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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)
v3
2
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)
v3
2
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)
v3
2
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)
v3
2
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)
v3
2
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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 - CBSE Sample Papers For Class 10
| 1. What is the CBSE exam pattern for Class 10 Mathematics? |  |
Ans. The CBSE exam pattern for Class 10 Mathematics includes a total of 80 marks. It is divided into two sections: Section A (objective type questions) and Section B (subjective type questions). Section A carries 20 marks and consists of multiple-choice questions, while Section B carries 60 marks and consists of long answer and short answer type questions.
| 2. How can I prepare effectively for the Class 10 Mathematics exam? | |
Ans. To prepare effectively for the Class 10 Mathematics exam, you can follow these tips:
- Understand the syllabus and exam pattern thoroughly.
- Create a study schedule and allocate specific time for each topic.
- Practice solving sample papers and previous year question papers to get familiar with the exam pattern.
- Identify your weak areas and focus on improving them.
- Take regular breaks during study sessions to avoid burnout.
- Seek help from teachers or classmates if you have any doubts or difficulties in understanding certain concepts.
| 3. Are there any important topics that I should focus on for the Class 10 Mathematics exam? | |
Ans. Yes, there are a few important topics that you should focus on for the Class 10 Mathematics exam. Some of these topics include:
- Real Numbers
- Polynomials
- Pair of Linear Equations in Two Variables
- Quadratic Equations
- Triangles
- Circles
- Constructions
- Surface Areas and Volumes
- Statistics and Probability
| 4. How can I improve my problem-solving skills in Mathematics? | |
Ans. To improve your problem-solving skills in Mathematics, you can follow these strategies:
- Understand the concepts and formulas thoroughly.
- Practice solving a variety of problems from different sources.
- Break down complex problems into smaller, manageable steps.
- Look for patterns and similarities in different types of problems.
- Discuss and solve problems with peers or study groups.
- Analyze your mistakes and learn from them.
- Seek guidance from teachers or tutors if you need additional help.
| 5. What are some effective time management tips for the Class 10 Mathematics exam? | |
Ans. Here are some effective time management tips for the Class 10 Mathematics exam:
- Create a study schedule and allocate specific time for each topic or chapter.
- Prioritize topics based on their weightage and difficulty level.
- Set realistic goals and deadlines for completing each topic.
- Practice solving sample papers within the given time duration to improve speed and accuracy.
- Break down your study sessions into shorter, focused sessions to avoid fatigue.
- Take short breaks during study sessions to refresh your mind.
- Review and revise the topics regularly to retain the concepts effectively.
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