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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
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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.
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