }{ x } =7\), then find the value of \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \)

Question 8.
Find the angle which is two times its supplementary angle.

Question 9.
Which of the following points lie on the x-axis? Show the position of other points also.
A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (-1, 0), F (0, -1) G (4, 0), H (0, 7)

Question 10.
Answer the following:
(i) A point lies on y-axis, then which coordinate is zero?
(ii) If a point is at a distance of 2 units from y-axis and 3 units from x-axis. Write the coordinates.

Question 11.
Into a circular drum of radius 4.2 m and height 3.5 m, how many full bags of wheat can be emptied if the space required for wheat on each bag is 2.1 cubic m? (Take π = 3.14)

Question 12.
Find the median of first 11 multiples of 3.

Question 13.
Simplify

OR
If 5^2x-1 – (25)^x-1 = 2500, then find the value of x.

Question 14.
If 3x + y + z = 0, show that 27x³ + y³ + z³ = 9xyz.
OR
If a + b = c, then show that b² + ac = c² – ab.

Question 15.
In the given figure, ∠ABC = 60°, ∠BCE = 25°, ∠DCE = 35° and ∠CEF = 145°. Prove that AB || EF.

Question 16.
Find the value of x and y, if (x + 4, 3y – 2) = (7, -5).

Question 17.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
OR
Show that the diagonals of a square are equal and bisect each other at right angle.

Question 18.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

OR
In the following figure, AB is equal to the radius of the circle. Find the value of ∠AMB, if ‘O’ is the centre of the circle.

Question 19.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Question 20.
An isosceles triangle with base 48 cm has area 240 sq cm. Find the remaining two sides of the triangle.

Question 21.
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the
(i) Inside surface area of the dome
(ii) Volume of the air inside the dome.

Question 22.
A jar contains 3000 white, black and red beads. The beads are thoroughly mixed and a sample of 60 is taken. The sample is found to contain 17 white beads, 32 black beads and 11 red beads. Estimate the number of beads of each colour in the jar.

SECTION-D

Question 23.
If 2^a = 3^b = 6^c, then show that \(c=\frac { ab }{ a+b }\)

Question 24.
If both (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors px² + 5x + 8, show that \(\frac { p }{ r }=1\)

Question 25.
In figure below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Question 26.
Bisectors of angle A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.Prove that the angles of a triangle DEF are 90° – \(\frac { 1 }{ 2 }\)A, 90° – \(\frac { 1 }{ 2 }\)B, and 90° – \(\frac { 1 }{ 2 }\)C

Question 27.
A cylinder is within the cube touching all the vertical faces. A cone is inside the cylinder. If their heights are same with the same base. Find the ratio of their volumes.

Question 28.
Construct a combined histogram and frequency polygon for the following frequency distribution.

Question 29.
ABC is a triangle right angled at C. A line through the mid¬point M of hypotenuse AB and parallel to BC intersect AC at D, show that:
(i) D is the mid-point of AC.
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac { 1 }{ 2 }\)AB

Question 30.
Kartikeya and Pallavi of class IX decided to collect Rs 25 for class cleanliness. Write it in linear equations in two variables. Also draw the graph. What values of both the students are depicted here?

Solutions

Solution 1.

Solution 2.
x – 3 = 0 => x = 3, p(x) = x³ + x² – 17x + a
For factor x – 3, remainder p(3) = 0
P(3) = (3)³ + (3)² – 17 x 3 + a = 0 => 27 + 9 – 51 + a = 0
a = 51 – 36 = 15 => a = 15

Solution 3.
∠P = ∠R (Isosceles A), ∠Q = 90°
∠P + ∠Q + ∠R = 180° => ∠P + ∠P + 90° = 180°
2 ∠P = 90° =>∠P = 45°

Solution 4.
ar (∆ADF) = \(\frac { 1 }{ 2 }\) ar (||gm ABCD) = ar (∆DCE)
ar (∆ADF) = ar (∆DCE)
ar (∆ADF) = 12 cm²
[In same base and between same parallels area of triangle is half the area of parallelogram.

Solution 5.
Volume = πh (R² – r²) cubic unit.

Solution 6.
Total students = 18 + 22 = 40, Boys = 18, girls = 22
P (Absent boys) = p (E) = \(\frac { 18 }{ 40 }\) = \(\frac { 9 }{ 20 }\)

Solution 7.

Solution 8.
Let angle be x. It supplementary angle = (180 – x)°
=> x = 2(180 – x)
=> x = 360 – 2x =>3x = 360° => x = 120°

Solution 9.
(i) Points lie on x-axis may be (a, 0) => B (1, 0), E (-1, 0), G (4, 0).
(ii) Points lie on y-axis may be (0, b) => C (0, 1), F (0, -1), H (0, 7)
(in) Point lies on origin => D(0, 0)
(iv) Point lies of I quadrant may be (x, y) = (1, 1).

Solution 10.
(i) If a point lies on y-axis, then its x-coordinate will be zero, i.e., it is represented by (0, b).
(ii) Coordinates of points be (2, 3).

Solution 11.
r = 4.2m
h = 3.5m

Complete or full bags of wheat = 92 bags.

Solution 12.
First 11 multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33
Median = Middle term of the observations
Total terms = n = 11

Median = 18 (6th term)

Solution 13.

Solution 14.
If a + b + c = 0 then a³ + b³ + c³ = 3abc
If a = 3x, b = y, c = z
then a + b + c = 3x + y + z = 0
and, a³ + b³ + c³ = 3 abc
(3x)³ + (y³) + (z)³ = 3 x (3x) x y x z
27 x³ + y³ + z³ = 9 xyz
OR
Since a + b = c
On squaring both sides (a + b)² = c²
a² + b² + 2 ab = c²
a² + b² + ab + ab = c²
(a² + ab) + b² = c² – ab
a (a + b) + b² = c² – ab
ac + b² = c² – ab [since a + b = c]
b² + ac = c² – ab

Solution 15.
∠BCD = ∠BCE + ∠DCE
= 25° + 35° = 60°

Solution 16.
[(x + 4), (3y – 2)] = (7, -5)
On comparing
x + 4 = 7 ⇒ x = 7 – 4 = 3 ⇒ x = 3
3y – 2 = -5 ⇒ 3y = -5 + 2 = -3
⇒ y = \(\frac { -3 }{ 3 }\) = -1
x = 3,y = -1

Solution 17.

Given: In quadrilateral ABCD
OA= OC and ∠AOB = ∠BOC = ∠COD
OB = OD = ∠DOA = 90°
To prove: ABCD is a rhombus
i.e. ABCD is a parallelogram having all sides equal.
Proof: In ∆AOD and ∆COD
OA = OC(Diagonals bisect each other)
∠AOD = ∠COD (Each 90°)
OD = OD (Common)
∴ ∆AOD ≅ ∆COD (By S.A.S)
=> AD = CD (By CPCT)
Similarly, AD = AB, and CD = BC
So, AB = BC = CD = DA
Since opposite sides of a quadrilateral ABCD are equal, so it is a parallelogram.
Here, all sides of the parallelogram are equal. So it is rhombus.
OR

Given: ABCD is a square and diagonals AC and BD intersect each other at O.
To Prove: AC = BD, OA = OC, OB = OD and ∠AOB = 90°
Proof: In ∆ABC and ∆DCB
AB = DC
∠ABC = ∠DCB
BC = BC
∴ ∆ABC ≅ ∆DCB
AC = DB
Hence the diagonals of a square are equal in length.
In ∆AOB and ∆COD ,
∠AOB = ∠COD
∠ABO = ∠CDO
AB = CD
∴ ∆AOB ≅ ∆COD
=> AO = CO and OB = OD
Hence diagonals of a square bisect each other.
Now in ∆AOB and ∆COB
OA = OC; AB = BC and BO = BO => ∴ ∆AOB ≅ ∆COB(SSS Congruence)
∠AOB = ∠COB (CPCT)
∠AOB = ∠COB
But ∠AOB + ∠COB = 180° (Linear pair)
2 ∠AOB = 180°
∠AOB = \(\frac { 180 }{ 2 }\) = 90°
∠AOB = 90°
Hence, the diagonals of a square bisect each other at right angle.

