Class 10 Mathematics (Standard) Term I : CBSE (Official) Marking Scheme with Solution (2021-22) | Mathematics (Maths) Class 10 PDF Download

 Page 1


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Standard) 
  SECTION A  
QN Correct 
Option 
HINTS/SOLUTION MAR
KS 
1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 
HCF(4,2) = 4:2 = 2:1 
 
1 
2 (a) 
 For lines to coincide: 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 
                so,  
5
15
 = 
7
21
 = 
-3
-k
 
                i.e. k= 9 
 1 
3 (b) 
By Pythagoras theorem  
The required distance   =v(200² + 150²) 
     = v(40000+ 22500)  = v(62500)  = 250m. 
So the distance of the girl from the starting point is 250m.   
1 
4 (d) 
Area of the Rhombus = 
1
2
 d1d2 =
1
2
 x 24 x 32= 384 cm².  
Using Pythagoras theorem  
side² = (
1
2
d1)² + (
1
2
d2)² = 12² +16² = 144 +256 =400 
                             Side = 20cm 
 Area of the Rhombus = base x altitude 
384 = 20 x altitude 
So altitude = 384/20 = 19.2cm 
1 
5 (a)  Possible outcomes are (HH), (HT), (TH), (TT) 
Favorable outcomes(at the most one head) are (HT), (TH), (TT) 
So probability of getting at the most one head =3/4 
1 
6 (d)  Ratio of altitudes = Ratio of sides for similar triangles 
So AM:PN = AB:PQ = 2:3 
1 
7 (b) 2sin
2
ß – cos
2
ß = 2  
Then 2 sin
2
ß – (1- sin
2
ß) = 2  
3 sin
2
ß =3 or sin
2
ß =1 
ß is 90? 
1 
8 (c)  Since it has a terminating decimal expansion,  
so prime factors of the denominator will be 2,5 
1 
9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 
will intersect.  
1 
10 (d) Distance of  point A(-5,6) from the origin(0,0) is 
 v(0 + 5)
2
+ (0 - 6)
2
 = v25 + 36 = v61 units 
1 
11 (b) 
a²=23/25, then a = v 23/5, which is irrational 
1 
12 (c)  LCM X HCF = Product of two numbers 
36 X 2 = 18 X x 
x = 4 
 
1 
13 (b) 
tan A= v3 = tan 60° so ?A=60°, Hence ?C = 30°.  
So cos A cos C- sin A sin C = (1/2)x (v3/2) - (v3/2)x (1/2) =0 
  
1 
14 (a) 1x +1x +2x =180°, x = 45°. 
  ?A , ?B  and ?C are 45°, 45° and 90°resp. 
sec A
cosec B
 –  
tan A
cot B
  =  
sec 45
cosec 45
 –  
tan 45
cot 45
  = 
v 2
v 2
 –  
1
1
   = 1-1=  0  
  
1 
Page 2


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Standard) 
  SECTION A  
QN Correct 
Option 
HINTS/SOLUTION MAR
KS 
1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 
HCF(4,2) = 4:2 = 2:1 
 
1 
2 (a) 
 For lines to coincide: 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 
                so,  
5
15
 = 
7
21
 = 
-3
-k
 
                i.e. k= 9 
 1 
3 (b) 
By Pythagoras theorem  
The required distance   =v(200² + 150²) 
     = v(40000+ 22500)  = v(62500)  = 250m. 
So the distance of the girl from the starting point is 250m.   
1 
4 (d) 
Area of the Rhombus = 
1
2
 d1d2 =
1
2
 x 24 x 32= 384 cm².  
Using Pythagoras theorem  
side² = (
1
2
d1)² + (
1
2
d2)² = 12² +16² = 144 +256 =400 
                             Side = 20cm 
 Area of the Rhombus = base x altitude 
384 = 20 x altitude 
So altitude = 384/20 = 19.2cm 
1 
5 (a)  Possible outcomes are (HH), (HT), (TH), (TT) 
Favorable outcomes(at the most one head) are (HT), (TH), (TT) 
So probability of getting at the most one head =3/4 
1 
6 (d)  Ratio of altitudes = Ratio of sides for similar triangles 
So AM:PN = AB:PQ = 2:3 
1 
7 (b) 2sin
2
ß – cos
2
ß = 2  
Then 2 sin
2
ß – (1- sin
2
ß) = 2  
3 sin
2
ß =3 or sin
2
ß =1 
ß is 90? 
1 
8 (c)  Since it has a terminating decimal expansion,  
so prime factors of the denominator will be 2,5 
1 
9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 
will intersect.  
1 
10 (d) Distance of  point A(-5,6) from the origin(0,0) is 
 v(0 + 5)
2
+ (0 - 6)
2
 = v25 + 36 = v61 units 
1 
11 (b) 
a²=23/25, then a = v 23/5, which is irrational 
1 
12 (c)  LCM X HCF = Product of two numbers 
36 X 2 = 18 X x 
x = 4 
 
