9 Maths - Sample Paper 1

Practice Paper – TERM II (2021 – 2022)
Class – IX
Mathematics (041)
Time: 2 hours Maximum Marks: 40
General Instructions:
1. The question paper consists of 14 questions divided into 3 sections A, B, C.
2. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in
two questions.
3. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in
one question.
4. Section C comprises of 4 questions of 4 marks each. An internal choice has been
provided in one question. It contains two case study-based questions.
5. Use of calculators is not permitted.
SECTION - A
Q. No.
1. A tank is in the form of a cuboid whose length is 20 metre. If 18 cubic metre water is
removed from it the water level goes down by 15 centimetre. Find the width of the tank.
Volume of removed water = Length X width X decrease in water level
⇒ 18 = 20 X width X 0.15
∴ width = 6 metre
OR
Find the volume of a sphere whose surface area is 616 sq. cm.
SA of sphere = 616 ⇒ 4 X 22/7 X r2 = 616
∴ r = 7 cm
Volume of sphere = 4/3 X 22/7 X (7)3
2. Find the value of the polynomial 2x + 5 at x = – 3.

2 (– 3) + 5 = –6 + 5 = –1
3. A joker’s cap is in the shape of a right circular cone with base radius 7 cm and height 24
cm. Find the area of sheet required for 10 such caps.
Slant height=
Area of sheet required for a cap = CSA of the cap = 22/7 X 7 X 25 = 550 cm2
Area of sheet required for 10 such cap = 10 X 550 = 5500 cm2
4. Prove that “A diagonal of a parallelogram divides it into two congruent triangles”.
Given: A parallelogram ABCD & BD is its diagonal.
To prove: ∆ABD ≅ ∆CDB
Proof: In ∆ABD and ∆CDB,
Proof: In ∆ABD and ∆CDB,
AB = CD [∵ opp. sides of IIgm]
DA = BC [∵ opp. sides of IIgm]
BD = DB [Common]
∴ ∆ABD ≅ ∆CDB (SSS rule) Hence Proved
5.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC =
30°. Find ∠BCD.
∠BDC = ∠BAC = 300 (∵ angles in same segment)
∠DBC + ∠BDC + ∠BCD = 1800 (∵ Angle sum property of ∆BCD)
⇒ 700 + 300 + ∠BCD = 1800
⇒ ∠BCD = 800
OR
Prove that equal chords of a circle subtend equal angles at the centre.
Given: AB and CD are two equal chords of a circle with centre O.
To Prove: ∠AOB = ∠COD
Construction: Join AO, BO, CO and DO.
Proof: In ∆AOB and ∆COD

AO = CO [radii of the same circle]
BO = DO [radii of the same circle]
AB = CD [Given]
∴ ∆AOB ≅ ∆COD (SSS rule)
So, ∠AOB = ∠COD (c.p.c.t.)
Hence Proved
6. If a sphere and a solid cube have the same surface area then find the ratio of their
volumes.
SECTION B
7. Find the factors of 2t2 – t – 10.
2t2 – t – 10
= 2t2 – 5t + 4t – 10
= t(2t – 5) + 2 (2t – 5)
= (2t – 5)(t + 2)
8. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA
respectively. Show that the quadrilateral PQRS is a rhombus.
ABCD is a rectangle.
∴ AB = DC & BC = AD
⇒ 1/2 AB = 1/2 DC & 1/2 BC = 1/2 AD
∴ AP = PB = DR = RC & BQ = QC = DS = AS
AS = BQ [Proved above]
∠SAP = ∠QBP [each 900]
AP = BP [Proved above]
∴∆SAP ≅ ∆QBP [SAS rule]
So, SP = PQ (c.p.c.t.) -------- ①
Similarly, PQ = QR -------- ②
QR = RS -------- ③
and RS = SP -------- ④
from ①, ②, ③ and ④, we get
PQ = QR = RS = SP
As four sides of PQRS are equal, PQRS is a rhombus.
OR
In parallelogram ABCD, two points P and Q are taken on the diagonal BD such that DP = BQ.
Show that APCQ is a parallelogram.
In ∆BCQ and ∆ DAP
BC = DA [opp. sides of IIgm]
∠CBQ = ∠ADP [Alternate interior angles]
BQ = DP [Given]
∴ ∆BCQ ≅ ∆ DAP [SAS rule]
So, CQ = AP [c.p.c.t.] -------- ①
Similarly, PC = QA -------- ②
As opp. sides of APCQ are equal So APCQ is a parallelogram.
9. Write (1/4a – 1/2b + 1)2 in the expanded form.
10. A cylindrical roller 1.5 m long has a diameter of 70 cm. How many revolutions will it make
to level a playground measuring 50 m X 33 m?
Radius of roller = 70/2 = 35 cm
Area covered in one revolution = 2 X 22/7 X 35/100 X 1.5 = 3.3 m2
Area of playground = 50 X 33 = 1650 m2 No. of revolution = 1650/3.3 = 500
SECTION C
11. Factorize the polynomial x3 – 6x2 + 11x – 6.
OR
Find the value of the following by using suitable identity:
(i) (999)3
= (1000 – 1)3
= (1000)3 – (1)3 – 3(1000)(1)(1000 – 1)
Using identity (a – b)3 = a3 – b3 – 3ab (a – b)
= 1000000000 – 1 – 2997000 = 997002999
(ii) 103 × 107
(100 + 3) X (100 + 7)
= (100)2 + (3 + 7) (100) + 3 X 7
Using identity (x + a) (x + b) = x2 + (a + b) x + ab
= 10000 + 2100 + 21
= 12121
12. Construct a triangle PQR in which QR = 6 cm, ∠Q = 600 and PR – PQ = 2 cm.
13.
Three girls Reshma, Salma and Mandeep are playing with each other standing at points R, S
and M respectively on the boundary of a circle of radius 5 m as shown in the figure. Reshma
throws a ball to Salma, Salma to Mandeep and Mandeep to Reshma. If the distance between
Reshma and Salma and between Salma and Mandeep is 6 meters each and O is the center of
the circle, then answer the following questions:
(i) Find the distance between Reshma and Mandeep.
(ii) Find the value of length OE.
14. Insurance is a means of protection from financial loss. It is a form of risk management,
primarily used to hedge against the risk of a contingent on uncertain loss. For insurance
awareness, an insurance company selected 2000 drivers at random in a particular city, find a
relationship between age and accidents. The data obtained is given in the following table:
Answer the following questions:

(i) Find the probability for a driver chosen at random from the city being 18 – 29 years of age
and having exactly 3 accidents in one year.
P(E) = 61/2000 or 0.0305
(ii) Find the probability for a driver chosen at random from the city having no accidents in
one year.
Favourable cases = 440 + 505 + 306 = 1251
P(E) = 1251/2000 or 0.6255

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Practice Paper – TERM II (2021 – 2022)

2. Find the value of the polynomial 2x + 5 at x = – 3.

Proof: In ∆AOB and ∆COD

Answer the following questions:

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