Solution 18.

Given: In trapezium ABCD, AB || DC and BC = AD
To Prove: ABCD is a cyclic quadrilateral.
Construction: Draw A ⊥ DC and BN ⊥ DC
Proof: In ∆AMD and ∆BNC
AD = BC (Given)
∠AMD = ∠BNC (Each 90°)
AM = BM
(Perpendicular distance between two parallel lines are equal)
∴ ∆AMD ≅ ∆BNC (By RHS congruency)
∴∠ADC = ∠BCD …(1) (CPCT)
∵ ∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° …(2)
=> ∠BAD + ∠BCD = 180° (∵ ∠ADC = ∠BCD)
=> This shows that opposite angles are supplementary.
=> ABCD is a cyclic quadrilateral.
OR

Given: O is the centre of circle and AB is equal to the radius.
Construction: Join OA, OB and OC.
To Find: ∠AMB
In ∆OAB, OA = OB = OC (Each equal to radius)
=> ∆OAB is an equilateral triangle.
=> ∠AOB = 60°
=> ∠AOB = \(\frac { 180 }{ 2 }\)= 90°
∠ACB = 30°
(Angle subtended by an arc at the centre is double that subtended at any part of the circle)
But ∠DAC = 90° (Angle in the semi-circle is 90°)
Now ∠CAM = 180°- 90° = 90° (Linear pair)
Now in ∆ (CAM)
∠A + ∠C + ∠M= 180°
90° + 30° + ∠M = 180°
∠M = 180° – 90° – 30° = 60°
∠AMB = 60°

Solution 19.
Steps of Construction:
1. Draw a line segment AB of 11 cm (XY + YZ + ZX =11 cm).
2. Construct an angle ∠PAB = 30° at point A and an angle ∠OBA of 90° at point B.
3. Bisect ∠PAB and ∠QBA. Let these bisectors intersect each other at point X.
4. Draw perpendicular bisector ST of AX and UV of BX.
5. Let ST intersects AB at Y and UV intersects AB at Z. Join XY, XZ.
Thus, ∆XYZ is the required triangle.

Solution 20.
Let two sides of Isosceles triangle = x cm
Area = 240 cm², Here, a = x, b = x, c = 48 cm

Hence two sides of Isosceles triangle = 26 cm.

Solution 21.
(i) Cost occured in white-washing the dome from inside = Rs 498.96
Cost of white-washing 1 m² = Rs 2

V = 523.908 m³ = 523.9 m³ approximately.
Hence, volume of air inside the dome is 523.9 m³.

Solution 22.
Sample sum = 17 + 32 + 11 = 60

Hence number of beads
(i) White = 850
(ii) Black = 1600
(iii) Red = 550

Solution 23.
2^a = 3^b ⇒ 2 = 3^b/a ….(1)
6^c = 3^b ⇒ 6 = 3^b/c …..(2)

Solution 24.
Let f(x) = px² + 5x + r. If (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors of polynomial f(x), then remainder must be zero.
If x – 2 = 0, => x = 2
Remainder = f(2) = p(2)² + 5(2) + r = 0
4p + 10 + r = 0
4p + r = – 10 …(1)

Solution 25.
∠BAD = ∠EAC (Given)
Adding ∠DAC on both sides
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE …(1)
In ∆BAC and ∆ADE
AC = AE (Given)
∠BAC = ∠DAE (Proved in Eqn. (1)]
=> AB = AD (Given)
∴ ∆ABC ≅ ∆DAE (By SAS congruency)
BC = DE (CPCT)
Hence, BC = DE .

Solution 26.
It is given that BF is the bisector of ∠B

Solution 27.
Let the length of each edge of the cube be ‘a’ units.
V1 = Volume of cube = a³ cubic units …(1)
Since a cylinder is within the cube and touches all the faces of the cube.
r = Radius of the base of the cylinder = \(\frac { a }{ 2 }\)
H = Height of the cylinder = a

Solution 28.
For frequency polygon 2 more class intervals (-10 – 0) and (50 – 60) are taken and classmarks of all class intervals are taken.
The class marks are -5, 5, 15, 25, 35,45, 55.
The graph is plotted on graph paper. Coordinates on both axes are given here in the graph paper.

Solution 29.
(i) In ∆ABC, M is the mid-point of AB and MD || BC so, D is the mid-point of AC.
(Converse .of mid-point theorem)
(ii) As DM || CB and AC is a transversal line for them
∠MDC + ∠DCB = 180° (Co-interior angles)
∠MDC + 90° = 180°
∠MDC = 90°
MD⊥AC
(iii) Join MC
In ∆AMD and ∆CMD AD = CD (D is mid-point of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (Common)

Solution 30.
Let Kartikaya collected = Rs x
Pallavi collected = Rs y
ATQ, Linear Equation in two variables
x + y = 25
(i) If x = 10,y = 25 – x = 25 – 10 = 15
(ii) If x = 15, y = 25 – 15 = 10


The graph is plotted.
Values: (i) Co-operation
(ii) Awareness about cleanliness,
(iii) Responsibility.

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CBSE Sample Papers for Class 9 Maths Paper 4 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 4

CBSE Sample Papers for Class 9 Maths Paper 4

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Simplify: 3^2/3 x 7^2/3.

Question 2.
When P(x) = x³ – ax² + x is divided by (x – a). Find the remainder.

Question 3.
On which axis does the point A(0, 4) lie?

Question 4.
Euclid divided his book “Elements” into how many chapters?

Question 5.
A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting head?

Question 6.
What is the total surface area (T.S.A.) of a cone whose radius is \(\frac { r }{ 2 }\) and slant height is 2l?

SECTION-B

Question 7.
Factorize: a(a – 1) – b(b – 1).

Question 8.
The angles of a triangle are in the ratio 2:3:7. Find the measure of each angle of triangle.

Question 9.
Find the angle at which the bisectors of any two adjacent angles of a parallelogram intersect.

Question 10.
Show that a median of a triangle divides it into two triangles of equal area.

Question 11.
Find the area of isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.

Question 12.
The mean of 40 numbers was found to be 38. Later on, it was detected that a number 56 was misread as 36. Find the correct mean of the (data) given numbers.
OR
Following data gives a number of children in 40 families.
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3,4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2,4, 3, 2, 1, 0, 5, 1, 2,4, 3, 4, 1, 6, 2, 2.
Represent it in the form of frequency distribution taking classes 0-2, 1-4, etc.

SECTION-C

Question 13.
Simplify

Question 14.
Find the product: (a-b-c) (a² + b² + c² + ab + ac – bc).
OR
The polynomial f(x) = x⁴ – 2x³ + 3x² – ax + b when divided by (x – 1) and (x + 1) leaves the remainder 5 and 19 respectively. Find the value of a and b. Hence find the remainder when fix) is divided by (x – 2).

Question 15.
Fill in the blanks:
(i) For the line 4x + 3y = 12, x-intercept =___ and y-intercept = ___
(ii) If x = 4, y = 3, is a solution of 2x + ky = 14, then k =___
(iii) If the point P(P, 4) lies on the line 3x + y = 10, then P =___

Question 16.
In the adjoining figure, ABCD is a parallelogram in which E and F are the mid points of the sides AB and CD respectively. Prove that the line segment CE and AF trisect the diagonal BD.

Question 17.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that
ar(∆AOD) = ar(∆BOC)

Question 18.
Construct an isosceles triangle whose base is 5 cm and whose vertical angle is 72°

Question 19.
Plot the points A(-5, 2), B(3, 2), C(-4, -3) and D(6, 0), E(0, 2), F(4, -4) on the graph paper. Write the name of quadrant also.

Question 20.
The total surface area of a cylinder is 462 cm². Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

Question 21.
A tent is in the form of a right circular cylinder, surmounted by a cone. The diameter of a cylinder is 24 m. The height of the cylindrical portion is 11m, while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.

Question 22.
Three coins are tossed simultaneously 150 times and it is found that 3 tails appeared 24 times; 2 tails appeared 45 times 1 tail appeared 72 times and no tail appear 9 times. If three coins are tossed simultaneously at random, find the probability of getting
(i) 3 tails,
(ii) 2 tails,
(iii) 1 tail.

