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Page 1
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80
Q.No. Section A Marks
1. D) -6,6 1
2. B) -5 1
3. D) From a point inside a circle only two tangents can be drawn. 1
4. A) 7 1
5. B) 20 cm 1
6.
A)
11
9
1
7. C) 140
?? 1
8.
B) 8?? 2
- 20
1
9.
C) 30
1
10.
B) isosceles and similar
1
11.
A) Irrational and distinct
1
12.
C)
3
v3
1
13.
B)
594
7
1
14.
B)
3
8
1
15.
B) (-4, 0)
1
16.
A) median
1
17.
C) (3,0)
1
18.
D)
3
26
1
19. B) 1
20. D) 1
Page 2
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80
Q.No. Section A Marks
1. D) -6,6 1
2. B) -5 1
3. D) From a point inside a circle only two tangents can be drawn. 1
4. A) 7 1
5. B) 20 cm 1
6.
A)
11
9
1
7. C) 140
?? 1
8.
B) 8?? 2
- 20
1
9.
C) 30
1
10.
B) isosceles and similar
1
11.
A) Irrational and distinct
1
12.
C)
3
v3
1
13.
B)
594
7
1
14.
B)
3
8
1
15.
B) (-4, 0)
1
16.
A) median
1
17.
C) (3,0)
1
18.
D)
3
26
1
19. B) 1
20. D) 1
2
Section B
21. (A)
(B)
480 = 2
5
x 3 x 5
720 = 2
4
x 3
2
x 5
LCM (480,720) = 2
5
x 3
2
x 5 = 1440
HCF (480, 720) = 2
4
x 3x 5 = 240
OR
85 = 5x17, 238 = 2x7x17
HCF( 85, 238) = 17
17 = 85xm -238
m = 3
½
½
½
½
1
1
22.(A)
(B)
Total number of possible outcomes = 6x6=36
For a product to be odd, both the numbers should be odd.
Favourable outcomes are (7,7) (7,9) (7,11) (9,7) (9,9) (9, 11) (11,7) (11,9)
(11,11)
no. of favourable outcomes = 9
P (product is odd) =
9
36
or
1
4
OR
Total number of three-digit numbers = 900.
Numbers with hundredth digit 8 & and unit’s digit 5 are 805,815,
825,....,895
Number of favourable outcomes = 10
P(selecting one such number) =
10
900
or
1
90
½
1
½
½
1
½
23.
2 (
v3
2
)
2
- (
1
v3
)
2
(v2)
2
=
7
12
1 ½
½
24 Let the required point be (x,0)
v(8 - ?? )
2
+ 25 = v41
=> (8 - ?? )
2
= 16
=> 8 - x =?4
=> x = 4 , 12
Two points on the x-axis are (4,0) & (12,0).
½
½
1
Page 3
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80
Q.No. Section A Marks
1. D) -6,6 1
2. B) -5 1
3. D) From a point inside a circle only two tangents can be drawn. 1
4. A) 7 1
5. B) 20 cm 1
6.
A)
11
9
1
7. C) 140
?? 1
8.
B) 8?? 2
- 20
1
9.
C) 30
1
10.
B) isosceles and similar
1
11.
A) Irrational and distinct
1
12.
C)
3
v3
1
13.
B)
594
7
1
14.
B)
3
8
1
15.
B) (-4, 0)
1
16.
A) median
1
17.
C) (3,0)
1
18.
D)
3
26
1
19. B) 1
20. D) 1
2
Section B
21. (A)
(B)
480 = 2
5
x 3 x 5
720 = 2
4
x 3
2
x 5
LCM (480,720) = 2
5
x 3
2
x 5 = 1440
HCF (480, 720) = 2
4
x 3x 5 = 240
OR
85 = 5x17, 238 = 2x7x17
HCF( 85, 238) = 17
17 = 85xm -238
m = 3
½
½
½
½
1
1
22.(A)
(B)
Total number of possible outcomes = 6x6=36
For a product to be odd, both the numbers should be odd.
