Class 10 Exam > Class 10 Notes > CBSE Sample Papers For Class 10 > Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1
Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 | 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)
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
Read More
|
303 docs|7 tests
|
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.
Related Exams
About this Document
|
4.97/5 Rating |
|
Dec 26, 2024 Last updated |
Document Description: Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 for Class 10 2024 is part of CBSE Sample Papers For Class 10 preparation.
The notes and questions for Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 have been prepared according to the Class 10 exam syllabus. Information about Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 covers topics
like and Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1.
Introduction of Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 in English is available as part of our CBSE Sample Papers For Class 10
for Class 10 & Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 | CBSE Sample Papers For Class 10
Description
Full syllabus notes, lecture & questions for Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 | 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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1
In this doc you can find the meaning of Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 defined & explained in the simplest way possible. Besides explaining types of
Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 theory, EduRev gives you an ample number of questions to practice Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 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
Previous Year Questions with Solutions
,Extra Questions
,practice quizzes
,mock tests for examination
,ppt
,shortcuts and tricks
,video lectures
,Viva Questions
,Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 | CBSE Sample Papers For Class 10
,study material
,Summary
,MCQs
,Free
,Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 | CBSE Sample Papers For Class 10
,Objective type Questions
,past year papers
,Important questions
,Semester Notes
,Exam
,Sample Paper
,Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 | CBSE Sample Papers For Class 10
;
Additional Information about Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 for Class 10 Preparation
Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 Free PDF Download
The Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 now and kickstart your journey towards success in the Class 10 exam.
Importance of Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1
The importance of Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 Notes
Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1.
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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 Notes on EduRev are your ultimate resource for success.
Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 Class 10 Questions
The "Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 on the App
Students of Class 10 can study Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1,
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: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1 is prepared as per the latest Class 10 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup