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

Board	CBSE
Class	IX
Subject	Maths
Sample Paper Set	Paper 4
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 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.
• The question paper consists of 30 questions divided into 4 sections A, B, C and D. Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions of 4 marks each.
• There is no overall choice. However, internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

Section A

Question 1.
It is known that, if x + y = 10, then x + y + z = 10 + z. Which Euclid’s axiom illustrates this statement?

Question 2.
If √3 = 1.73205, then what is the value of

Question 3.
If the mean of the observations m, m + 3, m + 5, m + 7, m + 10 is 9, then what is the value of m?

Question 4.
Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

Question 5.
The radius of a solid hemisphere is 7 cm. Find its total curved surface area.

Question 6.
Look at the given figure and answer the following questions

(i) What is the abscissa of P?
(ii) What is the ordinate of Q?
(iii) What are the coordinates of origin?
(iv) Which point is having abscissa as (-3)?

Section B

Question 7.
For what value of m, the polynomial x³ – 2mx² + 16 is divisible by (x + 2)?

Question 8.
Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m².

Question 9.
In the given figure, AE – AD and CE = BD. Prove that ∆AEB ≅ ∆ADC.

Question 10.
The relative humidity (in %) of a certain city for a month of 31 days was as follows
98.1 98.6 99.2 90.3 86.5 95.3 92.9
96.3 94.2 95.1 95.1 89.2 92.3 97.1
93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3
96.1 92.1 89.0
Find the probability of relative humidity less than 94% on any of these days.

Question 11.
In the given figure, BD = DC and ∠CBD = 40°. Find ∠BAC.

Question 12.
A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius with sand. She empties the bucket on the ground and makes a conical heap of the sand. If the height of the conical heap is 24 cm, then find its radius.

Section C

Question 13.
The point (2, 3) lies on the graph of the linear equation 3x – (a – 1) y = 2a – 1. If the same point also lies on the graph of the linear equation 5x + (1 – 2a) y = 3b, then find the value of b.

Question 14.
In the given figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 18 m, AE = 10 m and CF = 12 m, then find the length of AD.

Question 15.
ABCD is a rectangle and P,Q,R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.

Question 16.
S is any point in the interior of ∆PQR. Show that SQ + SR < PQ + PR.

Question 17.
(i) The given table shows the weights of 12 students

Find the mean weight.
(ii) If we add two students of same weight, then whether the mean weight will be different.
Or
Find the missing frequencies in the following frequency distribution, if it is known that the mean of the distribution is 1.46.

Question 18.
In the given figure two circles whose centres O and O’ intersect at P. Through P, a line parallel to OO’ intersecting the circles at C and D is drawn as shown. Prove that CD = 2OO’.

Or
If O is the circumcentre of a ∆ABC and OD ⊥ BC, then prove that ∠BOD = ∠B AC.

Question 19.
Construct a ∆ABC whose perimeter is 8.2 cm and the base angles are 45° and 60°.

Question 20.
Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.
Or
In the adjoining figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (∆APQ).

Question 21.
Find five rational number between -5 and -6.
OR
If both a and b are rational numbers, then find a and b from the following.

Question 22. From the given figure, write

(i) the ordinate of the points A and C.
(ii) the point identified by the coordinates (1, 1).
(iii) the perpendicular distance of the point G from the X-axis.
(iv) the coordinates of the points B and F.
(v) the perpendicular distance of the point I from the Y-axis.
(vi) the points whose perpendicular distance from Y-axis is 2 units.

Section D

Question 23.
In quadrilateral ABDC, AB = AC and AD is the bisector of ∠BAC. Anju has to prove that ∆ABD ≅ ∆ACD, which as follows

Here, AB = AC [given]
∠B AD = ∠CAD
[∵ AD is bisector of ∠BAC]
and ∠B = ∠C [∵corresponding angles of equal sides are equal]
∴ ∆ABD ≅ ∆ACD [by ASA congruence rule]
Further Anju shows this proof to his classmate Amita and she finds that there is some errors in the proof.
(i) Write the correct proof.
(ii) What is the mistake in Anju’s proof ?
(iii) Which value is depicted from this action ?

Question 24.
The cost of a shirt of a particular brand is Rs 1000. Write a linear equation, when the cost of x shirts is Rs y. Draw the graph of this equation and find the cost of 12 such shirts from the graph.

Question 25.
The maximum temperature is Celsius of some cities on a day are given below

(i) Find the probability that maximum temperature lies between 16° C and 30 ° C
(ii) Find the probability that minimum temperature is less than 25°C.
Or
A company selected 3000 households at random and surveyed them to find out a relationship between income level and the number of television sets in a home.
The information so obtained is listed in the following table

Find the probability
(i) of a household earning Rs (10000-14999) and have exactly one television.
(ii) of a household earning Rs 25000 and more per month and owing 2 televisions.
(iii) of a household not having any television.

