CBSE Class 10 Marking Scheme 2022 for Maths Basic Term 2

𝑴𝒂𝒓𝒌𝒊𝒏𝒈 𝑺𝒄𝒉𝒆𝒎𝒆
Mathematics –Basic(241)
Class- X Session- 2021-22
TERM II

Q.N. HINTS/SOLUTION Marks
1 2
3𝑥 − 7𝑥 − 6 = 0
⇒ 3𝑥 2 − 9𝑥 + 2𝑥 − 6 = 0 1/2
⇒ 3𝑥(𝑥 − 3) + 2(𝑥 − 3) = 0
⇒ (𝑥 − 3)(3𝑥 + 2) = 0 1/2
2
∵ 𝑥 = 3, − 1
3
OR
Since the roots are real and equal, ∴ 𝐷 = 𝑏 2 − 4𝑎𝑐 = 0
⇒k2 – 4×3×3 = 0 (∵ 𝑎 = 3, 𝑏 = 𝑘, 𝑐 = 3) 1
⇒k2 = 36
1/2 +1/2
⇒k = 6 𝑜𝑟 −6
2 Let 𝑙 be the side of the cube and L, B, H be the dimensions of the cuboid
Since 𝑙 3 = 64 𝑐𝑚3 ∴ 𝑙 = 4 𝑐𝑚 1/2
Total surface area of cuboid is 2[𝐿𝐵 + 𝐵𝐻 + 𝐻𝐿], Where L=12, B=4 and H=4
1/2
=2(12 × 4 + 4 × 4 + 4 × 12) 𝑐𝑚2 = 224𝑐𝑚2 1
3 Runs scored Frequency Cumulative Frequency
0-20 4 4
20-40 6 10
40-60 5 15
60-80 3 18
1/2
80-100 4 22

Total frequency (N) = 22
𝑁
2
= 11; So 40-60 is the median class. 1/2

𝑁
( )−𝑐𝑓
2
Median = 𝑙 + 𝑓
×ℎ 1/2
11−10
= 40 + x 20
5
= 44 runs 1/2

4 The common difference is 9 - 4=5 1
If the first term is 6 and common difference is 5, then new AP is,
6, 6+5, 6+10…
=6,11,16…. 1
5 ∵ Mode = 38.
∴ The modal class is 30-40. 1/2

𝑓 −𝑓
Mode = 𝑙 + 2𝑓 1−𝑓 −𝑓
0
×ℎ
1 0 2 1/2

16−12 1/2
=30 + x 10 = 38
32−12−𝑥

4
x 10 =8
20−𝑥
8(20-x) = 40
20-x= 5
1/2
X= 15
6
X D Y

.

E N

∵XY is the tangent to the circle at the point D
1/2
∴ OD  XY   ODX = 900   EDX = 900
Also, MN is the tangent to the circle at E
∴ OE  MN  OEN = 900   DEN = 900 1/2
⇒  EDX =  DEN (𝑒𝑎𝑐ℎ 900 ).
which are alternate interior angles.
1
∴ XY  MN
OR

D R C
∵Tangent segments drawn from
an external point to a circle are equal
S Q
∴ BP=BQ
CR=CQ
DR=DS
AP=AS A P B

⇒BP+CR+DR+AP = BQ+CQ+DS+AS 1

⇒ AB+DC = BC+AD

∴ AD= 10-7= 3 cm
1
Section-B

7 First Term of the AP(a) = 5
Common difference (d) = 8-5=3

Last term = 𝑎40 = a+(40-1) d
= 5 + 39 × 3 = 122 1

Also 𝑎31 = 𝑎 + 30𝑑 = 5 + 30 × 3 = 95
1
𝑛
Sum of last 10 terms = 2 (𝑎31 + 𝑎40 )
10
= (95 + 122)
2
= 5 × 217 = 1085 1

8
Let, AB be the tree broken at C,
B
Also let 𝐴𝐶 = 𝑥
𝐴𝐶
In ∆ CAD, sin300 = 8m 1
𝐷𝐶
1 𝑥
⇒2=
8
⇒𝑥 =4𝑚 C 1/2
1/2
⇒the length of the tree is = 8+4 =12m
8m
x

30 ∘ 1(correct
D A Fig.)
OR

Let AB and CD be two poles of height h meters also let P be a point between them on
the road which is x meters away from foot of first pole AB, PD= (80-x) meters.

