CBSE Class 10 Marking Scheme 2022 for Maths Standard

Marking Scheme
Class- X Session- 2021-22
TERM 1
Subject- Mathematics (Standard)
SECTION A
QN Correct HINTS/SOLUTION MAR
Option KS
1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 1
HCF(4,2) = 4:2 = 2:1

2 (a) 𝑎₁ 𝑏₁ 𝑐₁ 1
For lines to coincide: 𝑎₂ = 𝑏₂ = 𝑐₂
5 7 −3
so, 15 = 21 = −k
i.e. k= 9
3 (b) By Pythagoras theorem 1
The required distance =√(200² + 150²)
= √(40000+ 22500) = √(62500) = 250m.
So the distance of the girl from the starting point is 250m.
4 (d) 1 1 1
Area of the Rhombus = 2 d₁d₂ = 2 x 24 x 32= 384 cm².
Using Pythagoras theorem
1 1
side² = (2d₁)² + (2d₂)² = 12² +16² = 144 +256 =400
Side = 20cm
Area of the Rhombus = base x altitude
384 = 20 x altitude
So altitude = 384/20 = 19.2cm
5 (a) Possible outcomes are (HH), (HT), (TH), (TT) 1
Favorable outcomes(at the most one head) are (HT), (TH), (TT)
So probability of getting at the most one head =3/4
6 (d) Ratio of altitudes = Ratio of sides for similar triangles 1
So AM:PN = AB:PQ = 2:3
7 (b) 2sin2β – cos2β = 2 1
Then 2 sin2β – (1- sin2β) = 2
3 sin2β =3 or sin2β =1
β is 90ᵒ
8 (c) Since it has a terminating decimal expansion, 1
so prime factors of the denominator will be 2,5
9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 1
will intersect.
10 (d) Distance of point A(-5,6) from the origin(0,0) is 1
√(0 + 5)2 + (0 − 6)2 = √25 + 36 = √61 units
11 (b) a²=23/25, then a = √23/5, which is irrational 1
12 (c) LCM X HCF = Product of two numbers 1
36 X 2 = 18 X x
x=4

13 (b) tan A= √3 = tan 60° so ∠A=60°, Hence ∠C = 30°. 1
So cos A cos C- sin A sin C = (1/2)x (√3/2) - (√3/2)x (1/2) =0

14 (a) 1x +1x +2x =180°, x = 45°. 1
∠A , ∠B and ∠C are 45°, 45° and 90°resp.
sec A tan A sec 45 tan 45 √2 1
– = – = – = 1-1= 0
cosec B cot B cosec 45 cot 45 √2 1

15 (d) total distance 176 1
Number of revolutions= = 22
circumference 2 X 7 X 0.7
= 40
16 (b) perimeter of ∆ABC BC 1
=
perimeter of ∆DEF EF
7.5 2
= . So perimeter of ∆DEF = 15cm
perimeter of ∆DEF 4

17 (b) Since DE∥ BC, ∆ABC ~ ∆ADE ( By AA rule of similarity) 1
AD DE 3 DE
So = i.e. = . So DE = 6cm
AB BC 7 14
18 (a) Dividing both numerator and denominator by cosβ, 1
4 𝑠𝑖𝑛𝛽−3 cos 𝛽 4 𝑡𝑎𝑛𝛽−3 3−3
= = =0
4 sin 𝛽+3 cos 𝛽 4 tan 𝛽+3 3+3

19 (d) 𝑎₁ 𝑏₁ 𝑐₁ 1
-2(–5x + 7y = 2) gives 10x – 14y = –4. Now = 𝑏₂ = 𝑐₂ = -2
𝑎₂
20 (a) Number of Possible outcomes are 26 1
Favorable outcomes are M, A, T, H, E, I, C, S
probability = 8/26 = 4/13
SECTION B
21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y, 1
ATQ 81x + 81y = 1215
Or x+y = 15
which gives four co prime pairs-
1,14
2,13
4,11
7, 8

22 (c) Required Area is area of triangle ACD = ½(6)2 1
= 6 sq units
23 (b) tan α + cot α = 2 gives α=45°. So tan α = cot α = 1 1
tan20α + cot 20α = 120 + 120 = 1+1 = 2
24 (a) Adding the two given equations we get: 348x + 348y = 1740. 1
So x +y =5
25 (c) LCM of two prime numbers = product of the numbers 1
221= 13 x 17.
So p= 17 & q= 13
⸫3p - q= 51-13 =38
26 (a) Probability that the card drawn is neither a king nor a queen 1
52−8
= 52
= 44/52 = 11/13
27 (b) Outcomes when 5 will come up at least once are- 1
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6)
Probability that 5 will come up at least once = 11/36

28 (c) 1+ sin2α = 3 sinα cos α 1
sin2α + cos2α + sin2α = 3 sinα cos α
2 sin2α - 3sinα cos α + cos2α = 0
(2sinα -cos α)( sinα- cosα) =0
⸫cotα = 2 or cotα = 1

29 (a) Since ABCD is a parallelogram, diagonals AC and BD bisect each other, ∴ mid 1
point of AC= mid point of BD

𝑥+1 6+2 3+4 5+𝑦
( 2 , 2 )=( 2 , 2 )
Comparing the co-ordinates, we get,
𝑥+1 3+4
= 2 . So, x= 6
2
6+2 5+𝑦
Similarly, 2 = . So, y= 3
2
∴(x, y) = (6,3)