SECTION-D

Question 23.

Question 24.
If the polynomials (2x³ + ax² + 3x – 5) and (x³ + x² – 2x +a) leave the same remainder when divided by (x – 2). Find the value of a. Also, find the remainder in each case.

Question 25.
Two men starts from point A and B respectively, 42 km apart. One walks from A to B at 4 km/hr and another walks from B to A at a certain uniform speed. They meet each other after 6 hours. Find the speed of the second man.

Question 26.
In a ∆ABC, ∠B > ∠C. If AM is the bisector of ∠BAC and AN ⊥ BC, prove that ∠MAN = \(\frac { 1 }{ 2 }\) (∠B – ∠C).

Question 27.
In the given figure, ABCD is a square, M is the mid-point of AB and PQ ⊥ CM meets AD at P and CB produced at Q. Prove that
(i) PA = BQ,
(ii) CP = AB + PA

Question 28.
Prove that the sum of either pair of the opposite angles of a cyclic quadrilateral is 180°.
OR
Prove that the opposite angles of a cyclic quadrilateral are supplementary.

Question 29.
A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 15 cm and diameter 49 cm to cover its whole surface. Find the length and volume of the wire. If the specific gravity of copper be 9 g per cubic cm, find the weight of the wire.

Question 30.
15 students of Govt. School spend the following number of hours in a month for doing cleaning in their street 25, 15, 20, 20, 9, 20, 25, 15, 7, 13, 20, 12, 10, 15, 8.
(i) Find mean, median and mode from above data.
(ii) Which value is depicted from above data?

Solutions

Solution 1.
3^2/3 x 7^2/3 = (3 x 7)^2/3 = (21)^2/3 

Solution 2.
P(x) = x³ – ax² + x, x – a = 0 ⇒ x = a
When P(x) is divided by (x – a), we get
Remainder = P(+a) = a³ – a(+a)² + (+a) = a³ – a³ + a = a
Remainder = P(a) = a
⇒ Remainder = a

Solution 3.
y-axis ⇒ coordinates of y-axis is (0, b).

Solutions 4.
13 chapters.

Solution 5.

Solution 6.

Solution 7.
a(a – 1) – b(b – 1) = a² – a – b² + b = a² – b² – (a – b) = (a – b)(a + b) – (a – b)
= (a – b) (a + b – 1)

Solution 8.
Let the angles of the given triangles measure (2x)°, (3x)° and (7x)°.
2x + 3x + 7x = 180° [By angle sum property of A]
12x = 180° ⇒ x = 15°
Hence angles are
2x = 30°
3x = 45°
7x =105°

Solution 9.

∠A + ∠B = 180°
⇒ \(\frac { 1 }{ 2 }\)∠A + \(\frac { 1 }{ 2 }\)∠B = 90°
In ∆AOB
\(\frac { 1 }{ 2 }\)∠A + \(\frac { 1 }{ 2 }\)∠B + ∠AOB = 180°
⇒ 90° + ∠AOB = 180°
⇒ ∠AOB = 90°

Solution 10.

Given: A ∆ABC in which AD is median.
To Prove: ar(∆ABD) = ar(∆ADC)
Construction: Draw AL ⊥ BC
Proof: BD = DC
(AD is median so D is midpoint of BC)

=> ar(∆ABD) = ar(∆ADC) [∵ ar(∆) = \(\frac { 1 }{ 2 }\) x base x height]
Hence, a median of a triangle divides it into two triangles of equal area.

Solution 11.
a = 13,b = 13,c = 24

Hence, area of the triangle (∆) = 60 cm²

Solution 12.
Calculated mean of 40 numbers = 38
∴ Calculated sum of these numbers = 38 x 40 = 1520
Correct sum of these numbers = [1520 – (wrong item) + correct item]
= 150 – 36 + 56 = 1540
The correct mean = \(\frac { 1540 }{ 40 }\) = 38.5
Hence, correct mean = 38.5

Lowest data = 0, highest data = 6

Solution 13.

Solution 14.
(a – b – c) (a² + b² + c² + ab + ac – bc)
= [a + (-b) + (-c)] [a² + (-b)² + (-c)² – a(-b) – (-b) (-c) – (-c)a]
= a³ + (-b)³ + (-c)³ – 3a (-b)(-c)
= a³ – b³ – c³ – 3abc (a – b – c) (a² + b² + c² + ab + ac – bc)
= a³ – b³ – c³ – 3abc

OR
f(x) = x⁴ – 2x³ + 3x² – ax + b
If f(x), divided by (x – 1), let remainder be R₁ = 5.
x – 1 = 0, x = 1

Solution 15.
(i) For the line 4x + 3y = 12, x-intercept = 3 and y-intercept = 4, i.e., Ans = 3, 4
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = 2.
=> Putting x = 4 and y = 3 in eqn. 2x + ky = 14
2 x 4 + k x 3 = 14 =>3k = 14 – 8 = 6
3k = 6 => k = 2
(iii) If the point P(P, 4) lies on the line 3x +y = 10, then P = 2
=> Putting x = P and y = 4 in eqn. 3x + y = 10
3P + 4 = 10 =>3P = 6 => P = 2.

Solution 16.

Let BD be intersected by CE and AF at P and Q respectively. AB || DC and AB = DC [opposite sides of || gm]
=> AE = FC and \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) DC => AE || FC and AE = FC
=> AECF is a || gm => AF || CE => EP || AQ and FQ || CP.
In ∆BAQ, E is the mid point of AB.
and EP || AQ so, P is the mid point of BQ
BP = PQ [By converse of midpoint theorem]
Again in ∆DPC, F is mid point of DC and FQ || CP. So Q is the mid point of DP.
=> PQ = QD [By converse of mid point theorem]
∴ BP = PQ = QD
Hence, CE and AF trisect the diagonal BD.

Solution 17.

∆ACD and ∆BCD being on the same base DC and between the same parallels DC and AB, then
ar(∆ACD) = ar(∆BCD)
Subtracting ar(∆DOC) on both sides
ar (∆ACD) – ar(∆DOC) = ar(∆BCD) – ar(∆DOC)
ar(∆AOD) = ar(∆BOC)

Solution 18.

Steps of Construction:
1. Draw a line segment BC = 5 cm
2. Make ∠CBX = 72°, below the line segment
3. Make ∠XBY = 90°
4. Draw the right bisector PQ of BC, intersecting BY at O.
5. With O as centre and radius OB, draw a circle, intersecting PQ at A.
6. Join AB and AC.
Then ∆ABC is the required isosceles triangle in which AB = AC.

Solution 19.
Point A(-5, 2) lies in => II quadrant
B(3, -2) lies in => I quadrant
C(-4, -3) lies in => III quadrant
D(6, 0) lies in => x-axis (abscissa)
E(0, 2) lies in => y-axis (ordinate)
F(4, -4) lies in => IV quadrant

Solution 20.
C.S.A. of cylinder = \(\frac { 1 }{ 3 }\) x (T.S.A. of cylinder)
= \(\frac { 1 }{ 3 }\) x 462 = 154 cm²
(T.S.A. of cylinder – C.S.A. of cylinder) = 462 – 154 = 308 cm²
2πr(h + r) – 2πrh = 308

Solution 21.

Radius of the cylinder = \(\frac { 24 }{ 2 }\) = 12
m = R
Height, H = 11 m
Curved surface area of the cylindrical portion
= 2πRH sq. units
= (2π x 12 x 11) m²
= (264 π) m²
Radius of the cone = r = 12m,
height of the cone = (16 – 11) = 5m

Hence, the area of the canvas required for tent = 1320 m².

Solution 22.
Total number of trials = 150
Number of times 3 tails appeared = 24
Number of times 2 tails appeared = 45
Number of times 1 tail appeared = 72
Number of times 0 tail appeared = 9

Solution 23.

Solution 24.
Let f(x) = 2x³ + ax² + 3x – 5 and g(x) = x³ + x² – 2x + a
When f(x) is divided by (x – 2), then remainder = f(2) [x – 2 = 0 => x = 2]
f(2) = 2(2)³ + a(2)² + 3(2) – 5 = 16 + 4a – 5 + 6 = 17 + 4a
When g(x) is divided by (x – 2), remainder = g(2) [x – 2 = 0 => x = 2]
g(2) = (2)³ + (2)² -2 x 2 + a = 8 + a
But f(2) = g(2)
17 + 4a = 8 + a
3a = -9 => a = -3
Remainder in each case = 8 + (-3) = 8 – 3 = 5

Solution 25.
Let the speed of another man be x km/h.
Average speed of both men = (x + 4) km/h.

=> 6x = 18 =>x = \(\frac { 18 }{ 6 }\) = 3
x = 3 km/h

Solution 26.
Given: In ∆ABC, in which ∠B > ∠C, AN ⊥ BC and AM is the bisector of ∠A. Also find the angle MAN if ∠B = 65° and ∠C = 30°.
To Prove: ∠MAN = \(\frac { 1 }{ 2 }\) (∠B – ∠C)
Proof: Since AM is the bisector of ∠A

Solution 27.

In ∆PAM and ∆QBM
AM = BM (M is the mid point)
∠PAM = ∠QBM (each 90°)
∠PMA = ∠BMQ (V.O.A)
∴ ∆AMP ≅ ∆BMQ (ASA congruency)
∆AMP ≅ ∠BMQ (CPCT)
PA = BQ …(i)
and MP = MQ
Now join PC. Again in ∆CMP and ∆CMQ
PM = MQ (Proved)
∠CMP = ∠CMQ (each 90°)
CM = CM (Common)
∆CMP = ∆CMQ (RHS congruency)
CP = CQ (CPCT)
CP = CQ = BC + BQ = AB + PA[BC = AB, PA = BQ]
CP = AB + PA

Solution 28.
Given: A cyclic quadrilateral ABCD
To Prove: ∠A + ∠C = 180°
∠B + ∠D = 180°
Construction: Join AC and BD

Solution 29.
Length of the cylinder = 15 cm
Radius = 24.5 cm, diameter of the wire = 0.6 cm

Solution 30.
Data in ascending order 7, 8, 9, 10, 12, 13, 15, 15, 15, 20, 20, 20, 20, 25, 25 => n = 15.

Median =15
Mode = Maximum number of observations = 20 (4 times)
(ii) Value => Social work.

We hope the CBSE Sample Papers for Class 9 Maths Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 4, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 3 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 3

CBSE Sample Papers for Class 9 Maths Paper 3

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 3 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
Write a rational number between √2 and √3

Question 2.
Find the remainder when P(x) = x³ – 3x² + 4x + 32 is divided by (x + 2).

Question 3.
What is the equation of y-axis?

Question 4.
If 3∠A = 4∠B = 6∠C, then ∠A : ∠B : ∠C = ?

Question 5.
In which quadrant point (x, y) lies if x < 0, y > 0?

Question 6.
The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3, then find the ratio of their volumes.

SECTION-B

Question 7.
Factorize: x(x – y)³ + 3x²y (x -y).

Question 8.
The base of an Isosceles triangle measure 24 cm and its area is 192 cm². Find its perimeter.

Question 9.
The runs scored by two batsmen in a cricket match is 164. Write a linear equation in two variables x and y. Also write a solution of this equation.

Question 10.
12 packets of salts, each marked 2 kg, actually contained the following weights (in kg) of salt.
1.950, 2.020, 2.060, 1.980, 2.030, 1.970
2.040, 1.990, 1.985, 2.025, 2.000, 1.980
Out of these packets one packet is chosen at random. What is the probability that the chosen packet contains more than 2 kg of salt?

Question 11.
The percentage of marks obtained by a student in six unit test are given below.

The unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?

Question 12.
Evaluate (104)³ using suitable identity.

SECTION-C

Question 13.
If x = 7 + 4√3 , then find the value of \(\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) \)
OR
Find the value of √4.3 geometrically and represent √4.3 on a given real line.

Question 14.
a³ – b³ + 1 + 3ab

Question 15.
In the figure, AB || CD. Prove that ∠p + ∠q – ∠r = 180°.

Question 16.
In the given figure, ABC is a triangle in which AB = AC. Side BA is produced to D, such that AB = AD. Prove that ∠BCD = 90°.

Question 17.
Mid Point Theorem: Prove that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
OR
Prove that parallelograms on the same base and between same parallels are equal in area.

Question 18.
In the given figure O is the centre of circle, prove that ∠x + ∠y = ∠z

Question 19.
Construct a ∆ABC in which BC = 6 cm, ∠B = 60° and sum of other two sides is 9 cm [i.e., AB + AC = 9 cm].

Question 20.
Plot the points A(3, 3), B(2, 4), C(5, 5), D(0, 2), E(3, -3) and F(-5, -5) on a graph paper. Which of these points are the mirror images in
(i) x-axis,
(ii) y-axis?

Question 21.
A cylindrical metallic pipe is 14 cm long. The difference between the outside and inside surface is 44 cm². If the pipe is made up of 99 cubic cm of metal, find the outer and inner radii of the pipe.

Question 22.
The total surface area of a solid cylinder is 231 cm² and its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.

SECTION-D

Question 23.
Express \(0.6+0.\overline { 7 } +0.\overline { 47 } \) in the form \(\frac { p }{ q }\), wherep and q are integers and q ≠ 0.

Question 24.
If both (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors of polynomial Px² + 5x + r show that \(\frac { P }{ r }\) = 1.

Question 25.
A pharmacist needs to strengthen a 15% alcoholic solution to one of 32% alcohol. How much pure alcohol should be added to 800 ml of 15% solution?

Question 26.
In the figure, POQ is a line, Ray OR is perpendicular to line PQ, OS is another ray lying between rays OP and OR. Prove that
∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS)

Question 27.
In the given figure, C is the mid point of AB. If ∠DCA = ∠ECB and ∠DBC = ∠EAC, prove that
(i) ∆DBC ≅ ∆EAC
(ii) DC = EC

Question 28.
In the adjoining figure, D, E, F are mid points of the sides BC, CA and AB of ∆ABC. If BE and DF intersect at X while CF and DE intersect at Y, prove that XY = \(\frac { 1 }{ 4 }\)BC

Question 29.
Draw a histogram to represent the following frequency distribution:

Question 30.
Kartikeya wanted to make a temporary shelter for street dogs by making a box like structure with tarpaulin that covers all the four sides and the top of the house. How much tarpaulin would be referred to make the shelter of height 2.5 m with base dimensions 4m x 3m. Assuming stitching margin is negligible. Which values are depicted in this question?

Solutions

Solution 1.
√2 = 1.414, √3 = 1.732
Rational number between √2 and √3 = \(\frac { \sqrt { 2 } +\sqrt { 3 } }{ 2 } \) = 1.573 ~ 1.6 upto one place of decimal = \(\frac { 8 }{ 5 }\)

Solution 2.
P(x) = x³ – 3x² + 4x + 32,x + 2 = 0 => x = -2
When P(x) is divided by (x + 2), we get
Remainder = P(-2) = [(-2)³ – 3(-2)² + 4(-2) + 32] = -8 – 12 – 8 + 32 = 4
Remainder = 4

Solution 3.
x = 0.

Solution 4.
Let 3∠A = 4∠B = 6∠C = k°

Solution 5.
If x < 0, y > 0 => x = -ve and y = +ve => P(x, y) lies in quadrant II because it is (-, +).

Solution 6.
Let their radius be 2x cm and 3x cm and their height be 5y cm and 3y cm respectively and volume of cylinder = πr²h

Solution 7.
x(x – y)³ + 3x²y (x – y) = x(x – y) [(x – y)² + 3xy] = x(x – y) [(x² + y² – 2xy + 3xy)]
= x (x – y) (x² + y² + xy).

Solution 8.
AL ⊥ BC

Hence perimeter of Isosceles triangle = 20 + 20 + 24 = 64 cm.

Solution 9.
Both Batsmen make runs = x + y = 164
So, linear equation = x + y = 164
If x = 100, then y = 164 – 100 = 64
One solution is

Solution 10.
Total number of packets of salts = 12
Number of packets containing more than 2 kg of salt = 5.

Solution 11.
Number of test in which student gets more than 60% marks = 2.
Total number of test = 6

Solution 12.
(104)³ = (100 + 4)³ = (100)³ + (4)³ + 3(100)(4)(100 + 4)
= 1000000 + 64 + 1200 (100 + 4)
= 1000000 + 64 + 120000 + 4800 = 1124864

Solution 13.

Steps of Construction:
Draw a line segment AB = 4.3 units and extend it to C such that BC = 1 unit.
Find the mid point O of AC.
Draw a semicircle with O as centre and OA as radius.
Draw BD ⊥ AC intersecting the semicircle at D.
Then BD = √4.3 units
with B as centre and BD as radius, draw an arc meeting with AC produced at E. Then BE = BD = √4.3 units.

Solution 14.
a³ – b³ + 1 + 3ab = a³ + (-b)³ + 1 + 3ab = a³ + (-b)³ + (1)³ – 3 x a x (-b) x 1
a³ – b³ + 1 + 3ab = (a – b + 1) (a² + b² + 1 + ab + b – a)
= (a – b + 1) (a² + b² + ab – a + b + 1)
a³ – b³ + 1 + 3 ab = (a – b + 1) (a² + b² + ab – a + b + 1)

Solution 15.

AB || CD.
Through F, draw PFQ || AB || CD
So, q = ∠1 + ∠2
AB || PF and EF is transversal line
∠p° + ∠1 = 180° …(i)
Again
PF || CD and GF is transversal line
∠2 = ∠r [Alternate Interior angles] …(ii)

Adding (i) and (ii)
∠p + ∠1 + ∠2 = 180° + ∠r
∠p + (∠1 + ∠2) = 180° + ∠r
∠p + ∠q = 180° + ∠r
∠p + ∠q – ∠r = 180°

Solution 16.

AB = AC => ∠ABC = ∠ACB
AC = AD => ∠ADC = ∠ACD
∴ ∠ABC + ∠ADC = ∠ACB + ∠ACD = ∠BCD
∠DBC + ∠BDC = ∠BCD
[∵ ∠ABC = ∠DBC, ∠ADC = ∠BDC]
Adding ∠BCD on both sides
∠DBC + ∠BDC + ∠BCD = 2∠BCD
180° = 2 ∠BCD[Angle sum property of A]
=> ∠BCD = 90°
Hence, ∠BCD = 90°.

Solution 17.

Given: A ∆ABC in which D and E are the mid points of AB and AC respectively, DE is joined.
To prove: DE || BC and DE = \(\frac { 1 }{ 2 }\)BC
Construction: Draw CF || BA, meeting DE produced in F.
Proof: In ∆AED and ∆CEF
AE = CF (given)
∠AED = ∠CED (V.O.A.)
∠DAE = ∠FCE (alternate interior angles) BD || CF
∴ ∆AED ≅ ∆CEF (By ASA congruency)
=> AD = CF (CPCT)
But AD = BD
=> BD = CF and BD || CF (by construction)
∴ BCFD is a parallelogram.∴DF || BC and DF = BC
=> DE || BC [∵ DF || BC]
DE = \(\frac { 1 }{ 2 }\) DF = \(\frac { 1 }{ 2 }\) BC [∵ DF = BC]
DE || BC and DE = \(\frac { 1 }{ 2 }\) BC
OR
Given: Two parallelograms ABCD and ABEF on the same base AB and between the same parallel lines AB and FC.

Solution 18.
In ∆ACF, side CF is produced to B
∴ ∠y = ∠1 + ∠3 …(i)
[exterior angle = sum of interior opposite angles]
In ∆AED, side ED is produced to B
∠1 + ∠x = ∠4 …(ii)
Adding eq. (i) and (ii)
∠1 + ∠x + ∠y = ∠1 + ∠3 + ∠4
⇒∠x + ∠y = ∠3 + ∠4 [∵∠4 = ∠3 angles in the same segment ]
⇒∠x + ∠y = 2∠3 [∵∠AOB = 2∠ACB]
∠x + ∠y = ∠z
Hence ∠x + ∠y = ∠z

Solution 19.
Steps of Construction:
(i) Draw BC = 6 cm
(ii) Construct ∠CBX = 60°
(iii) Along BX, set off BP = 9 cm
(iv) Join CP
(v) Draw the perpendicular bisector of CP to intersect BP at A.
(vi) Join AC, then ∆ABC is the required triangle.

Solution 20.
(i) (3, -3) is the mirror image of (3, 3) in x-axis.
(ii) No mirror image in y-axis.

Solution 21.
Let the outer and inner radii be R cm and r cm respectively.
h = 14 cm, outer surface area = 2πRh = 2 x \(\frac { 22 }{ 7 }\) x R x 14 = (88R) cm²

Solution 22.
TSA of cylinder = 2πrh + 2πr² = 231 …(1)
CSA of cylinder = 2πrh = \(\frac { 2 }{ 3 }\) of (TSA of cylinder)

Solution 23.

Solution 24.
Let f(x) = Px² + 5x + r
If (x – 2) is a factor of P(x), then remainder = 0
x – 2 = 0 => x = 2
Remainder = f(2) = P(2)² + 5x² + r = 0
4P + 10 + r = 0
4P + r = -10 ..(i)

Solution 25.
Let x ml of the pure alcohol be added to be 15% solution to get 32% alcoholic solution. New volume of alcoholic solution = (800 + x) ml
Quantity of pure alcohol in (800 + x) ml of 32% solution = Quantity of pure alcohol in 800 ml of 15% solution + x ml of pure alcohol

Solution 26.
∠QOS = ∠ROQ + ∠ROS …(i)
∠POS = ∠POR – ∠ROS …(ii)
Subtracting eqn. (i) by (ii)
∠QOS – ∠POS = ∠ROQ – ∠POR + 2∠ROS
∠QOS – ∠POS = (∠ROQ – ∠ROQ) + 2∠ROS (∠ROQ = ∠POR = 90°)
=> 2∠ROS = ∠QOS – ∠POS
=> ∠ROS = \(\frac { 1 }{ 2 }\) [∠QOS – ∠POS]

Solution 27.

(i) ∠DCA = ∠ECB
=> ∠DCA + ∠DCE = ∠ECB + ∠DCE
=> ∠ACE = ∠BCD
Now in AAEC and ABCD
∠DBC = ∠EAC (given)
AC = BC (C is the mid point)
∠ACE = ∠BCD (Proved)
∴ ADBC ≅ AEAC (ASA congruency)
(ii) ADBC ≅ AEAC
DC = EC (CPCT)

Solution 28.

In ∆ABC, F and E are the mid-points of AB and AC respectively.
∴FE || BC and FE = \(\frac { 1 }{ 2 }\) BC = BD
∴FE || BD and FE = BD [∵FE||BC => FE||BD]
=> BDEF is a parallelogram whose diagonals BE and DF intersect each other at X.
∴X is the midpoint of DF.
Similarly, Y is the mid point of DE.
Thus, in ∆DEF, X and Y are the midpoints of DF and DE respectively.
So, XY || FE

Solution 29.
In the given frequency distribution, class sizes are different. So, to calculate the adjusted frequency for each class, Minimum class size = 5

Solution 30.
Required tarpaulin = Area of shelter = Curved surface area of cuboid + Top area
= 2 (l + b) x h + lb
= 2(4 + 3) x 2.5 + 4 x 3
= 35 + 12
= 47 m²
Values: (i) Care for animal, (ii) Kindness.

We hope the CBSE Sample Papers for Class 9 Maths Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 3, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 2 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 2

CBSE Sample Papers for Class 9 Maths Paper 1

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 2 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.
If √2 = 1.414 , then \(\sqrt { \frac { (\sqrt { 2 } -1) }{ (\sqrt { 2 } +1) } } =\) ?

Question 2.
If \(\frac { x }{ y } +\frac { y }{ x } =-1\), where x ≠ 0,y ≠ 0, then find the value of x³ – y³.

Question 3.
Who is known as the “Father of Geometry”?

Question 4.
Find the area of the ∆OAB with 0(0, 0). A(4, 0) and B(0, 6).

Question 5.
Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface area.

Question 6.
In 50 tosses of a coin, tail appears 32 times. If a coin is tossed at random, what is the probability of getting a head?