1 
13 (b) 
tan A= v3 = tan 60° so ?A=60°, Hence ?C = 30°.  
So cos A cos C- sin A sin C = (1/2)x (v3/2) - (v3/2)x (1/2) =0 
  
1 
14 (a) 1x +1x +2x =180°, x = 45°. 
  ?A , ?B  and ?C are 45°, 45° and 90°resp. 
sec A
cosec B
 –  
tan A
cot B
  =  
sec 45
cosec 45
 –  
tan 45
cot 45
  = 
v 2
v 2
 –  
1
1
   = 1-1=  0  
  
1 
15 (d) 
Number of revolutions= 
total distance 
circumference
 = 
176 
2 X
22
7
X 0.7
  
                                                             = 40 
1 
16 (b) 
perimeter of ?ABC 
perimeter of ?DEF 
 =  
BC
EF
  
7.5 
perimeter of ?DEF 
 = 
2
4
 . So  perimeter of ?DEF = 15cm 
 
1 
17 (b) Since DE? BC, ?ABC ~ ?ADE ( By AA rule of similarity) 
So 
AD
AB
 = 
DE
BC
  i.e. 
3
7
 = 
DE
14
. So DE = 6cm 
 1 
18 (a) Dividing both numerator and denominator by cosß,  
 4 ???????? -3 cos ?? 4 sin ?? +3 cos ?? = 
 4 ???????? -3
4 tan ?? +3
 = 
3-3
3+3
 =0 
 
 1 
19 (d) 
-2(–5x + 7y = 2) gives 10x – 14y = –4. Now 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 = -2 
1 
20 (a) 
Number of Possible outcomes are 26 
Favorable outcomes are M, A, T, H, E, I, C, S 
probability = 8/26 = 4/13 
1 
  SECTION B  
21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y,  
ATQ 81x + 81y = 1215  
Or  x+y = 15 
which gives four co prime pairs-  
1,14 
2,13 
4,11 
7, 8 
 
1 
22 (c) Required Area is area of triangle ACD = ½(6)2  
                                                               = 6 sq units 
1 
23 (b)  tan a + cot a = 2 gives a=45°. So tan a = cot a = 1  
tan
20
a + cot 
20
a = 1
20
 + 1
20
 = 1+1 = 2 
1 
24 (a) 
 Adding the two given equations we get: 348x + 348y = 1740.  
So x +y =5 
1 
25 (c)  LCM of two prime numbers = product of the numbers 
221= 13 x 17.  
So p= 17 & q= 13 
?3p - q= 51-13 =38 
1 
26 (a) Probability that the card drawn is neither a king nor a queen      
              = 
 52-8
52
 
               = 44/52 = 11/13 
1 
27 (b)  Outcomes when 5 will come up at least once are- 
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6) 
Probability that 5 will come up at least once = 11/36 
 
1 
28 (c)  1+ sin
2
a = 3 sina cos a 
sin
2
a + cos
2
a + sin
2
a = 3 sina cos a 
2 sin
2
a  - 3sina cos a + cos
2
a = 0 
(2sina -cos a)( sina- cosa) =0 
?cota = 2 or cota = 1   
 
1 
29 (a)  Since ABCD is a parallelogram, diagonals AC and BD bisect each other, ? mid 
point of AC= mid point of BD 
1 
Page 3


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Standard) 
  SECTION A  
QN Correct 
Option 
HINTS/SOLUTION MAR
KS 
1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 
HCF(4,2) = 4:2 = 2:1 
 
1 
2 (a) 
 For lines to coincide: 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 
                so,  
5
15
 = 
7
21
 = 
-3
-k
 
                i.e. k= 9 
 1 
3 (b) 
By Pythagoras theorem  
The required distance   =v(200² + 150²) 
     = v(40000+ 22500)  = v(62500)  = 250m. 
So the distance of the girl from the starting point is 250m.   
1 
4 (d) 
Area of the Rhombus = 
1
2
 d1d2 =
1
2
 x 24 x 32= 384 cm².  
Using Pythagoras theorem  
side² = (
1
2
d1)² + (
1
2
d2)² = 12² +16² = 144 +256 =400 
                             Side = 20cm 
 Area of the Rhombus = base x altitude 
384 = 20 x altitude 
So altitude = 384/20 = 19.2cm 
1 
5 (a)  Possible outcomes are (HH), (HT), (TH), (TT) 
Favorable outcomes(at the most one head) are (HT), (TH), (TT) 
So probability of getting at the most one head =3/4 
1 
6 (d)  Ratio of altitudes = Ratio of sides for similar triangles 
So AM:PN = AB:PQ = 2:3 
1 
7 (b) 2sin
2
ß – cos
2
ß = 2  
Then 2 sin
2
ß – (1- sin
2
ß) = 2  
3 sin
2
ß =3 or sin
2
ß =1 
ß is 90? 
1 
8 (c)  Since it has a terminating decimal expansion,  
so prime factors of the denominator will be 2,5 
1 
9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 
will intersect.  
1 
10 (d) Distance of  point A(-5,6) from the origin(0,0) is 
 v(0 + 5)
2
+ (0 - 6)
2
 = v25 + 36 = v61 units 
1 
11 (b) 
a²=23/25, then a = v 23/5, which is irrational 
1 
12 (c)  LCM X HCF = Product of two numbers 
36 X 2 = 18 X x 
x = 4 
 
