CBSE Class 10 Sample Paper 2023 Solutions for Maths Basic

Class- X
Mathematics Basic (241)
Marking Scheme SQP-2022-23
Time Allowed: 3 Hours Maximum Marks: 80

Section A

1 (c) a3b2 1

2 (c) 13 km/hours 1

3 (b) -10 1

4 (b) Parallel. 1

5 (c) k = 4 1

6 (b) 12 1

7 (c) ∠B = ∠D 1

8 (b) 5 : 1 1

9 (a) 25° 1

10 √3 1
(a)
2

11 (c) √3 1

12 (b) 0 1

13 (b) 14 : 11 1

14 (c) 16 : 9 1

15 (d) 147π cm2 1

16 (c) 20 1

17 (b) 8 1

18 3 1
(a)
26

19 (d) Assertion (A) is false but Reason (R) is true. 1

20 (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation 1
of Assertion (A).

Section B
For a pair of linear equations to have infinitely many solutions :
21 a1 b1 c1 k 3 k−3 ½
= = ⇒ = =
a2 b2 c2 12 k k
𝑘 3
= ⇒ k2 = 36 ⇒ k = ± 6 ½
12 𝑘
3 𝑘−3
Also, = ⇒ k2 – 6k = 0 ⇒ k = 0, 6. ½
𝑘 𝑘
Therefore, the value of k, that satisfies both the conditions, is k = 6. ½
(i) In ΔABD and ΔCBE
22
∠ADB = ∠CEB = 90º ½
∠ABD = ∠CBE (Common angle)
⇒ ΔABD ~ ΔCBE (AA criterion) ½

(ii) In ΔPDC and ΔBEC
∠PDC = ∠BEC = 90º ½
∠PCD = ∠BCE (Common angle)
⇒ ΔPDC ~ ΔBEC (AA criterion) ½

In ΔABC, DE || AC
BD/AD = BE/EC .........(i) (Using BPT) ½
In ΔABE, DF || AE
BD/AD = BF/FE ........(ii) (Using BPT) ½
From (i) and (ii)
BD/AD = BE/EC = BF/FE ½
BF BE
Thus, FE = EC ½

Let O be the centre of the concentric circle of radii 5 cm
23 and 3 cm respectively. Let AB be a chord of the larger circle
touching the smaller circle at P
Then AP = PB and OP⊥AB ½
Applying Pythagoras theorem in △OPA, we have
OA2=OP2+AP2 ⇒ 25 = 9 + AP2 ½
⇒ AP2 = 16 ⇒ AP = 4 cm ½
∴ AB = 2AP = 8 cm ½

24 (1 + sinθ)(1 − sinθ) (1 – sin2 θ) ½
Now, = (1
(1 + cosθ)(1 − cosθ) – cos2 θ)

cos2 θ cosθ 2
= =( ) ½
sin2 θ sinθ
½
= cot 2 θ
7 2 49
=( ) = ½
8 64

25 1
Perimeter of quadrant = 2r + 4 × 2 π r ½
1 22
⇒ Perimeter = 2 × 14 + 2 × 7 × 14 ½
⇒ Perimeter = 28 + 22 =28+22 = 50 cm 1
Area of the circle = Area of first circle + Area of second circle
⇒ πR2 = π (r1)2 + π (r1)2 ½
⇒ πR2 = π (24)2 + π (7)2 ⇒ πR2 = 576π +49π ½