Favourable outcomes are (7,7) (7,9) (7,11) (9,7) (9,9) (9, 11) (11,7) (11,9)
(11,11)
no. of favourable outcomes = 9
P (product is odd) =
9
36
or
1
4
OR
Total number of three-digit numbers = 900.
Numbers with hundredth digit 8 & and unit’s digit 5 are 805,815,
825,....,895
Number of favourable outcomes = 10
P(selecting one such number) =
10
900
or
1
90
½
1
½
½
1
½
23.
2 (
v3
2
)
2
- (
1
v3
)
2
(v2)
2
=
7
12
1 ½
½
24 Let the required point be (x,0)
v(8 - ?? )
2
+ 25 = v41
=> (8 - ?? )
2
= 16
=> 8 - x =?4
=> x = 4 , 12
Two points on the x-axis are (4,0) & (12,0).
½
½
1
3
25.
AB = v(3 + 5)
2
+ (0 - 6)
2
= 10
BC = v(9 - 3)
2
+ (8 - 0)
2
= 10
AC = v(9 + 5)
2
+ (8 - 6)
2
= 10v2
Since AB = BC, therefore ?? ABC is isosceles
½
½
½
½
Section C
26.(A)
(B)
Since D, E, F are the mid points of BC, CA, AB respectively
Therefore, EF||BC, DF||AC, DE||AB
BDEF is a parallelogram
? 1= ? 2 & ? 3 = ? 4
?? FBD ~ ?? DEF
Also, DCEF is a parallelogram
? 3= ? 6 & ? 1 = ? 2 ( proved above)
?? DEF ~ ?? ABC
OR
Since PQ//BC therefore ?? APR ~ ?? ABD
=>
????
????
=
????
????
…….. (i)
1
1
1
1
A
P
R
Q
B
D
C
1
2
3
4
5
6
Page 4
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80
Q.No. Section A Marks
1. D) -6,6 1
2. B) -5 1
3. D) From a point inside a circle only two tangents can be drawn. 1
4. A) 7 1
5. B) 20 cm 1
6.
A)
11
9
1
7. C) 140
?? 1
8.
B) 8?? 2
- 20
1
9.
C) 30
1
10.
B) isosceles and similar
1
11.
A) Irrational and distinct
1
12.
C)
3
v3
1
13.
B)
594
7
1
14.
B)
3
8
1
15.
B) (-4, 0)
1
16.
A) median
1
17.
C) (3,0)
1
18.
D)
3
26
1
19. B) 1
20. D) 1
2
Section B
21. (A)
(B)
480 = 2
5
x 3 x 5
720 = 2
4
x 3
2
x 5
LCM (480,720) = 2
5
x 3
2
x 5 = 1440
HCF (480, 720) = 2
4
x 3x 5 = 240
OR
85 = 5x17, 238 = 2x7x17
HCF( 85, 238) = 17
17 = 85xm -238
m = 3
½
½
½
½
1
1
22.(A)
(B)
Total number of possible outcomes = 6x6=36
For a product to be odd, both the numbers should be odd.
Favourable outcomes are (7,7) (7,9) (7,11) (9,7) (9,9) (9, 11) (11,7) (11,9)
(11,11)
no. of favourable outcomes = 9
P (product is odd) =
9
36
or
1
4
OR
Total number of three-digit numbers = 900.
Numbers with hundredth digit 8 & and unit’s digit 5 are 805,815,
825,....,895
Number of favourable outcomes = 10
P(selecting one such number) =
10
900
or
1
90
½
1
½
½
1
½
23.
2 (
v3
2
)
2
- (
1
v3
)
2
(v2)
2
=
7
12
1 ½
½
24 Let the required point be (x,0)
v(8 - ?? )
2
+ 25 = v41
=> (8 - ?? )
2
= 16
=> 8 - x =?4
=> x = 4 , 12
Two points on the x-axis are (4,0) & (12,0).