Question 26.
If a^x = c^q – b and c^y = a^z = d, then prove that xy = qz.
Or
If 5^2x-1 – 25^x-1 = 2500 then find the value of x

Question 27.
Factorise x³ – 3x² – 9x – 5.

Question 28.
Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank, which is 4.2 m in diameter and 4.5 m in high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank?

Question 29.

Question 30.
Two hemispherical domes are to be painted as shown in the following figure. If the circumferences of the bases of the domes are 17.6 cm and 70.4 cm respectively, then find the cost of painting at the rate of Rs 10 per cm².

Solutions

Solution 1.
Euclid’s second axiom illustrates that, if equals are added to equals, then wholes are equal.

Solution 2.

Solution 3.

Solution 4.
Given, OA = 3 cm and OD = 2 cm

We know that diagonals of a parallelogram bisect each other.
∴ AC = 2 OA = 6cm
and BD = 2 OD = 4cm

Solution 5.
Given, radius of solid hemisphere (r) = 7 cm
∴ Total curved surface area of hemisphere = 3πr²
= 3 x \(\frac { 22 }{ 7 }\) x 7 x 7
= 3 x 154
= 462 cm²

Solution 6.
(i) The abscissa of P is 4.
(ii) The ordinate of Q is 3.
(iii) The coordinates of origin are (0, 0).
(iv) The point A is having abscissa as (-3).

Solution 7.
Let f(x) = x³ – 2mx² + 16 ..(i)
Also, let f(x) is divisible by (x + 2).
Then, f(-2) = 0
= (-2)³-2m(-2)²+16 = 0
[put x = -2 in Eq. (i)]
-8-8m+16 = 0
8m = 8
m = 1

Solution 8.
Let ABC be a triangular field of the sides- AB = 50 m, BC = 65 m and CA = 65 m

Solution 9.
Given, AE = AD …(i)
and CE = BD …(ii)
On adding Eqs. (i) and (ii), we get
AE + CE = AD + BD
=> AC = AB
Now, in ∆AEB and ∆ADC, we have
AE = AD [given]
AB = AC [proved above]
and ∠A = ∠A [common angle]
∴ ∆AEB ≅ ∆ADC [by SAS congruence rule]
Hence proved.

Solution 10.
Given, total number of days = 31
∴ Total number of trials = 31
Let E be the event of getting a day when the relative humidity less than 94%.
∴ Number of trials in which E happened = 13

Solution 11.
In ∆BDC, BD = DC [given]
∴ ∠DCB = ∠CBD = 40° ,..(i)
[∵ angles opposite to equal sides are equal]
Also, ∠CBD + ∠BDC + ∠DCB = 180°
[∵ sum of angles of a triangle is 180°]
∴ 40° + ∠BDC + 40° = 180° [from Eq. (i)]
=> ∠BDC = 180°- 80° = 100°
As A, B, D and C are four points on a circle.
Therefore, ABDC forms a cyclic quadrilateral.
Hence, ∠A + ∠D = 180°
[∵ in a cyclic quadrilateral, sum of two opposite angles is equal to two right angles, i.e. 180°]
=> ∠A + 100° = 180° [∵∠D = ∠BDC = 100°]
∴∠A = 80°

Solution 12.
Given, height of cylindrical bucket, H = 32 cm
Radius of cylindrical bucket, R = 18 cm
∴Volume of sand = Volume of a cylindrical bucket

Solution 13.
Given, point (2, 3) lies on the line, 3x-(a-1)y = 2a-1
So, the point (2, 3) is the solution of
3x – (a – 1)y = 2a -1
3×2-(a-1)×3 = 2a-1 [put x = 2,y = 3]
=> 6 – 3a + 3 = 2a – 1
=> -3a -2a = -1-3-6
=> -5a = -10
=> a = 2
Also, (2, 3) is the solution of 5x + (1 – 2a) y = 3b.
∴5 × 2 + (1 – 2a) × 3 = 3b [put x = 2,y = 3]
On putting the value of a = 2, we get
5 × 2 + (1 – 2 × 2) × 3 = 3b
=> 10 – 9 = 3b => 1 = 3b =>\(\frac { 1 }{ 3 }\) = b
Hence, the value of b is \(\frac { 1 }{ 3 }\).

Solution 14.
Given, AE = 10 m, CF = 12 m and AB = 18 m
We know that,
Area of parallelogram = Base x Altitude

Solution 15.
Given, ABCD is a rectangle.

