Students can access the CBSE Sample Papers for Class 9 Maths with Solutions and marking scheme Set 5 will help students understand the difficulty level of the exam.
CBSE Sample Papers for Class 9 Maths Set 5 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This question paper contains 38 questions. All questions are compulsory.
- The Question paper is divided into Five sections – Sections A, B, C, D, and E.
- In Section A, questions numbers 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 numbers 21 to 25 are very short answer (VSA) type questions of 2 marks each.
- In Section C, questions numbers 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 the calculator is not allowed.
Section-A
Consists of Multiple Choice Type questions of 1 mark each.
Question 1.
If \(\sqrt{2}\) = 1.4142….. then \(\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\) is equal to
(A) 2.4142
(B) 5.8282
(C) 0.4142
(D) 0.1718
Answer:
(C) 0.4142
Explanation:
After rationalisation, we get
\(\sqrt{2}\) – 1 = 1.4142 – 1 = 0.4142
Question 2.
A soap manufacturer makes fragrant and non-fragrant liquid soaps. The liquid soaps are filled in plastic bottles and packed in equal size cartons for transportation. Each carton contains 50 bottles. A carton is checked randomly. Which of the following cannot be the number of fragrant and non-fragrant liquid bottles in the carton?
(A) (5, 45)
(B) (15, 35)
(C) (20, 30)
(D) (30, 40)
Answer:
(D) (30, 40)
Explanation:
Let Number of fragrant bottles = x
Number of non-fragrant bottles = y
Given: Total number of bottles = 50
So, x + y = 50
In (A) 5 + 45 = 50
(B) 15 + 35 = 50
(C) 20 + 30 = 50
(D) 30 + 40 = 70
Thus from above we can see that (D) is not the number of fragrant and non-fragrant bottle.
Question 3.
A number Y is greater than a number X and another number Z < 0. Which of the following relations can be true for a unique value of Z?
(A) X × Z = Y × Z
(B) X ÷ Z = Y ÷ Z
(C) X – Z = Y
(D) X + Z = Y
Answer:
(C) X – Z = Y
Question 4.
The figure below shows the side view of a shopping trolley. The metal plate is fixed on the side by the storekeeper for advertisement.
What is the shape of the metal plate?
(A) Square
(B) Rectangle
(C) Rhombus
(D) Parallelogram
Answer:
(D) Parallelogram
Question 5.
The facts or information collected with a definite purpose is called
(A) Class-mark
(B) Class-size
(C) Data
(D) Class-intervals
Answer:
(C) Data
Question 6.
Five friends Anchal, Amisha, Mahi, Vishu and Sahar are living in a hostel. At the end of every month, they calculate the expenses on food and shopping. The table given below shows their monthly expenses for the month of November.
Name | Anchal | Amisha | Mahi | Vishu | Sahar |
Expenditure (in ₹) | 3000 | 5000 | 6000 | 4500 | 7000 |
Which graphical representation method would best represent the data given?
(A) Histogram
(B) Bar Graph
(C) Frequency Polygon
(D) Ogive
Answer:
(B) Bar Graph
Question 7.
The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is
(A) ₹ 2.00
(B) ₹ 2.16
(C) ₹ 2.48
(D) ₹ 3.00
Answer:
(B) ₹ 2.16
Explanation:
Let a = 6 cm, b = 8 cm, c = 10 cm
Cost of painting 24 cm2 = ₹ 0.09 × 24 = ₹ 2.16.
Question 8.
If the height and the radius of cone is tripled, then the ratio of volume of new cone to that of original cone will be
(A) 27 : 1
(B) 9 : 1
(C) 3 : 1
(D) 6 : 1
Answer:
(A) 27 : 1
Explanation:
Let h and r be the height and radius of original cone
and let h’ and r’ be the height and radius of new cone.
Given, h’ = 3h and r’ = 3r
Volume of original cone, V = \(\frac{1}{3} \pi r^2 h\)
Volume of new cone, V’ = \(\frac{1}{3} \pi\left(r^{\prime}\right)^2 h^{\prime}\)
Hence, the ratio of new cone to the original cone is 27 : 1.
Question 9.
In the given figure, O is the centre of the circle and PA = PB.
Value of ∠OPA is
(A) 60°
(B) 90°
(C) 80°
(D) 100°
Answer:
(B) 90°
Explanation:
Line drawn through the centre of circle to bisect a chord is perpendicular to the chord.
Given, PA = PB; OP ⊥ AB
Hence, ∠OPA = 90°
Question 10.
Which statement is incorrect about the parallelogram?
