Students can access the CBSE Sample Papers for Class 9 Maths with Solutions and marking scheme Set 2 will help students understand the difficulty level of the exam.
CBSE Sample Papers for Class 9 Maths Set 2 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This question paper contains 38 questions. All questions are compulsory.
- Question paper is divided into Five sections – Sections A, B, C, D, and E.
- In Section A, questions number 1 to 18 are multiple-choice questions (MCQs) and questions number 19 and 20 are Assertion-Reason based questions of 1 mark each.
- In Section B, questions number 21 to 25 are very short answer (VSA) type questions of 2 marks each.
- In Section C, questions number 26 to 31 are short answer (SA) type questions carrying 3 marks each.
- In Section D, questions number 32 to 35 are long answer (LA) type questions carrying 5 marks each.
- In Section E, questions number 36 to 38 are case-based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks questions in each case study.
- There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C and 2 questions in Section D.
- Draw neat figures wherever required. Take π = \(\frac{22}{7}\) wherever required if not stated.
- Use of calculator is not allowed.
Section-A
Consists of Multiple Choice Type questions of 1 mark each.
Question 1.
A forest ranger keeps track of bears in his area. He plotted their location on a graph. The origin represents the ranger7s control room’s location. To access and maintain equipment, Road x and Road y have been laid and paved inside the forest. They pass through the control room.
One unit on the graph paper represents 1 km. Which bear is nearest to a paved road?
(A) Bear 389
(B) Bear 415
(C) Bear 425
(D) Bear 467
Answer:
(B) Bear 415
Explanation:
According to the graph ‘Bear 415’ is nearest to a paved road as compared to other Bear’s.
Question 2.
Ravi planted a red maple tree sapling. The height of the sapling is 0.25 m. The average growth rate of the height of a red maple tree is 0.27 m per year. The average life of a red maple tree is 80-100 years. Ravi estimated that his tree will grow up to 27 m. What is the likely reason behind his estimation? Which of the following equations represents the height (h) of the red maple tree after ‘t’ years of planting?
(A) h = 0.25 + 0.27
(B) h = 0.25t + 0.27
(C) h = 0.25 + 0.27t
(D) h = 0.25 + 27t
Answer:
(C) h = 0.25 + 0.27t
Explanation:
Original height of sapling = 0.25 m
Average growth rate per year = 0.27 m
Height of Red maple tree after t years,
h = 0.25 + 0.27t
Question 3.
Two lines intersect at a point P. Which of the following is true for the distance between the two lines as they travel beyond point P?
(A) The distance becomes constant.
(B) The distance increases continuously.
(C) The distance decreases continuously.
(D) The distance increases and decreases depending upon the intersection point.
Answer:
(B) The distance increases continuously.
Explanation:
As we can see from figure given below, the distance between the two intersecting lines Increases as they travel beyond the point.
Question 4.
Angle which is one-fifth of its complement is
(A) 25°
(B) 45°
(C) 15°
(D) 55°
Answer:
(C) 15°
Explanation:
Let the angle be x, then
By given condition, x = \(\frac{1}{5}\)(90° – x)
⇒ 5x = 90° – x
⇒ x = 15°
Question 5.
The sum of the opposite angles of a cyclic quadrilateral is
(A) 90°
(B) 180°
(C) 100°
(D) 360°
Answer:
(B) 180°
Explanation:
According to the quadrilateral’s property
Sum of opposite angles = 180°
Question 6.
\(\frac{1}{\sqrt{9}-\sqrt{8}}\) is equal to
(A) \(\frac{1}{2}(3-2 \sqrt{2})\)
(B) \(\frac{1}{3+2 \sqrt{2}}\)
(C) 3 – 2√2
(D) 3 + 2√2
Answer:
(D) 3 + 2√2
Explanation:
Question 7.
What is the degree of the polynomial (x3 + 5) (4 – x5)?
(A) 5
(B) 8
(C) 2
(D) 7
Answer:
(B) 8
Explanation:
Degree of x3 + 5 = 3
Degree of 4 – x5 = 5
Now, (x3 + 5)(4 – x5) = 4x3 – x8 + 20 – 5x5 = -x8 – 5x5 + 4x3 + 20
∴ Degree of (x3 + 5)(4 – x5) = 8
Question 8.
Which of the following is not the linear equation in two variables?
