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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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Importance of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25)
The importance of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25) cannot be overstated, especially for Class 10 aspirants.
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Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25) Notes
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25) Notes offer in-depth insights into the specific topic to help you master it with ease.
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Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25) Class 10 Questions
The "Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25) Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the
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The content of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2024-25) is prepared as per the latest Class 10 syllabus.
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