Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme with Solutions (2025-26)
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Page 1 of 11
MATHEMATICS BASIC – Code No. 241
MARKING SCHEME
CLASS - X (2025 - 26)
SECTION - A
Q. No. Answer Marks
1. Answer – D
As, 2025 = 3
4
× 5
4
So, the exponent of 3 in the prime factorization of 2025 is 4
1
2. Answer – B
On subtracting first equation from second equation, we get
2025 ?? + 2024 ?? - 2024 ?? - 2025 ?? = -1 - 1 ? (?? - ?? ) = -??
1
3. Answer – D
As, ?? (?? ) = k(x + 2)(x - 5) ? ?? (?? ) = k(?? 2
- 3x - 10), k ? 0
Since k can be any real number. So, there are Infinitely many such polynomials.
1
4. Answer – C
On simplification, given equations reduce to
(?? ) ?? 2
+ 2?? - 2 = 0 (?????????????????? ???????????????? )
(?? ) 2?? 2
- 3?? - 1 = 0 (?????????????????? ???????????????? )
(C) 3?? + 1 = 0 (?????? ?? ?????????????????? ???????????????? )
(D) 4?? 2
+ ?? = 0 (?????????????????? ???????????????? )
1
5. Answer – A
As, 2(?? + 10) = (3?? + 2) + 2?? ? ?? = 6
1
6. Answer – B
As,
50(51)
2
= 25?? ? ?? = 51
1
7.
Answer – D
Distance between the given points =
v
(
1
2
-
v3
2
)
2
+ (
1
2
+
v3
2
)
2
= v2
1
8. Answer – C
We know that, for the coordinates of a mirror image of a point in x-axis,
abscissa remains the same and ordinate will be of opposite sign of the ordinate
of given point. So, the Mirror image of the point (-3, 5) about x-axis is (-3, -5).
1
9. Answer – B
As, ?ABC ~ ?EFD ? ?A = ?E
1
Page 2
Page 1 of 11
MATHEMATICS BASIC – Code No. 241
MARKING SCHEME
CLASS - X (2025 - 26)
SECTION - A
Q. No. Answer Marks
1. Answer – D
As, 2025 = 3
4
× 5
4
So, the exponent of 3 in the prime factorization of 2025 is 4
1
2. Answer – B
On subtracting first equation from second equation, we get
2025 ?? + 2024 ?? - 2024 ?? - 2025 ?? = -1 - 1 ? (?? - ?? ) = -??
1
3. Answer – D
As, ?? (?? ) = k(x + 2)(x - 5) ? ?? (?? ) = k(?? 2
- 3x - 10), k ? 0
Since k can be any real number. So, there are Infinitely many such polynomials.
1
4. Answer – C
On simplification, given equations reduce to
(?? ) ?? 2
+ 2?? - 2 = 0 (?????????????????? ???????????????? )
(?? ) 2?? 2
- 3?? - 1 = 0 (?????????????????? ???????????????? )
(C) 3?? + 1 = 0 (?????? ?? ?????????????????? ???????????????? )
(D) 4?? 2
+ ?? = 0 (?????????????????? ???????????????? )
1
5. Answer – A
As, 2(?? + 10) = (3?? + 2) + 2?? ? ?? = 6
1
6. Answer – B
As,
50(51)
2
= 25?? ? ?? = 51
1
7.
Answer – D
Distance between the given points =
v
(
1
2
-
v3
2
)
2
+ (
1
2
+
v3
2
)
2
= v2
1
8. Answer – C
We know that, for the coordinates of a mirror image of a point in x-axis,
abscissa remains the same and ordinate will be of opposite sign of the ordinate
of given point. So, the Mirror image of the point (-3, 5) about x-axis is (-3, -5).
1
9. Answer – B
As, ?ABC ~ ?EFD ? ?A = ?E
1
Page 2 of 11
10. Answer – B
As, ?ABC~?PQR ?
AB
PQ
=
BC
QR
=
AC
PR
=
1
2
? PQ = 6 cm, QR = 8 cm
Perimeter of the triangle PQR (in cm) = 6 + 8 +10 = 24
1
Question given for Visually impaired candidates
Answer – B
The solution is same as above.