1 
13 (b) 
tan A= v3 = tan 60° so ?A=60°, Hence ?C = 30°.  
So cos A cos C- sin A sin C = (1/2)x (v3/2) - (v3/2)x (1/2) =0 
  
1 
14 (a) 1x +1x +2x =180°, x = 45°. 
  ?A , ?B  and ?C are 45°, 45° and 90°resp. 
sec A
cosec B
 –  
tan A
cot B
  =  
sec 45
cosec 45
 –  
tan 45
cot 45
  = 
v 2
v 2
 –  
1
1
   = 1-1=  0  
  
1 
15 (d) 
Number of revolutions= 
total distance 
circumference
 = 
176 
2 X
22
7
X 0.7
  
                                                             = 40 
1 
16 (b) 
perimeter of ?ABC 
perimeter of ?DEF 
 =  
BC
EF
  
7.5 
perimeter of ?DEF 
 = 
2
4
 . So  perimeter of ?DEF = 15cm 
 
1 
17 (b) Since DE? BC, ?ABC ~ ?ADE ( By AA rule of similarity) 
So 
AD
AB
 = 
DE
BC
  i.e. 
3
7
 = 
DE
14
. So DE = 6cm 
 1 
18 (a) Dividing both numerator and denominator by cosß,  
 4 ???????? -3 cos ?? 4 sin ?? +3 cos ?? = 
 4 ???????? -3
4 tan ?? +3
 = 
3-3
3+3
 =0 
 
 1 
19 (d) 
-2(–5x + 7y = 2) gives 10x – 14y = –4. Now 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 = -2 
1 
20 (a) 
Number of Possible outcomes are 26 
Favorable outcomes are M, A, T, H, E, I, C, S 
probability = 8/26 = 4/13 
1 
  SECTION B  
21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y,  
ATQ 81x + 81y = 1215  
Or  x+y = 15 
which gives four co prime pairs-  
1,14 
2,13 
4,11 
7, 8 
 
1 
22 (c) Required Area is area of triangle ACD = ½(6)2  
                                                               = 6 sq units 
1 
23 (b)  tan a + cot a = 2 gives a=45°. So tan a = cot a = 1  
tan
20
a + cot 
20
a = 1
20
 + 1
20
 = 1+1 = 2 
1 
24 (a) 
 Adding the two given equations we get: 348x + 348y = 1740.  
So x +y =5 
1 
25 (c)  LCM of two prime numbers = product of the numbers 
221= 13 x 17.  
So p= 17 & q= 13 
?3p - q= 51-13 =38 
1 
26 (a) Probability that the card drawn is neither a king nor a queen      
              = 
 52-8
52
 
               = 44/52 = 11/13 
1 
27 (b)  Outcomes when 5 will come up at least once are- 
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6) 
Probability that 5 will come up at least once = 11/36 
 
1 
28 (c)  1+ sin
2
a = 3 sina cos a 
sin
2
a + cos
2
a + sin
2
a = 3 sina cos a 
2 sin
2
a  - 3sina cos a + cos
2
a = 0 
(2sina -cos a)( sina- cosa) =0 
?cota = 2 or cota = 1   
 
1 
29 (a)  Since ABCD is a parallelogram, diagonals AC and BD bisect each other, ? mid 
point of AC= mid point of BD 
1 
(
?? +1
2
 , 
6+2
2
 ) = (
3+4
2
 , 
5+?? 2
 )  
Comparing the co-ordinates, we get,  
?? +1
2
 = 
3+4
2
 .  So, x= 6 
Similarly, 
6+2
2
 = 
5+?? 2
.  So, y= 3 
?(x, y) = (6,3) 
 
30 (c)   ?ACD ~? ABC( AA )  
? 
AC
AB
= 
AD
AC
 (CPST) 
8/AB = 3/8 
This gives AB = 64/3 cm.  
So BD = AB – AD = 64/3 -3 = 55/3cm. 
1 
31 (d) Any point (x, y) of perpendicular bisector will be equidistant from A & B. 
? v(?? - 4)
2
+ (?? - 5)
2
= v(?? + 2)
2
+ (?? - 3)
2
 
