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.
• 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.

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