CBSE Class 10 Question Paper 2024 Solution Maths Standard

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Secondary School Examination, 2024
MATHEMATICS PAPER CODE 30/4/1
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. It’s 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 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 totalled up and written on the left-
hand margin and encircled. This may be followed strictly.
8 If a question does not have any parts, marks must be awarded on the left-hand margin and
encircled. This may also be followed strictly.

1

9 In Q1-Q20, if a candidate attempts the question more than once (without cancelling the previous
attempt), marks shall be awarded for the first attempt only and the other answer scored out
with a note “Extra Question”.
10 In Q21-Q38, 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”.
11 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
12 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.
13 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). This is in view of the reduced
syllabus and number of questions in question paper.
14 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 totalling of marks awarded to an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totalling on the title page.
● Wrong totalling 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.
15 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.
16 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.
17 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for
spot Evaluation” before starting the actual evaluation.
18 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the
title page, correctly totalled and written in figures and words.
19 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.

2

MARKING SCHEME
MATHEMATICS (Subject Code–041)
(PAPER CODE: 30/4/1)

Q. No. EXPECTED OUTCOMES/VALUE POINTS Marks
SECTION A
This section comprises Multiple Choice Questions (MCQs) of 1 mark each

1.

Sol. (C) a – b 1
2.

Sol. (B) 13 1
3.

Sol. (B) 4 1
4.

Sol. (C) -1 1
5.

Sol. (B) 1:4 1

3

6.

Sol. (D) 150° 1
7.

Sol. (A) 12 1
8.

Sol. 9 1
(D) 36
9.

Sol. (C) 3π:1 1
10.

Sol. 1
(A) 2√2 1
11.

Sol. (D) – 15x + 9y = 5 1

4

12.

Sol. (B) 10 1
13.

Sol. (B) 7.5 cm 1
14.

Sol. (B) 45° 1
15.

Sol. (A) 96° 1

5

16.

Sol. (B) (0,-1) 1
17.

Sol. (A) -3 1
18.

Sol. (C) 8.4 cm 1

19.

Sol. (C) Assertion (A) is true but Reason (R) is false 1

6

20.

Sol. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct 1
explanation of the Assertion (A).
SECTION B
This section comprises Very Short Answer (VSA) type questions of 2 marks
each.
21.

Sol. 9 = 32
12 = 22 × 3 1
15 = 3 × 5
L.C.M = 22 × 32 × 5 = 180 1
Three bells will toll together after 180 min.
22. (a)

Sol. Angle subtended in 5 min. = 30° ½
30 22
Area described by minute hand = 360 × 7 × 14 × 14 1
154 ½
= 3 cm2 or 51.33 cm2 approx.

OR
22. (b)

Sol. 22 60
Length of arc = 2 × 7 × 42 × 360 1½

= 44 cm ½

7

23. (a)

Sol. 1 2 2 2
5( ) +4( ) −(1)2
2 √3
2 1½
1 2 √3
( ) +( )
2 2

67
= 12 ½
OR
23.(b)

Sol. sin (A – B) = sin 30°
A – B = 30° --------(i) ½
cos (A+B) = cos 60°
A + B = 60° ---------(ii) ½
Solving (i) and (ii)
A = 45° , B = 15° 1
24.

Sol.

Take a point P on circumference and join AP & BP. ½

8

1
∠APB = × 145° = 72.5° ½
2

∠APB + ∠ACB = 180° ½

 ∠ACB = 107.5° or x = 107.5° ½

25.

Sol. ∵ ∆AHK ∼∆ABC (given)
𝐻𝐾 𝐴𝐾
 𝐵𝐶 = 𝐴𝐶
6.4 8.0
 3.2 = 𝐴𝐶 1

 AC = 4 cm 1
SECTION C
This section comprises Short Answer (SA) type questions of 3 marks each.
26.

Sol. sin θ−cos θ+1
L.H.S = sin θ+cos θ−1

Divide Numerator and Denominator by cos θ.
tan θ−1+sec θ 1
= tan θ+1−sec θ
tan θ−1+sec θ
= (tan θ−sec θ)+(sec2θ−tan2θ) 1
tan θ−1+sec θ
= (sec θ−tan θ) (tan θ+sec θ−1) ½
1
= sec θ−tan θ = R.H.S ½

9

27.(a)

Sol. Total number of outcomes = 8
7 1
(i) P (at least one head) = 8
3
(ii) P (exactly 2 tails) = 8 1
4 1
(iii) P (at most one tail) = 8 or 2 1
OR
27.(b)

Sol. Total outcomes = 90
30 1 1
(i) P (2 digit number less than 40) = 90 or 3
8 4
(ii) P (a number divisible by 5 and greater than 50) = 90 or 45 1
9 1
(iii) P (a perfect square number) = 90 or 10 1

28

Sol. Let number of ₹50 notes = x
and number of ₹100 notes = y
1
Here x + y = 25 ------(i)
50x + 100y = 2000 or x + 2y = 40 -------(ii) 1

10

Solving eq.(i) and eq.(ii), we get
x = 10 and y = 15 ½+½
Therefore 10 notes of ₹50 and 15 notes of ₹100 are received.
29.(a)

Sol. P(x) = 4𝑥 2 + 4𝑥 − 3
= (2x + 3) (2x – 1) 1
−3 1
 Zeroes of the polynomial are 2 , 2 1
−3 1 −3+1 −4 −(coefficient of x)
Sum of Zeroes = 2 + 2 = = -1= 4 = (coefficient of x2) ½
2
−3 1 −3 constant term
Product of Zeroes = ×2 = = coefficient of x2 ½
2 4

OR
29(b).

