Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Basic) - 1
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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 - Mathematics (Maths) Class 10
| 1. What is the structure of the CBSE Class 10 Mathematics Sample Question Paper? |  |
Ans. The CBSE Class 10 Mathematics Sample Question Paper typically consists of various sections including multiple-choice questions, short answer questions, and long answer questions. It is designed to assess a student's understanding of key mathematical concepts and their ability to apply these concepts to solve problems.
| 2. How can students effectively prepare for the CBSE Class 10 Mathematics exam? | |
Ans. Students can prepare effectively by reviewing the syllabus, practicing previous years' question papers, and utilizing sample question papers. Additionally, focusing on understanding concepts rather than rote memorization, attending revision classes, and solving problems regularly can greatly enhance their preparation.
| 3. What types of questions can be expected in the CBSE Class 10 Mathematics exam? | |
Ans. The exam generally includes a variety of question types such as objective-type questions, very short answer questions, short answer questions, and long answer questions. Topics covered may include algebra, geometry, statistics, and probability, among others.
| 4. Are there any specific marking schemes for the CBSE Class 10 Mathematics exam? | |
Ans. Yes, the marking scheme for the CBSE Class 10 Mathematics exam typically allocates different marks to different types of questions. For example, multiple-choice questions may carry 1 mark each, while long answer questions may carry 4 or 5 marks each, depending on their complexity.
| 5. Where can students find solutions for the CBSE Class 10 Mathematics Sample Question Paper? | |
Ans. Students can find solutions for the CBSE Class 10 Mathematics Sample Question Paper in various resources such as official CBSE publications, educational websites, and study guides. Additionally, many coaching centers and online platforms offer detailed solutions and explanations for these papers.
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