∠A = ∠B = ∠C = ∠D = 90°
and AD = SC, AB = CD
Also, given P, Q, R and S are the mid-points of
AS, SC, CD and DA, respectively.
∴ PQ||AC and PQ = \(\frac { 1 }{ 2 }\)AC
and SR||AC and SR = \(\frac { 1 }{ 2 }\)AC
Thus, PQ = SR . ..(i)
Now, in ∆ASP and ∆SOP
AP = BP [vP is the mid-point of AB]
AS = BQ
[∵ AD = BC, so \(\frac { 1 }{ 2 }\) AD = \(\frac { 1 }{ 2 }\)BC => AS = BQ]
∠A = ∠B [given]
∴ ∆ASP ≅ ∆BQP [by SAS congruence rule]
∴ SP = PQ [by CPCT]… (ii)
Similarly, in ∆ RDS and ∆ RCQ,
SD = CQ,
DR = RC and ∠D = ∠C
∴∆ RDS ≅ ∆ RCQ [by SAS congruence rule]
∴ SR = RQ [by CPCT]… (iii)
From Eqs. (i), (ii) and (iii), it is clear that quadrilateral
PQRS is a rhombus.

Solution 16.
Given, S is any point in the interior of ∆ PQR.
Produce QS to intersect PR at T.

In ∆PQT.we have PQ+PT>QT
[∵ sum of any two sides of a triangle is greater than the third side]
PQ+PT>SQ+ST
In ∆TSR, we have
ST+TR>SR .(ii)
On adding Eqs. (i) and (ii). we get
PQ+PT+ST+TR>SQ+ST+SR
PQ+PT+TR>SQ+SR
PQ+PR>SQ+SR
or SQ+SR<PQ+PR
Hence proved.

Solution 17.
(i) Consider the table given below

Solution 18.
Given Given circles having centres 0 and 0′ intersect at P. Also CD || OO’.
To prove CD = 2OO’
Construction Draw OA and O’B perpendicular to CD from 0 and O’, respectively.

Proof OA⊥CD
OA bisects the chord.
Also, CP is perpendicular from the centre to the chord bisects the chord.
AP = \(\frac { 1 }{ 2 }\)CP => CP = 2AP …(i)
Similarly, O’B⊥PD
∴BP = \(\frac { 1 }{ 2 }\)PD => PD =2BP …(ii)
Now, CD = CP + DP = 2 AP + 2 BP
[from Eqs. (i) and (ii)]…(iii)
In quadrilateral ABO’O, OA = O’B
[two lines perpendicular to same line]
∴ ABO’O is parallelogram, AB = OO’
[∴ opposite sides of a parallelogram are equal]
CD = 2AB = 2OO’
Hence proved.
Or
Given OD⊥BC
To prove ∠BOD = ∠BAC
Proof In ∆ OBD and ∆ OCD,
OB = OC
OD = OD
∠ODB = ∠ODC
∆ OBD ≅ ∆ OCD
=> ∠BOD = ∠COD
But ∠BOC = 2∠B0D = 2∠BAC
=> ∠BOD = ∠BAC
Hence proved.

Solution 19.
Steps of construction
(i) First, draw PQ = 8.2 cm.
(ii) At point P, construct ∠RPQ = 60° and at point 0, construct ∠SQP = 45°.
(iii) Draw bisector of ∠RPQ and ∠SQP, which intersect each other at point A.
(iv) Join AP and AQ.
(v) Draw perpendicular bisector of AP which intersects PQ at point 6 and also draw perpendicular bisector of AQ which intersects PQ at point C.

(vi) Join AB and AC
Thus, we get the required ∆ABC.

Solution 20.
Let ABCD is a parallelogram. M and N are the mid-points of AB and DC, respectively.

To prove ar (parallelogram AMND) =ar (parallelogram MBCN)
Proof Since, ABCD is a parallelogram.
∴ AB = DC and AB||DC
[by property of parallelogram]
=> \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\)DC and AM||DN
[dividing both sides by 2]
=> AM = DN and AM||DN
[∵ M and N are the mid-points of AB and DC]
So, AMND is a parallelogram.
Similarly, we can prove that MBCN is also a parallelogram.
∴ Parallelograms AMND and MBCN are on same base AB and between the same parallel lines AB and DC.
∴ ar (parallelogram AMND) = ar (parallelogram MBCN)
Or
Given ABCDE is a pentagon, in which BP || AC and EQ || AD.
To prove ar (ABCDE) = ar (∆APQ)
Proof We know that triangles on the same base and between the same parallels are equal in area.
∴ar (∆ADE) = ar (∆ADQ) ,..(i)
[∵ ∆ADQ and ∆ADE lie on the same base AD and between the same parallels AD and EQ]
Similarly, ar (∆CAB) = ar (∆CAP) …(ii)
[∵ same base is AC and same parallels are AC and BP]
On adding Eqs. (i) and (ii), we get
ar (∆ADE) + ar (∆CAB)
= ar (∆ADQ) + ar (∆CAP)
On adding ar(∆CDA) both sides, we get
ar (∆ADE) + ar (∆CAB) + ar(∆CDA)
= ar (∆ADQ) + ar (∆CAP) + ar(∆CDA)
=> ar (ABCDE) = ar (∆APQ)
Hence proved.