1
11. Answer – A
From the figure, AE = 24 – r = AF. So, BF = 1 + r = 7 – r ? r = 3 cm
1
Question given for Visually Impaired candidates
Answer – B
As, PQ = PR = 24 cm
So, Area of Quadrilateral PQOR (in cm
2
) = 2 ×
1
2
× 24 × 10 = 240
1
12. Answer – B
As, cot
2
?? - cosec
2
?? = -?? , so it is NOT equal to Unity
1
13. Answer – C
As, Median class is 10-15. So, its upper limit is 15.
1
14. Answer – C
Since, 3 Median = Mode + 2 Mean. So, a = 3 & b = 2.
Thus, (2b + 3a) = 4 + 9 = 13
1
15. Answer – B
Radius (in cm) = v13
2
- 12
2
= 5
1
16. Answer – A
As, ?PAO= 90°. So, ?APO = 115° - 90° = 25°
1
Question given for Visually Impaired candidates
Answer – A
As, the chord is at a distance of 18 cm (more than the radius). So, the chord
will be at a distance of 5 cm on the opposite side of the centre. Thus, length
of the chord CD will be 2v13
2
- 5
2
= 24 ????
1
17. Answer – C
As, r1 : r2 = 3 : 4. So, the ratio of their areas = r1
2
: r2
2
= 9 :16
1
18. Answer – A
Since, the event is most unlikely to happen. Therefore, its probability is 0.0001
1
19. Answer – A
As, Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
1
Page 3
Page 1 of 11
MATHEMATICS BASIC – Code No. 241
MARKING SCHEME
CLASS - X (2025 - 26)
SECTION - A
Q. No. Answer Marks
1. Answer – D
As, 2025 = 3
4
× 5
4
So, the exponent of 3 in the prime factorization of 2025 is 4
1
2. Answer – B
On subtracting first equation from second equation, we get
2025 ?? + 2024 ?? - 2024 ?? - 2025 ?? = -1 - 1 ? (?? - ?? ) = -??
1
3. Answer – D
As, ?? (?? ) = k(x + 2)(x - 5) ? ?? (?? ) = k(?? 2
- 3x - 10), k ? 0
Since k can be any real number. So, there are Infinitely many such polynomials.
1
4. Answer – C
On simplification, given equations reduce to
(?? ) ?? 2
+ 2?? - 2 = 0 (?????????????????? ???????????????? )
(?? ) 2?? 2
- 3?? - 1 = 0 (?????????????????? ???????????????? )
(C) 3?? + 1 = 0 (?????? ?? ?????????????????? ???????????????? )
(D) 4?? 2
+ ?? = 0 (?????????????????? ???????????????? )
1
5. Answer – A
As, 2(?? + 10) = (3?? + 2) + 2?? ? ?? = 6
1
6. Answer – B
As,
50(51)
2
= 25?? ? ?? = 51
1
7.
Answer – D
Distance between the given points =
v
(
1
2
-
v3
2
)
2
+ (
1
2
+
v3
2
)
2
= v2
1
8. Answer – C
We know that, for the coordinates of a mirror image of a point in x-axis,
abscissa remains the same and ordinate will be of opposite sign of the ordinate
of given point. So, the Mirror image of the point (-3, 5) about x-axis is (-3, -5).
1
9. Answer – B
As, ?ABC ~ ?EFD ? ?A = ?E
1
Page 2 of 11
10. Answer – B
As, ?ABC~?PQR ?
AB
PQ
=
BC
QR
=
AC
PR
=
1
2
? PQ = 6 cm, QR = 8 cm
Perimeter of the triangle PQR (in cm) = 6 + 8 +10 = 24
1
Question given for Visually impaired candidates
Answer – B
The solution is same as above.
1
11. Answer – A
From the figure, AE = 24 – r = AF. So, BF = 1 + r = 7 – r ? r = 3 cm
1
Question given for Visually Impaired candidates
Answer – B
As, PQ = PR = 24 cm
So, Area of Quadrilateral PQOR (in cm
2
) = 2 ×
1
2
× 24 × 10 = 240
1
12. Answer – B
As, cot
2
?? - cosec
2
?? = -?? , so it is NOT equal to Unity
1
13. Answer – C
As, Median class is 10-15. So, its upper limit is 15.