Sol. Here α + β = – 1 and αβ = – 2 1
𝛼 β α2 + β2 (𝛼+𝛽)2 −2αβ
+α= = 1
𝛽 αβ αβ

(−1)2 −2(−2) 5 1
= = −2
−2

30.

Sol. Assuming
2−√3
to be a rational number.
5
2−√3 p
 = q ,where p and q are integers & q ≠0 ½
5
2q−5p
 √3 = q
1
Here RHS is rational but LHS is irrational. ½
Therefore our assumption is wrong. ½

11

2−√3
Hence is an irrational number. ½
5

31.

Sol. Correct figure ½

AP = AS -----(i)
BP = BQ -----(ii)
1
CR = CQ -----(iii)
DR = DS -----(iv)
Adding (i), (ii), (iii) & (iv)
AP + BP + CR + DR = AS + BQ + CQ + DS
1
 AB + CD = AD + BC
But ABCD is a parallelogram AB = CD and AD = BC
 2AB = 2AD or AB = AD ½

Hence, ABCD is a rhombus.

12

SECTION D
This section comprises Long Answer (LA) type questions of 5 marks each.
32.

Sol. Correct figure 1

Let AB and CD are two pillars of equal length h m and let P be the point on
road x m away from pillar CD.
In ∆CDP
h 1
tan 60° = √3 = x

⟹ h = √3 x ------(i) ½
In ∆ABP,
1 h 1
tan 30° = = 100−x
√3
100−𝑥
⟹h= -------(ii) ½
√3

Solving eq.(i) and eq.(ii)
x = 25 ½

13

and h = 25√3 = 25 × 1.732 = 43.3 ½
The length of each pillar is 43.3 m and the distance of the point on the road
from pillars is 75 m and 25 m respectively.
33.(a)

Sol. Correct figure 2

In ∆ABE and ∆CFB
∠EAB = ∠BCF 1
1
∠AEB = ∠CBF
1
 ∆ABE ∼ ∆CFB
OR
33.(b)

Sol. Correct figure 1

𝐴𝐵 𝐵𝐶 𝐴𝐷
∵ 𝑃𝑄 = 𝑄𝑅 = 𝑃𝑀
𝐴𝐵 2𝐵𝐷 𝐴𝐷 1
 𝑃𝑄 = 2𝑄𝑀 = 𝑃𝑀

14

𝐴𝐵 𝐵𝐷 𝐴𝐷
 𝑃𝑄 = 𝑄𝑀 = 𝑃𝑀 -------(i)

 ∆ABD ∼ ∆PQM 1
 ∠B = ∠Q -----(ii) 1
In ∆ABC and ∆PQR
𝐴𝐵 𝐵𝐶
= 𝑄𝑅
𝑃𝑄

∠B = ∠Q
 ∆ABC ∼ ∆PQR 1
34(a).

Sol. Let the original speed be x km/h
New speed = (x + 15) km/h ½
A.T.Q.
90 90 1 1½
- =
𝑥 𝑥+15 2

 𝑥 2 + 15𝑥 − 2700 = 0 1

(x + 60) (x - 45) = 0 1

x ≠ -60 , x = 45 1
The original speed of the train = 45km/h
OR
34(b).

Sol. For real and equal roots,
{−6(𝑐 + 1)}2 − 4(𝑐 + 1) × 3(𝑐 + 9) = 0 2
 12(c + 1) (2c – 6) = 0 2
c ≠ –1 So, c = 3 1

15

35.

Sol.
Age (in years) No. of patients (𝒇𝒊 ) Mid point (𝒙𝒊 ) 𝒙𝒊 𝒇𝒊
5 – 15 6 10 60
15 - 25 11 20 220
25 - 35 21 30 630
35 - 45 23 40 920
45 - 55 14 50 700
55 - 65 5 60 300
Total 80 2830
Correct table 1½
2830
 Mean = 80 ½
½
= 35.375
½
Modal class = (35 - 45)
23−21
 Mode = 35 + (2×23−21−14)×h 1

= 36.81 1
Therefore, mode and mean of given data are 36.81 years and 35.375 years
respectively.
SECTION E
This section comprises 3 case-study based questions of 4 marks each.
36. Case Study - 1

16

Sol. −3+1 0+4 1
(i) Mid point of FG is ( , 2 ) = (−1,2)
2

(ii) (a) AC = √(−1 − 3)2 + (−2 − 4)2 1
1
= √52 or 2√13
OR
1×3+3×3 1×2+3×4
(ii) (b) The coordinates of required point are ( , ) 1
1+3 1+3
7
i.e. (3, 2) 1

(iii) D(-2, -5) 1

17

37.

Sol. (i) Number on the first spot = 20 + 4 × 1 = 24 1
(ii) (a) 20 + 4n = 112 1
⟹ n = 23 1
OR
(ii) (b) d = 4 ½
10
𝑆10 = 2 [2 × 24 + 9 × 4] 1

= 420 ½

(iii) Number on the (n – 2)th spot = 20 + 4(n - 2)
= 12 + 4n 1

18

38.

Sol. (i) Volume of cuboidal carton = 30×32×15 ½
= 14400 cm3 ½
(ii)(a) Total surface area of milk packet = 2(15×8+8×5+5×15) 1
= 470 cm2 1
OR
30×32×15
(ii) (b) Number of milk packets in carton = 15×8×5 1

= 24 1
22
(iii) Capacity of the cup = 7 × 5 × 5 × 7 ½

= 550 𝑐𝑚3 or 550 ml ½

19