(A) 2x = 3
(B) 4 = 5x – 4y
(C) x2 + x = 1
(D) x – √2y = 3
Answer:
(C) x2 + x = 1
Explanation:
x2 + x = 1 is not linear as highest power is 2. Also, it is an equation in one variable.
Thus, it is not a linear equation in two variables.
Question 9.
A triangle whose all three sides are unequal is called
(A) Scalene triangle
(B) Isosceles triangle
(C) Equilateral triangle
(D) Right triangle
Answer:
(A) Scalene triangle
Question 10.
In parallelogram ABCD, AB = (2y – 3) and CD = 5 cm then value of y is
(A) 8 cm
(B) 4 cm
(C) 6 cm
(D) 10 cm
Answer:
(B) 4 cm
Explanation:
In ||gm ABCD
AB = CD (opposite sides of a ||gm are equal)
So, 2y – 3 = 5
⇒ 2y = 5 + 3
⇒ y = 4
Question 11.
The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, then its radius is
(A) 8 cm
(B) 12 cm
(C) 6 cm
(D) 15 cm
Answer:
(B) 12 cm
Explanation:
Let the radius of a cone, r = 4x
and slant height, l = 7x
CSA = 792 cm2
⇒ πrl = 792
⇒ \(\frac{22}{7}\) × 4x × 7x = 792
⇒ x2 = \(\frac{792 \times 7}{22 \times 4 \times 7}\)
⇒ x2 = 9
⇒ x = 3 cm
∴ radius = 4 × 3 = 12 cm
Question 12.
\(\frac{1}{\sqrt{2}}\) is a/an _______ number.
(A) Rational
(B) Irrational
(C) Fractional
(D) None of these
Answer:
(B) Irrational
Question 13.
In the given figure, AD || BC and ∠BCA = 40°. The measure of ∠DBC is equal to
(A) 40°
(B) 60°
(C) 50°
(D) 75°
Answer:
(A) 40°
Explanation:
∠BDA = ∠BCA = 40° (Angles in the same segment)
Now, since AD || BC,
∠DBC = ∠BDA (Alternate interior angles)
∴ ∠DBC = 40°
Question 14.
In the given figure, lines AB, CD, and EF meet at O. The value of x is
(A) 36°
(B) 90°
(C) 54°
(D) 18°
Answer:
(D) 18°
Explanation:
∠COF = 2x (vertically opposite angles)
∴ 3x + 2x + 5x = 180° (straight line angle)
⇒ 10x = 180°
⇒ x = 18°
Question 15.
The volume of a right circular cone with radius of 6 cm and height of 7 cm is
(A) 264 cm3
(B) 165 cm3
(C) 184 cm3
(D) 225 cm3
Answer:
(A) 264 cm3
Explanation:
Volume of right circular cone = \(\frac{1}{3} \pi r^2 h\)
= \(\frac{1}{3} \times \frac{22}{7} \times(6)^2 \times 7\)
= \(\frac{1}{3} \times \frac{22}{7} \times 36 \times 7\)
= 264 cm3
Question 16.
Euclid’s belong to the country
(A) Babylonia
(B) Egypt
(C) Greece
(D) India
Answer:
(C) Greece
Question 17.
A diagonal of a parallelogram divides it into two _______ triangles.
(A) Similar
(B) Congruent
(C) Equilateral
(D) Right angled
Answer:
(B) Congruent
Question 18.
Euclid’s divided his famous treatise ‘The Elements’ into
(A) 13 Chapters
(B) 12 Chapters
(C) 11 Chapters
(D) 9 Chapters
Answer:
(A) 13 Chapters
Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Question 19.
Assertion (A): The point of the form (a, -a) lies on the line x + y = 0.
Reason (R): Any point which satisfies the equation ax + by + c = 0 is the solution of the equation.
Answer:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Explanation:
In case of Assertion (A):
Let x = a and y = -a
x + y = a + (-a) = a – a = 0
∴ Assertion is true.
∴ In case of Reason (R):
Any pair (x, y) satisfies ax + by + c = 0 then (x, y) is the solution.
∴ Reason is true.
Both (A) and (R) are true and (R) is the correct explanation of (A).
Question 20.
Assertion (A): If the volumes of two spheres are in the ratio 27 : 8, then their surface areas are in the ratio 3 : 2.
