CBSE Sample Papers for Class 9 Maths Paper 4

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.
Find the value of {(23 + 2²)^2/3 + (140 – 19)^1/2}².

Question 2.
Find the value of 0.54 x 0.54 – 0.46 x 0.46

Question 3.
If ∆ PQR ≅ ∆ EFD then find
(i) ED
(ii) ∠ E

Question 4.
In the given figure, G is the centroid of ∆ ABC, such that GD = 3 cm and BC = 4 cm. Then find the area of ∆ ABC.

Question 5.
Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.

Question 6.
Find the class marks of the class 130 – 150

Question 7.
Find the value of

without actual calculation.

Question 8.
In the given figure ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.

Question 9.
Write the equation of x-axis, y-axis, coordinates of origin and coordinates of point P.

Question 10.
Plot the point A(4, 4) and B(-4, 4) and join OA, OB and BA. What figure do you obtain?

Question 11.
If V and S represent the volume and surface area of a cuboid, respectively of length l, breadth b and height h, then show that

Question 12.
Find the mode of 14,25,14,28,18, 17,18, 14, 23, 22,14,18.

SECTION-C

Question 13.
If (x² – 1) is a factor of polynomial ax⁴ + bx³ + cx² + dx + e. then prove that
a + c + e = b + d = 0.

Question 14.
If x + y + z = 1,xy + yz + zx = – 1 and xyz = -1, then find the value of x³ + y³ + z³.

Question 15.
Simplify \(\sqrt { 3-2\sqrt { 2 } } \)

Question 16.
Prove that the perimeter of a triangle is greater than the sum of their medians.

Question 17.
In the given figure, m and n are two mirrors placed parallel to each other. An incident ray AB strikes the mirror m at point B and then reflected to the mirror n along path BC and again reflects back along CD. Prove that AB || CD.

Question 18.
Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic.

Question 19.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and their side is 18 cm. Also verify it.

Question 20.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours. Each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Question 21.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the metal weighs 8.9 gm per cm³?

Question 22.
A fair die is thrown 120 times with the following frequencies of number divisible by 3 and not divisible by 3.
Divisible by 3 : 56
Not divisible by 3 : 64
Find the probability when the number is (i) divisible by 3 (ii) Not divisible by 3.

SECTION-D

Question 23.
If x + y + z = 10,xy + yz + zx = -15 and xyz = – 12, then find the values of x² + y² + z² and x³ + y³ + z³.

Question 24.
Prove that

Question 25.
Seema wants to invest Rs 20000 in two types of bond. She earns 12% on the first type and 15% on the second type. Find the investment in each if her total earning is Rs 2850. Find how much she invests at each type of bond. Write two values which are depicted here.

Question 26.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that
(i) ∆ AMC = ∆ BMD
(ii) ∠DBC is a right angle or ∠DBC = ∠ACB = 90°
(iii) ∆DCB ≅ ∆ACB
(iv) CM = \(\frac { 1 }{ 2 }\)AB

Question 27.
XY is a line parallel to side BC of a ∆ABC. If BE || AC and E and F respectively. Show that ar (∆ABE) = ar (∆ACF).

Question 28.
A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
OR
There is one and only one circle passing through three given non-collinear points.

Question 29.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and diameter of the graphite is 1 mm. If the length of the pencil is 14 m. Find the (i) Volume of graphite (ii) Weight of graphite (iii) Volume of wood (iv) Weight of pencil. If density of wood = 0.7 gm/cm³ and density of graphite = 2.1 gm/cm³.

Question 30.
If n observations x₁, x₂, x₃ …, x_n have sum of deviations -10 from 50 and sum of deviations 70 from 46, then find the value of n and mean of the observations.

Solutions

Solution 1.
(23 + 2²)^2/3 = (23 + 4)^2/3 = (27)^2/3 = 3^3×2/3 = 3² = 9
(140 – 19)^1/2 = (121)^1/2 = 11
{(23 + 2²)^2/3 + (140 – 19)^1/2}² = (9 + 11)² = 20² = 400

Solution 2.
0.54 x 0.54 – 0.46 x 0.46 = (0.54)² – (0.46)²
= (0.54 + 0.46) (0.54 – 0.46) = 1 x 0.08 = 0.08

Solution 3.
If ∆ PQR ≅ ∆ EFD
The ∠ P = ∠ E, ∠ Q = ∠ F, ∠ R = ∠ D and PQ = EF, QR = FD and PR = ED
so (i) ED = PR
(ii) ∠ E = ∠ P

Solution 4.
In ∆ GBC, BC = 4 cm and GD = 3 cm.
area (∆ GBC) = \(\frac { 1 }{ 2 }\) x Base x Altitude = \(\frac { 1 }{ 2 }\) x 4 x 3
= 6 cm²
ar (∆ ABC) = 3 x ar (∆ GBC) = 3 x 6 cm²
ar (∆ ABC) = 18 cm²

Solution 5.
l = 12m,b = 9m,h = 8m
The length of the longest rod = length of diagonal of the room

Hence the length of the longest rod that can be placed in the room is 17m

Solution 6.

