CBSE Class 10 Question Paper 2023 Solution Maths Basic

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2023
SUBJECT NAME MATHEMATICS (BASIC) (SUBJECT CODE 241) (PAPER CODE 430/1/1)

General Instructions: -
1 You are aware that evaluation is the most important process in the actual and correct
assessment of the candidates. A small mistake in evaluation may lead to serious problems
which may affect the future of the candidates, education system and teaching profession.
To avoid mistakes, it is requested that before starting evaluation, you must read and
understand the spot evaluation guidelines carefully.
2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the
examinations conducted, Evaluation done and several other aspects. Its’ leakage to
public in any manner could lead to derailment of the examination system and affect
the life and future of millions of candidates. Sharing this policy/document to
anyone, publishing in any magazine and printing in News Paper/Website etc may
invite action under various rules of the Board and IPC.”
3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not
be done according to one’s own interpretation or any other consideration. Marking
Scheme should be strictly adhered to and religiously followed. However, while
evaluating, answers which are based on latest information or knowledge and/or are
innovative, they may be assessed for their correctness otherwise and due marks be
awarded to them. In class-X, while evaluating two competency-based questions,
please try to understand given answer and even if reply is not from marking scheme
but correct competency is enumerated by the candidate, due marks should be
awarded.
4 The Marking scheme carries only suggested value points for the answers.
These are in the nature of Guidelines only and do not constitute the complete answer. The
students can have their own expression and if the expression is correct, the due marks
should be awarded accordingly.
5 The Head-Examiner must go through the first five answer books evaluated by each
evaluator on the first day, to ensure that evaluation has been carried out as per the
instructions given in the Marking Scheme. If there is any variation, the same should be
zero after deliberation and discussion. The remaining answer books meant for evaluation
shall be given only after ensuring that there is no significant variation in the marking of
individual evaluators.
6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be
marked. Evaluators will not put right (✓)while evaluating which gives an impression that
answer is correct and no marks are awarded. This is the most common mistake which
evaluators are committing.
7 If a question has parts, please award marks on the right-hand side for each part. Marks
awarded for different parts of the question should then be totaled up and written in the left-
hand margin and encircled. This may be followed strictly.
8 If a question does not have any parts, marks must be awarded in the left-hand margin and
encircled. This may also be followed strictly.
9 If a student has attempted an extra question, answer of the question deserving more
marks should be retained and the other answer scored out with a note “Extra Question”.
However, for MCQs (Q1 to Q20), only first attempt to be evaluated.
10 No marks to be deducted for the cumulative effect of an error. It should be penalized only

430/1/1 1 P.T.O.

once.
11 A full scale of marks __________(example 0 to 80/70/60/50/40/30 marks as given in
Question Paper) has to be used. Please do not hesitate to award full marks if the answer
deserves it.
12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours
every day and evaluate 20 answer books per day in main subjects and 25 answer books
per day in other subjects (Details are given in Spot Guidelines).
13 Ensure that you do not make the following common types of errors committed by the
Examiner in the past:-
● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right tick mark
is correctly and clearly indicated. It should merely be a line. Same is with the X for
incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it should
be marked as cross (X) and awarded zero (0)Marks.
15 Any un assessed portion, non-carrying over of marks to the title page, or totaling error
detected by the candidate shall damage the prestige of all the personnel engaged in the
evaluation work as also of the Board. Hence, in order to uphold the prestige of all
concerned, it is again reiterated that the instructions be followed meticulously and
judiciously.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines
for spot Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to
the title page, correctly totaled and written in figures and words.
18 The candidates are entitled to obtain photocopy of the Answer Book on request on
payment of the prescribed processing fee. All Examiners/Additional Head Examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out
strictly as per value points for each answer as given in the Marking Scheme.

430/1/1 2

430/1/1
MARKING SCHEME
MATHEMATICS (BASIC)
SECTION A

Ans. (c) 25 32 1

Ans. (b) 60 1

1
Ans. (d) 1
2

Ans. (b) 49 1

Ans. (d) 3 2 units 1
430/1/1 3 P.T.O.

Ans. (b) 4 1

Ans. (a) 0 1

Ans. (a) 4 : 7 1

430/1/1 4

6
Ans. (c) cm 1
5

Ans. (b) – 5, 6 1

Ans. (b) 1 1

Ans. (d) 30 1

Ans. (d) r3 1

430/1/1 5 P.T.O.