SECTION-B

Question 7.
If a + b + c = 0, then \(\left( \frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { c^{ 2 } }{ ab } \right) =?\)

Question 8.
Find the measure of an angle, if six times its complement is 12° less than twice its supplement.

Question 9.
If an angle of a parallelogram is two third of its adjacent angle, find the smallest angle of the parallelogram.

Question 10.
In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that \(\frac { ar(\Delta ABD) }{ ar(\Delta ADC) } =\frac { m }{ n } \)

Question 11.
The sides of a triangle are in the ratio 5:12:13 and its perimeter is 150 cm. Find the area of triangle.

Question 12.
Find the mean of the first six prime numbers.

SECTION-C

Question 13.
Rationalise the denominator of \(\frac { \left( \sqrt { 3 } +\sqrt { 2 } \right) }{ \left( \sqrt { 3 } -\sqrt { 2 } \right) } \)
OR
If a and b are rational numbers and \(\frac { \left( \sqrt { 11 } -\sqrt { 7 } \right) }{ \left( \sqrt { 11 } +\sqrt { 7 } \right) } =a-b\sqrt { 77 } \)
Find the value of a and b.

Question 14.
Factorize (p – q)³ + (q – r)³ + (r – p)³

Question 15.
Draw the graph of the equation 2x + 3y = 11. From your graph, find the value of y, when x = -2.

Question 16.
In a quadrilateral ABCD the line segment bisecting ∠C and ∠D meet at E.
Prove that ∠A + ∠B = 2∠CED.

Question 17.
In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, then prove that AC bisect BD.

Question 18.
Without plotting the given points on a graph paper, indicate the quadrants in which they lie
(a) ordinate = 6, abscissa = -3
(b) ordinate = -6, abscissa = 4
(c) abscissa = -5, ordinate = -7
(d) ordinate = 3, abscissa = 5
(e) (-6, 5)
(f) (2,-9)

Question 19.
A well of inner diameter 14 m is dug to a depth of 15 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of embankment so formed.

Question 20.
Find the number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 5 cm and diameter of 4.5 cm.

Question 21.
The table given below shows the ages of 75 teachers in a school.

Here 18-29 means from 18 to 29 including both. A teacher from the school is chosen at random. What is the probability that the teacher chosen is
(i) 40 years or more than 40 years old?
(ii) 49 years or less than 49 years old?
(iii) 60 years or more than 60 years old?

Question 22.
Construct a ∆ABC in which BC = 5.6 cm, ∠B = 30° and the difference between the other two sides is 3 cm.

SECTION D

Question 23.
If a and b are rational numbers and \(\frac { 4+3\sqrt { 5 } }{ 4-3\sqrt { 5 } } =a+b\sqrt { 5 } \) Find the value of a and b.

Question 24.
If x + y + z = 0, prove that x³ + y³ + z³ = 3xyz. Without actual calculation find the value of (-12)³ + 7³ + 5³ using above identity.

Question 25.
A taxi charges RS 20 for the first km and @ RS 12 per km for subsequent distance covered. Taking ‘the distance covered as x km and total fare RS y, write a linear equation depicting the relation in ,x and y
(i) Draw the graph between x and y.
(ii) From your graph find the taxi charges for covering 14 km.

Question 26.
In the given figure, AB || CD. Find the value of x.

Question 27.
In a ∆ABC, the bisectors of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90° + \(\frac { 1 }{ 2 }\)∠A

Question 28.
Prove that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Question 29.
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be \(\frac { 1 }{ 27 }\) of the volume of the given cone, at what height above the base, the section has been made?

Question 30.
A doctor suggests two ways for treatment of a particular disease one is by taking medicine only and other is by doing meditation and yoga.

(i) Draw frequency polygon for the given data on the same graph.
(ii) What is the importance of yoga and meditation in our life?

Solutions

Solution 1.

Solution 2.
\(\frac { x }{ y } +\frac { y }{ x } =-1\) ⇒ x² + y² = -xy
⇒ x² + y² + xy = 0
⇒ x³ – y³ = (x – y)(x² + y² + xy) = (x – y) x 0 = 0
⇒ x³ – y³ = 0

Solution 3.
Mathematician Euclid is known as “Father of Geometry’”.

Solution 4.
OA = 4 units, OB = 6 units
ar(∆OAB) = \(\frac { 1 }{ 2 }\) x OA x OB = \(\frac { 1 }{ 2 }\) x 4 x 6 = 12 sq. unit

Solution 5.
Volume of cube = (side)³, surface area of cube = 6(side)²
Let the volume of first cube = a³; and second cube = b³

Solution 6.
Probability of getting head

Solution 7.
a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc

Solution 8.
Let the measure of the required angle = x°
Measure of its complement = (90° – x).
Measure of its supplement = (180° – x)
6(90° – x) = 2(180° – x) – 12
⇒ 540° – 6x = 360° – 2x – 12 ⇒ 4x = 192°
⇒ x = 48°
Hence the measure of required angle = 48°.

Solution 9.
Let one angle of the parallelogram be x°.
We know that sum of two adjacent angles of a parallelogram is equal to 180°.

Solution 10.

Solution 11.
Let the sides of the triangle be 5x, 12x and 13x as a, b and c. Then
5x + 12x + 13x = 150 => 30x =150 => x = 5
∴ a = 25 cm, b = 60 cm and c = 65 cm
given 2s = 150 cm => s = 75 cm
(s – a) = 50, (s – b)= 15, (s – c) = 10

Solution 12.
The first six prime numbers are 2, 3, 5, 7, 11 and 13.

Solution 13.

Solution 14.
Putting p – q = x,q – r = y and r – p = z
x + y + z = p – q + q – r + r – p = O if x + y + z = 0 then x³ + y³ + z³ = 3xyz
∴ (p – q)³ + (q – r)³ + (r – p)³ = 3(p – q)(q – r)(r – p)

Solution 15.
2x + 3y = 11


Line AB is the required graph of 2x + 3y= 11.
Reading the Graph: Given x = -2. Take a point M on the x-axis such that OM = -2.
Draw PM, parallel to they-axis, meeting the line AB at P, it is PM = 5 = y.
So, when x = -2 then y = 5.

Solution 16.
Let CE and DE be the bisector of ∠C and ∠D respectively.

Solution 17.
We want to prove that AC bisect BD i.e., OB = OD
Let AC and BD intersect at O and BM = DN
Now ∴ ∆OND ≅ ∆OMB (By ASA)
=> OD = OB (CPCT)
[ ∵ ∠OND = ∠OMB (each 90°), DN = BM (given), ∠DON = ∠BOM (V.O.A.)]
=> AC bisect BD.

Solution 18.
(a) A(-3, 6) lies in quadrant II
(b) B(4, -6) lies in quadrant IV
(c) C(-5, -7) lies in quadrant III
(d) D(5, 3) lies in quadrant I
(e) E(-6, 5) lies in quadrant II
(f) F(2, -9) lies in quadrant IV

Solution 19.
For well: r = 7m, h = 15m = depth
Volume of the earth dug out = Volume of the well
= Volume of cylinder = πr²h

Solution 20.
Each coin is cylindrical in shape.

= 45 x 5 = 225
The number of coins required = 225

Solution 21.
Total number of teachers = 75
(i) Number of teachers who are 40 years or more than 40 years old = 35 + 10 = 45

(ii) Number of teachers who are 49 years or less than 49 years old = 5 + 25 + 35 = 65

(iii) Number of teachers who are 60 years or more = 0

Solution 22.
Steps of Construction:
(i) Draw BC = 5.6 cm
(ii) Construct ∠CBX = 30°
(iii) Set off BP = 3 cm
(iv) Join PC
(v) Draw the right bisector of PC, meeting BP produced at A.
(vi) Join AC. Then ∆ABC is the required triangle.

Solution 23.

