NAVODAYA VIDYALAYA SAMITI

MARKING SCHEME
CLASS X (2024-24)
MATHEMATICS BASIC (Code No.241)

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Cc c

Let A(-4, 0), B(4, 0), C(0, 3) are the given vertices.
1/2
Now, distance between A(-4, 0) and B(4, 0),
AB = √(x2-x1)2+(У2-y1)2

AB = √[4-(-4)]2+ (0-0)2
= √(4+4)2
1/2
= √64
= 8units
Distance between B(4, 0) and C(0, 3),
BC= √(0-4)2+(3-0)2
= √16+9 1/2
= √25
= 5units
Distance between A(-4, 0) and C(0, 3),
AC = = √[0-(-4)]²+(3-0)2
= √16+9
= √25
= 5units
** BC = AC
Hence, ∆ABC is an isosceles triangle because an isosceles triangle has two. 1/2
sides equal

OR

Since they are equidistant, PA = PB 1/2

Hence by applying the distance formula for PA = PB, we get
√(x - 7)² + (y - 1)² = √(x - 3)² + (y - 5)²
By squaring, we get 1/2
PA2 = PB2
(x - 7)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
x2 + 49 - 14x + y2 + 1 - 2y = x2 + 9 - 6x + y2 + 25 - 10y
-8x + 8y = -16 1/2
x-y=2
Thus, the relation between x and y is given by x - y - 2 = 0 1/2

1/2

Given:
TP and TQ are two tangents drawn from an external point T to the circle C
(O, r). 1/2
To prove: TP = TQ
Construction: Join OT.
Proof:
We know that a tangent to the circle is perpendicular to the radius through 1/2
the point of contact.
::. ∠OPT = ∠OQT = 90°
In ΔOPT and ΔOQT,
OT = OT...(Common)
OP = OQ ...(Radius of the circle)
ΔOPT = ΔOQT ...(90°)
.. ΔΟΡΤ = ΔΟQT ...(RHS congruence criterion)
⇒ TP = TQ ...(CPCT)
Hence, the lengths of the tangents drawn from an external point to a circle 1/2
are equal.
23
In the given AP, first term, a = 3 and common difference, d = (8-3 )= 5. 1/2
Let's its nth term be 83
Then, an = 83
⇒ a + (n-1)d = 83
⇒3+ (n-1)× 5 = 83
1
5n-2=83
⇒ 5n = 85
⇒ n = 17 1/2
Hence, the 17th term of the given AP is 83.

OR

Given arithmetic progression is 11,8,5,2.

First term a = 11 and common difference d = -3
If -150 is nth term, then we have
an = a + (n-1)d 1/2
-150 = 11+ (n-1)(-3)
We do not get n as integer by solving above expression. 1
Hence -150 is not a term in the given arithmetic progression series.

1/2
To evaluate (4/3) tan²60° + 3 cos²30° - 2sec²30° - (3/4) cot²60°,
We will first substitute the known values of trignomatic functions.

tan60° = √3, cos30° = √3/2, 1/2

sec30° = 2/√3, cot60° = 1/√3,

1/2
substituting these values in the above expression,
4/3(√3)2 + 3(√3/2)2 – 2(2/√3)2 - 3/4(1/√3)2
4/3*3 + 3*3/4 – 2*4/3 – 3/4*1/3
4 + 9/4 - 8/3 – ¼ 1/2
Common denominator = 12
(40+27-32-3)/12
40/12 1/2

10/3.
Using direct method to calculate mean :

No. of plants No. of houses xi fi.xi

0-2 1 1 1

2-4 2 3 6

4-6 1 5 5
1
6-8 5 7 35

8-10 6 9 54

10-12 2 11 22

12-14 3 13 39

Σfi=20 Σfi.xi=162

Mean=Σfi.xi/Σfi
=162/20
=8.1plants

Let us assume that 3 + √5 is a rational number.

⇒ 3 + √5 = p/q, where p and q are the integers and q ≠0. 1
⇒ - √5 = 3 - p/q
Since p, q and 3 are integers. So,
3 – p/q is a rational number. 1
But √5 is an irrational number.
This contradiction has arisen due to the wrong assumption that
3+√5 is a rational number.
Hence, 3 + √5 is an irrational number. 1
27

1/2

1/2

OR,
Consider the points be A(−3,10),B(6,−8),C(−1,6)
And we need to find the ratio between AC and CB
So, let the ratio be k:1
1/2
Therefore,
m1=k,m2=1
x1=−3,y1=10
1/2
x2=6,y2=−8
And x=−1,y=6
1

LHS 1/2
1/2
1/2

1/2

Hence, LHS = RHS. 1

 1

1/2

1/2

1
 Since ABCD is a parallelogram circumscribed in a circle.
30

AB=CD........(1)
BC=AD........(2) 1
DR=DS (Tangents on the circle from same point D)
CR=CQ(Tangent on the circle from same point C)
BP=BQ (Tangent on the circle from same point B )
AP=AS (Tangents on the circle from same point A)
Adding all these equations we get
DR+CR+BP+AP=DS+CQ+BQ+AS
(DR+CR)+(BP+AP)=(CQ+BQ)+(DS+AS)
CD+AB=AD+BC
Putting the value of equation 1 and 2 in the above equation we get 1
2AB=2BC
AB=BC...........(3)
From equation (1), (2) and (3) we get
AB=BC=CD=DA
∴ABCD is a Rhombus