Ans. (a) 7 cm 1

Ans. (c) SAS (Side – Angle – Side) Similarity 1

Ans. (b) 99 1

Ans. (a) 30 cm 1

430/1/1 6

Ans. (c) 24 cm 1

Ans. (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the
correct explanation of Assertion (A) 1

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

Solution:
430/1/1 7 P.T.O.

Let P(x, y) divide AB internally in the ratio 2 : 3
2  – 3  3 7 15
x= = =3 1
23 5

24 3– 1 5
y= = =1 1
23 5

Coordinates of the required point P are (3, 1)

OR

Solution: AB = 10 units  AB2 = 100

 (11 – 3)2 + (y + 1)2 = 100 1
y+1=6
1 1
 y = 5, – 7 +
2 2

Solution: tan2 60 – 2 cosec2 30 – 2 tan2 30

 3
2
2 2  1  1
= – 2(2) – 2   1
 3 2
17 1
=–
3 2

1
Solution: 92 = 2 2 23
2
1
510 = 2 3 5 17
2
1
HCF = 2
2
430/1/1 8

1
LCM = 2 2 3 5 17 23 = 23460
2

Solution: On solving the given equations and getting

22 8
x= and y = 1+1
5 5

OR

5 3 11 1 1 1 1
Solution: – = –  or – = –  1
10 6 22 2 2 2 2
1
 given pair of linear equations is inconsistent
2

Solution: In  ABC and  AMP,
1
 ABC =  AMP (90 each)
2
 BAC =  MAP (common) 1
By AA Similarity
1
 ABC   AMP
2
SECTION C

430/1/1 9 P.T.O.

Solution: LHS = sec  (1 – sin ) (sec  + tan )
1  1 sin  
= (1 – sin )    1
cos   cos  cos  
1  1  sin  
= (1 – sin )   1
cos   cos  
1 – sin 2  cos2  1 1
= = = 1 = RHS +
2 2
cos  cos  2 2

OR

1
1
1  sec  cos  = 1 + cos 
Solution: LHS = = 1
sec  1
cos 
(1 – cos ) (1  cos )
= 1
(1 – cos )
1 – cos2  sin 2 
= = = RHS 1
1 – cos  1 – cos 

Solution: AB = (4 – 1) 2  (2 – 7) 2 = 34
BC = (4  1)2  (2  1)2 = 34 2
CD = (– 4  1)2  (4  1)2 = 34
DA = (– 4 – 1)2  (4 – 7)2 = 34
 AB = BC = CD = DA
AC = (1  1)2  (7  1)2 = 68 1
BD = (4  4)2  (2 – 4)2 = 68
 AC = BD
Hence, ABCD is a square.

430/1/1 10

Solution:
1
(For Fig.)

Given: A circle with centre O and PQ, PR are
tangents to the circle from an external point P.
To Prove: PQ = PR
1
Construction: Join OP, OQ, OR
2
Proof : In  OPQ and  OPR
OP = OP (common)
OQ = OR (radii of the same circle)
 OQP =  ORP (each 90)
  POQ   POR (RHS congruence) 1
1
 PQ = PR
2

Solution: p(x) = x2 + 3x + 2
,  are its zeroes
1 1
 +  = – 3,  = 2 +
2 2
Now,
1
( + 1) + ( + 1) =  +  + 2 = – 3 + 2 = – 1
2
( + 1) ( + 1) =  + ( + ) + 1 = + 2 – 3 + 1 = 0 1
1
Required Polynomial is k(x2 + x) or x2 + x
2

Solution: Let us assume that 3+ 7 2 is a rational number.
p
3+ 7 2 = , p, q are integers and q  0 1
q
p  3q
 2 = 1
7q
RHS is rational but LHS is irrational
Our assumption is wrong 1
Hence, 3 + 7 2 is an irrational number
430/1/1 11 P.T.O.

