# CBSE Sample Papers for Class 9 Maths Paper 5

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

 Board CBSE Class IX Subject Maths Sample Paper Set Paper 5 Category CBSE Sample Papers

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.
(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 52x-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 2a = 3b = 6c, 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)
[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.

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
∠AMD = ∠BNC (Each 90°)
AM = BM
(Perpendicular distance between two parallel lines are equal)
∴ ∆AMD ≅ ∆BNC (By RHS congruency)
=> 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

(i) White = 850
(ii) Black = 1600
(iii) Red = 550

Solution 23.
2a = 3b ⇒ 2 = 3b/a ….(1)
6c = 3b ⇒ 6 = 3b/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 + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE …(1)
AC = AE (Given)
∠BAC = ∠DAE (Proved in Eqn. (1)]
∴ ∆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)
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