Class 10 Exam > Class 10 Notes > CBSE Sample Papers For Class 10 > Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23)
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) | CBSE Sample Papers For Class 10 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Class- X
Mathematics Basic (241)
Marking Scheme SQP-2022-23
Time Allowed: 3 Hours Maximum Marks: 80
Section A
1 (c) a
3
b
2
1
2 (c) 13 km/hours 1
3 (b) -10 1
4 (b) Parallel. 1
5 (c) k = 4 1
6 (b) 12 1
7 (c) ?B = ?D 1
8 (b) 5 : 1 1
9 (a) 25° 1
10
(a)
2
v3
1
11
(c) v3
1
12 (b) 0 1
13 (b) 14 : 11 1
14 (c) 16 : 9 1
15 (d) 147p cm
2
1
16 (c) 20 1
17 (b) 8 1
18
(a)
3
26
1
19 (d) Assertion (A) is false but Reason (R) is true. 1
Page 2
Class- X
Mathematics Basic (241)
Marking Scheme SQP-2022-23
Time Allowed: 3 Hours Maximum Marks: 80
Section A
1 (c) a
3
b
2
1
2 (c) 13 km/hours 1
3 (b) -10 1
4 (b) Parallel. 1
5 (c) k = 4 1
6 (b) 12 1
7 (c) ?B = ?D 1
8 (b) 5 : 1 1
9 (a) 25° 1
10
(a)
2
v3
1
11
(c) v3
1
12 (b) 0 1
13 (b) 14 : 11 1
14 (c) 16 : 9 1
15 (d) 147p cm
2
1
16 (c) 20 1
17 (b) 8 1
18
(a)
3
26
1
19 (d) Assertion (A) is false but Reason (R) is true. 1
20 (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation
of Assertion (A).
1
Section B
21
For a pair of linear equations to have infinitely many solutions :
a
1
a
2
=
b
1
b
2
=
c
1
c
2
?
k
12
=
3
k
=
k-3
k
?? 12
=
3
?? ? k
2
= 36 ? k = ± 6
Also,
3
?? =
?? -3
?? ? k
2
– 6k = 0 ? k = 0, 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6.
½
½
½
½
22
(i) In ?ABD and ?CBE
?ADB = ?CEB = 90º
?ABD = ?CBE (Common angle)
? ?ABD ~ ?CBE (AA criterion)
(ii) In ?PDC and ?BEC
?PDC = ?BEC = 90º
?PCD = ?BCE (Common angle)
? ?PDC ~ ?BEC (AA criterion)
[OR]
In ?ABC, DE || AC
BD/AD = BE/EC .........(i) (Using BPT)
In ?ABE, DF || AE
BD/AD = BF/FE ........(ii) (Using BPT)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus,
BF
FE
=
BE
EC
½
½
½
½
½
½
½
½
23
Let O be the centre of the concentric circle of radii 5 cm
and 3 cm respectively. Let AB be a chord of the larger circle
touching the smaller circle at P
Then AP = PB and OP?AB
Applying Pythagoras theorem in ?OPA, we have
OA
2
=OP
2
+AP
2
? 25 = 9 + AP
2
? AP
2
= 16 ? AP = 4 cm
? AB = 2AP = 8 cm
½
½
½
½
24
Now,
(1 + sin?)(1 - sin?)
(1 + cos?)(1 - cos?)
=
(1 – sin
2
?)
(1 – cos
2
?)
=
cos
2
?
sin
2
?
= (
cos?
sin?
)
2
= cot
2
?
= (
7
8
)
2
=
49
64
½
½
½
½
Page 3
Class- X
Mathematics Basic (241)
Marking Scheme SQP-2022-23
Time Allowed: 3 Hours Maximum Marks: 80
Section A
1 (c) a
3
b
2
1
2 (c) 13 km/hours 1
3 (b) -10 1
4 (b) Parallel. 1
5 (c) k = 4 1
6 (b) 12 1
7 (c) ?B = ?D 1
8 (b) 5 : 1 1
9 (a) 25° 1
10
(a)
2
v3
1
11
(c) v3
1
12 (b) 0 1
13 (b) 14 : 11 1
14 (c) 16 : 9 1
15 (d) 147p cm
2
1
16 (c) 20 1
17 (b) 8 1
18
(a)
3
26
1
19 (d) Assertion (A) is false but Reason (R) is true. 1
20 (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation
of Assertion (A).
