CBSE Class 10 Marking Scheme 2022 for Maths Standard Term 2

Marking Scheme
Class- X, Session- 2021-22
TERM II
Subject- Mathematics (Standard)
SECTION A
Q.No HINTS/SOLUTION MARKS
.
1 a = 6, d = 3 ; a25 = 6 + 24(3) = 78 1
a15 = 6 + 14(3) = 48 ; a 25 - a15 = 78 - 48 = 30 1
OR
1
7(𝑎 + 6𝑑) = 5(𝑎 + 4𝑑)
1
⇒2𝑎 + 22𝑑 = 0 ⇒ 𝑎 + 11𝑑 = 0 ⇒ 𝑡12 = 0

2 5mx² - 6mx + 9 = 0

b² - 4ac = 0 ⇒ (-6m)² - 4(5m)(9) = 0 1

⇒ 36m(m - 5) = 0

⇒ m = 0, 5 ; rejecting m=0, we get m = 5 1

3
A

r
2r
P O

B

let ∠𝐴𝑃𝑂 = 𝜃
𝑂𝐴 1 1/2
𝑆𝑖𝑛𝜃 = = ⇒ 𝜃 = 300
𝑂𝑃 2
1/2
⇒ ∠𝐴𝑃𝐵 = 2𝜃 = 600
1/2
Also ∠𝑃𝐴𝐵 = ∠𝑃𝐵𝐴 = 600 (∵ 𝑃𝐴 = 𝑃𝐵 )
1/2
⇒△ 𝐴𝑃𝐵 is equilateral
4 CSA (cone) = 𝜋𝑟𝑙 = 12320 1/2
22
× 56 × 𝑙 = 12320
7
1
𝑙 = 70 𝑐𝑚
ℎ = √702 − 562 = 42 cm 1/2

5 Modal class is 40 − 60, 𝑙 = 40, ℎ = 20, 𝑓1 =? , 𝑓0 = 10, 𝑓2 = 6 1/2
𝑓1 − 10 1/2
45 = 40 + 20 × [ ]
2𝑓1 − 10 − 6
1 𝑓1 −10
⇒ =
4 2𝑓1 −16

1
⇒2𝑓1 − 16 = 4𝑓1 − 40 ⇒ 𝑓1 = 12
6 Let the present age of Ritu be 𝑥 years
(𝑥 − 5)2 = 5𝑥 + 11 1
𝑥 2 − 15𝑥 + 14 = 0½
(𝑥 − 14)(𝑥 − 1) = 0 ⇒ 𝑥 = 1 𝑜𝑟 14 1/2

x = 14 years (rejecting 𝑥 = 1 as in that case Ritu’s age 5 years ago will be –ve ) 1/2

OR
9𝑥 2 − 6𝑝𝑥 + (𝑝2 − 𝑞 2 ) = 0
𝑎 = 9, 𝑏 = −6𝑝, 𝑐 = 𝑝2 − 𝑞 2 1/2
𝐷 = 𝑏 2 − 4𝑎𝑐 = (−6𝑝) 2 − 4(9)(𝑝2 − 𝑞 2 ) = 36𝑞 2
1/2
−𝑏 ± √𝐷 6𝑝 ± 6𝑞 𝑝 + 𝑞 𝑝−𝑞
𝑥= = = 𝑜𝑟 1
2𝑎 18 3 3
SECTION B
7 Distance (in m) 0-1 1–2 2-3 3-4 4-5
Number of
40 80 62 38 30
Students
𝑐𝑓 40 120 182 220 250
𝑛 250 1
= = 125 ⇒ 𝑚𝑒𝑑𝑖𝑎𝑛 𝑐𝑙𝑎𝑠𝑠 𝑖𝑠 2 − 3, 𝑙 = 2, ℎ = 1, 𝑐𝑓 = 120, 𝑓 = 62
2 2
𝑛
− 𝑐𝑓 1/2
𝑚𝑒𝑑𝑖𝑎𝑛 = 𝑙 + 2 ×𝑖
𝑓
5
=2+
62
129 5
= = 2 𝑚 𝑜𝑟 2.08𝑚 1
62 62
5
50% of students jumped below2 m and 50% above it. 1/2
62
8 Draw a circle of radius 4cm 1

