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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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Whether you're a beginner or aiming for advanced proficiency, Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme (2024-25) Notes on EduRev are your ultimate resource for success.
Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme (2024-25) Class 10 Questions
The "Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme (2024-25) Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
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Students of Class 10 can study Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme (2024-25) alongwith tests & analysis from the EduRev app,
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The content of Class 10 Mathematics (Standard): CBSE (Official) Marking Scheme (2024-25) is prepared as per the latest Class 10 syllabus.
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