(A) Consecutive angles are supplementary
(B) Opposite sides are parallel
(C) Diagonal bisects each other
(D) Diagonals are equal in length
Answer:
(D) Diagonals are equal in length
Question 11.
If the number of square centimetres in the surface area of a sphere is equal to the number of cubic cm in its volume, then the diameter of the sphere is
(A) 8 cm
(B) 6 cm
(C) 4 cm
(D) 10 cm
Answer:
(B) 6 cm
Explanation:
Given, Area of sphere = Volume of sphere
4πr2 = \(\frac{4}{3}\)πr3 [where r is the radius of sphere]
⇒ r = 3 cm [on solving]
∴ Diameter = 2r = 6 cm
Question 12.
Value of 710 × 810 is ________
(A) (15)10
(B) (56)10
(C) (15)20
(D) (56)20
Answer:
(B) (56)10
Explanation:
710 × 810 = 5610 [∵ am × bm = (ab)m]
Question 13.
Two consecutive angles of a parallelogram are in the ratio 1 : 3, then value of smaller angle is
(A) 35°
(B) 55°
(C) 75°
(D) 45°
Answer:
(D) 45°
Explanation:
Let the consecutive angles be x and 3x.
∴ x + 3x = 180° (∵ consecutive angles are supplementary)
⇒ 4x = 180°
⇒ x = 45°
∴ Smaller angle = x = 45°
Question 14.
An isosceles triangle has
(A) 3 sides equal
(B) 2 sides equal
(C) None of these sides equal
(D) All angles equal
Answer:
(B) 2 sides equal
Explanation:
An isosceles A has 2 sides equal.
Question 15.
When spherical ball of diameter 4.2 cm, is completely immersed in water, then the amount of water displaced by a solid will be
(A) 388.08 mL
(B) 3.8808 mL
(C) 38.808 mL
(D) 0.38808 mL
Answer:
(C) 38.808 mL
Explanation:
Amount of water displaced = Volume of solid spherical ball
r = \(\frac{4.2}{2}\) = 2.1 cm (given)
∴ Volume of solid spherical ball = \(\frac{4}{3} \pi(2.1)^3\)
= \(\frac{4}{3} \times \frac{22}{7} \times(2.1)^3 \mathrm{~cm}^3\)
= \(\frac{38808}{1000}\) cm3
∴ Amount of water displaced = 38.808 cm3 = 38.808 mL (∵ 1 cm3 = 1 mL)
Question 16.
Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, the lengths of AC and BD are
(A) AC = 6 cm, BD = 4 cm
(B) AC = 4 cm, BD = 6 cm
(C) AC = 8 cm, BD = 2 cm
(D) AC = 3 cm, BD = 2 cm
Answer:
(A) AC = 6 cm, BD = 4 cm
Explanation:
We know that the diagonals of a parallelogram bisect each other.
AC = 2 × OA = 2 × 3 cm = 6 cm
BD = 2 × OD = 2 × 2 cm = 4 cm
Therefore, AC = 6 cm and BD = 4 cm.
Question 17.
How many lines can be passed through two distinct points?
(A) 2
(B) 1
(C) 3
(D) 4
Answer:
(B) 1
Explanation:
Only one line passes through two distinct points.
Question 18.
In the given figure, A, B, C and D are the points on a circle such that ∠ACB = 40° and ∠DAB = 60°, the measure of ∠DBA is ________
(A) 80°
(B) 40°
(C) 60°
(D) 70°
Answer:
(A) 80°
Explanation:
∠ACB = ∠ADB [Angles in the same segment]
∴ ∠ADB = 40°
Now, in ∆ADB,
∠ADB + ∠DBA + ∠BAD = 180°
⇒ 40° + ∠DBA + 60° = 180°
⇒ ∠DBA = 80°
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): According to statistics more female children are born each year than male children in India.
Reason (R): In India, the death rate of a male child is higher than that of female child.
Answer:
(C) Assertion (A) is true, but Reason (R) is false.
Question 20.
Assertion (A): If the height of cone is 24 cm and diameter of base is 14 cm, then the slant height of cone is 25 cm.
Reason (R): If r be radius and h be the slant height of cone then the slant height l = \(\sqrt{h^2+r^2}\).
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): In cone h = 24 cm, r = \(\frac{14}{2}\) = 7 cm
∴ l = \(\sqrt{h^2+r^2}\)
⇒ l = \(\sqrt{24^2+7^2}\)
⇒ l = \(\sqrt{625}\)
⇒ l = 25 cm
∴ Assertion is true.
In case of Reason (R): l = \(\sqrt{h^2+r^2}\)
It is true and correct explanation of Assertion.