⇒ πR2 = 625π ⇒ R2 = 625 ⇒ R = 25 Thus, diameter of the circle = 2R = 50 cm. 1

Section C

26 Let us assume to the contrary, that √5 is rational. Then we can find a and b ( ≠ 0) such
𝑎
that √5 = (assuming that a and b are co-primes).
𝑏 1
So, a = √5 b ⇒ a2 = 5b2
Here 5 is a prime number that divides a2 then 5 divides a also
(Using the theorem, if a is a prime number and if a divides p2, then a divides p, where a is ½
a positive integer)
Thus 5 is a factor of a
Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c
½
We get (5c)2 = 5b2 ⇒ 5c2 = b2
This means 5 divides b2 so 5 divides b also (Using the theorem, if a is a prime number and
if a divides p2, then a divides p, where a is a positive integer).
Hence a and b have at least 5 as a common factor. ½
But this contradicts the fact that a and b are coprime. This is the contradiction to our
assumption that p and q are co-primes.
So, √5 is not a rational number. Therefore, the √5 is irrational. ½
6x2 – 7x – 3 = 0 ⇒ 6x2 – 9x + 2x – 3 = 0
27
⇒ 3x(2x – 3) + 1(2x – 3) = 0 ⇒ (2x – 3)(3x + 1) = 0 ½
⇒ 2x – 3 = 0 & 3x + 1 = 0
x = 3/2 & x = -1/3 Hence, the zeros of the quadratic polynomials are 3/2 and -1/3. ½

For verification
– coefficient of x 1
Sum of zeros = ⇒ 3/2 + (-1/3) = – (-7) / 6 ⇒ 7/6 = 7/6
coefficient of x2
constant
Product of roots = ⇒ 3/2 x (-1/3) = (-3) / 6 ⇒ -1/2 = -1/2 1
coefficient of x2
Therefore, the relationship between zeros and their coefficients is verified.
Let the fixed charge by Rs x and additional charge by Rs y per day
28 Number of days for Latika = 6 = 2 + 4
Hence, Charge x + 4y = 22
x = 22 – 4y ………(1) ½
Number of days for Anand = 4 = 2 + 2
Hence, Charge x + 2y = 16
x = 16 – 2y ……. (2) ½
On comparing equation (1) and (2), we get,

22 – 4y = 16 – 2y ⇒ 2y = 6 ⇒ y = 3 1
Substituting y = 3 in equation (1), we get,
x = 22 – 4 (3) ⇒ x = 22 – 12 ⇒ x = 10
Therefore, fixed charge = Rs 10 and additional charge = Rs 3 per day 1

AB = 100 km. We know that, Distance = Speed × Time.
AP – BP = 100 ⇒ 5x − 5y = 100 ⇒ x−y=20.....(i) ½
AQ + BQ = 100 ⇒ x + y = 100….(ii) ½
Adding equations (i) and (ii), we get,
x − y + x + y = 20 +100 ⇒ 2x = 120 ⇒ x = 60 1

Substituting x = 60 in equation (ii), we get, 60 + y = 100 ⇒ y = 40
1
Therefore, the speed of the first car is 60 km/hr and the speed of the second car
is 40 km/hr.

29 Since OT is perpendicular bisector of PQ.
Therefore, PR=RQ=4 cm ½
Now, OR = √𝐎𝐏 𝟐 − 𝐏𝐑𝟐 = √𝟓𝟐 − 𝟒𝟐 =3cm ½
Now, ∠TPR + ∠RPO = 90° (∵TPO=90°)
& ∠TPR + ∠PTR = 90° (∵TRP=90∘)
So, ∠RPO = ∠PTR ½
So, ⍙TRP ~ ⍙PRO [By A-A Rule of similar triangles] ½
TP RP
So, PO = RG
. ½
TP 4 20
⇒ 5 = 3 ⇒ TP = 3 cm ½
1
30 tan θ cot θ tan θ ½
LHS = + = 1 + tan θ
1−cot θ 1−tan θ 1− 1−tan θ
tan θ

tan2 θ 1
= +
tan θ−1 tan θ (1−tan θ)
½
tan3 θ−1
=
tan θ (tan θ−1)

(tanθ −1) (tan3 θ + tanθ+1 )
=
tan θ (tan θ−1) ½
(tan3 θ + tanθ+1 )
=
tan θ

= tanθ + 1 + sec = 1 + tanθ + secθ
½
sin θ cos θ
= 1+ +
cos θ sin θ
sin2 θ+ cos2 θ ½
=1+
sin θ cos θ