Solving we get -12x – 4y + 28=0 or 3x + y – 7=0  
 
1 
32 (b) 
 
cot ?? °
cot ?? °
=  
AC / BC
???? /????
 = CD/ BC = CD/ 2CD = ½ 
 
1 
33 (a)  The smallest number by which 1/13 should be multiplied so that its decimal 
expansion terminates after two decimal points is 13/100 as  
1
13
  x 
13
100
  = 
1
100
  = 
0.01 
Ans: 13/100 
 
1 
34 (b) 
 
 
1 
35 (a)  Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1 ? 
coordinates of P are (
9k -1
k+1
, 
8k+3
k+1
) 
Since P lies on the line x – y +2=0, then  
9k -1
k+1
 - 
8k+3
k+1
 +2 =0 
9k -1 -8k-3 +2k+2 =0 
which gives k=2/3 
 
1 
36 (c)  
 
 
1 
?ABE is a right triangle & FDGB is a 
square of side x cm 
?AFD ~? DGE( AA ) 
? 
AF
DG
= 
FD
GE
 (CPST) 
 
16 - x
x
= 
x
8 - x
 (CPST) 
128 = 24x or x = 16/3cm 
Shaded area = Area of semicircle +       
(Area of half square – Area of two 
quadrants) 
= Area of semicircle +(Area of half 
square – Area of semicircle)  
= Area of half square 
= ½ x 14 x14 = 98cm² 
 
 
  
Page 4


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Standard) 
  SECTION A  
QN Correct 
Option 
HINTS/SOLUTION MAR
KS 
1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 
HCF(4,2) = 4:2 = 2:1 
 
1 
2 (a) 
 For lines to coincide: 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 
                so,  
5
15
 = 
7
21
 = 
-3
-k
 
                i.e. k= 9 
 1 
3 (b) 
By Pythagoras theorem  
The required distance   =v(200² + 150²) 
     = v(40000+ 22500)  = v(62500)  = 250m. 
So the distance of the girl from the starting point is 250m.   
1 
4 (d) 
Area of the Rhombus = 
1
2
 d1d2 =
1
2
 x 24 x 32= 384 cm².  
Using Pythagoras theorem  
side² = (
1
2
d1)² + (
1
2
d2)² = 12² +16² = 144 +256 =400 
                             Side = 20cm 
 Area of the Rhombus = base x altitude 
384 = 20 x altitude 
So altitude = 384/20 = 19.2cm 
1 
5 (a)  Possible outcomes are (HH), (HT), (TH), (TT) 
Favorable outcomes(at the most one head) are (HT), (TH), (TT) 
So probability of getting at the most one head =3/4 
1 
6 (d)  Ratio of altitudes = Ratio of sides for similar triangles 
So AM:PN = AB:PQ = 2:3 
1 
7 (b) 2sin
2
ß – cos
2
ß = 2  
Then 2 sin
2
ß – (1- sin
2
ß) = 2  
3 sin
2
ß =3 or sin
2
ß =1 
ß is 90? 
1 
8 (c)  Since it has a terminating decimal expansion,  
so prime factors of the denominator will be 2,5 
1 
9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 
will intersect.  
1 
10 (d) Distance of  point A(-5,6) from the origin(0,0) is 
 v(0 + 5)
2
+ (0 - 6)
2
 = v25 + 36 = v61 units 
1 
11 (b) 
a²=23/25, then a = v 23/5, which is irrational 
1 
12 (c)  LCM X HCF = Product of two numbers 
36 X 2 = 18 X x 
x = 4 
 
1 
13 (b) 
tan A= v3 = tan 60° so ?A=60°, Hence ?C = 30°.  
So cos A cos C- sin A sin C = (1/2)x (v3/2) - (v3/2)x (1/2) =0 
  
1 
14 (a) 1x +1x +2x =180°, x = 45°. 
  ?A , ?B  and ?C are 45°, 45° and 90°resp. 
sec A
cosec B
 –  
tan A
cot B
  =  
sec 45
cosec 45
 –  
tan 45
cot 45
  = 
v 2
v 2
 –  
1
1
   = 1-1=  0  
  
1 
15 (d) 
Number of revolutions= 
total distance 
circumference
 = 
176 
2 X
22
7
X 0.7
  
                                                             = 40 
1 
16 (b) 
perimeter of ?ABC 
perimeter of ?DEF 
 =  
BC
EF
  
7.5 
perimeter of ?DEF 
 = 
2
4
 . So  perimeter of ?DEF = 15cm 
 
1 
17 (b) Since DE? BC, ?ABC ~ ?ADE ( By AA rule of similarity) 
So 
AD
AB
 = 
DE
BC
  i.e. 
3
7
 = 
DE
14
. So DE = 6cm 
 1 
18 (a) Dividing both numerator and denominator by cosß,  
 4 ???????? -3 cos ?? 4 sin ?? +3 cos ?? = 
 4 ???????? -3
4 tan ?? +3
 = 
3-3
3+3
 =0 
 