Draw OA and construct ∠𝐴𝑂𝐵 = 1200 1
Draw ∠𝑂𝐴𝑃 = ∠𝑂𝐵𝑃 = 900
PA and PB are required tangents 1
9 Runs Scored 0 - 40 40 - 80 80 - 120 120 - 160 160 - 200 TOTAL
Number of
12 20 35 30 23 120
Batsmen (𝑓𝑖 )

𝑥𝑖 20 60 100 140 180
𝑓𝑖 𝑥𝑖 240 1200 3500 4200 4140 13280 1𝟏𝟐
∑ 𝑓𝑖 𝑥𝑖 13280 1𝟏𝟐
𝑚𝑒𝑎𝑛 ( 𝑥̅ ) = = = 110.67 𝑟𝑢𝑛𝑠
∑ 𝑓𝑖 120

10
P
R

y
x

30° 60° 1
Q S
20m

𝑦
𝐼𝑛 ∆𝑃𝑄𝑆, 𝑡𝑎𝑛600 = ⇒ 𝑦 = 20√3m 1/2
20
𝑥 20
𝐼𝑛 ∆𝑅𝑆𝑄, 𝑡𝑎𝑛300 = ⇒𝑥= 𝑚 1/2
20 √3
20 40 40√3
𝑦 − 𝑥 = 20√3 − = = = 23.06𝑚 1
√3 √3 3

OR
P

A 60° Q
1
1.7 m 50m
B R

Let PR be the building and AB be the boy
𝑃𝑄
𝐼𝑛 ∆𝑃𝑄𝑅, 𝑡𝑎𝑛600 = ⇒ 𝑃𝑄 = 50√3𝑚 1
50
Height of the building = 𝑃𝑅 = (50√3 + 1.7)𝑚 = 88.2𝑚
1
SECTION C
11 Volume of shell = Volume of cylinder
4𝜋
⇒ [ 5³ − 3³] = 𝜋(7)2 ℎ 1𝟏𝟐
3
8 2
⇒ ℎ = = 2 𝑐𝑚 1
3 3

TSA of cylinder is
22 8 29 1276
= 2𝜋𝑟 (𝑟 + ℎ) = 2 × × 7 × (7 + ) = 44 × = 𝑐𝑚 2 𝑜𝑟 425.33𝑐𝑚 2
7 3 3 3 1𝟏𝟐

12

1

∠𝑂𝐴𝑃 + ∠𝑂𝐵𝑃 + ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 360𝑜 1𝟏𝟐
⇒ 900 + 900 + ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 360𝑜 (∵ 𝑇𝑎𝑛𝑔𝑒𝑛𝑡 ⊥ 𝑟𝑎𝑑𝑖𝑢𝑠 )
1𝟏𝟐
⇒ ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 1800
OR

Let ∠𝑃𝑇𝑄 = 𝜃
𝑇𝑃𝑄 is an isosceles triangle.
1 𝜃
∠𝑇𝑃𝑄 = ∠𝑇𝑄𝑃 = (180𝑜 − 𝜃 ) = 90° − 1𝟏𝟐
2 2
∠𝑂𝑃𝑇 = 90𝑜
𝜃 𝜃
∠𝑂𝑃𝑄 = ∠𝑂𝑃𝑇 − ∠𝑇𝑃𝑄 = 90𝑜 − (90𝑜 − ) = 1𝟏𝟐
2 2
1
∠𝑂𝑃𝑄 = ∠𝑃𝑇𝑄
2
2∠𝑂𝑃𝑄 = ∠𝑃𝑇𝑄
1
13 Case Study-1
i) T
30°

240m 1

30
° R
P 𝑥

240
In ∆𝑃𝑇𝑅, tan 30𝑜 = ⇒ 𝑥 = 240√3 𝑚 1
𝑥

ii) Distance of boat from tower= 240 √3 − 240(√3 − 1) = 240𝑚 1
Let the angle of depression = 𝜃
240
𝑡𝑎𝑛𝜃 = = 1 ⇒ 𝜃 = 45𝑜 1
240

14 i) 3000, 3005, 3010, ...,3900. 1
𝑎𝑛 = 𝑎 + (𝑛 − 1) 𝑑
3900 = 3000 + (n - 1)5
⇒ 900 = 5𝑛 − 5 ⇒ 5𝑛 = 905 ⇒ 𝑛 = 181
1
Minimum number of days of practice = 𝑛 − 1 = 180 𝑑𝑎𝑦𝑠
𝑛
ii) 𝑆𝑛 = (𝑎 + 𝑙) 1
2
181
= × (3000 + 3900) = 624450 pushups 1
2