1
Section B
21
For a pair of linear equations to have infinitely many solutions :
a
1
a
2
=
b
1
b
2
=
c
1
c
2
?
k
12
=
3
k
=
k-3
k
?? 12
=
3
?? ? k
2
= 36 ? k = ± 6
Also,
3
?? =
?? -3
?? ? k
2
– 6k = 0 ? k = 0, 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6.
½
½
½
½
22
(i) In ?ABD and ?CBE
?ADB = ?CEB = 90º
?ABD = ?CBE (Common angle)
? ?ABD ~ ?CBE (AA criterion)
(ii) In ?PDC and ?BEC
?PDC = ?BEC = 90º
?PCD = ?BCE (Common angle)
? ?PDC ~ ?BEC (AA criterion)
[OR]
In ?ABC, DE || AC
BD/AD = BE/EC .........(i) (Using BPT)
In ?ABE, DF || AE
BD/AD = BF/FE ........(ii) (Using BPT)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus,
BF
FE
=
BE
EC
½
½
½
½
½
½
½
½
23
Let O be the centre of the concentric circle of radii 5 cm
and 3 cm respectively. Let AB be a chord of the larger circle
touching the smaller circle at P
Then AP = PB and OP?AB
Applying Pythagoras theorem in ?OPA, we have
OA
2
=OP
2
+AP
2
? 25 = 9 + AP
2
? AP
2
= 16 ? AP = 4 cm
? AB = 2AP = 8 cm
½
½
½
½
24
Now,
(1 + sin?)(1 - sin?)
(1 + cos?)(1 - cos?)
=
(1 – sin
2
?)
(1 – cos
2
?)
=
cos
2
?
sin
2
?
= (
cos?
sin?
)
2
= cot
2
?
= (
7
8
)
2
=
49
64
½
½
½
½
25
Perimeter of quadrant = 2r +
1
4
× 2 p r
? Perimeter = 2 × 14 +
1
2
×
22
7
× 14
? Perimeter = 28 + 22 =28+22 = 50 cm
[OR]
Area of the circle = Area of first circle + Area of second circle
? pR
2
= p (r1)
2
+ p (r1)
2
? pR
2
= p (24)
2
+ p (7)
2
? pR
2
= 576p +49p
? pR
2
= 625p ? R
2
= 625 ? R = 25 Thus, diameter of the circle = 2R = 50 cm.
½
½
1
½
½
1
Section C
26
Let us assume to the contrary, that v5 is rational. Then we can find a and b ( ? 0) such
that v5 =
?? ?? (assuming that a and b are co-primes).
So, a = v5 b ? a
2
= 5b
2
Here 5 is a prime number that divides a
2
then 5 divides a also
(Using the theorem, if a is a prime number and if a divides p
2
, then a divides p, where a is
a positive integer)
Thus 5 is a factor of a
Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c
We get (5c)
2
= 5b
2
? 5c
2
= b
2
This means 5 divides b
2
so 5 divides b also (Using the theorem, if a is a prime number and
if a divides p
2
, then a divides p, where a is a positive integer).
Hence a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime. This is the contradiction to our
assumption that p and q are co-primes.
So, v5 is not a rational number. Therefore, the v5 is irrational.
1
½
½
½
½
27
6x
2
– 7x – 3 = 0 ? 6x
2
– 9x + 2x – 3 = 0
? 3x(2x – 3) + 1(2x – 3) = 0 ? (2x – 3)(3x + 1) = 0
? 2x – 3 = 0 & 3x + 1 = 0
x = 3/2 & x = -1/3 Hence, the zeros of the quadratic polynomials are 3/2 and -1/3.
For verification
Sum of zeros =
– coefficient of x
coefficient of x
2
? 3/2 + (-1/3) = – (-7) / 6 ? 7/6 = 7/6
Product of roots =
constant
coefficient of x
2
? 3/2 x (-1/3) = (-3) / 6 ? -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.