 1 
19 (d) 
-2(–5x + 7y = 2) gives 10x – 14y = –4. Now 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 = -2 
1 
20 (a) 
Number of Possible outcomes are 26 
Favorable outcomes are M, A, T, H, E, I, C, S 
probability = 8/26 = 4/13 
1 
  SECTION B  
21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y,  
ATQ 81x + 81y = 1215  
Or  x+y = 15 
which gives four co prime pairs-  
1,14 
2,13 
4,11 
7, 8 
 
1 
22 (c) Required Area is area of triangle ACD = ½(6)2  
                                                               = 6 sq units 
1 
23 (b)  tan a + cot a = 2 gives a=45°. So tan a = cot a = 1  
tan
20
a + cot 
20
a = 1
20
 + 1
20
 = 1+1 = 2 
1 
24 (a) 
 Adding the two given equations we get: 348x + 348y = 1740.  
So x +y =5 
1 
25 (c)  LCM of two prime numbers = product of the numbers 
221= 13 x 17.  
So p= 17 & q= 13 
?3p - q= 51-13 =38 
1 
26 (a) Probability that the card drawn is neither a king nor a queen      
              = 
 52-8
52
 
               = 44/52 = 11/13 
1 
27 (b)  Outcomes when 5 will come up at least once are- 
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6) 
Probability that 5 will come up at least once = 11/36 
 
1 
28 (c)  1+ sin
2
a = 3 sina cos a 
sin
2
a + cos
2
a + sin
2
a = 3 sina cos a 
2 sin
2
a  - 3sina cos a + cos
2
a = 0 
(2sina -cos a)( sina- cosa) =0 
?cota = 2 or cota = 1   
 
1 
29 (a)  Since ABCD is a parallelogram, diagonals AC and BD bisect each other, ? mid 
point of AC= mid point of BD 
1 
(
?? +1
2
 , 
6+2
2
 ) = (
3+4
2
 , 
5+?? 2
 )  
Comparing the co-ordinates, we get,  
?? +1
2
 = 
3+4
2
 .  So, x= 6 
Similarly, 
6+2
2
 = 
5+?? 2
.  So, y= 3 
?(x, y) = (6,3) 
 
30 (c)   ?ACD ~? ABC( AA )  
? 
AC
AB
= 
AD
AC
 (CPST) 
8/AB = 3/8 
This gives AB = 64/3 cm.  
So BD = AB – AD = 64/3 -3 = 55/3cm. 
1 
31 (d) Any point (x, y) of perpendicular bisector will be equidistant from A & B. 
? v(?? - 4)
2
+ (?? - 5)
2
= v(?? + 2)
2
+ (?? - 3)
2
 
Solving we get -12x – 4y + 28=0 or 3x + y – 7=0  
 
1 
32 (b) 
 
cot ?? °
cot ?? °
=  
AC / BC
???? /????
 = CD/ BC = CD/ 2CD = ½ 
 
1 
33 (a)  The smallest number by which 1/13 should be multiplied so that its decimal 
expansion terminates after two decimal points is 13/100 as  
1
13
  x 
13
100
  = 
1
100
  = 
0.01 
Ans: 13/100 
 
1 
34 (b) 
 
 
1 
35 (a)  Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1 ? 
coordinates of P are (
9k -1
k+1
, 
8k+3
k+1
) 
Since P lies on the line x – y +2=0, then  
9k -1
k+1
 - 
8k+3
k+1
 +2 =0 
9k -1 -8k-3 +2k+2 =0 
which gives k=2/3 
 
1 
36 (c)  
 
 
1 
?ABE is a right triangle & FDGB is a 
square of side x cm 
?AFD ~? DGE( AA ) 
? 
AF
DG
= 
FD
GE
 (CPST) 
 
16 - x
x
= 
x
8 - x
 (CPST) 
128 = 24x or x = 16/3cm 
Shaded area = Area of semicircle +       
(Area of half square – Area of two 
quadrants) 
= Area of semicircle +(Area of half 
square – Area of semicircle)  
= Area of half square 
= ½ x 14 x14 = 98cm² 
 
 
  
37 (d) 
                                                          Let O be the 
center of the circle. OA = OB = AB =1cm.  
So ?OAB is an equilateral triangle and ? ?AOB =60° 
 Required Area= 8x Area of one segment with r=1cm, ?= 60° 
                        = 8x(
60
360
 x p x 1²- 
v3
4
 x 1²) 
                        = 8(p/6 - v3/4)cm²  
 
1 
38 (b) 
Sum of zeroes = 2 + ½ = -5/p 
i.e. 5/2 = -5/p . So p= -2 
Product of zeroes = 2x ½ = r/p 
                     i.e. r/p = 1 or r = p = -2 
 