Reason (R): Volume of sphere = \(\frac{4}{3} \pi r^3\)
Surface area of a sphere = 4πr2
Answer:
(D) Assertion (A) is false, but Reason (R) is true.
Explanation:
∴ Assertion is false but Reason is true as volume of sphere = \(\frac{4}{3} \pi r^3\) and Surface area = 4πr2
Hence, Assertion (A) is false but Reason (R) is true.
Section-B
Consists of 5 questions of 2 marks each.
Question 21.
Find the value of k, if x – 2 is a factor of f(x) = x2 + kx + 2k.
Answer:
Given, (x – 2) is a factor of f(x).
∴ f(2) = 0
⇒ (2)2 + k(2) + 2k = 0
⇒ 4 + 2k + 2k = 0
⇒ 4 + 4k = 0
⇒ k = -1
Question 22.
Karan marks his city on the map as point A.
Savita says, ‘A dot is dimensionless, so your city is also dimensionless.’ Why is Savita wrong? Justify your answer.
Answer:
A dot in the map is for representational purpose.
Dot is used only to show the location of the city, not its area.
∴ Savita is wrong.
Question 23.
Represent the following frequency distribution by means of a histogram.
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Number of Students | 7 | 11 | 9 | 13 | 16 | 4 |
Answer:
Question 24.
Sanya has a triangular piece of land. She wants to divide it into four equal areas. Suggest a way to do so?
OR
Does joining four distinct points always produce a quadrilateral? Justify your answer.
Answer:
Sanya can find mid-points of the sides of the triangular region and create a smaller triangular region by connecting them, In this way, the triangular region can be divided into four triangles of equal area.
OR
No, it is not possible as there can be three cases.
When all the points are collinear, the resulting figure is a line.
When three points are collinear out of four, the resulting figure is a triangle.
When no three points out of four are collinear, the resulting figure is a quadrilateral.
Question 25.
Factorize: 8a3 + 8b3
OR
Factorize: 8x3 – (2x – y)3
Answer:
8a3 + 8b3 = (2a)3 + (2b)3
= (2a + 2b)[(2a)2 + (2b)2 – (2a) × (2b)] [∵ a3 + b3 = (a + b)(a2 + b2 – ab)]
= 2(a + b) × 4(a2 + b2 – ab)
= 8(a + b)(a2 + b2 – ab)
OR
8x3 – (2x – y)3 = (2x)3 – (2x – y)3
= [2x – (2x – y)][(2x)2 + (2x – y)2 + 2x(2x – y)] [Since, (a3 – b3) = (a – b)(a2 + b2 + ab)]
= y[4x2 + 4x2 + y2 – 4xy + 4x2 – 2xy]
= y[12x2 + y2 – 6xy]
Section-C
Consists of 6 questions of 3 marks each.
Question 26.
Irrational numbers can provide more precision on measuring scale. What can be the possible arguments in favour and against this statement?
Answer:
Irrational numbers are non-terminating with more number of decimals so precision on measuring scale can be more. But they are non-terminating, so fixing their exact location on a measuring scale is not possible.
Question 27.
A polynomial is expressed as: p(x) = x3 + x2 – x – 1. At what values of x is the polynomial p(x) = 0?
Answer:
p(x) = x3 + x2 – x – 1
Let x = 1,
then p(1) = 13 + 12 – 1 – 1
= 1 + 1 – 1 – 1
= 0
Now take x = -1
then, p(-1) = (-1)3 + (-1)2 – (-1) – 1
= -1 + 1 + 1 – 1
= 0
Hence, value of x = 1, -1
Question 28.
(i) Find values of a and b, if two ordered pairs (a – 3, -6) and (4, a + b) are equal.
(ii) Find in which quadrant point (a, b) lies.
OR
If the coordinates of a point A are (-2, 9) which can also be expressed as (1 + x, y2) and y > 0, then find in which quadrant do the following points lie: P(y, x), S(2x, -3y)
Answer:
(i) Here, two ordered pairs are equal.
∴ a – 3 = 4; a + b = 6
a = 4 + 3; 7 + b = -6 (substituting value of ‘a’)
a = 7; b = -6 – 7 = -13
Hence a = 7 and b = -13
(ii) Clearly point (7, -13) lies in IV quadrant.