Solution 7.
Let 7.83 = a, 1.17 = b, then a x a – b x b = a² – b² = (a + b) (a – b)
(a – b) = 7.83 -1.17 = 6.66

Solution 8.
∵ Ray QP and RP exist on line ST.
∴ ∠ SQP + ∠ PQR = 180° ⇒ ∠ PQS + ∠ PQR
and ∠SRP + ∠PRT = 180° ⇒ ∠PRQ + ∠PRT
=> ∠ PQS + ∠ PQR = ∠ PRQ + ∠ PRT
=> ∠ PQS + ∠ PQR = ∠ PQR + ∠ PRT [ ∵ ∠ PRQ = ∠ PQR Given]
=> ∠ PQS = ∠ PRT

Solution 9.
(i) Equation of x-axis is y = 0.
(ii) Equation of y-axis is x = 0.
(iii) Coordinates of origin = (0, 0).
(iv) Coordinates of point P = (x, y)

Solution 10.
The points are plotted on the graph paper. The figure is a ∆AOB,

Solution 11.
Volume of the cuboid = V = Ibh …(1)
Surface area of the cuboid = S = 2(lb + bh + hl) …(2)

Solution 12.
The given numbers are 14,25,14, 28, 18,17, 18, 14, 23, 22, 14 and 18.
Arranging the data in ascending order:
14, 14,14, 14, 17, 18, 18, 18, 22, 23,25, 28
M₀ = Mode = Maximum number of repeated value of the data
= 14 (Since 14 is occur 4 times)
M₀ = 14

Solution 13.
Let P(x) = ax⁴ + bx³ + cx² + dx + e
∵ (x² – 1) is a factor of P(x)
=> (x + 1) (x – 1) are factors of P(x).
=> (x + 1) and (x – 1) are factors of P(x).
=> P(- 1) = 0 and P(1) = 0

P(-1) = a(-1)⁴ + b(-1)³ + c(-1)² + d(-1) + e = 0
=> a – b + c – d + e = 0
P(1) = a(1)⁴ + b(1)³ + c(1)² + d(1) + c = 0
=> a + b + c + d + e = 0
Adding eq. (1) and (2)
2 (a + c + e) = 0 => a + c + e = 0
Subtracting eq. (2) by eq. (1)
2(b + d) = 0 => b + d = 0
=> a + c + e = b + d = 0

Solution 14.
We know that
x³ + y³ + z³ – 3 xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
=> x³ + y³ + z³ – 3xyz => (x + y + z) (x² + y² + z² + 2xy + 2yz + 2zx – 3xy – 3yz – 3zx)
(Subtracting and adding 2xy + 2yz + 2zx)
=> x³ + y³ + z³ – 3xyz = (x + y + z) {(x + y + z)² – 3(xy + yz + zx)}
=> x³ + y³ + z³ – 3 x (-1)= 1 x {(1)² – 3 x (-1)}
[Putting x + y + z = 1; xy + yz + zx = -1; xyz = -1]
=> x³ + y³ + z³ + 3 = 4
=> x³ + y³ + z³ = 4 – 3
x³ + y³ + z³ = 1

Solution 15.

Solution 16.
Given: In ∆ ABC, AD, BE and CF are medians.
To prove: AB + BC + AC > AD + BE + CF
Proof: We know that the sum of two sides of a triangle is always greater than the twice the median drawn on third side.
i. e. AB + AC > 2AD …(1)
AB + BC > 2BE …(2)
BC + AC > 2CF
Adding eq. (1), (2) and (3)
(AB + AC) + (AB + BC) + (BC + AC) > 2 AD + 2 BE + 2 CF
=> 2(AB + BC + AC) > 2(AD + BE + CF)
=> AB + BC + AC> AD + BE + CF.

Solution 17.
Given: Two plane mirrors m and n are in such a way that m || n. In incident ray AB after reflection, reflects along BC and CD. BM and CN are normal respectively on mirror m and n.
To prove: AB || CD
Proof: ∵ BM ⊥ m, CN ⊥ n. and m||n
CN ⊥ m => BM||CN

Solution 18.