30 (c) ∆ACD ~∆ ABC( AA ) 1
AC AD
∴ = (CPST)
AB AC
8/AB = 3/8
This gives AB = 64/3 cm.
So BD = AB – AD = 64/3 -3 = 55/3cm.
31 (d) Any point (x, y) of perpendicular bisector will be equidistant from A & B. 1
∴ √(𝑥 − 4)2 + (𝑦 − 5)2 = √(𝑥 + 2)2 + (𝑦 − 3)2
Solving we get -12x – 4y + 28=0 or 3x + y – 7=0

32 (b) cot 𝑦 ° AC/ BC 1
= = CD/ BC = CD/ 2CD = ½
cot 𝑥 ° 𝐴𝐶/𝐶𝐷

33 (a) The smallest number by which 1/13 should be multiplied so that its decimal 1
1 13 1
expansion terminates after two decimal points is 13/100 as 13 x 100 = 100 =
0.01
Ans: 13/100

34 (b) 1
∆ABE is a right triangle & FDGB is a
square of side x cm
∆AFD ~∆ DGE( AA )
AF FD
∴ = (CPST)
DG GE
16 − x x
= (CPST)
x 8−x
128 = 24x or x = 16/3cm
35 (a) Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1 ∴ 1
9k −1 8k+3
coordinates of P are ( k+1 , k+1 )
9k −1 8k+3
Since P lies on the line x – y +2=0, then - k+1 +2 =0
k+1
9k -1 -8k-3 +2k+2 =0
which gives k=2/3

36 (c) 1
Shaded area = Area of semicircle +
(Area of half square – Area of two
quadrants)
= Area of semicircle +(Area of half
square – Area of semicircle)
= Area of half square
= ½ x 14 x14 = 98cm²

37 (d) 1
.o

Let O be the
center of the circle. OA = OB = AB =1cm.
So ∆OAB is an equilateral triangle and ∴ ∠AOB =60°
Required Area= 8x Area of one segment with r=1cm, ѳ= 60°
60 √3
= 8x(360 x π x 1²- 4 x 1²)
= 8(π/6 - √3/4)cm²

38 (b) Sum of zeroes = 2 + ½ = -5/p 1
i.e. 5/2 = -5/p . So p= -2
Product of zeroes = 2x ½ = r/p
i.e. r/p = 1 or r = p = -2

39 (c) 2πr =100. So Diameter = 2r =100/π = diagonal of the square. 1
side√2 = diagonal of square = 100/ π
∴ side = 100/√2π = 50√2/π

40 (b) 3x+y = 243 = 35 1
So x+y =5-----------------------------------(1)
243x-y = 3
(35) x-y = 31
So 5x -5y =1--------------------------------(2)
𝑎₁ 𝑏₁
Since : 𝑎₂ ≠ 𝑏₂ , so unique solution

SECTION C
41 (c) Initially, at t=0, Annie’s height is 48ft 1
So, at t =0, h should be equal to 48
h(0) = -16(0)² + 8(0) + k = 48
So k = 48

42 (b) When Annie touches the pool, her height =0 feet 1
i.e. -16t² + 8t + 48 =0 above water level
2t² - t -6 =0
2t² - 4t +3t -6 =0
2t(t-2) +3(t-2) =0
(2t +3) (t-2) =0
i.e. t= 2 or t= -3/2
Since time cannot be negative , so t= 2seconds

43 (d) t= -1 & t=2 are the two zeroes of the polynomial p(t) 1
Then p(t)=k (t- -1)(t-2)
= k(t +1)(t-2)
When t = 0 (initially) h₁ = 48ft
p(0)=k(0²- 0 -2)= 48
i.e. -2k = 48
So the polynomial is -24(t²- t -2) = -24t² + 24t + 48.

44 (c) A polynomial q(t) with sum of zeroes as 1 and the product as -6 is given by 1
q(t) = k(t² - (sum of zeroes)t + product of zeroes)
= k(t² -1t + -6) ………..(1)
When t=0 (initially) q(0)= 48ft

q(0)=k(0²- 1(0) -6)= 48
i.e. -6k = 48 or k= -8
Putting k = -8 in equation (1), reqd. polynomial is -8(t² -1t + -6)
= -8t² + 8t + 48

45 (a) When the zeroes are negative of each other, 1
sum of the zeroes = 0
So, -b/a = 0
(k-3)
- −12 = 0
k−3
+ 12 = 0
k-3 = 0,
i.e. k = 3.

46 (a) Centroid of ΔEHJ with E(2,1), H(-2,4) & J(-2,-2) is 1
2+−2+ −2 1+4+ −2
( , ) = (-2/3, 1)
3 3

47 (c) If P needs to be at equal distance from A(3,6) and G(1,-3), such that A,P and G 1
are collinear, then P will be the mid-point of AG.
3+1 6+ −3
So coordinates of P will be ( 2 , 2 ) = (2, 3/2)

48 (a) Let the point on x axis equidistant from I(-1,1) and E(2,1) be (x,0) 1
then √(𝑥 + 1)2 + (0 − 1)2 = √(𝑥 − 2)2 + (0 − 1)2
x2 + 1 + 2x +1 = x2 + 4 - 4x +1
6x = 3
So x = ½ .
∴ the required point is (½, 0)

49 (b) Let the coordinates of the position of a player Q such that his distance from 1
K(-4,1) is twice his distance from E(2,1) be Q(x, y)
Then KQ : QE = 2: 1
2 X 2+1 X−4 2 X 1+1 X 1
Q(x, y) = ( , )
3 3
= (0,1)

50 (d) Let the point on y axis equidistant from B(4,3) and C(4,-1) be (0,y) 1
then √(4 − 0)2 + (3 − 𝑦)2 = √(4 − 0)2 + (𝑦 + 1)2
16 + y2 + 9 - 6y = 16 + y2 + 1 + 2y
-8y = -8
So y = 1 .
∴ the required point is (0, 1)