1
14. Answer – C
Since, 3 Median = Mode + 2 Mean. So, a = 3 & b = 2.
Thus, (2b + 3a) = 4 + 9 = 13
1
15. Answer – B
Radius (in cm) = v13
2
- 12
2
= 5
1
16. Answer – A
As, ?PAO= 90°. So, ?APO = 115° - 90° = 25°
1
Question given for Visually Impaired candidates
Answer – A
As, the chord is at a distance of 18 cm (more than the radius). So, the chord
will be at a distance of 5 cm on the opposite side of the centre. Thus, length
of the chord CD will be 2v13
2
- 5
2
= 24 ????
1
17. Answer – C
As, r1 : r2 = 3 : 4. So, the ratio of their areas = r1
2
: r2
2
= 9 :16
1
18. Answer – A
Since, the event is most unlikely to happen. Therefore, its probability is 0.0001
1
19. Answer – A
As, Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
1
Page 3 of 11
20. Answer – D
Q1. Since events given in Assertion are not equally likely, so probability of getting
two heads is not
1
3
.
Thus, Assertion (A) is false but reason (R) is true.
1
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
21 (A). It can be observed that,
2 × 5 × 7 × 11 + 11 × 13 = 11 × (70 + 13) = 11 x 83
which is the product of two factors other than 1. Therefore, it is a composite
number.
OR
1
1
21 (B). The smallest number which is divisible by any two numbers is their LCM.
So, Number which is divisible by both 306 and 657 = LCM (306, 657)
Since, 306 = 2
1
× 3
2
× 17
1
and 657 = 3
2
× 73
LCM (306, 657) = 2
1
× 3
2
× 17
1
× 73 = 22338
½
1
½
22. As, P(3, a) lies on the line L, so 3 + ?? = 5 ? ?? = 2
Now, the radius of the circle = CP = v3
2
+ 2
2
= v13 ??????????
Question given for Visually Impaired candidates
Diameter of the circle = Distance between (0,0) and (6,8) = v6
2
+ 8
2
= 10
Radius of the circle = ½ (Diameter of the circle) = 5 units
1
1
1½
½
23. Sum of the zeroes = 2 – 3 = -(?? + 1) ? ?? = 0
Product of the zeroes = – 6 = ?? ? ?? = -6
Hence, ?? = 0 & ?? = -6
1
1
24.
Discriminant, D = 16 + 12v2 > 0
As, Discriminant is positive. So, Roots are real and distinct.
1
1
25 (A).
Q2. 2 sin30°tan60°- 3 cos
2
60°sec
2
30° = 2 (
1
2
) (v3) - 3 (
1
2
)
2
(
2
v3
)
2
Q3. = v3 - 1
OR
1½
½
25 (B).
As , sinx . cosx (tanx+ cotx ) = sinx . cosx(
sinx
cosx
+
cosx
sinx
).
= sinx . cosx (
1
cosx .sinx
)
= 1 (Constant)
Since, the value of sinx. cosx (tanx + cotx) is constant, so its equal 1 for all
angles.
½
1½
Page 4
Page 1 of 11
MATHEMATICS BASIC – Code No. 241
MARKING SCHEME
CLASS - X (2025 - 26)
SECTION - A
Q. No. Answer Marks
1. Answer – D
As, 2025 = 3
4
× 5
4
So, the exponent of 3 in the prime factorization of 2025 is 4
1
2. Answer – B
On subtracting first equation from second equation, we get
2025 ?? + 2024 ?? - 2024 ?? - 2025 ?? = -1 - 1 ? (?? - ?? ) = -??
1
3. Answer – D
As, ?? (?? ) = k(x + 2)(x - 5) ? ?? (?? ) = k(?? 2
- 3x - 10), k ? 0
Since k can be any real number. So, there are Infinitely many such polynomials.
