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Page 1
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS BASIC (Code No.241)
TIME: 3 hours MAX.MARKS: 80
Q. No. Section A Marks
1. B) 90 1
2. A) consistent with unique solution 1
3. D) 7 1
4.
C) 2 v?? 2
+ ?? 2
1
5. D) 145
°
1
6. B) 15 cm 1
7.
A)
5
4
1
8.
B) ?? ??????
1
9.
C) 3780
1
10.
B) 40
1
11.
D) 52
°
1
12.
B) 5 cm
1
13.
A) cos 60
°
1
14.
(C) 3?? ?? 2
1
15.
D) 4
1
16.
B) real and equal
1
17.
C) 30 - 40
1
18.
D) 25?? 2
- 5?? - 2
1
19. A) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
20. C) Assertion (A) is true but reason (R) is false.
1
Section B
Page 2
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS BASIC (Code No.241)
TIME: 3 hours MAX.MARKS: 80
Q. No. Section A Marks
1. B) 90 1
2. A) consistent with unique solution 1
3. D) 7 1
4.
C) 2 v?? 2
+ ?? 2
1
5. D) 145
°
1
6. B) 15 cm 1
7.
A)
5
4
1
8.
B) ?? ??????
1
9.
C) 3780
1
10.
B) 40
1
11.
D) 52
°
1
12.
B) 5 cm
1
13.
A) cos 60
°
1
14.
(C) 3?? ?? 2
1
15.
D) 4
1
16.
B) real and equal
1
17.
C) 30 - 40
1
18.
D) 25?? 2
- 5?? - 2
1
19. A) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
20. C) Assertion (A) is true but reason (R) is false.
1
Section B
2
21 (A). ???? 2
= ???? 2
? (?? - 4)
2
+ (?? - 3)
2
= (?? - 3)
2
+ (?? - 4)
2
? ?? = ?? ???? ?? - ?? = 0
1
1
OR
21 (B). AB = 6 cm = AC
OC = v36 - 9 = 3v3 cm
Point C is (3v3, 0)
½
1
½
22. Correct figure
AM = 4 cm
OM = v????
2
- ????
2
= v5
2
- 4
2
= 3 cm
½
½
1
23 (A).
12
2
[2 × 20 + 11?? ]=900
? ?? = 10
Also ?? 12
= 20 + 11 × 10 = 130
½
1
½
OR
23 (B). Putting ?? = 1, ?? 1
= ?? = 6 - 1
2
= 5 . . . . . . . . . . . . . (?? )
Putting ?? = 2, ?? 2
= 2?? + ?? = 6 × 2 - 2
2
= 8 . . . . . . . . . . . . . (???? )
Solving (i) & (ii) ?? = -2
½
1
½
24.
?????? (?? - ?? ) =
1
2
? ?? - ?? = 30
°
………..(i)
?????? (?? + ?? ) =
1
2
? ?? + ?? = 60
°
………..(ii)
Solving (i) & (ii) to get ?? = 45
°
, ?? = 15
°
½
½
½+½
25.
Modal class is 15-20.
???????? = 15 + 5 × (
15-6
2×15-6-10
)
= 18.21(approx.)
Class 5-10 10-15 15-20 20-25 25-30 30-35
Frequency 5 6 15 10 5 4
½
1
½
Section-C
O
A
M
B
Page 3
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS BASIC (Code No.241)
TIME: 3 hours MAX.MARKS: 80
Q. No. Section A Marks
1. B) 90 1
2. A) consistent with unique solution 1
3. D) 7 1
4.
C) 2 v?? 2
+ ?? 2
1
5. D) 145
°
1
6. B) 15 cm 1
7.
A)
5
4
1
8.
B) ?? ??????
1
9.
C) 3780
1
10.
B) 40
1
11.
D) 52
°
1
12.
B) 5 cm
1
13.
A) cos 60
°
1
14.
(C) 3?? ?? 2
1
15.
D) 4
1
16.
B) real and equal
1
17.
C) 30 - 40
1
18.
