Download Central Board of Secondary Education (CBSE) Class 10th (10th Board Exam) Maths Basic Marking Scheme 2021 Model Paper
Mathematics-Basic (241)
Marking Scheme SQP-2020-21
Max. Marks: 80 Duration:3hrs
1
1
156 = 22 x 3 x 13
2
Quadratic polynomial is given by x2 - (a +b) x +ab
1
x2 -2x -8
3
HCF X LCM =product of two numbers
?
LCM (96,404) = 96 404 = 96 404
(96,404)
4
?
LCM = 9696
OR
Every composite number can be expressed (factorized) as a product
1
of primes, and this factorization is unique, apart from the order in
which the factors occur.
4
x ? 2y =0
3x + 4y -20 =0
1 -2
?
3
4
As, 1 1
is one condition for consistency.
2
2
Therefore, the pair of equations is consistent.
?
5
1
1
6
= 60?
Area of sector = r2
360?
?
A = 60? X 22 X (6)2 cm2
360?
7
A = 1 X 22 X36 cm2
6
7
= 18.86cm2
?
OR
Another method-
Horse can graze in the field which is a circle of radius 28 cm.
?
So, required perimeter = 2r= 2.(28) cm
=2 x 22 X (28)cm
7
= 176 cm
?
7
By converse of Thale's theorem DE II BC
ADE = ABC = 70?
?
Given BAC = 50?
ABC + BAC +BCA =180? (Angle sum prop of triangles)
700 + 500 + BCA = 180?
BCA = 180? - 120? = 60?
?
OR
EC = AC ? AE = (7- 3.5) cm = 3.5 cm
= 2 and =3.5 = 1
?
3
3.5
1
So,
Hence, By converse of Thale's Theorem, DE is not Parallel to BC.
?
8
Length of the fence =
= .5280 = 220 m
24/
?
So, length of fence = Circumference of the field
220m= 2 r=2 X 22 x r
7
So, r = 220 7 m =35 m
2 22
?
9
Sol: tan 30 ? =
?
1/3 =
8
AB = 8 / 3 metres
?
Height from where it is broken is 8/3 metres
10
Perimeter = Area
1
2r = r2
r = 2 units
11
3 median = mode + 2 mean
1
12
8
1
13
1 1
is the condition for the given pair of equations to have unique ?
2
2
solution.
4
2
2
p
4
?
Therefore, for all real values of p except 4, the given pair of equations
wil have a unique solution.
OR
1
Here, = 2 = 1
2
4
2
1
= 3 = 1 and 1 = 5
2
6
2
2
7
1 = 1 5
2
2
7
?
1
1
1
=
is the condition for which the given system of equations
2
2
2
wil represent parallel lines.
So, the given system of linear equations wil represent a pair of parallel ?
lines.
14
No. of red bal s = 3, No.black bal s =5
?
Total number of bal s = 5 + 3 =8
Probability of red bal s =3
?
8
OR
Total no of possible outcomes = 6
There are 3 Prime numbers, 2,3,5.
?
?
So, Probability of getting a prime number is 3 = 1
6
2
15
?
tan 60? =
15
3 =
15
h = 153 m
?
16
1
1
17 i)
Ans : b)
1
Cloth material required = 2X S A of hemispherical dome
= 2 x 2 r2
= 2 x 2x 22 x (2.5)2 m2
7
= 78.57 m2
ii)
a) Volume of a cylindrical pil ar = r2h
1
iii)
b) Lateral surface area = 2x 2rh
1
= 4 x22 x 1.4 x 7 m2
7
= 123.2 m2
iv)
d) Volume of hemisphere =2 r3
1
3
= 2 22 (3.5)3 m3
3 7
= 89.83 m3
v)
b)
Sum of the volumes of two hemispheres of radius 1cm each= 2 x 2 13
3
?
Volume of sphere of radius 2cm = 4 23
3
2
2 x 13
So, required ratio is
3
=
4
1:8
?
2 3
3
18 i)
c) (0,0)
1
ii)
a) (4,6)
1
iii)
a) (6,5)
1
iv)
a) (16,0)
1
v)
b) (-12,6)
1
19 i)
c)
90?
1
ii)
b) SAS
1
iii)
b) 4 : 9
1
iv)
d) Converse of Pythagoras theorem
1
v)
a) 48 cm2
1
20 i)
d) parabola
1
ii)
a) 2
1
iii)
b) -1, 3
1
iv)
c) 2 - 2 - 3
1
v)
d) 0
1
21
Let P(x,y) be the required point. Using section formula
12+21
{
,12+21} = (x, y)
1
1+2
1+2
x = 3(8)+1(4) , y = 3(5)+1(-3)
3+1
3+1
x = 7 y= 3
1
(7,3) is the required point
OR
Let P(x, y) be equidistant from the points A(7,1) and B(3,5)
1
Given AP =BP. So, AP2 = BP2
(x-7)2 + (y-1)2 = (x-3)2 + (y-5)2
X2 -14x+49 +y2-2y +1 = x2 -6x +9+y2 -10y+25
1
x ? y =2
22
By BPT,
?
= ............(1)
?
Also, = .................(2)
By Equating (1) and (2) =
1
23
To prove: AB + CD = AD + BC.
1
Proof: AS = AP ( Length of tangents from an external point to a circle
are equal)
1
BQ = BP
CQ = CR
DS = DR
AS + BQ + CQ + DS = AP + BP + CR + DR
(AS+ DS) + ( BQ + CQ) = ( AP + BP) + (CR + DR)
AD + BC = AB +CD
24
For the correct construction
2
25
15 cot A =8, find sin A and sec A.
Cot A =8/15
1
=8/15
By Pythagoras Theorem
AC2 =AB2 +BC2
?
AC =(8)2 + (15)2
AC= 17x
?
Sin A = 15/17
Cos A =8/17
OR
By Pythagoras Theorem
QR = (13)2 - (12)2 cm
1
QR = 5cm
Tan P =5/12
Cot R =5/12
1
Tan P -Cot R =5/12 -5/12
= 0
26
9,17,25, .......
Sn = 636
a = 9
?
d = a2 -a1
= 17 ? 9 = 8
S
n = [ 2a + (n-1) d]
2
Sn = [ 2a + (n-1) d]
2
?
636 = [ 2x 9 + (n-1) 8]
2
1272 = n [ 18 + 8n -8]
1272 = n [10 +8n]
8n2 +10n -1272 =0
4n2 + 5n -636 =0
?
n = -?2-4
2
-5?52-4 4(-636)
n =
24
n =--5?101
8
n=96 n =-106
8
8
n=12 n = --53
4
?
n=12 (since n cannot be negative)
27
Let 3 be a rational number.
Then 3 = p/q HCF (p,q) =1
1
Squaring both sides
(3)2 = (p/q)2
3 = p2/ q2
3q2 = p2
3 divides p2 ? 3 divides p
3 is a factor of p
Take p = 3C
?
3q2 = (3c)2
3q2 = 9C2
3 divides q2 ? 3 divides q
?
3 is a factor of q
Therefore 3 is a common factor of p and q
It is a contradiction to our assumption that p/q is rational.
1
Hence 3 is an irrational number.
28
Required to prove -: PTQ = 2OPQ
1
Sol :- Let PTQ =
Now by the theorem TP = TQ. So, TPQ is an isosceles triangle
TPQ = TQP = ? (180? -)
1
= 90? - ?
OPT = 90?
?
OPQ =OPT -TPQ =90? -(90? - ? )
= ?
= ? PTQ
?
PTQ = 2OPQ
29
Let Meena has received x no. of 50 re notes and y no. of 100 re
1
notes.So,
50 x + 100 y =2000
x + y =25
multiply by 50
1
50x + 100y =2000
50 x + 50 y = 1250
- - -
50y =750
Y= 15
1
Putting value of y=15 in equation (2)
x+ 15 =25
x = 10
Meena has received 10 pieces 50 re notes and 15 pieces of 100 re
notes
30
(i)
10,11,12...90 are two digit numbers. There are 81
numbers.So,Probability of getting a two-digit number
1
= 81/90 =9/10
(i )
1, 4, 9,16,25,36,49,64,81 are perfect squares. So,
1
Probability of getting a perfect square number.
= 9/90 =1/10
(i i)
5, 10,15....90 are divisible by 5. There are 18 outcomes..
1
So,Probability of getting a number divisible by 5.
= 18/90 =1/5
OR
(i)
Probability of getting A king of red colour.
1
P (King of red colour) = 2/52 =1/26
(i )
Probability of getting A spade
1
P ( a spade) = 13/52 = 1/4
(i i) Probability of getting The queen of diamonds
1
P ( a the queen of diamonds) = 1/52
31
r1 = 6cm
r2 = 8cm
r3 = 10cm
1
Volume of sphere = 4/3 r3
Volume of the resulting sphere = Sum of the volumes of the smaller
spheres.
4/
3
3
3 r3 = 4/3 r 1
+ 4/3 r 2
+4/3 r 3 3
1
4/
3
3
3
3 r3 = 4/3 (r 1 + r2 + r3 )
r 3 = 63 + 83 + 103
r3 = 1728
r =
3 1728
r = 12 cm
1
Therefore, the radius of the resulting sphere is 12cm.
32
(sin A-cos A+1)/ (sin A+cosA-1) = 1/(sec A-tan A)
L.H.S. divide numerator and denominator by cos A
= (tan A-1+secA)/ (tan A+1-sec A)
1
= (tan A-1+secA)/(1-sec A + tan A)
We know that 1+tan2 A=sec 2A
1
Or 1=sec2 A-tan2 A = (sec A + tan A)(sec A ? tan A)
=( sec A + tan A-1)/[(sec A + tan A)(sec A-tan A)-(sec A-tan A)]
=( sec A + tan A-1)/(sec A-tan A)(sec A + tan A-1)
1
= 1/(sec A-tan A) , proved.
33
Given:-
Speed of boat =18km/hr
Distance =24km
Let x be the speed of stream.
?
Let t1 and t2 be the time for upstream and downstream.
As we know that,
speed= distance / time
time= distance / speed
?
For upstream,
Speed =(18-x) km/hr
Distance =24km
Time =t1
Therefore,
24
t
1 =
18-
For downstream,
Speed =(18+x)km/hr
Distance =24km
Time =t2
Therefore,
24
t
2 =
18+
Now according to the question-
t1=t2+1
24
= 24 + 1
18-
18+
?
24(18+)- 24 (18- )
= 1
(18-)(18+)
48x=(18-x)(18+x)
48x=324+18x-18x- x2
x2 +48x-324=0
x2+54x-6x-324=0
x(x+54)-6(x+54)=0
(x+54)(x-6)=0
?
x=-54 or x=6
Since speed cannot be negative.
x=-54 will be rejected
x=6
Thus, the speed of stream is 6km/hr.
1
OR
Let one of the odd positive integer be x
then the other odd positive integer is x+2
1
their sum of squares = x? +(x+2)?
= x? + x? + 4x +4
= 2x? + 4x + 4
Given that their sum of squares = 290
2x? +4x + 4 = 290
2x? +4x = 290-4 = 286
2x? + 4x -286 = 0
1
2(x? + 2x - 143) = 0
x? + 2x - 143 = 0
x? + 13x - 11x -143 = 0
x(x+13) - 11(x+13) = 0
(x -11)(x+13) = 0
(x-11) = 0 , (x+13) = 0
Therefore , x = 11 or -13
According to question, x is a positive odd integer.
Hence, We take positive value of x
1
So , x = 11 and (x+2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13 .
34
1
Let AB and CD be the multi-storeyed building and the building
respectively.
Let the height of the multi-storeyed building= h m and
the distance between the two buildings = x m.
AE = CD = 8 m [Given]
BE = AB ? AE = (h ? 8) m
and
AC = DE = x m [Given]
Also,
FBD = BDE = 30? ( Alternate angles)
FBC = BCA = 45? (Alternate angles)
?
Now,
In ACB,
1
In BDE,
1
From (i) and (i ), we get,
h =3h -83
3h ? h =83
h (3 -1) =83
h = 83
3-1
h= 83 x3+1
1
3-1
3+1
h-= 43 (3 +1)
h = 12 +43 m
Distance between the two building
?
OR
E
C
D
A
B
From the figure, the angle of elevation for the first position of the
bal oon EAD = 60? and for second position BAC = 30?.The
1
vertical distance
ED = CB = 88.2-1.2 =87m.
Let AD = x m and AB = y m.
Then in right ADE, tan60? =
3 =87
1
X =87 ..........(i)
3
In right ABC, tan 30? =
1 =87
3
Y = 873 ..........(i )
1
Subtracting(i) and (i )
y-x =873 -- -87
3
1
y-x =87 .2.3
3.3
y-x = 583 m
Hence, the distance travelled by the bal oon is equal to BD
y-x =583 m.
1
35
Let A be the first term and D the common difference of A.P.
Tp=a=A+(p-1)D=(A-D)+pD (1)
?
Tq=b=A+(q-1)D=(A-D)+qD ..(2)
?
Tr=c=A+(r-1)D=(A-D)+rD ..(3)
?
Here we have got two unknowns A and D which are to be eliminated.
We multiply (1),(2) and (3) by q-r,r-p and p-q respectively and add:
a (q-r) = (A ? D )(q-r) + D p(q-r)
?
b(r-p) = (A-D) (r-p) + Dq (r-p)
?
c(p-q) = (A-D) (p-q) + Dr (p-q)
?
a(q-r)+b(r-p)+c(p-q)
1
=(A-D)[q-r+r-p+p-q]+D[p(q-r)+q(r-p)+r(p-q)]
= (A ? D ) ( 0 ) + D [ pq-pr + qr ? pq + rp ? rq )
1
=0
36
Height (in cm)
f C.F.
below 140
4
4
140-145
7
11
1
145-150
18 29
150-155
11 40
155-160
6
46
160-165
5
51
N=51
N/2=51/2=25.5
As 29 is just greater than 25.5, therefore median class is 145-150.
( -)
Median= l + 2
X h
Here, l= lower limit of median class =145
?
C=C.F. of the class preceding the median class =11
h= higher limit - lower limit =150-145=5
f= frequency of median class =18
median=
( 25.5-11)
= 145 +
X 5
?
18
=149.03
Mean by direct method
1
f x
Height (in cm)
i fxi
below 140
4 137.5 550
140-145
7 142.5 997.5
145-150
18 147.5 2655
150-155
11 152.5 1677.5
155-160
6 157.5 945
1
5 162.5 812.5
160-165
Mean = ________
N
=7637.5/51
= 149.75
1
This post was last modified on 07 March 2021