Class 10 Mathematics (Basic): CBSE (Official) Marking Scheme with Solution (2022-23) | Mathematics (Maths) Class 10 PDF Download

 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