Class 9 Sample Paper 2022 Solution Maths Term 2

Solutions of Practice Paper
Class – IX
Mathematics (Code: 041)
Term – II (2021 – 2022)
Time Duration: 2 hrs. Maximum Marks: 40

Q. No. Value point/Hints
1 Volume of removed water = Length X width X decrease in water level
⇒ 18 = 20 X width X 0.15
∴ width = 6 metre
OR
SA of sphere = 616
⇒ 4X X r2 = 616
∴ r = 7 cm
Volume of sphere = X X (7)3
= 1437 cm3
2 2 (– 3) + 5 = –6 + 5 = –1
3 Slant height= (7) + (24) = 25 cm
Area of sheet required for a cap = CSA of the cap = X 7 X 25 = 550 cm2
Area of sheet required for 10 such cap = 10 X 550 = 5500 cm2
4 Given: A parallelogram ABCD & BD is its
diagonal.
To prove: ∆ABD ≅ ∆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)
5 ∠BDC = ∠BAC = 300 (∵ angles in same segment)
∠DBC + ∠BDC + ∠BCD = 1800 (∵ Angle sum property of ∆BCD)
⇒ 700 + 300 + ∠BCD = 1800
⇒ ∠BCD = 800
OR

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.)
6 Given, Surface area of sphere = Surface area of cube
4𝜋r2 = 6a2

∴ =

= = 𝜋 = 𝜋 = 𝜋X X =

Therefore, required ratio is √6 : √𝜋.
7 2t2 – t – 10
= 2t2 – 5t + 4t – 10
= t(2t – 5) + 2 (2t – 5)
= (2t – 5)(t + 2)
8 ABCD is a rectangle.
∴ AB = DC & BC = AD
⇒ AB = DC & BC = AD
∴ AP = PB = DR = RC &
BQ = QC = DS = AS
In ∆SAP and ∆QBP
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 ∆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 Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
𝑎− 𝑏+1 = 𝑎 + − 𝑏 + (1) + 2 𝑎 − 𝑏 + 2 − 𝑏 (1) + 2 (1) 𝑎

= 𝑎 + 𝑏 + 1 – ab – b + a
10 Radius of roller = = 35 cm
Area covered in one revolution = 2 X X X 1.5 = 3.3 m2
Area of playground = 50 X 33 = 1650 m2
No. of revolution = = 500
.
11 Let p(x) = x3 – 6x2 + 11x – 6
Possible factors of – 6 are ± 1, ± 2, ±3 , ±6 etc.
p(1) = (1)3 – 6(1)2 + 11(1) – 6 = 12 – 12 = 0
So, x – 1 is a factor of p(x).
By long division method,
p(x) = (x – 1) (x2 – 5x + 6)
= (x – 1) (x2 – 3x – 2x + 6) = (x – 1)(x – 2)(x – 3)
OR
(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 X 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 Correct Construction
13 (i) In ∆RSO,
s= = 8 cm
Using Herons formulae,

ar (RSO) = 8 X(8 − 6)X(8 − 5)X(8 − 5) =12 cm2
Also, ar(RSO) = X SO X RE
12 = X 5 X RE

RE = 4.8 cm
Therefore, RM = 2 X RE = 2 X 4.8 = 9.6 cm
(ii) In rt. angled ∆REO,
(OR)2 = (RE)2 + (EO)2 [By Pythagoras Theorem]
(5)2 = (4.8)2 + (EO)2
EO = √25 − 23.04 = 1.4 cm
14 (i) P (E) = or 0.0305
(ii) Favourable cases = 440 + 505 + 306 = 1251
P (E) = or 0.6255