Solution 24.
x + y + z = 0 => x + y = – z => (x+y)³ = (-z)³
=> x³ + y³ + 3xy (x + y) = -z³
=> x³ + y³ + 3xy (-z) = -z³
=> x³ + y³ – 3xyz = -z³
=> x³ + y³ + z³ – 3xyz = 0
=> x³ + y³ + z³ = 3xyz
Putting x = -12, y = 7, z = 5
We get x + y + z = -12 + 7 + 5 = -12 + 12 = 0 then x³ + y³ + z³ = 3xyz
=> (-12)³ + (7)³ + (5)³ = 3 x (-12) x 7 x 5
= -3 x 7 x 60
= -21 x 60
= -1260
(-12)³ + 7³ + 5³ = -1260.

Solution 25.
y = 20 + 12 (x – 1) = 20 + 12x – 12
=> y = 12x + 8

On the graph paper take distance along x-axis and fare (in Rs) along y-axis. We plot the points A(6, 80) and B(11, 140) on the graph paper. Join AB and produce it on both sides to obtain the required graph.
From the graph, when x = 14, we find y = Rs 176.

Solution 26.

Through E draw a line GEH || AB || CD
Now GE || AB and EA is transversal
∴ ∠GEA = ∠EAB = 50°
[Alternate Interior Angles]
Again EH || CD and EC is a transversal
∠HEC + ∠ECD = 180°
[Cointerior angles of the same side of transversal]
=> ∠HEC + 100° = 180°
=> ∠HEC = 80°
Now GEH is a straight angle
∠GEA + ∠AEC + ∠HEC = 180°
=> 50° + x° + 80° =180° => x = 50°

Solution 27.
In ∆ABC we have
∠A + ∠B + ∠C = 180°

Solution 28.
Given: A circle C(O, r) in which arc AB subtends ∠AOB at the centre and ∠ACB at any point C on the remaining part of the circle.
To prove: (i) and (ii) ∠AOB = 2 ∠ACB, when AB is a minor arc or a semicircle.
(iii) Reflex ∠AOB = 2∠ACB, when AB is a major arc.
Construction: Join AB and CO. Produce CO to a point D outside the circle.

We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.
∴ ∠AOD = ∠OAC + ∠OCA
∠BOD = ∠OBC + ∠OCB
[∵ OC = OA = r ,OC = OB = r]
∠AOD = 2∠OCA and ∠BOD = 2∠OCB
In Figure (i) and Figure (ii) i.e., Case (i) and Case (ii)
∠AOD + ∠BOD = 2∠OCA + 2∠OCB => ∠AOB = 2(∠OCA + ∠OCB)
=> ∠AOB = 2∠ACB
In Figure (iii) i.e., Case (iii)
∠AOD + ∠BOD = 2∠OCA + 2∠OCB
=> Reflex ∠AOB = 2 (∠OCA + ∠OCB) => Reflex ∠AOB = 2∠ACB

Solution 29.
Let the smaller cone have radius r and height h cm and let the radius of the given original cone be R cm.

Solution 30.
(i) We take two imagined classes, one at the beginning, namely (10-20) and the other at the end, namely (70-80) each with frequency zero.
With these two classes, we have the following frequency table:

Importance of Yoga and Meditation:
1. Meditation gives you an experience, an inner state, where what is you and what is your is separated.
2. By meditation life seems to be flowing without conflict and just feel more cohesive.
3. Yoga is important for us to achieve:
(i) Good Health
(ii) Success
(iii) Overall well being
(iv) For peace, joy and love
(v) For inner exploration.

We hope the CBSE Sample Papers for Class 9 Maths Paper 2 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 2, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 9 Maths Paper 1 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 1

CBSE Sample Papers for Class 9 Maths Paper 1

Students who are going to appear for CBSE Class 9 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Papers for Class 9 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

General Instructions:

• All questions are compulsory.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.

Question 2.
What is the degree of zero polynomial?

Question 3.
In ∆ABC, BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?

Question 4.
Write the signs of abscissa and ordinate of a point in quadrant (II).

Question 5.
The total surface area of a cube is 216 cm². Find its volume.

Question 6.
Find the slope of the line 2x + 3y – 4 = 0.

SECTION-B

Question 7.
If a + b + c = 0, then a³ + b³ + c³ = ?

Question 8.
Find four different solutions of 2x + y = 6

Question 9.
The base of an isosceles triangle is 16 cm and its area is 48 cm². Find the perimeter of a triangle.

Question 10.
If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 13. Find the value of x.

Question 11.
1500 families with 2 children each were selected randomly and the following data were recorded.

Out of these families, one family is selected at random. What is the probability that the selected family has
(i) 2 girls
(ii) no girl.

Question 12.
Two coins are tossed 1000 times and the outcomes are recorded as under.

A coin is thrown at random. What is the probability of getting
(i) at most one head ?
(ii) atleast one head ?

SECTION-C

Question 13.
If x = 3 + √8 , find the value of \(\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \)

Question 14.
If p = 2 – a, prove that a³ + 6ap + p³ – 8 = 0.

Question 15.
In the given figure, prove that x = α + β + γ.

Question 16.
If the bisector of the vertical angle of a triangle bisect the base, prove that the triangle is Isosceles.

Question 17.
The sides BA and DC of quad. ABCD are produced as shown in the given figure. Prove that x° + y° = a° + b°.

Question 18.
In an equilateral triangle, prove that the centroid and the circumcentre coincide.

Question 19.
Construct a Δ ABC whose perimeter is 14 cm and sides are in the ratio 2:3:4.

Question 20.
Plot the points A(-5, 2), B(-4, -3), C(3, -2) and D(6, 0) on a graph paper. Write the name of figure formed by joining them.

Question 21.
The diameter of a roller, 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground. Find the cost of levelling it at 75 paise per square metre.

Question 22.
The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost of painting the vessel all over at 14 paise per cm².

SECTION-D

Question 23.
Show that

Question 24.
Prove that x³ + y³ + z³ – 3xyz =(x + y + z) (x² + y² + z² – xy – yz – zx). Hence find the value of x³ + y³ + z³ if x + y + z = 0.

Question 25.
In a ∆ABC, the side AB and AC are produced to P and Q respectively. The bisector of ∠PBC and ∠QCB intersect at a point O. Prove that ∠BOC = 90° – \(\frac { 1 }{ 2 }\) ∠A.

Question 26.
In each of the following figures, AB || CD. Find the value of x in each case.

Question 27.
Draw the graph of the equation 3x + 2y = 12. At what points does the graph cut the x-axis and the y-axis.

Question 28.
A river 2m deep and 45 m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.
Now answer these questions.
1. Why we should preserve and not waste the water and other natural resources? Write one method to restore the water at your home.
2. Why it is necessary to clean all rivers in the country?

Question 29.
The mean of 25 observations is 36. If the mean of first 13 observations is 32 and that of last 13 observations is 39, find the 13th observation.

Question 30.
Plot the points A(2, 5), B(-2, 2) and C(4, 2) on a graph paper. Join AB, BC and AC. Calculate the area of ∆ABC.

Solutions

Solution 1.

Solution 2.
Degree of zero polynomial is not defined.

Solution 3.

∠A + ∠B = ∠ACD
⇒ ∠A + 40° = 120°
⇒ ∠ A = 80°

Solution 4.
The points in quadrant II are in the form (-, +).

Solution 5.
The Total Surface Area of cube (T.S.A.) = 6a² = 216
=> a² = 36 => a = √36 = 6 cm
Volume of cube = (a)³ = (6)³ = 216 cm³
Volume of cube = 216 cm³

Solution 6.
2x + 3y – 4 = 0 ⇒ 3y = – 2x + 4

Solution 7.
a + b + c = 0 ⇒ a + b = – c
⇒ (a + b)³ = (- c)³ ⇒ a³ + b³ + 3ab (a + b) = -c³
⇒ a³ + b³ + 3ab (- c) = -c³ ⇒ a³ + b³ + c³ = 3abc

Solution 8.
y = 6 – 2x
(i) If x = 1,y = 6 – 2 = 4
(ii) If x = 2,y = 6 – 4 = 2
(iii) If x = 3,y = 6 – 6 = 0
(iv) If x = 4,y = 6 – 8 = -2

Solution 9.
b =16 cm, area = 48 cm²

Perimeter of Isosceles triangle = 2a + b = 20 + 16 = 36 cm.