½
½
1
1
28
Let the fixed charge by Rs x and additional charge by Rs y per day
Number of days for Latika = 6 = 2 + 4
Hence, Charge x + 4y = 22
x = 22 – 4y ………(1)
Number of days for Anand = 4 = 2 + 2
Hence, Charge x + 2y = 16
x = 16 – 2y ……. (2)
On comparing equation (1) and (2), we get,
½
½
Page 4
Class- X
Mathematics Basic (241)
Marking Scheme SQP-2022-23
Time Allowed: 3 Hours Maximum Marks: 80
Section A
1 (c) a
3
b
2
1
2 (c) 13 km/hours 1
3 (b) -10 1
4 (b) Parallel. 1
5 (c) k = 4 1
6 (b) 12 1
7 (c) ?B = ?D 1
8 (b) 5 : 1 1
9 (a) 25° 1
10
(a)
2
v3
1
11
(c) v3
1
12 (b) 0 1
13 (b) 14 : 11 1
14 (c) 16 : 9 1
15 (d) 147p cm
2
1
16 (c) 20 1
17 (b) 8 1
18
(a)
3
26
1
19 (d) Assertion (A) is false but Reason (R) is true. 1
20 (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation
of Assertion (A).
1
Section B
21
For a pair of linear equations to have infinitely many solutions :
a
1
a
2
=
b
1
b
2
=
c
1
c
2
?
k
12
=
3
k
=
k-3
k
?? 12
=
3
?? ? k
2
= 36 ? k = ± 6
Also,
3
?? =
?? -3
?? ? k
2
– 6k = 0 ? k = 0, 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6.
½
½
½
½
22
(i) In ?ABD and ?CBE
?ADB = ?CEB = 90º
?ABD = ?CBE (Common angle)
? ?ABD ~ ?CBE (AA criterion)
(ii) In ?PDC and ?BEC
?PDC = ?BEC = 90º
?PCD = ?BCE (Common angle)
? ?PDC ~ ?BEC (AA criterion)
[OR]
In ?ABC, DE || AC
BD/AD = BE/EC .........(i) (Using BPT)
In ?ABE, DF || AE
BD/AD = BF/FE ........(ii) (Using BPT)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus,
BF
FE
=
BE
EC
½
½
½
½
½
½
½
½
23
Let O be the centre of the concentric circle of radii 5 cm
and 3 cm respectively. Let AB be a chord of the larger circle
touching the smaller circle at P
Then AP = PB and OP?AB
Applying Pythagoras theorem in ?OPA, we have
OA
2
=OP
2
+AP
2
? 25 = 9 + AP
2
? AP
2
= 16 ? AP = 4 cm
? AB = 2AP = 8 cm
½
½
½
½
24
Now,
(1 + sin?)(1 - sin?)
(1 + cos?)(1 - cos?)
=
(1 – sin
2
?)
(1 – cos
2
?)
=
cos
2
?
sin
2
?
= (
cos?
sin?
)
2
= cot
2
?
= (
7
8
)
2
=
49
64
½
½
½
½
25
Perimeter of quadrant = 2r +
1
4
× 2 p r
? Perimeter = 2 × 14 +
1
2
×
22
7
× 14
? Perimeter = 28 + 22 =28+22 = 50 cm
[OR]
Area of the circle = Area of first circle + Area of second circle
? pR
2
= p (r1)
2
+ p (r1)
2
? pR
2
= p (24)
2
+ p (7)
2
? pR
2
= 576p +49p
? pR
2
= 625p ? R
2
= 625 ? R = 25 Thus, diameter of the circle = 2R = 50 cm.
½
½
1
½
½
1
Section C
26
Let us assume to the contrary, that v5 is rational. Then we can find a and b ( ? 0) such
that v5 =
?? ?? (assuming that a and b are co-primes).
So, a = v5 b ? a
2
= 5b
2
Here 5 is a prime number that divides a
2
then 5 divides a also
(Using the theorem, if a is a prime number and if a divides p
2
, then a divides p, where a is
a positive integer)
Thus 5 is a factor of a
Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c
We get (5c)
2
= 5b
2
? 5c
2
= b
2
This means 5 divides b
2
so 5 divides b also (Using the theorem, if a is a prime number and
if a divides p
2
, then a divides p, where a is a positive integer).
Hence a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime. This is the contradiction to our
assumption that p and q are co-primes.
So, v5 is not a rational number. Therefore, the v5 is irrational.