D) 25?? 2
- 5?? - 2
1
19. A) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
20. C) Assertion (A) is true but reason (R) is false.
1
Section B
2
21 (A). ???? 2
= ???? 2
? (?? - 4)
2
+ (?? - 3)
2
= (?? - 3)
2
+ (?? - 4)
2
? ?? = ?? ???? ?? - ?? = 0
1
1
OR
21 (B). AB = 6 cm = AC
OC = v36 - 9 = 3v3 cm
Point C is (3v3, 0)
½
1
½
22. Correct figure
AM = 4 cm
OM = v????
2
- ????
2
= v5
2
- 4
2
= 3 cm
½
½
1
23 (A).
12
2
[2 × 20 + 11?? ]=900
? ?? = 10
Also ?? 12
= 20 + 11 × 10 = 130
½
1
½
OR
23 (B). Putting ?? = 1, ?? 1
= ?? = 6 - 1
2
= 5 . . . . . . . . . . . . . (?? )
Putting ?? = 2, ?? 2
= 2?? + ?? = 6 × 2 - 2
2
= 8 . . . . . . . . . . . . . (???? )
Solving (i) & (ii) ?? = -2
½
1
½
24.
?????? (?? - ?? ) =
1
2
? ?? - ?? = 30
°
………..(i)
?????? (?? + ?? ) =
1
2
? ?? + ?? = 60
°
………..(ii)
Solving (i) & (ii) to get ?? = 45
°
, ?? = 15
°
½
½
½+½
25.
Modal class is 15-20.
???????? = 15 + 5 × (
15-6
2×15-6-10
)
= 18.21(approx.)
Class 5-10 10-15 15-20 20-25 25-30 30-35
Frequency 5 6 15 10 5 4
½
1
½
Section-C
O
A
M
B
3
26.
Let v5 be a rational number.
? v5 =
p
q
, where q?0 and p & q are coprime.
5q
2
= p
2
? p
2
is divisible by 5
? p is divisible by 5----- (i)
? p = 3a, where ‘a’ is a postive integer
25a
2
= 5q
2
? q
2
= 5a
2
?q
2
is divisible by 5
? q is divisible by 5 ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are coprime.
? v5 is an irrational number.
½
1
1
½
27(A). Let the required point on the y axis be P(0,y).
Let AP : PB be k : 1
Therefore,
-?? +4
?? +1
= 0
?k=4
Therefore, required ratio is 4:1
& ?? =
8-5
5
=
3
5
Hence point of intersection is (0,
3
5
).
½
1
½
½
½
OR
27 (B). Let the line 4?? + ?? = 4 intersects AB at ?? (?? 1
, ?? 1
) such that AP: PB=?? :1
?? 1
=
3?? -2
?? +1
and ?? 1
=
5?? -1
?? +1
(?? 1
, ?? 1
) lies on 4?? + ?? = 4
Therefore, 4(
3?? -2
?? +1
)+(
5?? -1
?? +1
)=4
? ?? =1
Required ratio is 1:1
1
½
1
½
Page 4
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS BASIC (Code No.241)
TIME: 3 hours MAX.MARKS: 80
Q. No. Section A Marks
1. B) 90 1
2. A) consistent with unique solution 1
3. D) 7 1
4.
C) 2 v?? 2
+ ?? 2
1
5. D) 145
°
1
6. B) 15 cm 1
7.
A)
5
4
1
8.
B) ?? ??????
1
9.
C) 3780
1
10.
B) 40
1
11.
D) 52
°
1
12.
B) 5 cm
1
13.
A) cos 60
°
1
14.
(C) 3?? ?? 2
1
15.
D) 4
1
16.
B) real and equal
1
17.
C) 30 - 40
1
18.
D) 25?? 2
- 5?? - 2
1
19. A) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
20. C) Assertion (A) is true but reason (R) is false.
1
Section B
2
21 (A). ???? 2
= ???? 2
? (?? - 4)
2
+ (?? - 3)
2
= (?? - 3)
2
+ (?? - 4)
2
? ?? = ?? ???? ?? - ?? = 0
1
1
OR
21 (B). AB = 6 cm = AC
OC = v36 - 9 = 3v3 cm
Point C is (3v3, 0)
½
1
½
22. Correct figure
AM = 4 cm
OM = v????
2
- ????
2
= v5
2
- 4
2
= 3 cm
½
½
1
23 (A).
12
2
[2 × 20 + 11?? ]=900
? ?? = 10
Also ?? 12
= 20 + 11 × 10 = 130
½
1
½
OR
23 (B). Putting ?? = 1, ?? 1
= ?? = 6 - 1
2
= 5 . . . . . . . . . . . . . (?? )
Putting ?? = 2, ?? 2
= 2?? + ?? = 6 × 2 - 2
2
= 8 . . . . . . . . . . . . . (???? )
Solving (i) & (ii) ?? = -2
½
1
½
24.
?????? (?? - ?? ) =
1
2
? ?? - ?? = 30
°
………..(i)
?????? (?? + ?? ) =
1
2
? ?? + ?? = 60
°
………..(ii)
Solving (i) & (ii) to get ?? = 45
°
, ?? = 15
°
½
½
½+½
25.
Modal class is 15-20.
???????? = 15 + 5 × (
15-6
2×15-6-10
)
= 18.21(approx.)
Class 5-10 10-15 15-20 20-25 25-30 30-35
Frequency 5 6 15 10 5 4
½
1
½
Section-C
O
A
M
B
3
26.
Let v5 be a rational number.
? v5 =
p
q
, where q?0 and p & q are coprime.
5q
2
= p
2
? p
2
is divisible by 5
? p is divisible by 5----- (i)
? p = 3a, where ‘a’ is a postive integer
25a
2
= 5q
2
? q
2
= 5a
2
?q
2
is divisible by 5
? q is divisible by 5 ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are coprime.
? v5 is an irrational number.
½
1
1
½
27(A). Let the required point on the y axis be P(0,y).
Let AP : PB be k : 1
Therefore,
-?? +4
?? +1
= 0
?k=4
Therefore, required ratio is 4:1
& ?? =
8-5
5
=
3
5
Hence point of intersection is (0,
3
5
).
½
1
½
½
½
OR
27 (B). Let the line 4?? + ?? = 4 intersects AB at ?? (?? 1
, ?? 1
) such that AP: PB=?? :1
?? 1
=
3?? -2
?? +1
and ?? 1
=
5?? -1
?? +1
(?? 1
, ?? 1
) lies on 4?? + ?? = 4
Therefore, 4(
3?? -2
?? +1
)+(
5?? -1
?? +1
)=4
? ?? =1
Required ratio is 1:1
1
½
1
½
4
28.
LHS= (
1
???????? - ???????? )(
1
???????? - ???????? )
=
1-?????? 2
?? ???????? ×
1-??????
2
?? ????????
=
??????
2
?? ???????? ×
?????? 2
?? ????????
=???????? ????????
RHS =
???????? ???????? ?????? 2
?? +??????
2
??
=???????? ???????? = LHS
½
1
½
1
29.
Class x frequency(f)
?? =
?? - 25
10
????
0-10 5 6 -2 -12
10-20 15 10 -1 -10
20-30 25 15 0 0
30-40 35 9 1 9
40-50 45 10 2 20
? ?? =50
? ???? = 7
???????? = 25 + 10 × (
7
50
)
= 26.4
Correct
table
?? ?? ??
1
½
30 (A).
(i) ?? ?????? ? ????????
??????? = ???????
Or OP bisects ???
(ii)?? ?????? ? ????????
?AQ=QB and ??????? = ???????
AB is a straight line
therefore ??????? = ??????? = 90
°
Hence OP is right bisector of AB
1
1
1
OR
30 (B). Correct Given, to prove, figure and construction
Correct proof
1
2
Page 5
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS BASIC (Code No.241)
TIME: 3 hours MAX.MARKS: 80
Q. No. Section A Marks
1. B) 90 1
2. A) consistent with unique solution 1
3. D) 7 1
4.
C) 2 v?? 2
+ ?? 2
1
5. D) 145
°
1
6. B) 15 cm 1
7.
A)
5
4
1
8.
B) ?? ??????
1
9.
C) 3780
1
10.
B) 40
1
11.
D) 52
°
1
12.
B) 5 cm
1
13.
A) cos 60
°
1
14.
(C) 3?? ?? 2
1
15.
D) 4
1
16.
B) real and equal
1
17.
C) 30 - 40
1
18.