½
½
1
3
25.
AB = v(3 + 5)
2
+ (0 - 6)
2
= 10
BC = v(9 - 3)
2
+ (8 - 0)
2
= 10
AC = v(9 + 5)
2
+ (8 - 6)
2
= 10v2
Since AB = BC, therefore ?? ABC is isosceles
½
½
½
½
Section C
26.(A)
(B)
Since D, E, F are the mid points of BC, CA, AB respectively
Therefore, EF||BC, DF||AC, DE||AB
BDEF is a parallelogram
? 1= ? 2 & ? 3 = ? 4
?? FBD ~ ?? DEF
Also, DCEF is a parallelogram
? 3= ? 6 & ? 1 = ? 2 ( proved above)
?? DEF ~ ?? ABC
OR
Since PQ//BC therefore ?? APR ~ ?? ABD
=>
????
????
=
????
????
…….. (i)
1
1
1
1
A
P
R
Q
B
D
C
1
2
3
4
5
6
4
?? AQR ~ ?? ACD
=>
????
????
=
????
????
…….. (ii)
Now,
????
????
=
????
????
……….(iii)
Using (i), (ii) & (iii),
????
????
=
????
????
But, BD = DC
=> PR = RQ or AD bisects PQ
1
1
27. Let the numbers be x and 18-x.
1
?? +
1
18-?? =
9
40
=> 18×40 = 9x(18-?? )
=> ?? 2
-18 ?? + 80=0
=> (?? -10)(?? -8)=0
=> ?? =10, 8.
=> 18-?? =8, 10
Hence two numbers are 8 and 10.
½
1
1
½
28.
From given polynomial ?? + ?? =
5
6
, ???? =
1
6
?? 2
+ ?? 2
= (
5
6
)
2
- 2 x
1
6
=
13
36
And ?? 2
?? 2
= (
1
6
)
2
=
1
36
?? 2
-
13
36
?? +
1
36
? Required polynomial is 36?? 2
-13 ?? +1
1
1
½
½
29. (???????? + ?????? ?? )
2
+ (???????? - ?????? ?? )
2
= 2 ( ?????? 2
?? + ?????? 2
?? ) = 2
=> (1)
2
+ (???????? - ?????? ?? )
2
= 2
=> (???????? - ?????? ?? )
2
= 1
=> ???????? - ?????? ?? = ? 1
1 ½
1
½
30.(A)
(B)
Angle described by minute hand in 5 min = 30°.
length of minute hand =18 cm = r.
Area swept by minute hand in 35 minutes
=(
22
7
x18x18×
30
360
) x 7
= 594 ???? 2
.
OR
Area of minor segment = Ar. Sector OAB- Ar. ?? OAB
=
90
360
x
22
7
× 14x14 -
v3
4
x 14x14
= 69.23 cm
2
2
1
2
1
Page 5
1
Marking Scheme
Class X Session 2024-25
MATHEMATICS STANDARD (Code No.041)
TIME: 3 hours MAX.MARKS: 80
Q.No. Section A Marks
1. D) -6,6 1
2. B) -5 1
3. D) From a point inside a circle only two tangents can be drawn. 1
4. A) 7 1
5. B) 20 cm 1
6.
A)
11
9
1
7. C) 140
?? 1
8.
B) 8?? 2
- 20
1
9.
C) 30
1
10.
B) isosceles and similar
1
11.
A) Irrational and distinct
1
12.
C)
3
v3
1
13.
B)
594
7
1
14.
B)
3
8
1
15.
B) (-4, 0)
1
16.
A) median
1
17.
C) (3,0)
1
18.