Hence, A and R are true and R is correct explanation of A.
Section-B
Consists of 5 questions of 2 marks each.
Question 21.
Write ordinates of the following points:
(3, 4), (4, 0), (0, 4), (5, -3)
Answer:
4, 0, 4, -3
Question 22.
Find the point at which the equation 3x – 2y = 6 meets the x-axis.
Answer:
On x-axis, y co-ordinate is zero.
So, put y = 0 in 3x – 2y = 6, we get
3x – 0 = 6
∴ x = 2
∴ 3x – 2y = 6 meets the x-axis at (2, 0).
Question 23.
In the given figure, ∆ABC and ∆ADC are equilateral triangles on common base AC, each side of triangles being 2a units. Vertices A and C lies on X-axis, vertices B and D lies on Y-axis. O is the mid-point of AC and BD. Find the co-ordinates of the point B.
Answer:
Since, ∆ABC is an equilateral triangle with side 2a units, therefore
AB = BC = CA = 2a units
O is the mid-point of AC, then
OA = \(\frac{1}{2}\)AC
= \(\frac{1}{2}\)(2a)
= a units
Now, in right ∆AOB
OB = \(\sqrt{A B^2-O A^2}\)
= \(\sqrt{(2 a)^2-a^2}\)
= \(\sqrt{4 a^2-a^2}\)
= \(\sqrt{3 a^2}\)
= a√3 units
Thus, co-ordinates of B are (0, a√3).
Question 24.
Find any two irrational numbers between 0.1 and 0.12.
OR
Find an irrational number between \(\frac{1}{7}\) and \(\frac{2}{7}\), when it is given that \(\frac{1}{7}=0 . \overline{142857}\).
Answer:
Required two irrational numbers are 0.10100100010000….. and 0.10200200020000………
OR
Given, \(\frac{1}{7}\) = 0.142857142857…….
\(\frac{2}{7}\) = 0.285714285714……….
Hence, required irrational number lies between 0.142857……….0.285714
It can be 0.142858 or 0.20203.
Question 25.
Find the value of k, so that polynomial x3 + 3x2 – kx – 3 has one factor as x + 3.
OR
Find the value of the polynomial: p(x) = x3 – 3x2 – 2x + 6 at x = \(\sqrt{2}\).
Answer:
Let f(x) = x3 + 3x2 – kx – 3
Since (x + 3) is a factor of f(x).
Then, f(-3) = 0
⇒ (-3)3 + 3(-3)2 – k(-3) – 3 = 0
⇒ -27 + 27 + 3k – 3 = 0
⇒ 3k – 3 = 0
⇒ k = 1
OR
Given, p(x) = x3 – 3x2 – 2x + 6
Then, p(√2) = (√2)3 – 3(√2)2 – 2(√2) + 6
= 2√2 – 6 – 2(√2) + 6
= 0
Section-C
Consists of 6 questions of 3 marks each.
Question 26.
Hard plastic square-shaped sheets are available.
The side length of sheet is as per requirement.
The price of a sheet is z per square meter.
Anuj requires two sheets – a smaller sheet with side length x m and a larger sheet with side length y m. He has two choices:
Choice 1 – buy two separate sheets of side lengths x m and y m.
Choice 2 – buy a single sheet with side length (x + y) m.
What is the difference in price between the two choices?
Answer:
Price of two sheets in choice 1 = z(x2 + y2)
Price of two sheets in choice 2 = z(x + y)2
Difference in price between the two choices = z(x + y)2 – z(x2 + y2)
= zx2 + zy2 + 2xyz – zx2 – zy2
= 2xyz
Question 27.
Evaluate \(\sqrt{5+2 \sqrt{6}}+\sqrt{8-2 \sqrt{15}}\).
Answer:
Question 28.
Find the value of \(\frac{4}{(216)^{-\frac{2}{3}}}-\frac{1}{(256)^{-\frac{3}{4}}}\).
OR
Simplify \(\left(\frac{5^{-1} \times 7^2}{5^2 \times 7^{-4}}\right)^{\frac{7}{2}} \times\left(\frac{5^{-2} \times 7^3}{5^3 \times 7^{-5}}\right)^{-\frac{5}{2}}\)
Answer:
OR
Question 29.
Given below is the figure of a circle with centre O. The measure of ∠BOC = 88°.
Priya claims, “The length of OB is equal to the length of OC.”
Siya and Aditi provide different justifications for Priya’s claim.
Siya says, “OB and OC are radii of the same circle.”
Aditi says, “OC is the base of ∠BOC.”
Who has given the correct justification for Priya’s claim?
