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

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Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Basic) 
 
Q.
N. 
CORRECT 
OPTION 
HINTS/SOLUTION 
1 (d) P(perfect Square)=5/45=1/9 
2 (c) length of the arc= O /360° (2pr)=(60°/360°)x2x(22/7)x21=22cm 
3 (a) TanO = sinO/cosO = sinOxsecO = xy 
4 (d) The lines are parallel hence No solution 
5 (b) P(even composite no) =2/6=1/3 
6 (a) Let the cost of one chair=Rs. x 
Let the cost of one table=Rs. y 
8x+5y=10500 
5x+3y=6450 
Solving the above equations 
Cost of each chair= x= Rs. 750 
7 (c) CosO=I-cos
2
O=sin
2
O 
Therefore Sin
2
O+sin
4
O=cosO+cos
2
O=1 
8 (a) Terminating 
9 (c ) 2
3
x3
3
 
10 (c) 1
st
 No. x 2
nd
 No. = HCF X LCM 
12960=18 X LCM 
LCM=720 
11 (c) AE/AC=DE/BC=a/a+b=x/y 
X=ay/(a+b) 
12 (d)  (2x4+1x1)/3 , (2x6+1x3)/3 
=(3,5) 
13 (c) 3825=3
2
x5
2
x17 
14 (d) AB
2
=AD
2
+BD
2
 
AB=5cm 
AC
2
=AB
2
+CB
2
 
AC=13 cm 
Cot ?? =CB/AB=12/5 
15 (a) x+y=12 
X-y=8 
Solving the above equations 
X=10,y=2 
16 (d) AB
2
=AC
2
+AC
2
 
=AC
2
+BC
2
 
Hence, angle C=90° 
17 (d)   Let the zeroes be a and b 
Then, a=-1 , a+b=-(-7)/1 
Hence, b=7+1=8 
18 (a) P(same no on each die)=6/36=1/6 
19 (b)  (2,6)=((3p-2)/2, (4+2q)/2) 
3p-2=4, 4+2q=12 
P=2, q = 4 hence p+q = 6 
20 (c ) 147/120= 49/40=49/2
3
x5 
Page 2


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Basic) 
 
Q.
N. 
CORRECT 
OPTION 
HINTS/SOLUTION 
1 (d) P(perfect Square)=5/45=1/9 
2 (c) length of the arc= O /360° (2pr)=(60°/360°)x2x(22/7)x21=22cm 
3 (a) TanO = sinO/cosO = sinOxsecO = xy 
4 (d) The lines are parallel hence No solution 
5 (b) P(even composite no) =2/6=1/3 
6 (a) Let the cost of one chair=Rs. x 
Let the cost of one table=Rs. y 
8x+5y=10500 
5x+3y=6450 
Solving the above equations 
Cost of each chair= x= Rs. 750 
7 (c) CosO=I-cos
2
O=sin
2
O 
Therefore Sin
2
O+sin
4
O=cosO+cos
2
O=1 
8 (a) Terminating 
9 (c ) 2
3
x3
3
 
10 (c) 1
st
 No. x 2
nd
 No. = HCF X LCM 
12960=18 X LCM 
LCM=720 
11 (c) AE/AC=DE/BC=a/a+b=x/y 
X=ay/(a+b) 
12 (d)  (2x4+1x1)/3 , (2x6+1x3)/3 
=(3,5) 
13 (c) 3825=3
2
x5
2
x17 
14 (d) AB
2
=AD
2
+BD
2
 
AB=5cm 
AC
2
=AB
2
+CB
2
 
AC=13 cm 
Cot ?? =CB/AB=12/5 
15 (a) x+y=12 
X-y=8 
Solving the above equations 
X=10,y=2 
16 (d) AB
2
=AC
2
+AC
2
 
=AC
2
+BC
2
 
Hence, angle C=90° 
17 (d)   Let the zeroes be a and b 
Then, a=-1 , a+b=-(-7)/1 
Hence, b=7+1=8 
18 (a) P(same no on each die)=6/36=1/6 
19 (b)  (2,6)=((3p-2)/2, (4+2q)/2) 
3p-2=4, 4+2q=12 
P=2, q = 4 hence p+q = 6 
20 (c ) 147/120= 49/40=49/2
3
x5 
Three decimal places 
21 (d ) Perimeter of protractor=Circumference of semi-circle + 2 x radius 
=pr+2r 
 