Solution 21.

Solution 22.
(i) Coordinates of points A and C are (-3, 1) and (-2, 0), respectively.
Hence, ordinate of points A and C are 1 and 0.
(ii) (1,1) represents the point D.
(iii) Coordinates of point G is (5, 4).
Hence, the perpendicular distance of the point G from the X-axis is 5 units.
(iv) Coordinates of points B and F are (-4, – 3) and (5,0) respectively.
(v) Coordinates of point / is (-2, -2).
Hence, the perpendicular distance of the point l from the Y-axis is -2 units.
(vi) There are two points C and l whose distance from Y-axis is 2 units.

Solution 23.
(i) Given In ∆ABC, AB = AC and AD is the bisector of ∠BAC.
To prove ∆ABD ≅ ∆ACD
Proof In ∆ABD and ∆ACD,
AB = AC
∠BAD = ∠CAD
and AD = AD
∴ ∆ABD ≅ ∆ACD
[by SAS congruence rule]
(ii) Anju has used the result ∠B = ∠C for proving ∆ABD ≅ ∆ACD, but it is possible only when we consider ∆ABC.
(iii) The value depicted from this action, is cooperative learning among students without any bias.

Solution 24.
Given, cost of x shirts = Rs y
∴ Cost of 1 shirt = Rs \(\frac { y }{ x }\)
According to the question,
\(\frac { y }{ x }\) = 1000 => 1000x – y = 0 ,..(i)
which is the required linear equation.
Eq. (i) can be written as y = 1000x.
When x = 1 then y = 1000 x 1 = 1000
When x =2, then y = 1000 x 2 = 2000
Thus, we have the following table

Now, plot the points (1,1000) and (2,2000) on the graph paper and join them by a line to get required graph.

For the cost of 12 shirts from the graph. draw a line parallel to Y-axis from a point x = 12 on X-axis, which intersects the graph at point P.
Draw a perpendicular from point P on Y-axis, which intersects the Y-axis at y = 12000.
Thus, the cost of 12 shirts is Rs 2000.

Solution 25.
∵ Total number of cities = 1 + 2 + 18 + 21 + 19 + 18 = 79
(i) Let E1 be the event, which show maximum temperature lies between 16° C and 30° C. Number of cities = 2 + 18 + 21 = 41
∴ Required probability, P(E1) = \(\frac { 41 }{ 79 }\)
(ii) Let E2 be the event, which show minimum temperature less than 25° C
Number of cities = 1 + 2 + 18 = 21
∴ Required probability, P(E2) = \(\frac { 21 }{ 79 }\)
Or
∴Total number of households selected by the company = 3000
(i) The number of households earning
Rs (10000 -14999) per month and having exactly one television =240
P40 fi
∴ Required probability = \(\frac { 240 }{ 3000 }\) = \(\frac { 8 }{ 100 }\) = 0.08
(ii) The number of households earning ? 25000 and more per month and owing 2 televisions = 750
∴ Required probability = \(\frac { 750 }{ 3000 }\) = 0.25
(iii) The number of households not having any television = 20 +10 = 30
∴ Required probability = \(\frac { 30 }{ 3000 }\) = \(\frac { 3 }{ 300 }\) = \(\frac { 1 }{ 100 }\)

Solution 26.

Solution 27.
Let p(x) = x³ – 3x² – 9x – 5
Here, -5 is the constant term in p(x)and the factors of -5 are ±1 and ±5.
On putting x = 1 in p(x), we get
p(1) = (1)³ – 3(1)² – 9(1) – 5
= 1 – 3 – 9 – 5
= -16≠0
∴ (x -1) is not a factor of p(x).
On putting x =-1 in p(x), we get
P(-1) = (-1)³ – 3(-1)² – 9(-1) – 5
= -1-3+9-5
= -9+9=0
∴ x -(-1) = (x + 1) is a factor of p(x).
i.e. x +1 is a factor of x³ – 3x² – 9x – 5.
Now, we divide x³ – 3x² – 9x – 5 by long division method.

i.e. p(x) = (x+1)(x² – 4x – 5)
Now, x² – 4x – 5 = x² + x – 5x – 5
[by splitting middle term]
= x(x + 1)-5(x +1)
= (x + 1)(x – 5)
∴ p(x) = x³ – 3x² – 9x – 5 = (x + 1)(x + 1)(x – 5)

Solution 28.
(i) We have, cylindrical petrol storage tank having diameter,
d = 4.2 m

Solution 29.

Solution 30.
Let r be the radius of the base of hemispherical dome
with circumference = 17.6 cm
∴ 2πr = 17.6 [∵ 2πr = circumference of base of hemispherical dome]

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