Solution: In  ABE, DF  AE (given), hence by BPT
BD BF 1
= _________ (i) 1
DA FE 2
In  ABC, DE  AC (given), hence by BPT
BD BE
= _________ (ii) 1
DA EC
From (i) and (ii)
BF BE 1
=
FE EC 2

OR

Solution: In  AOB and  COD,
AO CO OA OB
=  =
BO OD OC OD
 AOB =  COD (vertically opp. angles)
  AOB   COD (SAS Similarity) 2
  CAB= ACD (or  DBA=  BDC)
But, these are alternate interior angles
 AB  CD  ABCD is a trapezium 1
430/1/1 12

SECTION D

Note: There is an error in the question, so full marks to be awarded to the
Candidate, who attempted.
OR

Solution: Let age of father = x years
and age of son = (45 – x) years 1
Five years ago, age of father = (x – 5) years
Age of son = (40 – x) years 1
A. T. Q., (x – 5) (40 – x) = 124 1
x2 – 45x + 324 = 0
(x – 36) (x – 9) = 0 1
1
x = 36, x = 9 (rejected)
2
1
 Father’s age = 36 years and son’s age = 9 years
2

Solution: Radius of hemispherical bowl = radius of cylinder = 7 cm
Height of cylinder = 13 – 7 = 6 cm 1
Inner surface area of the vessel = 2rh + 2r2
22
= 2r(h + r) = 2 7(6 + 7)
7
= 44 13 = 572 cm2 2
2
Volume of the vessel = r2h + r3
3
2
= r2(h + r)
3
22 14
= 7 7 (6 + )
7 3
4928
= cm3 or 1642.67 cm3 2
3
430/1/1 13 P.T.O.

Solution:
Daily Exp. (₹) No. of household xi fixi
(fi)
100 – 150 4 125 500
150 – 200 5 175 875
200 – 250 12 225 2700
250 – 300 2 275 550
300 – 350 2 325 650
25 5275
–  f i xi 5275 1
Mean x = = = 211 1
 fi 25 2
1
Mode: Modal Class = 200 – 250
2
l = 200, f1 = 12, f0 = 5, f2 = 2, h = 50
 f1 – f0 
Mode = l +   h
 1
2f – f 0 – f 2
 12 – 5 
= 200 +   50 1
 24 – 5 – 2 
3750 1
= or 220·59
17 2

Solution:

430/1/1 14

h 1
In  ABC, tan 60 = h= 3x 1+
x 2
h 1 h 1
In  ABD, tan 30 =  = 1+
20  x 3 20  x 2
3 h = 20 + x
 
3 3 x = 20 + x
1
 x = 10
2
h= 3 x = 10 3
1
 Height of tower = 10 3 m or 17·3 m
2
OR

Solution:

h 1
In  APC, = tan 45  h = x 1+
x 2
4000 4000 1
In  APC, = tan 60  x = 1+
x 3 2
4000
 h=x=
3
4000 1
Distance between the aeroplanes = 4000 –
3 2
 1 
= 40001  
 3
5080 1
= m or 1693.33 m (approx.)
3 2
(Note: ½ mark to be deducted for not using 3 =1.73)

430/1/1 15 P.T.O.

SECTION E

Solution: a = 2, d = 3
(i) Number of pots in the 10th row
= a10 = a + 9d = 29 1
(ii) a5 – a2 = (a + 4d) – (a + d) = 3d = 9 1
n
(iii) Sn = 100  [2(2) + (n – 1)3] = 100 1
2
2
3n + n – 200 = 0  (3n + 25) (n – 8) = 0
25
 n = 8 (n = – rejected), 1
3
OR
12
(iii) S12 = [2(2) + 11(3)] 1
2
= 222 1

430/1/1 16

Solution: (i) Area of square ABCD = (40)2 = 1600 cm2 1
22
(ii) Area of circle = r2 = 10 10
7
2200
= cm2 or 314.28 cm2 1
7
1 2200
(iii) Area of 4 quadrants = 4( r2) = cm2 1
4 7
 2200 2200 
Remaining area = 1600 –   
 7 7 

430/1/1 17 P.T.O.

4400 6800
= 1600 – = cm2 or 971.43 cm2 1
7 7
OR
1 2 2200
(iii) Area of 4 quadrants = 4( r ) = cm2 1
4 7
Combined area of circle + 4 quadrants
2200 2200 4400
= + = cm2 or 628.57 cm2 1
7 7 7

430/1/1 18

21
Solution: (i) P(type O) = 1
50
1
(ii) No. of people with AB type blood group = 50 – (21 + 22 + 5) = 2
2
2 1 1
P(type AB) = or
50 25 2
21  2 23
(iii) P(neither type A nor type B) = = 1+1
50 50
OR
21  22  5 24
(iii) P(type A or type B or type O) = = 1+1
50 25

430/1/1 19 P.T.O.