1 
39 (c)  2pr =100. So Diameter = 2r =100/p = diagonal of the square. 
sidev2 = diagonal of square = 100/ p 
? side = 100/v2p = 50v2/p 
 
1 
40 (b) 3
x+y 
= 243 = 3
5 
So x+y =5-----------------------------------(1) 
 243
x-y 
= 3 
(3
5
)
 x-y 
= 3
1 
So 5x -5y =1--------------------------------(2) 
Since : 
?? 1
?? 2
 ? 
?? 1
?? 2
 , so unique solution 
 
1 
  SECTION C  
41 (c)  Initially, at t=0, Annie’s height is 48ft 
So, at t =0, h should be equal to 48 
h(0) = -16(0)² + 8(0) + k = 48 
So k = 48  
 
1 
42 (b) 
When Annie touches the pool, her height =0 feet 
i.e. -16t² + 8t + 48 =0 above water level 
   2t² - t -6 =0 
2t² - 4t +3t -6 =0 
2t(t-2) +3(t-2) =0 
(2t +3) (t-2) =0 
i.e. t= 2 or t= -3/2 
Since time cannot be negative , so t= 2seconds    
 
1 
43 (d) t= -1 & t=2 are the two zeroes of the polynomial p(t) 
 Then p(t)=k (t- -1)(t-2) 
         = k(t +1)(t-2)  
When t = 0 (initially) h1 = 48ft 
p(0)=k(0²- 0 -2)= 48 
  i.e. -2k = 48  
So the polynomial is -24(t²- t -2) = -24t² + 24t + 48.    
   
1 
44 (c) A polynomial q(t) with sum of zeroes as 1 and the product as -6  is given by  
q(t) = k(t² - (sum of zeroes)t + product of zeroes) 
       = k(t² -1t + -6)   ………..(1) 
  When t=0 (initially) q(0)= 48ft 
1 
.o 
Page 5


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Standard) 
  SECTION A  
QN Correct 
Option 
HINTS/SOLUTION MAR
KS 
1 (b) Least composite number is 4 and the least prime number is 2. LCM(4,2) : 
HCF(4,2) = 4:2 = 2:1 
 
1 
2 (a) 
 For lines to coincide: 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 
                so,  
5
15
 = 
7
21
 = 
-3
-k
 
                i.e. k= 9 
 1 
3 (b) 
By Pythagoras theorem  
The required distance   =v(200² + 150²) 
     = v(40000+ 22500)  = v(62500)  = 250m. 
So the distance of the girl from the starting point is 250m.   
1 
4 (d) 
Area of the Rhombus = 
1
2
 d1d2 =
1
2
 x 24 x 32= 384 cm².  
Using Pythagoras theorem  
side² = (
1
2
d1)² + (
1
2
d2)² = 12² +16² = 144 +256 =400 
                             Side = 20cm 
 Area of the Rhombus = base x altitude 
384 = 20 x altitude 
So altitude = 384/20 = 19.2cm 
1 
5 (a)  Possible outcomes are (HH), (HT), (TH), (TT) 
Favorable outcomes(at the most one head) are (HT), (TH), (TT) 
So probability of getting at the most one head =3/4 
1 
6 (d)  Ratio of altitudes = Ratio of sides for similar triangles 
So AM:PN = AB:PQ = 2:3 
1 
7 (b) 2sin
2
ß – cos
2
ß = 2  
Then 2 sin
2
ß – (1- sin
2
ß) = 2  
3 sin
2
ß =3 or sin
2
ß =1 
ß is 90? 
1 
8 (c)  Since it has a terminating decimal expansion,  
so prime factors of the denominator will be 2,5 
1 
9 (a) Lines x=a is a line parallel to y axis and y=b is a line parallel to x axis. So they 
will intersect.  
1 
10 (d) Distance of  point A(-5,6) from the origin(0,0) is 
 v(0 + 5)
2
+ (0 - 6)
2
 = v25 + 36 = v61 units 
1 
11 (b) 
a²=23/25, then a = v 23/5, which is irrational 
1 
12 (c)  LCM X HCF = Product of two numbers 
36 X 2 = 18 X x 
x = 4 
 
1 
13 (b) 
tan A= v3 = tan 60° so ?A=60°, Hence ?C = 30°.  
So cos A cos C- sin A sin C = (1/2)x (v3/2) - (v3/2)x (1/2) =0 
  
1 
14 (a) 1x +1x +2x =180°, x = 45°. 
  ?A , ?B  and ?C are 45°, 45° and 90°resp. 
sec A
cosec B
 –  
tan A
cot B
  =  
sec 45
cosec 45
 –  
tan 45
cot 45
  = 
v 2
v 2
 –  
1
1
   = 1-1=  0  
  
1 
15 (d) 
Number of revolutions= 
total distance 
circumference
 = 
176 
2 X
22
7
X 0.7
  