1
½
½
½
½
27
6x
2
– 7x – 3 = 0 ? 6x
2
– 9x + 2x – 3 = 0
? 3x(2x – 3) + 1(2x – 3) = 0 ? (2x – 3)(3x + 1) = 0
? 2x – 3 = 0 & 3x + 1 = 0
x = 3/2 & x = -1/3 Hence, the zeros of the quadratic polynomials are 3/2 and -1/3.
For verification
Sum of zeros =
– coefficient of x
coefficient of x
2
? 3/2 + (-1/3) = – (-7) / 6 ? 7/6 = 7/6
Product of roots =
constant
coefficient of x
2
? 3/2 x (-1/3) = (-3) / 6 ? -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.
½
½
1
1
28
Let the fixed charge by Rs x and additional charge by Rs y per day
Number of days for Latika = 6 = 2 + 4
Hence, Charge x + 4y = 22
x = 22 – 4y ………(1)
Number of days for Anand = 4 = 2 + 2
Hence, Charge x + 2y = 16
x = 16 – 2y ……. (2)
On comparing equation (1) and (2), we get,
½
½
22 – 4y = 16 – 2y ? 2y = 6 ? y = 3
Substituting y = 3 in equation (1), we get,
x = 22 – 4 (3) ? x = 22 – 12 ? x = 10
Therefore, fixed charge = Rs 10 and additional charge = Rs 3 per day
[OR]
AB = 100 km. We know that, Distance = Speed × Time.
AP – BP = 100 ? 5x - 5y = 100 ? x-y=20.....(i)
AQ + BQ = 100 ? x + y = 100….(ii)
Adding equations (i) and (ii), we get,
x - y + x + y = 20 +100 ? 2x = 120 ? x = 60
Substituting x = 60 in equation (ii), we get, 60 + y = 100 ? y = 40
Therefore, the speed of the first car is 60 km/hr and the speed of the second car
is 40 km/hr.
1
1
½
½
1
1
29
.
Since OT is perpendicular bisector of PQ.
Therefore, PR=RQ=4 cm
Now, OR = v???? ?? - ????
?? = v?? ?? - ?? ?? =3cm
Now, ?TPR + ?RPO = 90° (?TPO=90°)
& ?TPR + ?PTR = 90° (?TRP=90°)
So, ?RPO = ?PTR
So, ?TRP ~ ?PRO [By A-A Rule of similar triangles]
So,
TP
PO
=
RP
RG
?
TP
5
=
4
3
? TP =
20
3
cm
½
½
½
½
½
½
30
LHS =
tan?
1-cot ?
+
cot ?
1-tan?
=
tan?
1-
1
tan ?
+
1
tan ?
1-tan?
=
tan
2
?
tan?-1
+
1
tan? (1-tan?)
=
tan
3
?-1
tan? (tan?-1)
=
(tan ? -1) (tan
3
? + tan ?+1 )
tan? (tan?-1)
=
(tan
3
? + tan ?+1 )
tan?
= tan ?+ 1 + sec = 1 + tan ?+ sec ?
= 1 +
sin ?
cos ?
+
cos ?
sin ?
= 1 +
sin
2
?+ cos
2
?
sin ? cos ?
½
½
½
½
½
Page 5
Class- X
Mathematics Basic (241)
Marking Scheme SQP-2022-23
Time Allowed: 3 Hours Maximum Marks: 80
Section A
1 (c) a
3
b
2
1
2 (c) 13 km/hours 1
3 (b) -10 1
4 (b) Parallel. 1
5 (c) k = 4 1
6 (b) 12 1
7 (c) ?B = ?D 1
8 (b) 5 : 1 1
9 (a) 25° 1
10
(a)
2
v3
1
11
(c) v3
1
12 (b) 0 1
13 (b) 14 : 11 1
14 (c) 16 : 9 1
15 (d) 147p cm
2
1
16 (c) 20 1
17 (b) 8 1
18
(a)
3
26
1
19 (d) Assertion (A) is false but Reason (R) is true. 1
20 (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation
of Assertion (A).
1
Section B
21
For a pair of linear equations to have infinitely many solutions :
a
1
a
2
=
b
1
b
2
=
c
1
c
2
?
k
12
=
3
k
=
k-3
k
?? 12
=
3
?? ? k
2
= 36 ? k = ± 6
Also,
3
?? =
?? -3
?? ? k
2
– 6k = 0 ? k = 0, 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6.