D) 25?? 2
- 5?? - 2
1
19. A) Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A)
1
20. C) Assertion (A) is true but reason (R) is false.
1
Section B
2
21 (A). ???? 2
= ???? 2
? (?? - 4)
2
+ (?? - 3)
2
= (?? - 3)
2
+ (?? - 4)
2
? ?? = ?? ???? ?? - ?? = 0
1
1
OR
21 (B). AB = 6 cm = AC
OC = v36 - 9 = 3v3 cm
Point C is (3v3, 0)
½
1
½
22. Correct figure
AM = 4 cm
OM = v????
2
- ????
2
= v5
2
- 4
2
= 3 cm
½
½
1
23 (A).
12
2
[2 × 20 + 11?? ]=900
? ?? = 10
Also ?? 12
= 20 + 11 × 10 = 130
½
1
½
OR
23 (B). Putting ?? = 1, ?? 1
= ?? = 6 - 1
2
= 5 . . . . . . . . . . . . . (?? )
Putting ?? = 2, ?? 2
= 2?? + ?? = 6 × 2 - 2
2
= 8 . . . . . . . . . . . . . (???? )
Solving (i) & (ii) ?? = -2
½
1
½
24.
?????? (?? - ?? ) =
1
2
? ?? - ?? = 30
°
………..(i)
?????? (?? + ?? ) =
1
2
? ?? + ?? = 60
°
………..(ii)
Solving (i) & (ii) to get ?? = 45
°
, ?? = 15
°
½
½
½+½
25.
Modal class is 15-20.
???????? = 15 + 5 × (
15-6
2×15-6-10
)
= 18.21(approx.)
Class 5-10 10-15 15-20 20-25 25-30 30-35
Frequency 5 6 15 10 5 4
½
1
½
Section-C
O
A
M
B
3
26.
Let v5 be a rational number.
? v5 =
p
q
, where q?0 and p & q are coprime.
5q
2
= p
2
? p
2
is divisible by 5
? p is divisible by 5----- (i)
? p = 3a, where ‘a’ is a postive integer
25a
2
= 5q
2
? q
2
= 5a
2
?q
2
is divisible by 5
? q is divisible by 5 ----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are coprime.
? v5 is an irrational number.
½
1
1
½
27(A). Let the required point on the y axis be P(0,y).
Let AP : PB be k : 1
Therefore,
-?? +4
?? +1
= 0
?k=4
Therefore, required ratio is 4:1
& ?? =
8-5
5
=
3
5
Hence point of intersection is (0,
3
5
).
½
1
½
½
½
OR
27 (B). Let the line 4?? + ?? = 4 intersects AB at ?? (?? 1
, ?? 1
) such that AP: PB=?? :1
?? 1
=
3?? -2
?? +1
and ?? 1
=
5?? -1
?? +1
(?? 1
, ?? 1
) lies on 4?? + ?? = 4
Therefore, 4(
3?? -2
?? +1
)+(
5?? -1
?? +1
)=4
? ?? =1
Required ratio is 1:1
1
½
1
½
4
28.
LHS= (
1
???????? - ???????? )(
1
???????? - ???????? )
=
1-?????? 2
?? ???????? ×
1-??????
2
?? ????????
=
??????
2
?? ???????? ×
?????? 2
?? ????????
=???????? ????????
RHS =
???????? ???????? ?????? 2
?? +??????
2
??
=???????? ???????? = LHS
½
1
½
1
29.
Class x frequency(f)
?? =
?? - 25
10
????
0-10 5 6 -2 -12
10-20 15 10 -1 -10
20-30 25 15 0 0
30-40 35 9 1 9
40-50 45 10 2 20
? ?? =50
? ???? = 7
???????? = 25 + 10 × (
7
50
)
= 26.4
Correct
table
?? ?? ??
1
½
30 (A).
(i) ?? ?????? ? ????????
??????? = ???????
Or OP bisects ???
(ii)?? ?????? ? ????????
?AQ=QB and ??????? = ???????