                                                             = 40 
1 
16 (b) 
perimeter of ?ABC 
perimeter of ?DEF 
 =  
BC
EF
  
7.5 
perimeter of ?DEF 
 = 
2
4
 . So  perimeter of ?DEF = 15cm 
 
1 
17 (b) Since DE? BC, ?ABC ~ ?ADE ( By AA rule of similarity) 
So 
AD
AB
 = 
DE
BC
  i.e. 
3
7
 = 
DE
14
. So DE = 6cm 
 1 
18 (a) Dividing both numerator and denominator by cosß,  
 4 ???????? -3 cos ?? 4 sin ?? +3 cos ?? = 
 4 ???????? -3
4 tan ?? +3
 = 
3-3
3+3
 =0 
 
 1 
19 (d) 
-2(–5x + 7y = 2) gives 10x – 14y = –4. Now 
?? 1
?? 2
 = 
?? 1
?? 2
 = 
?? 1
?? 2
 = -2 
1 
20 (a) 
Number of Possible outcomes are 26 
Favorable outcomes are M, A, T, H, E, I, C, S 
probability = 8/26 = 4/13 
1 
  SECTION B  
21 (c) Since HCF = 81, two numbers can be taken as 81x and 81y,  
ATQ 81x + 81y = 1215  
Or  x+y = 15 
which gives four co prime pairs-  
1,14 
2,13 
4,11 
7, 8 
 
1 
22 (c) Required Area is area of triangle ACD = ½(6)2  
                                                               = 6 sq units 
1 
23 (b)  tan a + cot a = 2 gives a=45°. So tan a = cot a = 1  
tan
20
a + cot 
20
a = 1
20
 + 1
20
 = 1+1 = 2 
1 
24 (a) 
 Adding the two given equations we get: 348x + 348y = 1740.  
So x +y =5 
1 
25 (c)  LCM of two prime numbers = product of the numbers 
221= 13 x 17.  
So p= 17 & q= 13 
?3p - q= 51-13 =38 
1 
26 (a) Probability that the card drawn is neither a king nor a queen      
              = 
 52-8
52
 
               = 44/52 = 11/13 
1 
27 (b)  Outcomes when 5 will come up at least once are- 
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6) 
Probability that 5 will come up at least once = 11/36 
 
1 
28 (c)  1+ sin
2
a = 3 sina cos a 
sin
2
a + cos
2
a + sin
2
a = 3 sina cos a 
2 sin
2
a  - 3sina cos a + cos
2
a = 0 
(2sina -cos a)( sina- cosa) =0 
?cota = 2 or cota = 1   
 
1 
29 (a)  Since ABCD is a parallelogram, diagonals AC and BD bisect each other, ? mid 
point of AC= mid point of BD 
1 
(
?? +1
2
 , 
6+2
2
 ) = (
3+4
2
 , 
5+?? 2
 )  
Comparing the co-ordinates, we get,  
?? +1
2
 = 
3+4
2
 .  So, x= 6 
Similarly, 
6+2
2
 = 
5+?? 2
.  So, y= 3 
?(x, y) = (6,3) 
 
30 (c)   ?ACD ~? ABC( AA )  
? 
AC
AB
= 
AD
AC
 (CPST) 
8/AB = 3/8 
This gives AB = 64/3 cm.  
So BD = AB – AD = 64/3 -3 = 55/3cm. 
1 
31 (d) Any point (x, y) of perpendicular bisector will be equidistant from A & B. 
? v(?? - 4)
2
+ (?? - 5)
2
= v(?? + 2)
2
+ (?? - 3)
2
 
Solving we get -12x – 4y + 28=0 or 3x + y – 7=0  
 
1 
32 (b) 
 
cot ?? °
cot ?? °
=  
AC / BC
???? /????
 = CD/ BC = CD/ 2CD = ½ 
 
1 
33 (a)  The smallest number by which 1/13 should be multiplied so that its decimal 
expansion terminates after two decimal points is 13/100 as  
1
13
  x 
13
100
  = 
1
100
  = 
0.01 
Ans: 13/100 
 
1 
34 (b) 
 
 
1 
35 (a)  Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k:1 ? 
coordinates of P are (
9k -1
k+1
, 
8k+3
k+1
) 
Since P lies on the line x – y +2=0, then  
9k -1
k+1
 - 
8k+3
k+1
 +2 =0 
9k -1 -8k-3 +2k+2 =0 
which gives k=2/3 
 
1 
36 (c)  
 
 
1 
?ABE is a right triangle & FDGB is a 
square of side x cm 
?AFD ~? DGE( AA ) 
? 
AF
DG
= 
FD
GE
 (CPST) 
 
16 - x
x
= 
x
8 - x
 (CPST) 
128 = 24x or x = 16/3cm 
Shaded area = Area of semicircle +       
(Area of half square – Area of two 
quadrants) 
= Area of semicircle +(Area of half 
square – Area of semicircle)  
= Area of half square 
= ½ x 14 x14 = 98cm² 
 