½
½
½
½
22
(i) In ?ABD and ?CBE
?ADB = ?CEB = 90º
?ABD = ?CBE (Common angle)
? ?ABD ~ ?CBE (AA criterion)
(ii) In ?PDC and ?BEC
?PDC = ?BEC = 90º
?PCD = ?BCE (Common angle)
? ?PDC ~ ?BEC (AA criterion)
[OR]
In ?ABC, DE || AC
BD/AD = BE/EC .........(i) (Using BPT)
In ?ABE, DF || AE
BD/AD = BF/FE ........(ii) (Using BPT)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus,
BF
FE
=
BE
EC
½
½
½
½
½
½
½
½
23
Let O be the centre of the concentric circle of radii 5 cm
and 3 cm respectively. Let AB be a chord of the larger circle
touching the smaller circle at P
Then AP = PB and OP?AB
Applying Pythagoras theorem in ?OPA, we have
OA
2
=OP
2
+AP
2
? 25 = 9 + AP
2
? AP
2
= 16 ? AP = 4 cm
? AB = 2AP = 8 cm
½
½
½
½
24
Now,
(1 + sin?)(1 - sin?)
(1 + cos?)(1 - cos?)
=
(1 – sin
2
?)
(1 – cos
2
?)
=
cos
2
?
sin
2
?
= (
cos?
sin?
)
2
= cot
2
?
= (
7
8
)
2
=
49
64
½
½
½
½
25
Perimeter of quadrant = 2r +
1
4
× 2 p r
? Perimeter = 2 × 14 +
1
2
×
22
7
× 14
? Perimeter = 28 + 22 =28+22 = 50 cm
[OR]
Area of the circle = Area of first circle + Area of second circle
? pR
2
= p (r1)
2
+ p (r1)
2
? pR
2
= p (24)
2
+ p (7)
2
? pR
2
= 576p +49p
? pR
2
= 625p ? R
2
= 625 ? R = 25 Thus, diameter of the circle = 2R = 50 cm.
½
½
1
½
½
1
Section C
26
Let us assume to the contrary, that v5 is rational. Then we can find a and b ( ? 0) such
that v5 =
?? ?? (assuming that a and b are co-primes).
So, a = v5 b ? a
2
= 5b
2
Here 5 is a prime number that divides a
2
then 5 divides a also
(Using the theorem, if a is a prime number and if a divides p
2
, then a divides p, where a is
a positive integer)
Thus 5 is a factor of a
Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c
We get (5c)
2
= 5b
2
? 5c
2
= b
2
This means 5 divides b
2
so 5 divides b also (Using the theorem, if a is a prime number and
if a divides p
2
, then a divides p, where a is a positive integer).
Hence a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime. This is the contradiction to our
assumption that p and q are co-primes.
So, v5 is not a rational number. Therefore, the v5 is irrational.
1
½
½
½
½
27
6x
2
– 7x – 3 = 0 ? 6x
2
– 9x + 2x – 3 = 0
? 3x(2x – 3) + 1(2x – 3) = 0 ? (2x – 3)(3x + 1) = 0
? 2x – 3 = 0 & 3x + 1 = 0
x = 3/2 & x = -1/3 Hence, the zeros of the quadratic polynomials are 3/2 and -1/3.
For verification
Sum of zeros =
– coefficient of x
coefficient of x
2
? 3/2 + (-1/3) = – (-7) / 6 ? 7/6 = 7/6
Product of roots =
constant
coefficient of x
2
? 3/2 x (-1/3) = (-3) / 6 ? -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.
½
½
1
1
28
Let the fixed charge by Rs x and additional charge by Rs y per day
Number of days for Latika = 6 = 2 + 4
Hence, Charge x + 4y = 22
x = 22 – 4y ………(1)
Number of days for Anand = 4 = 2 + 2
Hence, Charge x + 2y = 16
x = 16 – 2y ……. (2)
On comparing equation (1) and (2), we get,
½
½
22 – 4y = 16 – 2y ? 2y = 6 ? y = 3
Substituting y = 3 in equation (1), we get,
x = 22 – 4 (3) ? x = 22 – 12 ? x = 10
Therefore, fixed charge = Rs 10 and additional charge = Rs 3 per day
[OR]
AB = 100 km. We know that, Distance = Speed × Time.