1
4. Answer – C
On simplification, given equations reduce to
(?? ) ?? 2
+ 2?? - 2 = 0 (?????????????????? ???????????????? )
(?? ) 2?? 2
- 3?? - 1 = 0 (?????????????????? ???????????????? )
(C) 3?? + 1 = 0 (?????? ?? ?????????????????? ???????????????? )
(D) 4?? 2
+ ?? = 0 (?????????????????? ???????????????? )
1
5. Answer – A
As, 2(?? + 10) = (3?? + 2) + 2?? ? ?? = 6
1
6. Answer – B
As,
50(51)
2
= 25?? ? ?? = 51
1
7.
Answer – D
Distance between the given points =
v
(
1
2
-
v3
2
)
2
+ (
1
2
+
v3
2
)
2
= v2
1
8. Answer – C
We know that, for the coordinates of a mirror image of a point in x-axis,
abscissa remains the same and ordinate will be of opposite sign of the ordinate
of given point. So, the Mirror image of the point (-3, 5) about x-axis is (-3, -5).
1
9. Answer – B
As, ?ABC ~ ?EFD ? ?A = ?E
1
Page 2 of 11
10. Answer – B
As, ?ABC~?PQR ?
AB
PQ
=
BC
QR
=
AC
PR
=
1
2
? PQ = 6 cm, QR = 8 cm
Perimeter of the triangle PQR (in cm) = 6 + 8 +10 = 24
1
Question given for Visually impaired candidates
Answer – B
The solution is same as above.
1
11. Answer – A
From the figure, AE = 24 – r = AF. So, BF = 1 + r = 7 – r ? r = 3 cm
1
Question given for Visually Impaired candidates
Answer – B
As, PQ = PR = 24 cm
So, Area of Quadrilateral PQOR (in cm
2
) = 2 ×
1
2
× 24 × 10 = 240
1
12. Answer – B
As, cot
2
?? - cosec
2
?? = -?? , so it is NOT equal to Unity
1
13. Answer – C
As, Median class is 10-15. So, its upper limit is 15.
1
14. Answer – C
Since, 3 Median = Mode + 2 Mean. So, a = 3 & b = 2.
Thus, (2b + 3a) = 4 + 9 = 13
1
15. Answer – B
Radius (in cm) = v13
2
- 12
2
= 5
1
16. Answer – A
As, ?PAO= 90°. So, ?APO = 115° - 90° = 25°
1
Question given for Visually Impaired candidates
Answer – A
As, the chord is at a distance of 18 cm (more than the radius). So, the chord
will be at a distance of 5 cm on the opposite side of the centre. Thus, length
of the chord CD will be 2v13
2
- 5
2
= 24 ????
1
17. Answer – C
As, r1 : r2 = 3 : 4. So, the ratio of their areas = r1
2
: r2
2
= 9 :16
1
18. Answer – A
Since, the event is most unlikely to happen. Therefore, its probability is 0.0001
1
19. Answer – A
As, Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
1
Page 3 of 11
20. Answer – D
Q1. Since events given in Assertion are not equally likely, so probability of getting
two heads is not
1
3
.
Thus, Assertion (A) is false but reason (R) is true.
1
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
21 (A). It can be observed that,
2 × 5 × 7 × 11 + 11 × 13 = 11 × (70 + 13) = 11 x 83
which is the product of two factors other than 1. Therefore, it is a composite
number.
OR
1
1
21 (B). The smallest number which is divisible by any two numbers is their LCM.
So, Number which is divisible by both 306 and 657 = LCM (306, 657)
Since, 306 = 2
1
× 3
2
× 17
1
and 657 = 3
2
× 73
LCM (306, 657) = 2
1
× 3
2
× 17
1
× 73 = 22338
½
1
½
22. As, P(3, a) lies on the line L, so 3 + ?? = 5 ? ?? = 2
Now, the radius of the circle = CP = v3
2
+ 2
2
= v13 ??????????
Question given for Visually Impaired candidates
Diameter of the circle = Distance between (0,0) and (6,8) = v6
2
+ 8
2
= 10
Radius of the circle = ½ (Diameter of the circle) = 5 units
1
1
1½
½
23. Sum of the zeroes = 2 – 3 = -(?? + 1) ? ?? = 0
Product of the zeroes = – 6 = ?? ? ?? = -6
Hence, ?? = 0 & ?? = -6
1
1
24.