D)
3
26
1
19. B) 1
20. D) 1
2
Section B
21. (A)
(B)
480 = 2
5
x 3 x 5
720 = 2
4
x 3
2
x 5
LCM (480,720) = 2
5
x 3
2
x 5 = 1440
HCF (480, 720) = 2
4
x 3x 5 = 240
OR
85 = 5x17, 238 = 2x7x17
HCF( 85, 238) = 17
17 = 85xm -238
m = 3
½
½
½
½
1
1
22.(A)
(B)
Total number of possible outcomes = 6x6=36
For a product to be odd, both the numbers should be odd.
Favourable outcomes are (7,7) (7,9) (7,11) (9,7) (9,9) (9, 11) (11,7) (11,9)
(11,11)
no. of favourable outcomes = 9
P (product is odd) =
9
36
or
1
4
OR
Total number of three-digit numbers = 900.
Numbers with hundredth digit 8 & and unit’s digit 5 are 805,815,
825,....,895
Number of favourable outcomes = 10
P(selecting one such number) =
10
900
or
1
90
½
1
½
½
1
½
23.
2 (
v3
2
)
2
- (
1
v3
)
2
(v2)
2
=
7
12
1 ½
½
24 Let the required point be (x,0)
v(8 - ?? )
2
+ 25 = v41
=> (8 - ?? )
2
= 16
=> 8 - x =?4
=> x = 4 , 12
Two points on the x-axis are (4,0) & (12,0).
½
½
1
3
25.
AB = v(3 + 5)
2
+ (0 - 6)
2
= 10
BC = v(9 - 3)
2
+ (8 - 0)
2
= 10
AC = v(9 + 5)
2
+ (8 - 6)
2
= 10v2
Since AB = BC, therefore ?? ABC is isosceles
½
½
½
½
Section C
26.(A)
(B)
Since D, E, F are the mid points of BC, CA, AB respectively
Therefore, EF||BC, DF||AC, DE||AB
BDEF is a parallelogram
? 1= ? 2 & ? 3 = ? 4
?? FBD ~ ?? DEF
Also, DCEF is a parallelogram
? 3= ? 6 & ? 1 = ? 2 ( proved above)
?? DEF ~ ?? ABC
OR
Since PQ//BC therefore ?? APR ~ ?? ABD
=>
????
????
=
????
????
…….. (i)
1
1
1
1
A
P
R
Q
B
D
C
1
2
3
4
5
6
4
?? AQR ~ ?? ACD
=>
????
????
=
????
????
…….. (ii)
Now,
????
????
=
????
????
……….(iii)
Using (i), (ii) & (iii),
????
????
=
????
????
But, BD = DC
=> PR = RQ or AD bisects PQ
1
1
27. Let the numbers be x and 18-x.
1
?? +
1
18-?? =
9
40
=> 18×40 = 9x(18-?? )
=> ?? 2
-18 ?? + 80=0
=> (?? -10)(?? -8)=0
=> ?? =10, 8.
=> 18-?? =8, 10
Hence two numbers are 8 and 10.
½
1
1
½
28.
From given polynomial ?? + ?? =
5
6
, ???? =
1
6
?? 2
+ ?? 2
= (
5
6
)
2
- 2 x
1
6
=
13
36
And ?? 2
?? 2
= (
1
6
)
2
=
1
36
?? 2
-
13
36
?? +
1
36
? Required polynomial is 36?? 2
-13 ?? +1
1
1
½
½
29. (???????? + ?????? ?? )
2
+ (???????? - ?????? ?? )
2
= 2 ( ?????? 2
?? + ?????? 2
?? ) = 2
=> (1)
2
+ (???????? - ?????? ?? )
2
= 2
=> (???????? - ?????? ?? )
2
= 1
=> ???????? - ?????? ?? = ? 1
1 ½
1
½
30.(A)
(B)
Angle described by minute hand in 5 min = 30°.
length of minute hand =18 cm = r.
Area swept by minute hand in 35 minutes
=(
22
7
x18x18×
30
360
) x 7
= 594 ???? 2
.