ℎ
In ∆ABP, 𝑡𝑎𝑛60𝑜 = 𝑥 ⇒ ℎ = 𝑥√3 .…(1)
1

ℎ 80−𝑥 1/2
In ∆CDP, 𝑡𝑎𝑛 30𝑜 = 80−𝑥 ⇒ ℎ = √3
….(2)
80 − 𝑥
𝑥√3 = [∵ 𝐿𝐻𝑆(1) = 𝐿𝐻𝑆(2), 𝑠𝑜 𝑒𝑞𝑢𝑎𝑡𝑖𝑛𝑔 𝑅𝐻𝑆]
√3
⇒ 3𝑥 = 80 − 𝑥 ⇒ 4𝑥 = 80 ⇒ 𝑥 = 20𝑚
So, 80 − 𝑥 = 80 − 20 = 60𝑚 1/2
Hence the point is 20m from one pole and 60 meters from the other pole.

A C

h h
1(correct
60° 30° Fig.)
B D
x P 80-x

9 PA = PB (Tangent segments drawn to a circle from an external point are equal)

∴ In ∆𝐴𝑃𝐵,  PAB =  PBA
Also,  APB = 600 1
In ∆𝐴𝑃𝐵, sum of three angles is 1800.

Therefore,  PAB +  PBA = 1800 -  APB= 1800 – 600 = 1200.
∴ PAB =  PBA = 600 (∵ PAB =  PBA) 1
∵ ∆𝐴𝑃𝐵 is an equilateral triangle.
So, 𝐴𝐵 = 6𝑐𝑚 1

10 Let the three consecutive multiples of 5 be 5x, 5x+5, 5x+10.
Their squares are (5𝑥)2 , (5𝑥 + 5)2 and(5𝑥 + 10)2 .
(5𝑥)2 + (5𝑥 + 5)2 + (5𝑥 + 10)2 = 725 1
⇒25𝑥 2 + 25𝑥 2 + 50x + 25 + 25𝑥 2 + 100x + 100 = 725
⇒ 75𝑥 2 + 150𝑥 − 600 = 0
⇒ 𝑥 2 + 2𝑥 − 8 = 0
⇒ (𝑥 + 4)(𝑥 − 2) = 0
⇒ 𝑥 = −4, 2 1
⇒ 𝑥 = 2 (ignoring –ve value)
So the numbers are 10, 15 and 20 1
Section-C

11

A

P O´ O

B

Draw two concentric circles with center O and radii 3cm and 7cm respectively. 1
Join OP and bisect it at 𝑂′ , so 𝑃𝑂′ = 𝑂′ 𝑂 1
1
Construct circle with center 𝑂′ and radius 𝑂′ 𝑂
1
Join PA and PB

A OR

P 60° 120° O

B
Draw a circle of radius 6cm
Draw OA and Construct ∠ 𝐴𝑂𝐵 = 1200
1
Draw ∠ 𝑂𝐴𝑃 = ∠ 𝑂𝐵𝑃 = 900 1
PA and PB are required tangents 1
6
Join OP and apply tan∠𝐴𝑃𝑂 = tan 30° = 𝑃𝐴
1
⇒ Length of tangent = 6√3 cm

12
Converting the cumulative frequency table into exclusive classes, we get:
Age No of passengers(fi) xi fi xi

0-10 14 5 70
10-20 30 15 450
20-30 38 25 950
30-40 52 35 1820 2
40-50 50 45 2250
50-60 61 55 3355
60-70 42 65 2730
70-80 13 75 975
∑ 𝑓𝑖 = 300 ∑ 𝑓𝑖 𝑥𝑖 =12600

∑ 𝑓𝑖 𝑥𝑖 12600
Mean age = 𝑥̅ = = 1
∑ 𝑓𝑖 300

𝑥̅ = 42 1

13 (i) The ship is nearer to the lighthouse as its angle of depression is greater.
𝐴𝐵
In ∆ ACB, tan 600 = 𝐵𝐶
A 1
40
√3 = 𝐵𝐶
300
40 40√3 600
∴ BC = √3 = m
3

1

300 60°
D B C

(ii)
𝐴𝐵
In ∆ ADB, tan 300 = 𝐵𝐷
1 40
⇒ = 𝐷𝐵
√3
∴ DB = 40√3𝑚 1
60
Time taken to cover this distance = (2000 × 40√3) minutes
60√3 1
= 100 = 2.076 minutes

14 (i) Let 𝑟1 𝑎𝑛𝑑 𝑟2 be respectively the radii of apples and oranges
∵ 2𝑟1 : 2𝑟2 = 2: 3 ⇒ 𝑟1 : 𝑟2 = 2: 3 1/2

2 2
𝑟1 2 2 2 1
4𝜋𝑟1 : 4𝜋𝑟2 = ( ) = ( ) = 4: 9 1
𝑟2 3 2

(ii) Let the height of the drum be h.
Volume of the drum = volume of the cylinder + volume of the sphere
4 1
π32h = (π32 × 8 + 3 π33 ) 𝑐𝑚3
⇒ ℎ = (8 + 4)𝑐𝑚
⇒ ℎ = 12𝑐𝑚
1