OR
Here, A(-2, 9) can also be expressed as (1 + x, y2)
∴ (-2, 9) = (1 + x, y2) where y > 0
∴ 1 + x = -2, y2 = 9
x = -3, y = √9
x = -3, y = 3 (∵ y > 0)
P(y, x) = (3, -3), it lies in the IV quadrant.
S(2x, -3y) = [2 × (-3), (-3 × 3)]
= (-6, -9), it lies in III quadrant .
Question 29.
A joker’s cap is in the from of right circular cone of base radius of 7 cm and slant height 25 cm. Find the area of sheet required for 10 such caps.
OR
A zoo is in the shape of an isosceles trapezium. It is divided into three zones – Zone 1, Zone 2 and Zone 3. Animals are kept without cages in Zone 1. Zone 2 is for visitors and Zone 3 is reserved for park authorities.
To avoid the entry of animals in Zones 2 and 3, a 1.8 km long wired fencing is installed. “The area reserved for animals is twice the area reserved for the zoo authorities.” Do you have enough information to support this statement? Explain your answer.
Answer:
l = 25 cm, r = 7 cm
∴ l2 = h2 + r2
⇒ h2 = l2 – r2
⇒ h2 = (25)2 – (7)2
⇒ h = 24 cm
Area required = C.S.A. of cone = πrl
= \(\frac{22}{7}\) × 7 × 25
= 550 cm2
Area required to make 10 such caps = 10 × 550 = 5500 cm2 = 0.55 m2
OR
No, we don’t have enough information regarding area under zone 1, 2 and 3.
As, Area reserved under Zone 1 = Area reserved under zone 2 + 3
But it doesn’t means that the area reserved under Zone 2 and 3 are equal and their sum is equal to the area of zone 1.
Question 30.
A tile is made by joining the vertices of four equilateral triangles. The side length of the triangles is 15 cm. What is the area of the tile?
Answer:
Sides of triangle a = 15 cm, b = 15 cm, c = 15 cm
Thus, area of tile = 225√3 sq. cm.
Question 31.
The game of billiards is played with balls placed on a rectangular table. One ball is struck with the end of a stick, called a cue. The ball bounces into other balls and reflects off the sides of the table. In a real game, the ball may spin, but for mathematical purposes, it is considered that the ball travels in a straight line with the same reflection and incidence angles.
On a billiard table ABCD, the ball placed at O is struck with the cue. What is the value of ∠a + ∠d?
Answer:
According to figure
MX is normal and OM is incident ray.
∴ ∠AMX = 90°
(∵ Incident ray makes a right angle with normal)
OM is bisector
∴ ∠a = 45° …..(i)
Similarly ∠d = 45° ……(ii)
(∵ YN is normal and PN is bisector)
Adding equations (i) and (ii)
∴ ∠a + ∠d = 90°
Section-D
Consists of 4 questions of 5 marks each.
Question 32.
If a = \(\frac{2^{x-1}}{2^{x-2}}\), b = \(\frac{2^{-x}}{2^{x+1}}\) and a – b = 0, find the value of x.
OR
If x = \(\frac{\sqrt{5}+1}{\sqrt{5}-1}\) and y = \(\frac{\sqrt{5}-1}{\sqrt{5}+1}\), then find the value of x2 + y2.
Answer:
Question 33.
PQ and RS are two parallel chords of a circle whose centre is O and radius is 10 cm. If PQ = 16 cm and RS = 12 cm, find the. distance between PQ and RS if
(i) On the same side of the centre O.
(ii) On opposite side of the centre.
Answer:
Given, OP = OR = 10 cm (Radii of same circle)
PQ = 16 cm
RS = 12 cm
Draw OL ⊥ PQ and OM ⊥ RS
Since perpendicular from the centre to the chord bisects the chord.
∴ PL = LQ = \(\frac{1}{2}\)PQ = 8 cm
RM = MS = \(\frac{1}{2}\)RS = 6 cm
In right triangle OLP,
OP2 = OL2 + PL2 (By Pythagoras theorem)
⇒ 100 = OL2 + 64
⇒ OL = \(\sqrt{100-64}\)
⇒ OL = \(\sqrt{36}\)
⇒ OL = 6 cm
In right triangle OMR,
OR2 = OM2 + RM2 (By Pythagoras theorem)
⇒ 100 = OM2 + 36
⇒ OM = \(\sqrt{100-36}\)
⇒ OM = \(\sqrt{64}\)
⇒ OM = 8 cm
(i) If PQ and RS lie on same side of centre O.