Discriminant, D = 16 + 12v2 > 0
As, Discriminant is positive. So, Roots are real and distinct.
1
1
25 (A).
Q2. 2 sin30°tan60°- 3 cos
2
60°sec
2
30° = 2 (
1
2
) (v3) - 3 (
1
2
)
2
(
2
v3
)
2
Q3. = v3 - 1
OR
1½
½
25 (B).
As , sinx . cosx (tanx+ cotx ) = sinx . cosx(
sinx
cosx
+
cosx
sinx
).
= sinx . cosx (
1
cosx .sinx
)
= 1 (Constant)
Since, the value of sinx. cosx (tanx + cotx) is constant, so its equal 1 for all
angles.
½
1½
Page 4 of 11
Section –C
[This section comprises of solution short answer type questions (SA) of 3 marks each]
26.
To prove that (v2 - v5) is an irrational number, we will use the contradiction
Method.
Let, if possible, v2 - v5 = ?? , where ?? is any rational number (Clearly ?? ? 0)
so , v2 = ?? + v5 ? 2 = (?? + v5 )
2
? 2 = ?? 2
+ 5 + 2v5??
? -?? 2
- 3 = 2v5??
?
-?? 2
-3
2?? = v5 ……(1)
(Note: v5 is an irrational number, as the square root of any prime number is
Always an irrational number)
In equation (1), LHS is a rational number while RHS is an irrational number
but an irrational number cannot be equal to a rational number.
So, our assumption is wrong.
Thus, (v2 - v5) is an irrational number.
1
1
1
27 (A).
Area of land
(in hectares)
No. of
families
1 – 3 20
3 – 5 45 f0
Modal class 5 – 7 80 f1
7 – 9 55 f2
9 – 11 40
11 – 13 12
? Modal class = 5 – 7 , l = 5, h = 2
Mode = l + (
?? 1
-?? 0
2?? 1
-?? 0
-?? 2
)h = 5 + (
80-45
2(80)-45-55
) 2 = 6.166…..
Q4. Hence, modal agriculture holdings of the village is 6.17 hectare (approx.)
OR
1
2
Page 5
Page 1 of 11
MATHEMATICS BASIC – Code No. 241
MARKING SCHEME
CLASS - X (2025 - 26)
SECTION - A
Q. No. Answer Marks
1. Answer – D
As, 2025 = 3
4
× 5
4
So, the exponent of 3 in the prime factorization of 2025 is 4
1
2. Answer – B
On subtracting first equation from second equation, we get
2025 ?? + 2024 ?? - 2024 ?? - 2025 ?? = -1 - 1 ? (?? - ?? ) = -??
1
3. Answer – D
As, ?? (?? ) = k(x + 2)(x - 5) ? ?? (?? ) = k(?? 2
- 3x - 10), k ? 0
Since k can be any real number. So, there are Infinitely many such polynomials.
1
4. Answer – C
On simplification, given equations reduce to
(?? ) ?? 2
+ 2?? - 2 = 0 (?????????????????? ???????????????? )
(?? ) 2?? 2
- 3?? - 1 = 0 (?????????????????? ???????????????? )
(C) 3?? + 1 = 0 (?????? ?? ?????????????????? ???????????????? )
(D) 4?? 2
+ ?? = 0 (?????????????????? ???????????????? )
1
5. Answer – A
As, 2(?? + 10) = (3?? + 2) + 2?? ? ?? = 6
1
6. Answer – B
As,
50(51)
2
= 25?? ? ?? = 51
1
7.
Answer – D
Distance between the given points =
v
(
1
2
-
v3
2
)
2
+ (
1
2
+
v3
2
)
2
= v2
1
8. Answer – C
We know that, for the coordinates of a mirror image of a point in x-axis,
abscissa remains the same and ordinate will be of opposite sign of the ordinate
of given point. So, the Mirror image of the point (-3, 5) about x-axis is (-3, -5).
1
9. Answer – B
As, ?ABC ~ ?EFD ? ?A = ?E
1
Page 2 of 11
10. Answer – B
As, ?ABC~?PQR ?