22 (c) 0= P( E) =1 
23 (b) CD/BD=BD/AD 
BD
2
=CDXAD=6X3 
BD=3v2 cm 
24 (b ) 3/6=5/k ?K=10 
25 (d ) C1/C2=2pr/2pR 
2p/4p=2pr/2pR  
   r/R=1/2 
A1/A2=pr
2
/pR
2
=(r/R)
2
=(1/2)2=1/4 
A2=4A1 
26 (d) sinO=a/b 
H
2
=P
2
+B
2
 
 b
2
=a
2
+B
2
 
B=v(b
2
-a
2
) 
tanO=P/B=a/v(b
2
-a
2
) 
27 (a) x+y=2sin
2
O+2cos
2
O+1 
=2(sin
2
O+ cos
2
O)+1 
=2+1=3 
28 (b) 2pr- r=37 
     r{2x(22/7)-1}=37 
      r=37x7/37 
      r=7 
circumference=2x(22/7)x7=44cm 
29 (c) 1 = 1 
2 = 2 × 1 
3 = 3 × 1 
4 = 2 × 2 
5 = 5 × 1 
6 = 2 × 3 
7 = 7 × 1 
8 = 2 × 2 × 2 
9 = 3 × 3 
10 = 2 × 5 
So, LCM of these numbers = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 
Hence, least number divisible by all the numbers from 1 to 10 is 2520 
30 ( c) LCM 0f 4,7,14=28 
Bells will they ring together again at 6:28 AM 
31 (b) Let age of Father=x Years 
Let age of son = y years 
x+y = 65 
2(x-y)=50 
Solving the above equations 
Father’s Age =x = 45 years 
32 ( c) (tanOcosecO)
2
-(sinOsecO)
2
 
       =tan
2
Ocosec
2
O-sin
2
Osec
2
O 
       =(sin
2
O/cos
2
O)x1/ sin
2
O - sin
2
Ox1/cos
2
O 
        =(1- sin
2
O)/ cos
2
O= cos
2
O/ cos
2
O         =1 
33 ( d) A1/A2=(P1/P2)
2
=(26/39)
2
 
Page 3


Marking Scheme  
Class- X Session- 2021-22 
TERM 1 
Subject- Mathematics (Basic) 
 
Q.
N. 
CORRECT 
OPTION 
HINTS/SOLUTION 
1 (d) P(perfect Square)=5/45=1/9 
2 (c) length of the arc= O /360° (2pr)=(60°/360°)x2x(22/7)x21=22cm 
3 (a) TanO = sinO/cosO = sinOxsecO = xy 
4 (d) The lines are parallel hence No solution 
5 (b) P(even composite no) =2/6=1/3 
6 (a) Let the cost of one chair=Rs. x 
Let the cost of one table=Rs. y 
8x+5y=10500 
5x+3y=6450 
Solving the above equations 
Cost of each chair= x= Rs. 750 
7 (c) CosO=I-cos
2
O=sin
2
O 
Therefore Sin
2
O+sin
4
O=cosO+cos
2
O=1 
8 (a) Terminating 
9 (c ) 2
3
x3
3
 
10 (c) 1
st
 No. x 2
nd
 No. = HCF X LCM 
12960=18 X LCM 
LCM=720 
11 (c) AE/AC=DE/BC=a/a+b=x/y 
X=ay/(a+b) 
12 (d)  (2x4+1x1)/3 , (2x6+1x3)/3 
=(3,5) 
13 (c) 3825=3
2
x5
2
x17 
14 (d) AB
2
=AD
2
+BD
2
 
AB=5cm 
AC
2
=AB
2
+CB
2
 
AC=13 cm 
Cot ?? =CB/AB=12/5 
15 (a) x+y=12 
X-y=8 
Solving the above equations 
X=10,y=2 
16 (d) AB
2
=AC
2
+AC
2
 
=AC
2
+BC
2
 
Hence, angle C=90° 
17 (d)   Let the zeroes be a and b 
Then, a=-1 , a+b=-(-7)/1 
Hence, b=7+1=8 
18 (a) P(same no on each die)=6/36=1/6 
19 (b)  (2,6)=((3p-2)/2, (4+2q)/2) 
3p-2=4, 4+2q=12 
P=2, q = 4 hence p+q = 6 
20 (c ) 147/120= 49/40=49/2
3
x5 
Three decimal places 
21 (d ) Perimeter of protractor=Circumference of semi-circle + 2 x radius 
=pr+2r 
 
22 (c) 0= P( E) =1 
23 (b) CD/BD=BD/AD 
BD
2
=CDXAD=6X3 
BD=3v2 cm 
24 (b ) 3/6=5/k ?K=10 
25 (d ) C1/C2=2pr/2pR 
2p/4p=2pr/2pR  
   r/R=1/2 
A1/A2=pr
2
/pR
2
=(r/R)
2
=(1/2)2=1/4 
A2=4A1 
26 (d) sinO=a/b 
H
2
=P
2
+B
2
 
 b
2
=a
2
+B
2
 
B=v(b
2
-a
2
) 
tanO=P/B=a/v(b
2
-a
2
) 
27 (a) x+y=2sin
2
O+2cos
2
O+1 
=2(sin
2
O+ cos
2
O)+1 
=2+1=3 
28 (b) 2pr- r=37 
     r{2x(22/7)-1}=37 
      r=37x7/37 
      r=7 
circumference=2x(22/7)x7=44cm 
29 (c) 1 = 1 
2 = 2 × 1 
3 = 3 × 1 
4 = 2 × 2 
5 = 5 × 1 
6 = 2 × 3 
7 = 7 × 1 
8 = 2 × 2 × 2 
9 = 3 × 3 
10 = 2 × 5 
So, LCM of these numbers = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 
Hence, least number divisible by all the numbers from 1 to 10 is 2520 
30 ( c) LCM 0f 4,7,14=28 
Bells will they ring together again at 6:28 AM 
31 (b) Let age of Father=x Years 
Let age of son = y years 
x+y = 65 
2(x-y)=50 
Solving the above equations 
Father’s Age =x = 45 years 
32 ( c) (tanOcosecO)
2
-(sinOsecO)
2
 