AP – BP = 100 ? 5x - 5y = 100 ? x-y=20.....(i)
AQ + BQ = 100 ? x + y = 100….(ii)
Adding equations (i) and (ii), we get,
x - y + x + y = 20 +100 ? 2x = 120 ? x = 60
Substituting x = 60 in equation (ii), we get, 60 + y = 100 ? y = 40
Therefore, the speed of the first car is 60 km/hr and the speed of the second car
is 40 km/hr.
1
1
½
½
1
1
29
.
Since OT is perpendicular bisector of PQ.
Therefore, PR=RQ=4 cm
Now, OR = v???? ?? - ????
?? = v?? ?? - ?? ?? =3cm
Now, ?TPR + ?RPO = 90° (?TPO=90°)
& ?TPR + ?PTR = 90° (?TRP=90°)
So, ?RPO = ?PTR
So, ?TRP ~ ?PRO [By A-A Rule of similar triangles]
So,
TP
PO
=
RP
RG
?
TP
5
=
4
3
? TP =
20
3
cm
½
½
½
½
½
½
30
LHS =
tan?
1-cot ?
+
cot ?
1-tan?
=
tan?
1-
1
tan ?
+
1
tan ?
1-tan?
=
tan
2
?
tan?-1
+
1
tan? (1-tan?)
=
tan
3
?-1
tan? (tan?-1)
=
(tan ? -1) (tan
3
? + tan ?+1 )
tan? (tan?-1)
=
(tan
3
? + tan ?+1 )
tan?
= tan ?+ 1 + sec = 1 + tan ?+ sec ?
= 1 +
sin ?
cos ?
+
cos ?
sin ?
= 1 +
sin
2
?+ cos
2
?
sin ? cos ?
½
½
½
½
½
= 1 +
1
sin ? cos ?
= 1 + sec?cosec ?
[OR]
sin ? + cos ? = v3 ? (sin ? + cos ? )
2
= 3
? sin
2
?+ cos
2
? + 2sin ? cos ? = 3
? 1 + 2sin ? cos ? = 3 ? 1 sin ? cos ? = 1
Now tan? + cot? =
sin ?
cos ?
+
cos ?
isn ?
=
sin
2
?+ cos
2
?
sin ? cos ?
=
1
sin ? cos ?
=
1
1
= 1
½
½
½
½
½
½
½
31
(i) P(8 ) =
5
36
(ii) P(13 ) =
0
36
= 0
(iii) P(less than or equal to 12) = 1
1
1
1
Section D
32
Let the average speed of passenger train = x km/h.
and the average speed of express train = (x + 11) km/h
As per given data, time taken by the express train to cover 132 km is 1 hour less than the
passenger train to cover the same distance. Therefore,
132
?? -
132
?? +11
= 1
?
132 (?? +11-?? )
?? (?? +11)
= 1 ?
132 ?? 11
?? (?? +11)
= 1
? 132 × 11 = x(x + 11) ? x
2
+ 11x – 1452 = 0
? x
2
+ 44x -33x -1452 = 0
? x (x + 44) -33(x + 44) = 0 ? (x + 44)(x – 33) = 0
? x = – 44, 33
As the speed cannot be negative, the speed of the passenger train will be 33 km/h and the
speed of the express train will be 33 + 11 = 44 km/h.
[OR]
Let the speed of the stream be x km/hr
So, the speed of the boat in upstream = (18 - x) km/hr
& the speed of the boat in downstream = (18 + x) km/hr
ATQ,
distance
upstream speed
-
distance
downstream speed
= 1
?