AB
PQ
=
BC
QR
=
AC
PR
=
1
2
? PQ = 6 cm, QR = 8 cm
Perimeter of the triangle PQR (in cm) = 6 + 8 +10 = 24
1
Question given for Visually impaired candidates
Answer – B
The solution is same as above.
1
11. Answer – A
From the figure, AE = 24 – r = AF. So, BF = 1 + r = 7 – r ? r = 3 cm
1
Question given for Visually Impaired candidates
Answer – B
As, PQ = PR = 24 cm
So, Area of Quadrilateral PQOR (in cm
2
) = 2 ×
1
2
× 24 × 10 = 240
1
12. Answer – B
As, cot
2
?? - cosec
2
?? = -?? , so it is NOT equal to Unity
1
13. Answer – C
As, Median class is 10-15. So, its upper limit is 15.
1
14. Answer – C
Since, 3 Median = Mode + 2 Mean. So, a = 3 & b = 2.
Thus, (2b + 3a) = 4 + 9 = 13
1
15. Answer – B
Radius (in cm) = v13
2
- 12
2
= 5
1
16. Answer – A
As, ?PAO= 90°. So, ?APO = 115° - 90° = 25°
1
Question given for Visually Impaired candidates
Answer – A
As, the chord is at a distance of 18 cm (more than the radius). So, the chord
will be at a distance of 5 cm on the opposite side of the centre. Thus, length
of the chord CD will be 2v13
2
- 5
2
= 24 ????
1
17. Answer – C
As, r1 : r2 = 3 : 4. So, the ratio of their areas = r1
2
: r2
2
= 9 :16
1
18. Answer – A
Since, the event is most unlikely to happen. Therefore, its probability is 0.0001
1
19. Answer – A
As, Both assertion (A) and reason (R) are true and reason (R) is the correct
explanation of assertion (A).
1
Page 3 of 11
20. Answer – D
Q1. Since events given in Assertion are not equally likely, so probability of getting
two heads is not
1
3
.
Thus, Assertion (A) is false but reason (R) is true.
1
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
21 (A). It can be observed that,
2 × 5 × 7 × 11 + 11 × 13 = 11 × (70 + 13) = 11 x 83
which is the product of two factors other than 1. Therefore, it is a composite
number.
OR
1
1
21 (B). The smallest number which is divisible by any two numbers is their LCM.
So, Number which is divisible by both 306 and 657 = LCM (306, 657)
Since, 306 = 2
1
× 3
2
× 17
1
and 657 = 3
2
× 73
LCM (306, 657) = 2
1
× 3
2
× 17
1
× 73 = 22338
½
1
½
22. As, P(3, a) lies on the line L, so 3 + ?? = 5 ? ?? = 2
Now, the radius of the circle = CP = v3
2
+ 2
2
= v13 ??????????
Question given for Visually Impaired candidates
Diameter of the circle = Distance between (0,0) and (6,8) = v6
2
+ 8
2
= 10
Radius of the circle = ½ (Diameter of the circle) = 5 units
1
1
1½
½
23. Sum of the zeroes = 2 – 3 = -(?? + 1) ? ?? = 0
Product of the zeroes = – 6 = ?? ? ?? = -6
Hence, ?? = 0 & ?? = -6
1
1
24.
Discriminant, D = 16 + 12v2 > 0
As, Discriminant is positive. So, Roots are real and distinct.
1
1
25 (A).
Q2. 2 sin30°tan60°- 3 cos
2
60°sec
2
30° = 2 (
1
2
) (v3) - 3 (
1
2
)
2
(
2
v3
)
2
Q3. = v3 - 1
OR
1½
½
25 (B).
As , sinx . cosx (tanx+ cotx ) = sinx . cosx(
sinx
cosx
+
cosx
sinx
).
= sinx . cosx (
1
cosx .sinx
)
= 1 (Constant)
Since, the value of sinx. cosx (tanx + cotx) is constant, so its equal 1 for all
angles.
½
1½
Page 4 of 11
Section –C
[This section comprises of solution short answer type questions (SA) of 3 marks each]
26.
To prove that (v2 - v5) is an irrational number, we will use the contradiction
Method.