OR
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Answer:
Siya has given the correct justification for Priya’s claim.
As, length of OB and OC is equal because they are two radii of the same circle.
OR
According to the question,
OA = AB = OB
∴ ∆OAB is an equilateral triangle
∠AOB = 60°
∠ACB = \(\frac{1}{2}\)∠AOB
(Angle subtended by an arc at the circumference is half of the angle at the centre of circle)
⇒ ∠ACB = \(\frac{1}{2}\) × 60°
⇒ ∠ACB = 30°
∠ACB + ∠ADB = 180°
(Opposite angles of cyclic quadrilateral are supplementary)
⇒ ∠ADB = 180° – ∠ACB
⇒ ∠ADB = 180° – 30° = 150°
Question 30.
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 31.
Simplify \((\sqrt{x})^{-\frac{2}{3}} \sqrt{y^4} \div \sqrt{(x y)^{-\frac{1}{2}}}\)
Answer:
Section-D
Consists of 4 questions of 5 marks each.
Question 32.
In the figure, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
OR
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer:
According to the question, E and F are the midpoints of sides AB and CD.
∴ AE = \(\frac{1}{2}\)AB and CF = \(\frac{1}{2}\)CD
In the parallelogram opposite sides are equal, so
AB = CD
\(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD
∴ AE = CF
Again, AB || CD
\(\frac{1}{2}\)AB || \(\frac{1}{2}\)CD
So, AE || FC
Hence, AECF is a parallelogram.
In ∆ABP,
E is the mid-point of AB and EQ || AP.
∴ Q is the mid-point of BP. (By converse of mid-point theorem)
Similarly, P is the mid-point of DQ.
∴ DP = PQ = QB
∴ Line segments AF and EC trisect the diagonal BD.
Hence Proved.
OR
Given: ABCD is a quadrilateral P, Q, R and S are mid-points of sides AB, BC, CD and DA, respectively.
Construction: Join SP, PQ, QR, RS and AC.
Proof: In ∆DAC,
RS || AC
and RS = \(\frac{1}{2}\)AC (Mid-point theorem)
In ∆ABC,
PQ || AC
and PQ = \(\frac{1}{2}\)AC (Mid-point theorem)
From (i) and (ii), we get
RS || PQ and RS = PQ
∴ PQRS is a parallelogram.
Since diagonals of a parallelogram bisect each other.
∴ PR and QS bisect each other.
Question 33.
(a) Simplify: \(\frac{73 \times 73 \times 73+27 \times 27 \times 27}{73 \times 73-73 \times 27+27 \times 27}\)
(b) What is the degree of polynomial \(\sqrt{3}\)?
Answer:
(b) Degree of polynomial is 0.
Question 34.
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cone made of recycled cardboard. Each cone has a base diameter of 40 cm and height of 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 15 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}\) = 1.02 )
OR
A hemispherical dome, open at base is made from sheet of fibre. If the diameter of hemispherical dome is 80 cm and \(\frac{13}{170}\) of sheet actually used was wasted in making the dome, then find the cost of dome at the rate of ₹\(\frac{35}{100}\) cm2.
Answer:
Given, h = 1 m, radius = \(\frac{40}{2}\) = 20 cm = 0.2 m
and let slant height be l m.
∴ l2 = h2 + r2
⇒ l2 = 12 + 0.22
⇒ l = \(\sqrt{1+0.04}\)
⇒ l = \(\sqrt{1.04}\) (Since given \(\sqrt{1.04}\) = 1.02)
Curved surface area of cone will be painted = πrl
= (3.14 × 0.2 × 1.02) m2
= 0.64046 m2
Therefore, curved surface area of 50 cones = 50 × 0.64046 = 32.028 m2
∴ Cost of painting 32.028 m2 = ₹(15 × 32.028) = ₹ 384.336 = ₹ 384
OR
Given, Diameter = 80 cm
∴ r = 40 cm
C.S.A. of the dome = 2πr2
C.S.A. = 2 × \(\frac{1}{2}\) × 40 × 40
C.S.A. = \(\frac{70400}{7}\) cm2
Since, \(\frac{13}{170}\) of sheet was wasted,
Area of sheet wasted = \(\frac{13}{170} \times \frac{70400}{7}\) = \(\frac{915200}{1190}\) cm2
∴ Total area = \(\frac{70400}{7}+\frac{915200}{1190}\) = 10826.21 cm2
Cost of sheet per square metre = ₹\(\frac{35}{100}\)
∴ Total cost of sheet = \(\frac{35}{100}\) × 10826.21 = ₹ 3789.17
Question 35.
Two equilateral triangles on a straight line are shown below.