1
= 1 + sin θ cos θ = 1 + sec θ cosec θ
½
sin θ + cos θ = √3 ⇒ (sin θ + cos θ)2 = 3
½
⇒ sin2 θ + cos2 θ + 2sin θ cos θ = 3
⇒ 1 + 2sin θ cos θ = 3 ⇒ 1 sin θ cos θ = 1 ½

sin θ cos θ ½
Now tanθ + cotθ = +
cos θ isn θ
sin2 θ+ cos2 θ ½
=
sin θ cos θ
½
1 1
= = 1 =1
sin θ cos θ ½
31 5 1
(i) P(8 ) =
36
0 1
(ii) P(13 ) = =0
36
(iii) P(less than or equal to 12) = 1 1

Section D
Let the average speed of passenger train = x km/h.
32
and the average speed of express train = (x + 11) km/h
½
As per given data, time taken by the express train to cover 132 km is 1 hour less than the
passenger train to cover the same distance. Therefore,
132 132 1
− =1
𝑥 𝑥+11
132 (𝑥+11−𝑥) 132 𝑥 11
⇒ =1⇒ =1 ½
𝑥 (𝑥+11) 𝑥 (𝑥+11)

⇒ 132 × 11 = x(x + 11) ⇒ x2 + 11x – 1452 = 0
⇒ x2 + 44x -33x -1452 = 0 1

⇒ x (x + 44) -33(x + 44) = 0 ⇒ (x + 44)(x – 33) = 0 1

⇒ x = – 44, 33 ½

As the speed cannot be negative, the speed of the passenger train will be 33 km/h and the ½
speed of the express train will be 33 + 11 = 44 km/h.
Let the speed of the stream be x km/hr ½
So, the speed of the boat in upstream = (18 - x) km/hr
& the speed of the boat in downstream = (18 + x) km/hr ½
distance distance
ATQ, - =1
upstream speed downstream speed
24 24
⇒ − =1 1
18 − 𝑥 18 + 𝑥

1 1 18 + 𝑥−(18−𝑥)
⇒ 24 [ − ] = 1 ⇒ 24 [(18 ]=1
18 − 𝑥 18 + 𝑥 − 𝑥).(18 + 𝑥) 1
2𝑥 2𝑥
⇒ 24 [ ] = 1 ⇒ 24 [(18 ]=1
(18 − 𝑥).(18 + 𝑥) − 𝑥).(18 + 𝑥)
⇒ 48x = 324 - x2 ⇒ x2 + 48x - 324 = 0 1
⇒ (x + 54)(x - 6) = 0 ⇒ x = -54 or 6 ½
As speed to stream can never be negative, the speed of the stream is 6 km/hr. ½
Figure ½
33 Given, To prove, constructions 1½
Proof 2
Application ---- 1
1
34 Volume of one conical depression = 3 x π r 2 h ½
1 22 1½
= 3 x 7 x 0.52 x 1.4 cm3 = 0.366 cm3
Volume of 4 conical depression = 4 x 0.366 cm3
= 1.464 cm3 ½

Volume of cuboidal box = L x B x H ½

= 15 x 10 x 3.5 cm3 = 525 cm3 1½
Remaining volume of box = Volume of cuboidal box –
Volume of 4 conical depressions
½
= 525 cm3 − 1.464 cm3 = 523.5 cm3
1
Let h be height of the cylinder, and r the common radius of
the cylinder and hemisphere.
Then, the total surface area = CSA of cylinder + CSA of ½
hemisphere
= 2𝜋rh + 2𝜋r2 = 2𝜋 r (h + r) 2
22
= 2 x 7 x 30 (145 + 30) cm2 1
22
= 2x x 30 x 175 cm2 ½
7
= 33000 cm2 = 3.3 m2 1

35 Class Interval Number of policy holders (f) Cumulative Frequency (cf)
Below 20 2 2
20-25 4 6
25-30 18 24
30-35 21 45
35-40 33 78
40-45 11 89
45-50 3 92
50-55 6 98
1
55-60 2 100

n = 100 ⇒ n/2 = 50, Therefore, median class = 35 – 40,
Class size, h = 5, Lower limit of median class, l = 35,
frequency f = 33, cumulative frequency cf = 45
n
− cf
⇒Median = l + [ 2
]×h ½
f
50 − 45
⇒Median = 35 + [ ]×5 1½
33 1
25
= 35 + = 35 + 0.76
33
= 35.76 Therefore, median age is 35.76 years 1