OR
Area of minor segment = Ar. Sector OAB- Ar. ?? OAB
=
90
360
x
22
7
× 14x14 -
v3
4
x 14x14
= 69.23 cm
2
2
1
2
1
5
31. Let v3 be a rational number.
? v3 =
?? ?? , where q?0 and let p & q be co-prime.
3q
2
= p
2
? p
2
is divisible by 3 ? p is divisible by 3 ----- (i)
? p = 3a, where ‘a’ is some integer
9a
2
= 3q
2
? q
2
= 3a
2
?q
2
is divisible by 3 ? q is divisible by 3----- (ii)
(i) and (ii) leads to contradiction as ‘p’ and ‘q’ are co-prime.
½
1
1
½
Section D
32.(A)
(B)
x+2y=3, 2x-3y+8=0
Correct graph of each equation
Solution x=-1 and y=2
OR
Let car I starts from A with speed x km/hr and car Il Starts from B with
speed y km/hr (x>y)
Case I- when cars are moving in the same direction.
Distance covered by car I in 9 hours = 9x.
Distance covered by car II in 9 hours = 9y
Therefore 9 (x-y) = 180
=> x-y= 20 ……………. (i)
case II- when cars are moving in opposite directions.
Distance covered by Car I in 1 hour = x
Distance covered by Car II in 1 hour = y
Therefore x + y=180 ………….. (ii)
Solving (i) and (ii) we get, x=100 km/hr, y=80 km/hr.
2+2 = 4
1
2
2
1
33. Correct given, to prove, construction, figure
Correct proof
AR = AQ = 7cm
BP = BR = AB-AR = 3cm
CP =CQ = 5cm
BC = BP+PC = 3+5 = 8 cm
1
2
½
½
½
½
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FAQs on Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solutions (2024-25) - Mathematics (Maths) Class 10
| 1. What are the key components of the Class 10 Mathematics (Standard) CBSE marking scheme? |  |
Ans. The marking scheme for Class 10 Mathematics (Standard) typically includes various components such as the distribution of marks across different units, types of questions (like MCQs, short answer, and long answer questions), and the weightage given to each chapter. It is designed to assess students’ understanding of concepts, problem-solving skills, and application of mathematical principles.
| 2. How are the questions categorized in the CBSE Class 10 Mathematics exam? | |
Ans. Questions in the CBSE Class 10 Mathematics exam are generally categorized into different types: Very Short Answer Questions (1 mark), Short Answer Questions (2 marks), Long Answer Questions (3 marks), and Higher Order Thinking Skills (HOTS) questions (4 or 5 marks). Each category assesses different levels of understanding and skills, from basic recall to higher-order analytical abilities.
| 3. What is the importance of the Class 10 Mathematics marking scheme for students? | |
Ans. The marking scheme is crucial for students as it provides clear guidelines on how marks are allocated for various types of questions. Understanding the scheme helps students prioritize their study efforts, focus on high-weightage topics, and develop effective exam strategies. It also aids in self-assessment and identifying areas that require more practice.
| 4. How can students effectively prepare for the Class 10 Mathematics exam based on the marking scheme? | |
Ans. Students can prepare effectively by reviewing the marking scheme to understand the distribution of marks and types of questions. They should practice previous years' question papers, focus on solving sample papers aligned with the marking scheme, and emphasize topics with higher weightage. Regular revision and seeking help in challenging areas will also enhance their preparedness.
| 5. What changes have been made in the recent marking schemes for Class 10 Mathematics exams? | |
Ans. Recent changes in the marking schemes for Class 10 Mathematics exams may include adjustments in the weightage of different units, the introduction of new types of questions to assess higher-order thinking skills, and modifications in the evaluation criteria to better reflect students' understanding and application of mathematical concepts. Keeping updated with the latest syllabus and guidelines from the CBSE is essential for students.
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Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solutions (2024-25) Free PDF Download
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Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solutions (2024-25) Notes
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Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solutions (2024-25) Class 10 Questions
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The content of Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme with Solutions (2024-25) is prepared as per the latest Class 10 syllabus.
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