Given: In cyclic quadrilateral ABCD, a quadrilateral PQRS formed by angle bisectors of ∠ A, ∠ B, ∠ C and ∠ D.
To prove: PQRS is a cyclic quadrilateral.
i.e., ∠ APB + ∠ CRD= 180°
or ∠ PQR +∠ PSR = 180°
Proof: In A APB and A CRD, by angle sum properly of triangle
∠APB + ∠PAB + ∠PBA = 180°,
and ∠CRD + ∠RCD + ∠RDC = 180°


A pair of angles of quadrilateral PQRS is supplementary.
=> PQRS is a cyclic quadrilateral.

Solution 19.
Steps of construction:
1. Draw a base PQ = 12 cm.
2. Make an angle 90° at the point Q.
3. Cut off the line segment QS = 18 cm.
4. Draw perpendicular bisector AB of SP to intersect SQ at the point O. Join O, P.
5. ∆OPQ is the required triangle.
Verification: Points situated on perpendicular bisector of a line segment are at equal distance from both ends, i.e. OS = OP.
Now, OQ + OS = 18 cm (Given)
=> OQ + OP = 18 cm (∵ OS = OP)

Solution 20.

The sides of triangular shaped cloth of umbrella are a = 20 cm, b = 50 cm, c = 50 cm
If s is semi perimeter of piece of triangular shaped cloth.
2s = a + b + c
= 20 + 50 + 50 = 120cm

Solution 21.
Diameter of the metallic ball = d = 4.2 cm
Radius of the metallic ball = r = \(\frac { 4.2 }{ 2 }\) = 2.1 cm = \(\frac { 21 }{ 10 }\) cm

Weight of 1 cm³ of metal = 8.9 gms
Weight of 38.808 cm³ of metal = 38.808 x 8.9 = 345.3912 gm
= 345.39 gm
Hence the required weight of the metal = 345.39 gm.

Solution 22.
Total possible outcomes = 120

Solution 23.
We know that

Solution 24.

Solution 25.
Let Seema invest Rs x on first type of bonds and Rs (20000 – x) on the second type.


Investment on first type of bonds = Rs 5000
Investment on second type of bonds = Rs (20000 – 5000) = Rs 15,000.
Two values depicted here are
(i) Saving habit
(ii) Care for future
(iii) Help the country for development

Solution 26.

In ∆AMC and ∆BMD
AM = BM (given)
∠AMC = ∠BMD (vertically opposite angles)
MC = MD (given)
∆AMC ≅ ∆BMD (SAS congruency)
∆AMC ≅ ∆BMD

Solution 27.

Given: A ∆ABC in which XY || BC, BE || CA and CF || BA.
=> BE || CY and CF || BX
To prove: ar (∆ABE) = ar (∆ACF)
Proof: ||gm EBCY and ∆ABE are on the same base BE and between the same parallels BE & CA.
ar (∆ABE) = \(\frac { 1 }{ 2 }\) ar (||gm EBCY) …(1)
Again ||gm BCFX and ∆ACF are on the same base CF and between the same parallels CF and BA.
=> ar (∆ACF) = \(\frac { 1 }{ 2 }\) ar (||gm BCFX) …(2)
But ||gm EBCY and ||gm BCFX are on the same base BC and between the same parallels BC and EF.
=> ar (||gm EBCY) = ar (||gm BCFX) …(3)
From (1), (2) and (3)
ar(∆ABE) = ar(∆ACF)

Solution 28.

Given: Three non-collinear points P, Q and R.
To prove: A circle passes through these three points P, Q and R, and such circle is one and only one.
Construction: Join PQ and QR. Their perpendicular bisectors AL and BM intersect at O.
Join OP, OQ and OR.
Proof: ∵ Point O is at the perpendicular bisector of chord PQ.
∴ OP = OQ …(i)
Similarly O is at the perpendicular bisector of chord QR => OQ = OR …(ii)
From Eqn. (i) & (ii)
OP = OQ = OR = r (Let)
Now taking O as a centre and r radius if we draw a circle it will pass through all three points P, Q and R, i.e., P, Q and R exist on the circumference of the circle.
Now let another circle be (O’, s) which pass through points P, Q and R and perpendicular bisectors of PQ and QR i.e., AL and BM passes through the centre O’.
But the intersection point of AL and BM is O. i.e.
O’ and O coincide each other or O and O’ are the same point
∴ OP = r and OP’ = s and O and O’ are coincide => r = s.
=> C (O, r) = C (O’, s)
=> There is one and only one circle through which three non-collinear points P, Q and R pass.
=> There is one and only one circle passing through three given non-collinear points.

Solution 29.
(i) Given: Diameter of cylinder of graphite = 1mm = \(\frac { 1 }{ 10 }\) cm.
Radius of graphite cylinder = \(\frac { 1 }{ 20 }\)
Length of graphite cylinder = 14 cm

Solution 30.
Given

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