 
  
37 (d) 
                                                          Let O be the 
center of the circle. OA = OB = AB =1cm.  
So ?OAB is an equilateral triangle and ? ?AOB =60° 
 Required Area= 8x Area of one segment with r=1cm, ?= 60° 
                        = 8x(
60
360
 x p x 1²- 
v3
4
 x 1²) 
                        = 8(p/6 - v3/4)cm²  
 
1 
38 (b) 
Sum of zeroes = 2 + ½ = -5/p 
i.e. 5/2 = -5/p . So p= -2 
Product of zeroes = 2x ½ = r/p 
                     i.e. r/p = 1 or r = p = -2 
 
1 
39 (c)  2pr =100. So Diameter = 2r =100/p = diagonal of the square. 
sidev2 = diagonal of square = 100/ p 
? side = 100/v2p = 50v2/p 
 
1 
40 (b) 3
x+y 
= 243 = 3
5 
So x+y =5-----------------------------------(1) 
 243
x-y 
= 3 
(3
5
)
 x-y 
= 3
1 
So 5x -5y =1--------------------------------(2) 
Since : 
?? 1
?? 2
 ? 
?? 1
?? 2
 , so unique solution 
 
1 
  SECTION C  
41 (c)  Initially, at t=0, Annie’s height is 48ft 
So, at t =0, h should be equal to 48 
h(0) = -16(0)² + 8(0) + k = 48 
So k = 48  
 
1 
42 (b) 
When Annie touches the pool, her height =0 feet 
i.e. -16t² + 8t + 48 =0 above water level 
   2t² - t -6 =0 
2t² - 4t +3t -6 =0 
2t(t-2) +3(t-2) =0 
(2t +3) (t-2) =0 
i.e. t= 2 or t= -3/2 
Since time cannot be negative , so t= 2seconds    
 
1 
43 (d) t= -1 & t=2 are the two zeroes of the polynomial p(t) 
 Then p(t)=k (t- -1)(t-2) 
         = k(t +1)(t-2)  
When t = 0 (initially) h1 = 48ft 
p(0)=k(0²- 0 -2)= 48 
  i.e. -2k = 48  
So the polynomial is -24(t²- t -2) = -24t² + 24t + 48.    
   
1 
44 (c) A polynomial q(t) with sum of zeroes as 1 and the product as -6  is given by  
q(t) = k(t² - (sum of zeroes)t + product of zeroes) 
       = k(t² -1t + -6)   ………..(1) 
  When t=0 (initially) q(0)= 48ft 
1 
.o 
q(0)=k(0²- 1(0) -6)= 48 
  i.e. -6k = 48  or k= -8 
Putting k = -8 in equation (1),  reqd. polynomial is -8(t² -1t + -6) 
                                              = -8t² + 8t + 48  
                                                  
45 (a) When the zeroes are negative of each other,  
sum of the zeroes = 0 
So, -b/a = 0 
  - 
(k-3)
-12
 = 0 
+ 
k-3
12
 = 0 
k-3 = 0,  
i.e. k = 3.  
                          
1 
46 (a)  Centroid of ?EHJ  with E(2,1), H(-2,4) & J(-2,-2) is 
(
2+-2+ -2 
3
 , 
1+4+ -2 
3
) = (-2/3, 1)  
                      
1 
47 (c) If P needs to be at equal distance from A(3,6) and G(1,-3), such that A,P and G 
are collinear, then P will be the mid-point of AG.  
So coordinates of P will be (
3+1 
2
 , 
6+ -3 
2
) = (2, 3/2)   
                    
1 
48 (a) Let the point on x axis equidistant from I(-1,1) and E(2,1) be (x,0) 
then v(?? + 1)
2
+ (0 - 1)
2
= v(?? - 2)
2
+ (0 - 1)
2
     
x
2
 + 1 + 2x +1 = x
2
 + 4 - 4x +1   
6x = 3 
So x = ½ .  
 ? the required point is (½, 0)  
                           
1 
49 (b)  Let the coordinates of the position of a player Q such that his distance from 
K(-4,1)  is twice his distance from E(2,1) be Q(x, y)  
Then KQ : QE = 2: 1  
Q(x, y) = (
2 X 2+1 X-4 
3
 , 
2 X 1+1 X 1 
3
) 
         = (0,1)    
                  
1 
50 (d) Let the point on y axis equidistant from B(4,3) and C(4,-1) be (0,y) 
then v(4 - 0)
2
+ (3 - ?? )
2
= v(4 - 0)
2
+ (?? + 1)
2
     
 16 + y
2
 + 9 - 6y =  16 + y
2
 + 1 + 2y   
-8y = -8 
So y = 1 .  
 ? the required point is (0, 1)                            
1