Section E
1 Since the production increases uniformly by a fixed number every year, the
36
number of Cars manufactured in 1st, 2nd, 3rd, . . .,years will form an AP.
So, a + 3d = 1800 & a + 7d = 2600 ½
So d = 200 & a = 1200 ½
2 t12 = a + 11d ⇒ t30 = 1200 + 11 x 200 ½
⇒ t12 = 3400 ½
𝑛 10
3 Sn = 2 [2𝑎 + (𝑛 − 1)𝑑] ⇒ S10 = 2 [2 𝑥 1200 + (10 − 1) 200] ½
13
⇒ S10 = 2 [2 𝑥 1200 + 9 x 200] ½
⇒ S10 = 5 x [2400 + 1800 ] ½
⇒ S10 = 5 x 4200= 21000 ½
Let in n years the production will reach to 31200
𝑛 𝑛
Sn = [2𝑎 + (𝑛 − 1)𝑑] = 31200 ⇒ [2 𝑥 1200 + (𝑛 − 1)200] = 31200 ½
2 2
𝑛
⇒ 2 [2 x 1200 + (𝑛 − 1)200] = 31200 ⇒ 𝑛 [ 12 + (𝑛 − 1) ] = 312
½
⇒ n2 + 11n -312 = 0
⇒ n2 + 24n - 13n -312 = 0 ½
⇒ (n +24)(n -13) = 0
⇒ n = 13 or – 24. As n can’t be negative. So n = 13 ½
Case Study – 2
37

N

1 LB = √ (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 ⇒ LB = √ (0 − 5)2 + (7 − 10)2 ½
LB = √ (5)2 + (3)2 ⇒ LB = √25 + 9 LB = √34

Hence the distance is 150 √34 km ½

2 3 x 5 + 2 x 0 3 x 7 + 2 x 10
Coordinate of Kota (K) is ( , ) ½
3+2 3+2
15+0 21+20 41
½
=( , )= (3, 5 )
5 5

3 L(5, 10), N(2,6), P(8,6)
½
LN = √ (2 − 5)2 + (6 − 10)2 = √ (3)2 + (4)2 = √9 + 16 = √25 = 5 ½
NP = √ (8 − 2)2 + (6 − 6)2 = √ (4)2 + (0)2 = 4 ½

PL = √ (8 − 5)2 + (6 − 10)2 = √ (3)2 + (4)2 ⇒ LB = √9 + 16 = √25 = 5
as LN = PL ≠ NP, so ∆ LNP is an isosceles triangle. ½

Let A (0, b ) be a point on the y – axis then AL = AP
⇒ √ (5 − 0)2 + (10 − b)2 = √ (8 − 0)2 + (6 − b)2 ½
⇒ (5)2 + (10 − b)2 = (8)2 + (6 − b)2 ½
25
⇒ 25 + 100 − 20𝑏 + b2 = 64 + 36 − 12𝑏 + b2 ⇒ 8b = 25 ⇒ b = 8 ½
25 ½
So, the coordinate on y axis is (0, 8 )

Case Study – 3
38

1 PC ½
sin 60° =
PA
√3 18
⇒ = ⇒ PA = 12 √3 m ½
2 PA

2 PC ½
sin 30° =
PB
1 18
⇒ = ⇒ PB = 36 m ½
2 PB
PC 18
3 tan 60° = AC ⇒ √3 = AC ⇒ AC = 6 √3 m 1
PC 1 18
tan 30° = ⇒ = ⇒ CB = 18 √3 m
CB √3 CB ½
Width AB = AC + CB = 6 √3 + 18 √3 = 24 √3 m ½
RB = PC =18 m & PR = CB = 18 √3 m ½
QR 1 QR
tan 30° = ⇒ = ⇒ QR = 18 m 1
PR √3 18 √3
QB = QR + RB = 18 + 18 = 36m. Hence height BQ is 36m
½