       =tan
2
Ocosec
2
O-sin
2
Osec
2
O 
       =(sin
2
O/cos
2
O)x1/ sin
2
O - sin
2
Ox1/cos
2
O 
        =(1- sin
2
O)/ cos
2
O= cos
2
O/ cos
2
O         =1 
33 ( d) A1/A2=(P1/P2)
2
=(26/39)
2
 
       A1/A2=(2/3)
2
=4/9 
34 (a ) Let no of Cars=x 
       Let no of motorcycles=y 
       X+y=20 
       4x+2y=56 
Solving the above equations 
No of cars=x=8 
35 ( c) H
2
=P
2
+B
2
 
H
2
=15
2
+8
2
 
H=17m 
36 ( c)  (altitude)
2
=(side)
2
-(side/2)
2
 
                  =8
2
-4
2
= 64-16  =48 
Altitude=4v3 cm 
37 (d) P=3/9=1/3 
38 ( b) O/360°xpr
2
=1/6x pr
2
 
      O=60° 
39 ( d)  
 
Height of Vertical stick/Shadow of vertical stick=height of tower/shadow of tower 
20/10=Height of tower/50 
Height of tower=100 m 
40 (d) 37x+43y=123      ____(1) 
43x+37y=117      ____(2) 
Adding (1) and (2) 
X+y=3      ______(3) 
Subtracting (2) from (1) 
-x+y=1..............(4) 
Adding (3) and (4), 
2y=4 
y=2 
? x=1 
? solution is x=1 and y=2 
41 ( b) AB=v{(4-1)
2
+(0-4)
2
} 
             =v(3
2
+4
2
) 
         AB=5 units 
42 (a) (x-7)
2
+(y-1)
2
=(x-3)
2
+(y-5)
2
 
X
2
+49-14x+y
2
+1-2y=x
2
+9-6x+y
2
+25-10y 
Simplifying 
x-y=2 
43 (a ) 3x + y – 9 = 0 
Let R divide the line in ratio k:1 
R( 2k+1/k+1, 7k+3/k+1) 
3(2k+1/k+1)+( 7k+3/k+1)-9=0 
4k-3=0 
K=3/4 
3 : 4 
44 ( c) Distance of M from X-axis=v(2-2)
2
+(0-3)
2
=v9=3units 
45 (b)  ( (1+3)/2 , (4+5)/2) =  (4/2, 9/2) =   (2, 9/2) 
46 (c)  Cubic 
47 (d ) Four Zeroes as the curve intersects the x-axis at 4 points 
48 ( d) p?0 
49 (d) 3 Zeroes as the curve intersects the x-axis at 3 points 
50 ( c) -3,-1,2 
 
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FAQs on Class 10 Mathematics (Basic) Term I : CBSE (Official) Marking Scheme with Solution (2021-22) - Mathematics (Maths) Class 10

1. What is the marking scheme for Class 10 Mathematics (Basic) Term I exam according to CBSE?
Ans.The marking scheme for the Class 10 Mathematics (Basic) Term I exam typically comprises a total of 40 marks. It includes various sections, where questions may be divided into different types such as very short answer, short answer, and long answer questions. The distribution of marks can vary with some questions carrying more weight than others, usually based on their complexity.
2. How should students prepare for the Class 10 Mathematics (Basic) Term I exam?
Ans. Students should start by thoroughly reviewing the syllabus and understanding the topics covered. Practice previous years' question papers and sample papers to familiarize themselves with the exam pattern. Regular revision of concepts and solving numerical problems will enhance confidence. Utilizing study materials such as textbooks, online resources, and coaching classes can also be beneficial.
3. What types of questions are included in the Class 10 Mathematics (Basic) Term I exam?
Ans. The Class 10 Mathematics (Basic) Term I exam usually includes a variety of question types such as multiple-choice questions (MCQs), very short answer questions, short answer questions, and long answer questions. These questions may cover topics like algebra, geometry, statistics, and arithmetic, assessing both conceptual understanding and problem-solving skills.
4. Are there specific topics that students should focus on for the Mathematics (Basic) Term I exam?
Ans. Yes, students should focus on fundamental topics outlined in the CBSE syllabus. Key areas often include real numbers, polynomials, linear equations, triangles, quadrilaterals, statistics, and probability. Mastery of these concepts is crucial as they form the basis for many questions in the exam.
5. How can students manage their time effectively during the Mathematics (Basic) Term I exam?
Ans. Effective time management during the exam can be achieved by first reading through the entire question paper to gauge the difficulty level of questions. Students should allocate time based on the marks assigned to each question, ensuring they leave time to review their answers. Practicing under timed conditions beforehand can also help improve speed and efficiency during the actual exam.
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