What is the measure of ‘x’?
Answer:
In equilateral ΔABC, ∠2 = 60°
XY is a straight line
∴ ∠1 + ∠2 + ∠3 = 180° (angles on straight line are supplementary)
⇒ 75° + 60° + ∠3 = 180°
⇒ ∠3 = 180° – 135°
⇒ ∠3 = 45°
In equilateral ΔDEF, ∠5 = 60°
So, ∠4 + ∠5 + ∠6 = 180°
⇒ ∠4 = 180° – 125°
⇒ ∠4 = 55°
In ΔADM,
∠3 + ∠4 + ∠AMD = 180° (Angle sum property)
⇒ 45° + 55° + ∠AMD = 180°
⇒ ∠AMD = 180° – 100°
⇒ ∠AMD = 80°
∴ ∠NMC = ∠AMD = 80° (vertically opposite angles are equal)
In ΔMNC,
∠MCN = ∠ACB = 60° (angle of equilateral triangle)
Now, ∠MCN + ∠NMC + ∠CNM = 180°
⇒ 60° + 80° + x = 180°
⇒ x = 180° – 140°
⇒ x = 40°
Section-E
Cased-Based Subjective Questions.
Question 36.
Cleanliness drive is the way to raise awareness on the importance of cleanliness in one’s neighbourhood. Residents of a certain locality joined ‘Cleanliness drive’ together to clean their area. Participation of the women was 10 more than men. Taking x as number of women and y as number of men.
(i) Which mathematical concept is used here?
(ii) Write the suitable linear equation in two variables for above?
OR
If the number of women is double of the number of men then what is the number of women?
(iii) Find the number of women if number of men is 30.
Answer:
(i) ‘Linear equation in two variables’ is mathematical concept used here.
(ii) Here, Linear equation in two variables will be x = 10 + y
(iii) If number of men, y = 30,
Then, x = 10 + 30 = 40
Thus, number of women = 40
OR
We have, x = 10 + y
If x = 2y
Then, 2y = 10 + y
⇒ 2y – 2 = 10
⇒ 2y – y = 10
⇒ y = 10
∴ x = 2y = 2 × 10 = 20
Thus, number of women = 20.
Question 37.
Maths teacher of class 9th gave students coloured paper in the shape of quadrilateral and then ask the students to make parallelogram from it by using paper folding as shown in given figure:
(i) If ∠RSP = 30°, then find ∠RQP.
(ii) If ∠RSP = 50°, then find ∠SPQ.
OR
If SP = 3 cm, then measure of side RQ will be?
(iii) State property about the diagonals of a parallelogram?
Answer:
(i) In || gm PQRS,
∠RSP = 30°
∴ ∠RQP = ∠RSP = 30° (Opposite angles of a || gm are equal)
(ii) Adjacent angles of a parallelogram are supplementary,
Thus, ∠RSP + ∠SPQ = 180°
⇒ 50° + ∠SPQ = 180°
⇒ ∠SPQ = 180° – 50°
⇒ ∠SPQ = 130°
OR
In parallelogram opposite sides are equal,
Thus, RQ = SP
∴ RQ = 3 cm
(iii) Diagonals of a parallelogram bisects each other.
Question 38.
Shakshi prepared a Rangoli in triangular shape on Diwali. She makes a small triangle under a big triangle as shown in the figure.
Sides of big triangle are 25 cm, 26 cm and 28 cm. Also, ∆PQR is formed by joining midpoints of sides of ∆ABC. Use the above data to help her in resolving below doubts.
(i) Which theorem will be used to calculate sides of ∆PQR?
(ii) What is the semi-perimeter of ∆ABC?
OR
What is the length of RQ, PQ and QR?
(iii) If colourful rope is to be placed along the sides of small ∆PQR. What is the length of the rope?
Answer:
(i) Mid point theorem is used to calculate sides of ∆PQR.
Which states that Line joining the mid-point of any two sides of a triangle is parallel to the third side and is equal to half of it.
(ii) P = sum of all 3 sides = (25 + 26 + 28) cm = 79 cm
Now, semi-perimeter = \(\frac{1}{2}\) × P
= \(\frac{1}{2}\) × 79
= 39.5 cm
OR
Length of RQ = \(\frac{1}{2}\)BC
= \(\frac{1}{2}\) × 28
= 14 cm
Similarly PQ = \(\frac{1}{2}\) × 25 = 12.5 cm
And, PR = \(\frac{1}{2}\) × 26 = 13 cm
(iii) Length of rope = Perimeter of ∆PQR
= (12.5 + 13 + 14) cm
= 39.5 cm