Let, if possible, v2 - v5 = ?? , where ?? is any rational number (Clearly ?? ? 0)
so , v2 = ?? + v5 ? 2 = (?? + v5 )
2
? 2 = ?? 2
+ 5 + 2v5??
? -?? 2
- 3 = 2v5??
?
-?? 2
-3
2?? = v5 ……(1)
(Note: v5 is an irrational number, as the square root of any prime number is
Always an irrational number)
In equation (1), LHS is a rational number while RHS is an irrational number
but an irrational number cannot be equal to a rational number.
So, our assumption is wrong.
Thus, (v2 - v5) is an irrational number.
1
1
1
27 (A).
Area of land
(in hectares)
No. of
families
1 – 3 20
3 – 5 45 f0
Modal class 5 – 7 80 f1
7 – 9 55 f2
9 – 11 40
11 – 13 12
? Modal class = 5 – 7 , l = 5, h = 2
Mode = l + (
?? 1
-?? 0
2?? 1
-?? 0
-?? 2
)h = 5 + (
80-45
2(80)-45-55
) 2 = 6.166…..
Q4. Hence, modal agriculture holdings of the village is 6.17 hectare (approx.)
OR
1
2
Page 5 of 11
27 (B).
Class interval fi
xi
(Mid-
value)
di=
?? ?? -????
?? fi di
0-20 7 10 - 1 - 7
20-40 p 30 0 0
40-60 10 50 1 10
60-80 9 70 2 18
80-100 13 90 3 39
Total 39 + p 60
Assumed mean(A) = 30, Width of the interval (h) = 20
Mean= 30 +
60
39+p
× 20 = 54 ? 50 = 39 + p ? p = 11
2
1
28.
Tangents drawn to a circle from an external point are equal.
So, AP = AS, PB = BQ,
CR = CQ, DR = DS
On adding the above equations,
(AP+ PB ) + (CR+ RD ) = (AS + BQ ) + (CQ+ DS)
? AB+ CD = AD+ BC
?
AB + CD
AD + BC
= 1
1½
1
½
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FAQs on Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme with Solutions (2025-26) - Mathematics (Maths) Class 10
| 1. What is the marking scheme for Class 10 Mathematics (Basic) CBSE exams? |  |
Ans. The marking scheme for Class 10 Mathematics (Basic) typically includes a total of 100 marks, divided into various sections. The exam usually consists of multiple-choice questions, short answer questions, and long answer questions. For example, there might be around 20 multiple-choice questions worth 1 mark each, 6 short answer questions worth 2 marks each, and 8 long answer questions worth 3 marks each. Additionally, a section on internal assessment may contribute to the overall score.
| 2. How can students prepare effectively for the Class 10 Mathematics (Basic) exam? | |
Ans. To prepare effectively, students should start by understanding the syllabus and the exam pattern. Regular practice of sample papers and previous years' question papers is crucial. They should focus on key topics such as algebra, geometry, and statistics. Creating a study schedule, revising regularly, and seeking help for difficult concepts can significantly enhance understanding and retention. Joining study groups or tutoring can also be beneficial.
| 3. What are the important chapters to focus on for the Class 10 Mathematics (Basic) exam? | |
Ans. Important chapters typically include Number Systems, Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, and Statistics. Geometry topics such as Triangles, Circles, and Coordinate Geometry are also significant. Students should prioritize these chapters as they often carry more weight in terms of marks and are frequently tested in exams.
| 4. Are there any specific strategies to solve mathematics problems in the exam efficiently? | |
Ans. Yes, several strategies can help solve problems efficiently. First, read each question carefully and identify what is being asked. Break down complex problems into smaller, manageable parts. Use diagrams or sketches for geometry-related questions. For numerical problems, write down the formulas and apply them systematically. Time management is crucial, so practice solving problems within a set time limit to improve speed and accuracy.
| 5. How is internal assessment conducted in Class 10 Mathematics (Basic)? | |
Ans. Internal assessment in Class 10 Mathematics (Basic) usually comprises periodic tests, assignments, and practical work. Schools may conduct regular assessments throughout the academic year, and these scores contribute to the final grade. The internal assessment aims to evaluate students' understanding and application of mathematical concepts in real-world scenarios, promoting continuous learning and improvement.
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