Distance between PQ and RS = LM
= OM – OL
= 8 – 6
= 2 cm
(ii) If PQ and RS lie on opposite sides of centre O.
Distance between PQ and RS = LM
= OL + OM
= 6 + 8 cm
= 14 cm
Question 34.
Using factor theorem, show that (m – n), (n – p) and (p – m) are factors of m(n2 – p2) + n(p2 – m2) + p(m2 – n2)
Answer:
Let f = m(n2 – p2) + n(p2 – m2) + p(m2 – n2)
∴ f(m = n) = n(n2 – p2) + n(p2 – n2) + p(n2 – n2)
= n(n2 – p2) – n(n2 – p2) + 0
= 0
So, m – n is a factor of f.
Similarly, f(n = p) = 0 & f(p = m) = 0
∴ (m – n), (n – p) and (p – m) are factors of f.
Question 35.
In the figure below, BC = AC.
What is the measure of ∠BAD?
Answer:
BC = AC (Given)
∴ ∠CBA = ∠BAC = a° (Angles opposite to equal sides are equal)
So, ∠BAD = a° + x° ……(i)
In ∆ABD,
∠ABC + ∠ADB + ∠BAD = 180°
∴ a° + x° + (a° + x°) = 180°
Now, 2a° + 2x° = 180°
⇒ 2(a° + x°) = 180°
⇒ a° + x° = 90°
∴ ∠BAD = 90° [from (i)]
Section-E
Cased-Based Subjective Questions.
Question 36.
Prime Minister’s National Relief Fund (also called PMNRF in short) is the fund raised to provide support for people affected by natural and man-made disasters. Natural disasters that are covered under this include floods, cyclones, earthquakes, etc. Man-made disasters that are included are major accidents, acid attacks, riots, etc. Two friends Sita and Gita, together contributed ₹ 200 towards Prime Minister’s Relief Fund.
(i) How to represent the above situation in linear equations in two variables?
(ii) If Sita contributed ₹ 76, then how much was contributed by Gita?
OR
(i) If both contributed equally, then how much is contributed by each?
(ii) Write the standard form of linear equation if x = -5?
Answer:
(i) Let contribution given by Sita = ₹ x
And, contribution done by Gita = ₹ y
∴ Linear equation = x + y = 200
(ii) If x = ₹ 76, then 76 + y = 200
⇒ y = 200 – 76
⇒ y = ₹ 124
OR
If x = y, then x + x = 200
⇒ 2x = 200
⇒ x = 100
Thus, each contributed ₹ 100.
(iii) Since x = -5
⇒ x + 5 = 0
Thus, standard form of x = -5 is 1.x + 0.y + 5 = 0.
Question 37.
Ladli Scheme was launched by the Delhi Government in the year 2008. This scheme helps to make women strong and will empower a girl child. This scheme was started in 2008.
Read the above bar graph and answer the following questions:
(i) In which year the budget was minimum?
(ii) What is Bar-graph?
OR
Which scheme was launched by government and in which year the budget was maximum?
(iii) What was the difference in the budget in the year 2008-2009 and 2009-10?
Answer:
(i) According to graph,
In year 2007-08 the budget was minimum.
(ii) Bar graph is a pictorial representation of data.
It can be vertical or horizontal.
Heights of the bar depend on the values of the variable.
There is gap in between consecutive rectangles.
OR
Ladli scheme was launched by the government.
In year 2010-11 the budget was maximum.
(iii) Budget in the year 2008-09 = 9060 million
Budget in the year 2009-10 = 9160 million
Difference = 9160 – 9060 = 100 million
Question 38.
Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm.
(i) What will be the semi-perimeter of the highlighted triangle?
(ii) What is the area of the highlighted triangle?
OR
What will be the length of equal sides?
(iii) Which formula is used to calculate area of the triangle?
Answer:
OR
Let x cm be the length of equal sides of the isosceles triangle.
So, x + x + 4 = 20
⇒ 2x + 4 = 20
⇒ 2x = 20 – 4
⇒ 2x = 16
⇒ x = 8 cm
(iii) Heron’s formula is used to calculate area of triangle.