Solution 10.
Mean of the given observations

Solution 11.
Total number of families = 1500

Solution 12.
Total number of tosses = 266 + 540 + 194 = 1000
(i) At most one head = Number of times one head or no head appears
= 540+ 194 = 734
P(E1) = P(at most one head) = \(\frac { 734 }{ 1000 }\) =0.734
(ii) Atleast one head = Number of times one head or two head appears
= 540 + 266 = 806
P(E2) = P(atleast one head) = \(\frac { 806 }{ 1000 }\) = 0.806

Solution 13.
x + 3√8

Solution 14.
We have p = 2 – a => a + p + (-2) = 0
=> Putting a = x,p = y and (-2) = z.
we get a + p + (-2) = 0 =>x + y + z = 0
⇒a³ + p³ + (-2)³ = 3 × a × p × (-2) [If x + y + z = 0 then x³ + y³ + z³ = 3xyz]
⇒a³ + p³ + (-2)³ = -6ap
⇒a³ + 6ap + p³ – 8 = 0

Solution 15.

Join B and D and produce BD to E.
p + q = β and s + t = x
Side BD of A ABD is produced to E
∴ p + α = s ..(1)
Side BD of A CBD is produced to E.
∴ q + γ = t ..(2)
Adding (1) and (2)
(p + α) + (q + γ) = s + 1
(p + q) + (α) + (γ) = x
β + α + γ = x
⇒ x = α + β + γ

Solution 16.

Given: ∆ AABC in which AD is the bisector of ∠ A, which meets BC in D such that BD = DC.
To prove: AB = AC
Construction: Produce AD to E such that AD = DE and join EC.
Proof: In ∆ABD and ∆ECD
BD = DC (given)
AD = DE (by construction)
∠ADB = ∠EDC (V.O.A)
=> ∴ ∆ABD ≅ ∆ECD
∴ ∆ABC ≅ ∆ECD
=> AB = EC and ∠1 = ∠3 …(2) (CPCT)
Also ∠1 = ∠2 …(3) [∵AD bisects ∠A]
∴ ∠2 = ∠3 [using (2)and(3)]
=> EC = AC …(4) [side opposite to equal angles]
=> AB = AC [using (1)and(4)]
=> ∆ABC is Isosceles.

Solution 17.
We have ∠ A + b° = 180° (Linear pair)
=> ∠A = 180° – b° …(1)
Also ∠C + a° = 180° (Linear pair)
∠C = 180° – a° …(2)

Solution 18.

Let ∆ ABC be the given equilateral triangle and let its median AD,
BE and CF intersect at G.
Then G is the centroid of ∆ ABC
In ∆ BCE and ∆ CBF
BC = CB (common)
∠B = ∠C [each 60°]

This shows that G is the circumcentre of ∆ ABC
=> G is the centroid as well as circumcentre of ∆ ABC
Note: Centroid: The point of intersection of the medians of a triangle is called centroid.
The centroid of the triangle is the point located at \(\frac { 2 }{ 3 }\) of the distance from a vertex along the median.
Circumcentre: The centre of the circumcircle of the triangle is called the circumcentre.

Solution 19.
Steps of construction:
(i) Draw a line segment PQ =14 cm. Draw a ray PX making an acute angle with PQ and draw in the downward direction.
(ii) From P, mark set of (2 + 3 + 4) = 9 equal distance along PX.
(iii) Mark points L, M, N on PX such that PL = 2 cm, LM = 3 cm and MN = 4 cm.
(iv) Join NQ. Through L and M draw LB || NQ and MC || NQ. Cutting PQ at B and C respectively.
(v) With B as centre and radius BP draw an arc. With C as centre and radius CQ draw another arc, cutting the previous arc at A.
(vi) Join AB and AC. ∆ ABC is the required triangle.

Solution 20.
Points A(-5, 2), B(-4, -3), C (3, -2) and D(6, 0) are shown on graph paper.

The figure formed by joining them is called quadrilateral.

Solution 21.
Radius of roller = r = 42 cm, Length = h = 120 cm
Area covered by the roller in 1 revolution = C.S.A. of the roller = 2πrh sq. unit.

Solution 22.
Outer radius of the vessel R = 14 cm
Inner radius of the vessel = r = 10 cm
Area of the outer surface = 2πR² cm²
= (2π x 14 x 14) cm² = 392 π cm²
Area of the inner surface = 2πt² cm²
= (2π x 10 x 10) cm² = 200 π cm²
Area of the ring (shaded) at the top = π(R² – r²)
= π[( 14)² – (10)²] = π(14 + 10) (14 – 10)
= 96 π cm²
Total area to be painted = 392 π + 200 π + 96 π
= 688 π cm²

Solution 23.

Solution 24.
x³ + y³ + z³ – 3xyz = (x³ + y³) + z³ – 3xyz
x³ + y³ + z³ – 3xyz = [(x + y)³ – 3xy(x + y)] + z³ – 3xyz [let x + y = u]
= u³ – 3xyu + z³ – 3xyz
= u³ + z³ – 3xy (u + z) = (u³ + z³) – 3xy (u + z)
= (u + z) (u² – uz + z²) – 3xy (u + z) = (u + z) [u² + z² – uz – 3xy]
= (x + y + z) [(x + y)² + z² – (x + y)z – 3xy] [u = x + y]
= (x + y + z) [x² + y² + z² + 2xy – xz -yz – 3xy]
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
if x + y + z = 0
x³ + y³ + z³ – 3xyz = 0 × (x² + y² + z² – xy – yz – zx)
x³ + y³ + z³ – 3xyz = 0
x³ + y³ + z³ = 3xyz

Solution 25.
We have ∠B + ∠CBP = 180° (Linear pair)

Solution 26.
(i) Draw EF || AB || CD

=> ∠1 + ∠2 = x°
Now AB || EF and AE is transversal.
∠1 + ∠BAE = ∠1 + 104° = 180°
(co interior angles)
∠1 = 180°- 104° = 76°
Again EF || CD and EC is the transversal
∠2 + ∠ECD = ∠2 + 116° = 180°
=> ∠2 = 64°
Hence x = ∠1 + ∠2 = 76° + 64° = 140°
=> x = 140°
(ii) Draw EO || AB || CD

x° = ∠1 + ∠2
Now EO || AB and OB is transversal
∠1 + ∠ABO = 180° [Co-interior angles]
∠1 + 40°= 180°
=> ∠1 = 180°- 40° = 140° => ∠1 = 140°
Again EO || CD and DO is transversal
∠2 + ∠CDO = 180°[Co-interior angles]
∠2 + 35° = 180°
=> ∠2 = 180° – 35° = 145°
=> ∠2 = 145°
∠1 + ∠2 = (140° + 145°) = 285°
=> x = ∠1 + ∠2 = 285°
=> x = 285°

Solution 27.
Linear equation 3x + 2y = 12

The point A(4, 0) cuts on the x-axis and point B(0, 6) cuts on the y-axis.

Solution 28.

Volume of the water running into the sea per minute
= Volume of cuboid
= l x bx h
= 50 x 45 x 2
= 4500 m³
(i) We should preserve the water and other natural resources for ourself and for our next generation because they can not be polluted or finished very fast.
(ii) Method of restoring fresh water is by ‘rain water harvesting’ in home.
(iii) Our rivers are very polluted and causing the problems in ecosystem and becoming the shortage of fresh water at large scale. So, it is necessary to clean all rivers of our country.

Solution 29.
Mean of first 13 observations = 32
Sum of the first 13 observations = 32 x 13 = 416
Mean of last 13 observations =39
Sum of last 13 observations = 39 x 13 = 507
Mean of 25 observations = 36
Sum of all the 25 observations = 36 x 25 = 900
∴ the 13th observation = (416 + 507 – 900) = 23
Hence the 13th observation = 23

Solution 30.

From Graph:
BC = CE + BE = (4 – 0) + (0 – (-2))
Base = BC = 4 – (-2) = 4 + 2 = 6 units
Height = AD = OF – OE = 5 – 2 = 3 units
ar (∆ ABC) = \(\frac { 1 }{ 2 }\) x Base x height
= \(\frac { 1 }{ 2 }\) x 6 x 3 = \(\frac { 18 }{ 2 }\) = 9 sq units
ar (∆ ABC) = 9 sq. units.

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