AB is a straight line
therefore ??????? = ??????? = 90
°
Hence OP is right bisector of AB
1
1
1
OR
30 (B). Correct Given, to prove, figure and construction
Correct proof
1
2
5
31. Let the two-digit number be 10?? + ??
Therefore (10?? + ?? ) + (10?? + ?? ) = 99
? ?? + ?? = 9 ……….(i)
Also, ?? = 3 + ?? ……..(ii)
Solving (i) & (ii) to get ?? = 3 , ?? = 6
Therefore, required number is 63
½
½
½
½
½
½
Section D
32 (A). Let the number of books purchased be ??
Therefore, cost price of 1 book =
1920
??
Therefore
1920
?? -
1920
?? +4
= 24
? 1920 × 4 = 24?? (?? + 4)
or ?? 2
+ 4?? - 320 = 0
? (?? + 20)(?? - 16) = 0
? ?? = 16, ?? ? -20
Number of books bought=16
Price of each book =
1920
16
= ?120
1
1
1
1
1
OR
32 (B). Let the initial average speed of the train be ?? km/hr.
Therefore
132
?? +
140
?? +4
= 4
? 4?? 2
- 256?? - 528 = 0
or ?? 2
- 64?? - 132 = 0
? (?? - 66)(?? + 2) = 0
? ?? = 66, ?? ? -2
Initial average speed of train= 66 km/hr
Time taken to cover the distances separately=
132
66
&
140
70
i.e. 2 hours each
1
1
1
1
1
33. Correct Given, to prove, Construction and figure
Correct Proof
?? ?? × ?? =2
3
34.
(i) Perimeter of sector = 2?? +
2?????? 360
= 73.12
? 2(24) +
2×3.14×24×?? 360
= 73.12
? ?? = 60
°
(ii)Area of minor segment = (
3.14×24×24×60
360
-
1.73
4
× 24 × 24) ???? 2
= (301.44 - 249.12) ???? 2
= 52.32 ???? 2
1
1
2
1
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FAQs on Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme with Solutions (2024-25) - Mathematics (Maths) Class 10
| 1. What is the marking scheme for Class 10 Mathematics (Basic) in CBSE exams? |  |
Ans. The marking scheme for Class 10 Mathematics (Basic) generally includes a total of 100 marks, divided into various sections. There are typically two types of questions: objective type (multiple choice questions) and subjective type (short and long answer questions). The objective questions usually carry 1 mark each, while the subjective questions may carry 2, 3, or 4 marks depending on their complexity. This structure aims to assess students' understanding and application of mathematical concepts.
| 2. How can students prepare effectively for the Class 10 Mathematics (Basic) exam? | |
Ans. Students can prepare effectively by first familiarizing themselves with the syllabus and exam pattern. Regular practice of previous years' question papers and sample papers can help in understanding the question formats. Additionally, students should focus on key topics, revise regularly, and clarify doubts with teachers or peers. Utilizing study materials, such as textbooks and online resources, can also enhance understanding and retention of concepts.
| 3. What are some important topics that students should focus on for the Class 10 Mathematics (Basic) exam? | |
Ans. Important topics for Class 10 Mathematics (Basic) typically include Number Systems, Algebra, Geometry, Statistics, and Probability. Specific areas such as quadratic equations, triangles, circles, and surface areas and volumes are crucial. Mastery of these topics not only aids in scoring well in exams but also builds a strong foundation for higher-level mathematics.
| 4. Are there any specific tips for solving mathematical problems in the exam? | |
Ans. Yes, some effective tips include reading the questions carefully to understand what is being asked, breaking down complex problems into smaller, manageable parts, and showing all steps clearly to gain partial credit. Time management is also essential; students should allocate time wisely to different sections of the paper. Practicing mental math can speed up calculations during the exam.
| 5. How does the CBSE evaluate the performance of students in the Class 10 Mathematics (Basic) exam? | |
Ans. CBSE evaluates students based on their performance in the written exam, which includes both theoretical knowledge and practical application of mathematical concepts. The evaluation considers accuracy, clarity of reasoning, and methodology used in solving problems. Additionally, internal assessments, such as class tests and projects, may contribute to the overall performance evaluation, ensuring a comprehensive assessment of the student's understanding and skills in mathematics.
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Additional Information about Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme with Solutions (2024-25) for Class 10 Preparation
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The content of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme with Solutions (2024-25) is prepared as per the latest Class 10 syllabus.
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