OR

Consider the problem

Let us join point O to C
1
In ΔOPA and ΔOCA
OP=OC (Radii of the same circle)
AP=AC (Tangent from point A)
AO=AO (Common side)
ΔOPA≅ΔOCA (SSS congruence criterion)
Therefore, P↔C,A↔A,O↔O 1
∠POA=∠COA.........(1)
Similarly,
ΔQOB≅ΔOCB
∠QOB=∠COB.........(2)
Since, POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA+∠COA+∠COB+∠QOB=1800
So, from equation (1) and equation (2) 1
2∠COA+2∠COB=1800
∠COA+∠COB=900
∠AOB=900
(a) y = 2x - 2.
put x = 1 in equation.
⇒ y = 2(1) - 2.
⇒ y = 0.
Co-ordinate = (1,0).
put x = 0 in equation.
⇒ y = 2(0) - 2. 1
⇒ y = -2.
Co-ordinate = (0,-2).
(b) y = 4x - 4.
put x = 1 in equation.
⇒ y = 4(1) - 4. 1
⇒ y = o.
Co-ordinate = (1,0).
put x = 0 in equation.
⇒ y = 4(0) - 4.
⇒ y = -4.
Co-ordinate = (0,-4).
Red line represent the equation y = 2x - 2.
Blue line represent the equation y = 4x - 4.

Section - D
32 Let x be the base of the triangle, then the altitude will be (x−7).
By Pythagoras theorem, 1
X2+(x−7)2=(13)2
2x2−14x+49−169=0 1
2x2−14x−120=0
X2−7x−60=0
X2−12x+5x−60=0
(x−12)(x+5)=0
1
x=12,x=−5
Since the side of the triangle cannot be negative, so the base of the triangle
is 12cm and the altitude of the triangle will be 12−7=5cm. 2

OR

Let the length be l m and the breadth be b m.

Then the area would be lb = 400
Perimeter would be 2 l+b) = 80 1

lb =400
1
⇒2(l+b)=80
⇒l+b=40 1
∴b=40−l --(1)
Substituting (1) in Area, we get
⇒ l(40−l)=400
⇒40l−l2=400
⇒l2−40l+400=0
⇒(l−20)(l−20)=0 1
∴l=20
has equal roots, so it is possible to design the rectangle of given parameters.
⇒b=40−20=20
We now know that the length of the park is 20 m and the breadth of the park is
1
20m.
33 Basic Proportionality Theorem states that, if a line is parallel to a side of a
1
triangle which intersects the other sides into two distinct points then the line
divides those sides of the triangle in proportion.

Let ABC be the triangle.

The line l parallel to BC intersect AB at D and AC at E.
1

1
34

Area of major sector = Area of circle – area of minor sector

Area of circle = πr2 = (22/7)*(21)2 1

= 1386 cm2

∴ Area of major sector = (1386 – 231)cm2

1
= 1155 cm2
35
2

In △ABD, 1
tan300 =BD/AD
1/√3=75/AD
AD=75√3m
In △BCD 1
tan600 =BD/CD
√3=75/CD
CD=75/√3 m 1
AC=AD+CD=(75√3+75/√3)m
=300/1.73=173.41 m

OR

1
In ΔABE,
tan600 =h/AB
h=AB√3.... (i)
In ΔABC, 1
tan450 =7/AB
AB=7 ...(ii)
By equation (i) and equation (ii)
h=7√3 1
So, height of tower is =h+7=7(√3+1).
36
(i) Section - E
½
There are total of 12 face cards in a deck of 52 cards.
Therefore the probability of getting a face card is,
P(getting a face card) = number of face cards/total no of cards ½
= 12/52
= 3/13
(ii)
1
There are a total of 02 cards of red king in a deck of 52 cards
Therefore probability of getting a red king card is,
P(getting a red king card) = number of red king cards/total no of 1
cards
= 2/52
= 1/26
OR
1
There are total of 06 red face cards in a deck of 52 cards
Therefore the probability of getting a red face card is,
P(getting a red face card) = no of red face cards/total no of cards
1
= 6/52
= 3/26

(iii) 1/2
There is only one jack of hearts in a deck of 52 cards
Therefore probability of getting jack of hearts is 1/2
P(getting jack of hearts) = number of jack of hearts/total no of cards
= 1/52.
37
(i) Degree means the highest power of exponent in a polynomial.
1
In, ℎ (t)= 25t−5t²
Degree = 2.

(ii) Let the roots of polynomial be α and β.

-5t2 + 25t = 0
α + β = -b/a
½
= -25/(-5)
=5
α.β = c/a
=0
½
∴ α = 0, β = 5
Hence, 0 & 5 are the zeroes of the given polynomial.

(iii) value of given polynomial 25t-5t² at t=3.

Putting t=3, 1
h(3) = 25(3) – 5*(3)2
= 75-45
= 30.
∴ h(3) = 30. 1

OR

h(0) = 25(0) – 5*(0)2

= 0-0 1
=0

h(4) = 25(4) – 5*(4)2

= 100 – 80 1
= 20
38
(i) Height of hemispherical dome = Radius of hemispherical dome = 21m.

Volume of dome = 2/3 πr3 1

= (2/3)*(22/7)*(213)

= 19,404 m3

(ii) Formula for calculating the Volume of Sphere is

1
4/3 πr3

Cloth required to cover the hemispherical dome = surface area of

(iii)
hemispherical dome, 1
= 2 πr2

Given, radius of its base is 14m, ∴ = 2*(22/7)*(14)2

= 1232 m2
Cloth required to cover the hemispherical dome = 1232 m2 1

OR

volume of cuboid with dimensions 8m*6m*4m is: 1

volume of cuboid = l*b*h
= 8*6*4
= 192 m3 1