24
18 - ?? -
24
18 + ?? = 1
½
1
½
1
1
½
½
½
½
1
Read More
|
303 docs|7 tests
|
FAQs on Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) - CBSE Sample Papers For Class 10
| 1. What is the marking scheme for the CBSE Class 10 Mathematics (Basic) exam in 2022-23? |  |
Ans. The marking scheme for the CBSE Class 10 Mathematics (Basic) exam in 2022-23 is the official document provided by the CBSE board that outlines the distribution of marks for various sections, topics, and question types in the exam. It provides guidelines for how the exam papers will be evaluated and how marks will be awarded to students based on their performance.
| 2. How can I access the CBSE (Official) Marking Scheme for Class 10 Mathematics (Basic) exam in 2022-23? | |
Ans. The CBSE (Official) Marking Scheme for the Class 10 Mathematics (Basic) exam in 2022-23 can be accessed on the official website of the CBSE board. It is usually made available for download in a PDF format. Students and teachers can visit the CBSE website, navigate to the examination section, and look for the marking scheme specifically for the Class 10 Mathematics (Basic) exam.
| 3. What does the CBSE (Official) Marking Scheme for Class 10 Mathematics (Basic) exam include? | |
Ans. The CBSE (Official) Marking Scheme for the Class 10 Mathematics (Basic) exam includes the following information:
- Weightage of marks for each section, like Algebra, Geometry, Statistics, etc.
- Breakdown of marks for different types of questions, such as short answer type, long answer type, etc.
- Guidelines for marking schemes for each question type, including step-wise marking.
- Sample answers or marking rubrics for better understanding.
- Important instructions for examiners regarding evaluating the papers and awarding marks.
| 4. How can the CBSE (Official) Marking Scheme help students in preparing for the Class 10 Mathematics (Basic) exam? | |
Ans. The CBSE (Official) Marking Scheme can be highly beneficial for students in their exam preparation. It provides an insight into the exam pattern, weightage of marks, and the marking scheme followed by the examiners. By referring to the marking scheme, students can understand the distribution of marks for different topics and question types, which helps them prioritize their study accordingly. It also helps them understand the expectations of the examiners in terms of the depth of knowledge and the presentation of answers.
| 5. Are the marks awarded in the Class 10 Mathematics (Basic) exam based solely on the CBSE (Official) Marking Scheme? | |
Ans. No, the marks awarded in the Class 10 Mathematics (Basic) exam are not solely based on the CBSE (Official) Marking Scheme. While the marking scheme provides guidelines to the examiners, the final marks are also influenced by the quality and accuracy of the answers provided by the students. Examiners evaluate the answer scripts based on the marking scheme, but they also consider other factors like logical reasoning, presentation, and overall understanding of the concepts. Therefore, it is important for students to focus on comprehensive preparation and provide clear and concise answers to maximize their marks.
Related Exams
About this Document
|
4.85/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) for Class 10 2024 is part of CBSE Sample Papers For Class 10 preparation.
The notes and questions for Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) have been prepared according to the Class 10 exam syllabus. Information about Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) covers topics
like and Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23).
Introduction of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) in English is available as part of our CBSE Sample Papers For Class 10
for Class 10 & Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) in Hindi for CBSE Sample Papers For Class 10 course.
Download more important topics related with notes, lectures and mock test series for Class 10
Exam by signing up for free. Class 10: Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) | CBSE Sample Papers For Class 10
Description
Full syllabus notes, lecture & questions for Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) | CBSE Sample Papers For Class 10 - Class 10 | Plus excerises question with solution to help you revise complete syllabus for CBSE Sample Papers For Class 10 | Best notes, free PDF download
Information about Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23)
In this doc you can find the meaning of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) defined & explained in the simplest way possible. Besides explaining types of
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) theory, EduRev gives you an ample number of questions to practice Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) tests, examples and also practice Class 10
tests
|
Explore Courses for Class 10 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Objective type Questions
,Semester Notes
,Viva Questions
,MCQs
,study material
,Important questions
,practice quizzes
,Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) | CBSE Sample Papers For Class 10
,Exam
,Previous Year Questions with Solutions
,mock tests for examination
,Extra Questions
,Summary
,Free
,Sample Paper
,shortcuts and tricks
,past year papers
,video lectures
,ppt
,Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) | CBSE Sample Papers For Class 10
,Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) | CBSE Sample Papers For Class 10
;
Additional Information about Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) for Class 10 Preparation
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) Free PDF Download
The Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) is an invaluable resource that delves deep into the core of the Class 10 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) now and kickstart your journey towards success in the Class 10 exam.
Importance of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23)
The importance of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) cannot be overstated, especially for Class 10 aspirants.
This document holds the key to success in the Class 10 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) Notes
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) Notes on EduRev are your ultimate resource for success.
Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) Class 10 Questions
The "Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) 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.
Study Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) on the App
Students of Class 10 can study Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme (2022-23) is prepared as per the latest Class 10 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup