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7/30/2019 Math class xth
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CBSEContinuous and Comprehensive Evaluation (CCE)
Sample
Question Papers
Term 1 (April to September 2013)
OSWAAL BOOKS
Class
Mathematics
Solutions
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CONTENTS
l Solutions
q Sample Question Paper 6 1 - 12
q Sample Question Paper 7 13 - 24
q Sample Question Paper 8 25 - 35
q Sample Question Paper 9 36 - 47
q Sample Question Paper 10 48 - 60
ll
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Solutions | 1
1. (C),243
2 5 73 2 3 =
3
3 3 3
7
2 5 7 = 3 3
1
2 51
Hence, above rational numbers have terminating decimal expansion.
2. (B), Given, p( x) = x2 – 5 x + 6
Sum of zeroes = + = –F
H
GI
K
J 5
1
= 5
Product of zeros = =6
1= 6
Now + – 3 = 5 – 3(6)
= 5 – 18 = – 13. 1
3. (A), XY || QR so
PQ
XQ =PR YR
7
3 =
63
YR
YR =3 × 63
7= 27 cm 1
4. (A), Given, sec + tan = 7
(sec tan )(sec tan )
(sec tan )
= 7
sec tan
sec tan
2 2
= 7 1
sec tan = 7 1
sec – tan =1
7
5. (B), The required LCM = p3 1
6. (C), 4 x –3 y = 92 x + ky = 11
For unique solutiona
a
1
2=
b
b
1
2
c
c
1
2
4
2=
3
k
9
11
2 =–3
k
k =3
2
1
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2 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
7. (B), Given 5 tan =12
tan =12
5
So
13
3
sin=
13
3
12
13
F
H G
I
K J = 4. 1
8. (B), In the distribution,
Median class = 160 – 165
Hence Upper limit of median class = 165
and Modal class = 150 – 155
So Lower limit of Modal class = 150
Hence, Sum of upper limit of Median class and lower limit of Modal class
= 165 + 150 = 315. 1
Question number 9 to 14 carry two marks each.
9. No.
15 does not divide 175.
LCM is exactly divisible by their HCF. 1
Two numbers cannot have their HCF as 15 and LCM as 175. 1
10. Given, p ( x) = 2 x2 – x – 6
= 2 x2 – 4 x + 3 x = 6
= 2 x ( x – 2) + 3 ( x – 2)
= (2 x + 3) ( x – 2)
So zeroes are 2 and 3
2 ½
Sum of zeroes (SOZ) = 2 +F
H GI K J
3
2
1
2½
Product of zeroes (POZ) = 2 ×
3
23
SOZ =coeff. of
oeff. of
x
c x2
1
2
1
2
( )
POZ =constant
coeff. of 2 x
6
23 ½
Thus relationship is verified. ½
11. According to question AO = 20 cm, BO = 12 cm, PB = 18 cm
In AQO and BPO
AOP = BOP (V.O.s)
A = B = 90º
AQO ~ BPO 1
AQ
BP=
QO
PO=
AO
BO
AQ
18=
20
12
AQ = 30 cm. 1
B
C
A
12
5
13
P
B
O A
Q
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Solutions | 3
12. Given, 4cos = 11sin
cos =11
4sin
Now
11 7
11 7
cos sin
cos sin
=
1111
47
11 114
7
sin sin
sin sin
1
=
sin
sin
121
47
121
47
F H G
I K J
F H G
I K J
=121 28
121 28
93
149
–
1
13. p( x) = ax2 + bx + c
Let and1
be the zeroes of p( x), then
Product of zeroes, a ×
1
=
c
a 1
So, required condition is c = a. 1
14. Model class : 30 – 40. Here l = 30, f 1 = 45, f 2 = 12, f 0 = 30, h = 10 ½
Mode = l + f f
f f f 1 0
1 0 22
F
H GI
K J × h ½
= 30 +45 30
90 30 12
F H G
I K J × 10 ½
= 30 + 3125 = 33125 ½
Or
Class Interval Frequency
0-50 8
50-100 15
100-150 32
150-200 26
200-250 12
250-300 7
Total 100 2
15. In ABC and DEF AB
DE= 3 ;
AC
DF= 3; P1 = 3P2 BC = 3EF
A
B C
D
E F
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4 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
AB
DE=
AC
DF=
BC
EF= 3
ABC ~ DEF 1
ar ABC
ar ( DEF)
( )
= AB
DE
F H G
I K J
2
= (3)2 = 9 2
16. 3 x3 + 4 x2 + 5 x – 13 = (3 x + 10) g( x) + (16 x – 43) ½
3 4 11 30
3 10
3 2 x x x
x
= g ( x) ½
3 10 3 4 11 30
10
6 11
6 20
9 30
9 30
0
2 33 2
2
2
2
2 x x x x
x x
x x
x x
x
x
x x
3 3
(–) (–)
–
– –
( ) ( )
(–)(–)
(
Hence, g( x) = x2 – 2 x + 3. 2
17. L.C.M. of 48 and 60
2 | 48, 60
2 | 24, 30
3 | 12, 154, 5 1
48 = 24 × 3
60 = 22 × 3 × 5 ½
L.C.M. = 24 × 3 × 5
= 16 × 15 = 240 1
240 books are required ½
Or
(i) Required number of minutes is the LCM of 15 and 12.18 = 2 × 32
and 12 = 22 × 3 1
LCM of 18 and 12 = 22 × 32 = 36Hence, Ravish and Priya will meet again at the starting point after 36 minutes. 1(ii) LCM of numbers. ½(iiii) Healthy competition is necesssary for personal development and progrees. ½
18.2 2
2 2
sin 63 sin 7
sec 20 cot 70
+ 2 sin 36° sin 42° sec 48° sec 54° 1
=2 2
2 2
sin (90 27 ) sin 7
sec (90 70 ) cot 70
+ 2sin (90° – 54°) sec 54° sin 42° sec (90° – 42°) 1
=2 2
2 2
cos 27 sin 7
cosec 70 cot 70
+ 2cos 54° sec 54° sin 42° cosec 42°
=11 + 2 (1) = 3 1
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Solutions | 5
19. Given, p( x) = 3 x2 – 4 x – 7 and and are its zeroes.
sum of zeroes = + = –coeff. of
coeff. of 2
x
x= –
F H G
I K J
4
3
4
3
Product of zeroes =constant
coeff. of 2 x= –
7
3
7
3
F H G
I K J
1
for the new polynomial
Sum,1 1
=
4
37
3
4
7
Product,1 1
=
1 1
7
3
3
7
1
required quadratic polynomial = x2 – (sum of zeroes) x + product of zeroes
= x2 –F
H GI K J
F
H GI K J
4
7
3
7 x =
1
7(7 x2 + 4 x – 3)
= 7 x2 + 4 x – 3. 1
20. 7 x – 4 y = 49 ...(1)
on comparing with the equation a1 x + b1 y = 0
a1 = 7, b1 = –4, c1 = 49
Again 5 x – 6 y = 57 ...(2)
on comparing with the equation a2 x + b2 y = c2
a2 = 5, b2 = –6, c2 = 57
Sincea
a
1
2=
7
5and
b
b
1
2=
4
6
a
a
1
2
b
b
1
21
So, system has unique solution.
Multiply eqn. (1) by 5 and multiply eqn. (2) by 7 and subtract,
35 x – 20 y = 245
35 x – 42 y = 399 1
– + –
22 y = – 154
y = – 7
Put the value of y in eqn. (1)
5 x – 6(–7) = 57 5 x = 57 – 42 = 15
x = 3
Hence, x = 3 and y = –7. 1
Or
ax + by =a b
2
or 2ax + 2by = a + b ...(i)
and 3 x + 5 y = 4 ...(ii) 1
Multiplying (i) by 5 and (ii) by 2b, we get
10ax + 10 by = 5a + 5b
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6 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
6bx + 10 by = 8b
– – –
On subtracting x(10a – 6b ) = 5a – 3b 1
x =5 3
2 5 3
1
2
a b
a b
( )
Putting x =1
2in (1), we get y =
1
2
Hence x =1
2and y =
1
2 1
21. Classes :125 115
2
=
10
2= 5
115 – 5 = 110 ; 115 + 5 = 120
C.I. f c.f.
110 – 120 6 6
120 – 130 25 31
130 – 140 48 79
140 – 150 72 151
150 – 160 116 267
160 – 170 63 330 1
n = 330,n
2= 165 ½
median class is 150 – 160,
So, Median = 150 +165 151
116
F H G
I K J × 10 1
= 150 +140
116= 151.21 ½
22. In s ADE and ABC 1
A is common
D = B [corr. s]
ADE ~ ABC [AA]
AD AB
=DEBC
½
AB AD
=BCDE
½
ADDB =
75 ½
DB AD
+ 1 =57
+ 1 ½
AB AD
=127
=BCDE
½
In s BFC and EFDBFC = EFD [V.O. s)FBC = EFD [alt. s]
BFC ~ EFD [AA]
arEFDarBFC
=2
2
DE
BC=
2712
½
= 49144 ½
A
F
ED
B C
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Solutions | 7
Or
In TXM, We have
XM || BN
TBTX
=TNTM
...(i) 1
In TMC, we have
XM || CM
TX TC
=TNTM
...(ii) 1
From equation (i) and (ii), we get
TBTX
=TX TC
TX 2 = TB × TC 1
23. LHS. =sin cos
sin cos
sin cos
sin cos
=(sin cos ) (sin cos )
sin cos
2 2
2 2
=(sin cos ) – sin cos (sin cos ) sin cos
sin ( sin )
2 2 2 2
2 2
2 2
1
1
=1 1
12 2
sin sin 1
=2
2 12sin = RHS Proved 1
24.
C.I. f x u = i x ah fu
100-110 4 105 – 2 – 8
110-120 14 115 – 1 – 14
120-130 21 125 = a 0 0
130-140 8 135 1 8
140-150 3 145 2 6
Total 50 – 8 1½
x = a + fu
f
= 125 + 8
50
½
= 125 – 016
x = 12484 1
Question number 25 to 34 carry four marks each.
25. Let 3 – 5 is a rational number.
3 – 5 = p
q, q 0
5 = 3 – p
q
=3q p
q
T B X C
MN
A
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8 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
3q p
q
is rational number. 1
But 5 is a irrational number. 1
Since irrational number cannot be equal to rational number.
Our assumption is wrong. 1
3 – 5 is an irrational number. 1
26. Let the cost of one pencil be ` x and the cost of one chocolate be ` y
According to question,
2 x + 3 y = 11 ...(1)
x + 2 y = 7 ...(2)
Now, 2 x + 3 y = 11 x =11 3
2
y
y 1 3 5
x 4 1 – 2 1
and x + 2 y = 7 x = 7 – 2 y
y 0 1 3
x 7 5 1 1
–4 –3 –2 –1 1 2 3 4 5 6 7
y'
–1
–2
–3
1
2
3
5
4
x' x
y
6
7
8
0–5–6
–4
–5
–6
–7
8
(1, 3)
(4, 1)(5, 1)
(7, 0)
2 + 3 = 11 x y
x y+ 2 = 7
(–2, 5)
Putting the above points and drawing the lines joining them, we get the graph of the aboveequation. Clearly two lines interesect at point (1, 3).
Solution of eqns. (1) and (2) is x = 1 and y = 2
Cost of one pencil = ` 1
and cost of one chocolate = ` 3. 1
27. LHS =cos
tan
sin
sin cos
2 3
1
=cos
sin
cos
sin
sin cos
2 3
1
½
1
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Solutions | 9
=cos
cos sin–
sin
cos sin
3 3
1
=cos sin
cos sin
3 3
1
= (cos sin )(cos sin sin cos )(cos sin )
2 21
= 1 + sin cos = RHS Proved ½
28.
Classes f c.f.
5 – 10 2 2
10 – 15 12 14
15 – 20 2 16
20 – 25 4 20
25 – 30 3 23
30 – 35 4 27
35 – 40 3 30 2
Total f = 30 = N
SinceN
2= 15,
Median class is 15 – 20
Median = l +
N2
cf
f
× h 1
From table, l = 15, N = 30, c.f. = 14, f = 2, h = 5
Median = 15 +15 14
2
F H G
I K J × 5 = 15 + 25
= 175. 1
29. p( x) = 2 x2 + 5 x + k
Sum of zeroes = –coeff. of
coeff. of 2
x
x= + =
5
21
Product of zeroes =constant
coeff. of 2 x= =
k
2
According to question, 2 + 2 + = 214
1
(2 – 2 + =21
4[ ()2 = 2 + 2 + 2]
F
H GI K J
5
2
2
–k
2=
21
4
25
4
21
4 =
k
21
1 =k
2k = 2
Hence, k = 2 1
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10 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
Or
In cyclic quadrilateral ABCD, we have
A + C = 180º
and D = 180º 1
x + 7 + 3 y + 23 = 180
x + 3 y = 150 ..(1)1
and y + 8 + 4 x + 12 = 180
4 x + y = 160 ...(2)1
Solving equation (1) and (2), we get y = 40º, x = 30º
A = ( x + 7)º = 37º
B = ( y + 8)º = 48º
C = (3 y + 23)º = 143º
D = (4 x + 12)º = 132º 1
30. Given : ABC is a triangle in which DE || BC.
To prove :
AD
BD =
AE
CE
Constuction : Draw DN AE and EM AD, Join BE and CD. 1
Proof : In ADE, area ( ADE) =1
2× AE × DN ...(i)
In DEC,
area ( DCE) =1
2× CE × DN ...(ii)
By (i) / (ii)area ( ADE)
area ( DEC)
=
1
2
AE DN
1
2CE DN
area ( ADE)
area ( DEC)
= AE
CE...(iii) ½
Now in ADE, area ( ADE) =1
2× AD × EM ...(iv) ½
and in DEB, area (DEB) =1
2× EM × BD ...(v)
By (iv) / (v),area ( ADE)
area ( DEB)
=
1
2 AD EM
1
2BD EM
½
area ( ADE)
area ( DEB)
=
AD
BD ...(iv)½
DEB and DEC lies on the same base DE and between same parallel lines DE and BC.
area (DEB) = area (DEC)
From quation (iii),area ( ADE)
area ( DEB)
= AE
CE...(vii)
From equations (vi) and (vii), we get
AE
CE=
AD
BD Proved 1
Or
Let BD = DE = EC = x
A
M N
D E
B C
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Solutions | 11
BE = 2 x
BC = 3 x
AE2 = AB2 + BE2 = AB2 + 4 x2 1
AC2 = AB2 + BC2 = AB2 + 9 x2
AD2 = AB2 + BD23 = AB2 + x2 1
Now, 8AE2 = 8AB2 + 32 x2
and 3AC2 + 5AD2 = 3(AB2 + 9 x2) + 5 (AB2 + x2) 1
= AB2 + 27 x2 + 5AB2 + 5 x2
= 8AB2 + 32 x2
3AC2 + 5AD2 = 8AE2. Proved 1
31. Given x = r sin A.cos C
y = r sin A.sin C
z = r cos A
then x2 = r2sin2 A cos2C
z2 = r2cos2 A 1
y2 = r2sin2 A sin2C 1 x2 + y2 + z2 = r2sin2 A cos2C + r2sin2 Asin2C + r2cos2 A 1
= r2sin2 A (cos2C + sin2C) + r2cos2 A
= r2sin2 A + r2 cos2 A
= r2 (sin2 A + cos2 A)
= r2 proved 1
32. FEC GBD
EC = BD ...(1)1
It is given that
1 = 2
AE = AD ...(2)1
From eqns. (1) and (2), AE
EC=
AD
BD
DE || BC [converse of BP.T.]
1 = 3 and 2 = 4 1
Thus in ADE and ABC, A = A
1 = 3
2 = 4
So by AAA criterion of similarity, we have
ADE ~ ABC. Proved 1
33. cos ºsin º6525 = cos ºsin( º º )
cos ºcos º
6590 65
6565 = 1 1
tan º
cot º
20
70=
tan( º–70º )
cot º
cot º
cot º
90
70
70
70 = 1
sin 90º = 1
tan 5º tan 35º tan 60º tan 55º tan 85º = tan (90º – 85º) tan (90º – 55º) tan 55º tan 60º tan 85°
1
= cot 85º. tan 85º. cot 55º tan 55º. 3
= 1 × 1 × 3 = 3 1
Given expression = 1 – 1 – 1 + 3 = 3 – 1. 1
A
BF C
E1 2
G
D
3 4
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12 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
34.
Less than c.f. More than c.f.
10 8 0 35
20 15 10 2730 20 20 20
40 30 30 15
50 35 40 5 2
Scale
x-axis 10 units = 1 cm
y-axis 1 cm = 5 c.f.
35
30
25
20
15
10
5
10 20 30 40 50
CLASSES
C . F .
Scale
-axis 10 units = 1 cm-axis 1 cm = 5 cf x y y
Median =24
0
more than ogive
less than ogive
2
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Solutions | 13
1. (D), Since 16 = ± 4, so it is a rational number. 1
2. (B), Given, a = – 15 and b = 15
So, quadratic polynomial = ( x – ) ( x – ) = ( x – 15 ) ( x + 15 )
= x2 – ( 15 )2 = x2 – 15. 1
3. (A), In the figure AB || ED then ABC and DEC are similar. 1
4. (C), (1 + tan2 ) cos2 =2
2
sin1cos
cos2
=2 2
2
(sin cos )
cos
× cos2
= 1 1
5. (C),81
2 3 53 2 2 = 3 2
9
2 51
Hence, this rational number has a terminating decimal expansion. 1
6. (D), For a unique solution
a
a
1
2
b
b1
2
i.e.,3
6
2
k
– 4 k
i.e., all real numbers except – 4. 1
7. (D), Given tan = 1
Then5sin 4 cos5sin 4cos
=
cos (5tan 4)
cos (5tan 4)
=51 451 4
= 9 1
8. (B), The median of the data is 425. 1
9. a =9009
3003= 3 ½
b =1001
143= 7 ½
Since 143 = 11 × 13, so c = 11 or 13 ½d = 13 or 11. ½
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14 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
10. Given quadratic polynomial = 4 x2 + 7 x + 8
Sum of zeroes = –coeff. of
coeff. of 2
x
x= –
7
41
and Product of zeroes =constant
coeff. of 2
x
=8
4
= 2. 1
11. In 's ADE and ABC,
ADE = C [Given] ½
A = A [Common] ½
By AA criterion of similarity, ADE ~ ACB 1
12. x = 30°
sin 30° + cos y = 1
12
+ cos y = 1 ½
cos y =12
= cos 60° 1
y = 60° ½
13. Given, p( x) = 2 x2 – 9
2 x2 – 9 = 0
( 2 x)2 – (3)2 = 0
( 2 x + 3) ( 2 x – 3) = 0 1
x = –3
2and
3
2Hence, zeroes of the polynomials are
3
2and –
3
2. 1
14. Modal class : 20 – 30
Here, l = 20, f 1 = 40, f 0 = 24, f 2 = 36, h = 10 ½
Mode = l f f
f f f h
( )1 0
1 0 22½
= 20 +( – )
– –
40 24
80 24 36× 10 ½
= 20 +16 10
20
= 28 ½
Or
Classes 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
Frequency ( f i) 3 4 2 2 1 f i = 12
xi (Class marks) 35 45 55 65 75
f i xi 105 180 110 130 75 f i xi = 600
Mean, x–
=
x f
f
i i
i
600
12 = 50 1
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Solutions | 15
Mean weight of a worker = 50 kg. 1
Section - C
15. In ∆ ABC, given
A
BE D C
F
AB = AC ⇒ ∠ ABC = ∠ ACB 1
or ∠ ABD = ∠ECF
∠ ADB = ∠EFC, [Each 90º] 1
∴ ∆ ABD ~ ∆CEF. [AA Similarity] 1
16. If α, β are zeroes of the polynomial
p( y) = 6 y2 – 7 y + 2
α + β = –n
n 2
coeff of 76coeff of
y
y=
αβ =2
Constant term
Coefficient of y=
26
=13
Now,1 1
α β+ =
7613
α + β =αβ
=72
1
1
α×
1
β= 3 1
∴ Required polynomial = y2 –1 1 + α β y +
1 1·
α β
∴ = y2 –72
y + 3 = 21(2 7 6)
2y y− + 1
17. By Euclid’s division algorithm, 510 = 92 × 5 + 5092 = 50 × 1 + 42
50 = 42 × 1 + 8
42 = 8 × 5 + 2 18 = 2 × 4 + 0.
HCF (510, 92) = 2
92 = 22 × 23
510 = 2 × 3 × 5 × 17
LCM = 22 × 23 × 3 × 5 × 17 = 23460
HCF × LCM = 2 × 23460 = 46920 1
Product of 2 numbers = 510 × 92
⇒ HCF × LCM = Product two numbers. 1
Or
By Euclid’s division algorithm, for two positive integers a and b, we have
a = bq + r, 0 ≤ r < b
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16 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
Let b = 6 r = 0, 1, 2, 3, 4, 5
So a = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 1Clearly, a = 6q, 6q + 2, 6q + 4 are even, as they are divisible by 2.
But 6q + 1, 6q + 3, 6q + 5 are odd, as they are not divisible by 2. 1
Any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5. 1
18.sec ( º ) cot
(sin º sin º)
cos º tan º tan º
(sec º– cot º )
2 2
2 2
2 2 2
2 2
90
2 25 65
2 60 28 62
3 43 47
=(cos cot )
(sin º cos º)
tan º cot º
[sec º– tan º ]
ec 2 2
2 2
2 2
2 22 25 25
21
2
1
228 28
3 43 43
1
= 1
2 1
1
228
1
283
22
( )
tan ºtan º 1
=1
2
1
6 =1
3 1
19. x x x x x
x x x
x x
x x
x
x2 3 2
3 2
2
2
2 6 11 12
2
5 9 12
5 5 10
4 2
5
–
– –
– –
–
Quotient x – 5, remainder = 4 x – 2 1
Dividend = (Divisor × Quotient) + Remainder 1
= ( x2 – x + 2) ( x – 5) + 4 x – 2
= x3 – x2 + 2 x – 5 x2 + 5 x – 10 + 4 x – 2
= x3 – 6 x2 + 11 x – 12 Proved 1
20. Let1
y= a, the given equations become
4 x + 6a = 15 ...(1)
6 x – 8a = 14 ...(2) ½
Multiply eqn (1) by 4 and eqn (2) by 3 and adding
16 x + 24a = 60
18 x – 24a = 42
On adding 34 x = 102
x =102
34= 3 ½
Put the value of x in eqn (1)
4(3) + 6a = 15
6a = 15 – 12 = 3
a =3
6
=1
2
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Solutions | 17
a =1
y=
1
2 y = 2 1
Hence x = 3 and y = 2
Again y = px – 2 2 = p (3) –2 3 p = 4 p =4
3 1
Or
29 x + 41 y = 169 ...(1)
41 x + 29 y = 181 ...(2)
Adding equations (1) and (2), we get
70 x + 70 y = 350 1
x + y = 5 ...(3)
Subtracting eqn. (1) from eqn. (2), we get
–12 x + 12 y = –12
– x + y = –1 ...(4)
Adding equations (3) and (4), we get2 y = 4
y =4
2= 2 1
Putting this value of y in eqn (3), we get
x + 2 = 5 x = 3
Hence, x = 3 and y = 2 . 1
21. Let assumed mean, a = 35 and h = 10.
xi di= x a
hi
f i f idi
(Class Marks)5 – 3 5 – 15
15 – 2 13 – 26
25 – 1 20 – 20
35 0 15 0
45 1 7 7
55 2 5 10
Total f i = 65 f idi = – 44 1½
Mean, x–
= a +
f d
f
i i
i× h ½
= 35 +44
65× 10 = 35 – 676
x–
= 2824. 1
22. In ADB, AB2 = AD2 + BD2 [Pythagoras Theorem] ...(1)
A
B C
D3 x x
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18 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
In ADC, AC2 = AD2 + CD2 [Pythagoras Theorem] ...(2) 1
Subtracting eqn. (2) from eqn. (1), we get
AB2 – AC2 = BD2 –CD2 1
=3
4BC
1
4BC
2 2F H G
I K J
F H G
I K J =
BC
2
2
2 (AB2 – AC2) = BC2
2 (AB)2 = 2AC2 + BC2. 1
Or
According to question, BD = CD =BC
21
BC = 2BD
Using Pythagoras theorem in the right ABC, we have
AC2 = AB2 + BC2
= AB2 + 4BD2 1
= (AB2
+ BD2) + 3BD
2
AC2 = AD2 + 3CD2. 1
23. LHS = (coses – cot )2
=1
2
sin
cos
sin
F H G
I K J 1
=1
2F H G
I K J
cos
sin
=( cos )
sin
1 2
2
=( cos )( cos )
cos
1 1
1 2
1
=( cos )( cos )
( cos )( – cos )
1 1
1 1
=1
1
cos
cos
1
= R.H.S. Proved
24. (i) Here class intervals are not in inclusive form. So, we first convert them in inclusive formby subtracting 1/2 from the lower limit and adding 1/2 to the upper limit of each cases. whereh is the difference between the lower limit of a class and the upper limit of the preceding class.
The given frequency distribution in inclusive form is as follows : ½
Age (in years) No of cases
45 - 145 6
145 - 245 11
245 - 345 21
345 - 445 23
445 - 545 14
545 - 645 5
From table it is clear from table that the modal class is 345 - 445.
Here, l = 34.5, h = 10, f 1
= 23, f 0
= 21, f 2
= 14 ½
CB
A
D
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Solutions | 19
Now, Mode = l f f
f f f h+
–
2 –0
0
1
1 2–
Mode =23–21345+ 10
46–21–14
= 345 +2
11 10
= 345 + 181 = 3631 1
(ii) Mode of grouped data. ½
(iii) If we parctise habit of cleanliness we will be able to put disease at on arm’s length. ½
25. Let 5 be a rational number.
5 =
a
b , [a, b are co-prime integers and b 0]
a = b 5
a2 = 5b2
5 is a factor of a2 1
5 is a factor of a
Let a = 5c, [c is some integer]
25c2 = 5b2
5c2 = b2 1
5 is a factor of b2
5 is a factor of b
5 is a common factor of a, bBut this contradicts the fact that a, b are co-primes.
5 is irrational
Let 2 – 5 be rational 1
2 – 5 = a
2 – a = 5
2 – a is rational, so 5 is rational.
But 5 is not rational contradiction
2 – 5 is irrational. 1
26. 2 x – y = 1 y = 2 x – 1
x 0 1 3
y – 1 1 5
and x + 2 y = 13 y =13
2
x
x 1 3 5
y 6 5 4
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20 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
Putting the above points and drawing the lines joining them we get the graphs of above equations.
–4 –3 –2 –1 1 2 3 4 5 6 7
y'
–1
–2
–3
1
2
3
5
4
x' x
y
6
7
8
0–5–6
–4
–5
–6–7
8
2
–
= 1
x
y
(1, 6)
(3, 5)
(5, 4)
x y+ 2 = 13
(1, 1)
(0, –1)B
Clearly two lines intersect at point A (3, 5).
2
Hence, x = 3 and y = 5 is the solution of above equations. 1
ABC is the trianguler shaded region formed by the lines and the y-axis. 1
27. L.H.S. =tan sec
tan sec
1
1=
(tan sec ) (sec tan )
tan sec
2 2
1 [1 + tan2 = sec2 ] 1½
=(tan sec ) (sec tan )(sec tan )
tan sec
1
1
=(tan sec ) [ sec tan ]
tan sec
1
11
= tan + sec = RHS [Proved] ½
28.
CI f cf
0 – 10 5 5
10 – 20 x 5 + x
20 – 30 6 11 + x
30 – 40 y 11 + x + y
40 – 50 6 17 + x + y50 – 60 5 22 + x + y
Here from table, N = 22 + x + y = 40 1
x + y = 18 ...(1)
Since median = 31, median class is 30 – 40.
Median = l +
N c f
f
2F
H
GGG
I
K
J J J
. .
h
31 = 30 +20 11– ( )L
N
MO
Q
P x
y
× 10 1
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Solutions | 21
1 =( – )9 10 x
y
y = 90 – 10 x
From (1), 10 x + y = 90 ...(2)1
(On subtraction) x + y = 18
– – –9 x = 72
x =72
9= 8
From (1), y = 18 – 8 = 10. 1
29. According to question, and are zeroes of
p( x) = 6 x2 – 5 x + k
So, sum of zeroes + = –F
H GI K J
5
6
5
6...(1)
Product of zeroes = =k
6
– =1
6 [Given] ...(2) 1
Adding equation (1) and (2), we get 2 = 1
=1
21
Put the value of in equation (2), we get1
2– =
1
61
2
1
6 =
2
6=
1
3 1
=k
612
13
k = 1. 1
Or
Let the fraction be x
y, then according to the question
x
y
2
2=
9
11
11 x –9 y = –4 ...(1) 1
and x
y
3
3=
5
6 6 x – 5 y = –3 ...(2) 1
Now (1) and (2) 11 x –9 y + 4 = 0
and 6 x – 5 y + 3 = 0
Solving these equations by cross-multiplication method, we get x
(–9)( ) – (–5)( )3 4=
y
( )( ) – ( )( )4 6 11 3=
1
11 6( )(–5) – ( )(–9)
x
–27 20=
y
24 33
1
54– –55
x y
7 9
1
11
Hence x = 7, y = 9
Fraction =7
9 1
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22 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
30. Proof :
In quadrilateral ABCD, AO
BO=
CO
DO[Given]
AO
CO=
BO
DO...(1)
In ABD EO || AB (Construction)
AE
ED=
BO
DO[By BPT] ...(2)1
From eqns, (1) and (2), AE
ED=
AO
CO
In ADC, AE
ED=
AO
CO
EO || DC [Converse of BPT] 2
EO || AB [Constructions]
AB || DC
In quad. ABCD, AB || DC 1
ABCD is a trapezium. Proved
Or
(i) Let the initial position of the man be at O and his final position be B, Since the man goes to10 m due east and then 24 m due north. Therefore, AOB is a right triangle right angled at A suchthat OA = 10 m and AB = 24 m.
S
O
E
N B
A
W
24 m
10 m
By Pythagoras theorem, we have ½
OB2 = OA 2 + AB2
OB2 = (10)2 + (24)2
= 100 + 576 = 676
OB = 676 = 26 m 1
Hence, the man is at a distance of 26 m from the starting point.
(ii) Right angled triangle. ½
(iii) Slow and steady wins the race. ½
31. tan = AB
BC
1
5
In ABC, AC2 = AB2 + BC2 = 1 + 5 = 6 1
AC = 6
(i)cos sec
cos sec
ec
ec
2 2
2 2
=
( ) –
( )
66
5
66
5
66
5
66
5
22
2
2
F
H GI
K J
F
H GI
K J
=
24
36
2
3 1½
A B
CD
OE
A
B C
16
5
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Solutions | 23
(ii) sin2 +cos2 =1
6
5
6
2 2F H G
I K J
F
H GI
K J
=1
6
5
6 = 1. 1½
32. Proof BC = 2BD [ AD is the median]
A
B D C
P
Q M Rand QR 2 QM [PM is the median] 1
Given, AB
PQ=
AD
PM=
BC
QR
AB
PQ=
AD
PM=
2BD
2QM1
In triangles ABD and PQM, AB
PQ=
AD
PM=
BD
QM
ABD ~ PQM [SSS Similarity]
B = Q [By CPCT] 1
In ABC and PQR, AB
PQ=
BC
QR
and B = Q [ ABC ~ PQR] 1
33. LHS = (1 + cot A – cosec A) (1 + tan A + sec A)
= 11
11
F H G
I K J
F H G
I K J
cos
sin sin
sin
cos cos
A
A A
A
A A 1
=sin cos
sin
sin cos
cos
A A
A
A A
A
F H G
I K J
F H G
I K J
1 1
=(sin cos )
sin cos
A A
A A
2 211
=sin cos sin cos –
sin cos
2 2 2 1 A A A A
A A
1
= 1 2 sin cossin cos
A A – 1 A A
= 2 = RHS. 1
34.
Monthly Wages c.f.
More than 80 200
More than 100 180
More than 120 150
More than 140 130 2
More than 160 90
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24 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
x
y
Scale
-axis 2 cm = 10 units
-axis 1 cm = 10 units
x
y
Lower Lts
200
190180
170
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
080 100 120 140 160
More than ogive
(80, 200)
(100, 180)
(120, 150)
(140, 130)
(160, 90)
2
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Solutions | 25
Section - A
1. (C), The number of rational numbers lying between 2 and 3 is infinite. 1
2. (B), Given, Polynomial = x2 – 3
Sum of zeroes =coeff. of
coeff. of 2
x
x= 0
Product of zeroes =constant
coeff. of 2 x= – 3
Hence required answer 0, – 3. 1
3. (D), For similar triangles,
area of triangle 1
area of triangle 2= ( )
235
=925
1
4. (C), tan x = sin 45º cos 45º + cos 60º
=1
2
1
2
1
2
F H G
I K J F H G
I K J
+
=1
2
1
2+ = 1 = tan 45º
⇒ x = 45º. 1
5. (C), It is clear that 0·102003000...........is non-terminating non-repeating decimal, so it cannot
be expressed in the form of p
q. 1
6. (B), 4 x –5 y = 5
kx + 3 y = 3
For the condition of inconsistent
a
a
1
2=
b
c
c
c
1
1
1
2
≠
i.e.,4
k=
−≠
5
3
5
3
⇒4
k=
−5
3
⇒ k = −125
1
7. (B), tan 3A = cot (A – 26)
⇒ cot (90º – 3A) = cot (A – 26º)
⇒ 90º – 3A = A – 26º
⇒ 4A = 90º + 26º = 116º
⇒ Α =116
4
º= 29º. 1
Sample Question Paper–8
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26 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
8. (B), The modal class of the distribution is 50 – 60. 1
9. 6762
2 3381
3
7
7
1127
161
23
= 7 × 161
=161
7
2 × 3381 =
Composite number x = 6762. 1
10. Sum = – 2; product of zeroes = 1 1
p( x) = x2 – (sum of the zeroes) x + product of the zeroes
= x2 – (– 2) x + 1 ½
p( x) = x2 + 2 x + 1 ½
11.
AB || QR
PA
AQ PBBR
5125 – 5
6 x
½
x = 4 1
PR = 4 + 6 = 10 cm ½
12. LHS =1
1
cos
cos
A
A =
1
1
1
1
cos
cos
cos
cos
A
A
A
A 1
=( cos )
( cos )
( cos )
sin
1
1
12
2
2
2
A
A
A
A
=1 1
cos
sin sin
cos A
A A
A
sin A
= cosec A – cot A = RHS Proved. 1
1
P
A B
Q R
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Solutions | 27
13. x2 – x – 72 = x2 – 9 x + 8 x – 72
= ( x – 9) ( x + 8)
Zeroes of the given polynomial are 9 and – 8 1
Sum of the zeroesb
a
=(–1)1
= 1 = 9 – 8 ½
Product of the zeroesca
=721
= – 72 = – 72 = 9 × (– 8) ½
Hence Verified.
14.
Classes Frequency cf
135 – 140 4 4
140 – 145 18 22
145 – 150 7 29
150 – 155 11 40155 – 160 6 46
160 – 165 5 51 1
From above,
Median class is 145 – 150 ½
and Modal class is 140 – 145 ½
Or
More than Frequency
More than 75 5
More than 70 20
More than 65 34 1
More than 60 42
More than 55 48
More than 50 50 1
Question number 15 to 24 carry three marks each.
15. We have, PQR ~ PAB [P is common andPA
PQ=
PB
PR] 1
area PQR
area PAB
=PQ
PA
F H G
I K J
2
32
area PAB=
4
3
2k
k
F H G
I K J 1
area PAB = 18 cm2
area of AQRB = area of PQR – area of PAB = 32 – 18 = 14 cm2 1
P
Q R
3k
A B
k
4k
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28 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
16. p( x) = x2 + 7 x + 9
q( x) = x + 2
r( x) = – 1
g( x) = ?
p( x) = g( x).q( x) + r( x) ½
x2 + 7 x + 9 = g( x).( x + 2) – 1 ½
g( x) =x x
x
2 7 10
2
1
=( )( )
( )
x x
x
2 5
2½
g( x) = x + 5. ½
17. Let 3 be a rational number
3 =a
b. [a and b are integers and co-primes]
On squaring both the sides 3 = ab
2
2 1
3b2 = a2 a2 is divisible by 3 a is divisible by 3 ...(1)We can write a = 3c for some integer c a2 = 9c2
3b2 = 9c2
b2 = 3c2 1 b2 is divisible by 3 b is divisible by 3 ...(2)From (1) and (2), we get 3 is a factor of ‘a’ and ‘b’.Which is contradicting the fact that a and b are co-primes. Hence our assumption that 3 is
rational number is false. So 3 is irrational number. 1Or
154
+5
40=
15 25 5 254 25 40 25
1
=375100
+125
10001
= 375 + 0125 = 3875 1
18. cos + sin = 2 cos ½
L.H.S. sin = 2 cos – cos
= cos ( 2 – 1) × 2 12 1
=cos (2 1)
2 1
½
sin =cos
2 1
½
2 sin + sin = cos ½
cos – sin = 2 sin ½
= 2 sin
= R.H.S. Proved. ½
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Solutions | 29
19.
2 + 2 x
6 + 2 – 4 + 3 x x x3 2
– + –6 – 4 + 2 x x x3 2
6 – 6 + 3 x x2
– + –6 – 4 + 3 x x2
– 2 + 1 x
3 – 2 + 1 x x2
Quotient = 2 x + 2, Remainder is – 2 x + 1 1½
p( x) = g( x).q( x) + r( x)
= (3 x2 – 2 x + 1) (2 x + 2) + (– 2 x + 1) ½
= 6 x3 – 4 x2 + 2 x + 6 x2 – 4 x + 2 – 2 x + 1
= 6 x3 + 2 x2 – 4 x + 3 ½= p( x) Verified. ½
20. Let the sum of the ages of the 2 children be x and the age of the father be y years.
y = 2 x i.e. 2 x – y = 0 ...(1) 1
and 20 + y = x + 40
x – y = – 20 ...(2)
Subtracting (2) from (1), we get x = 20
From (1), y = 2 x = 2 × 20 = 40
y = 40 1
Hence, the age of the father = 40 years. 1Or
(i) Let the fixed charges of taxi be Rs x and the running charges be Rs. y per km. According to the question
x + 10 y = 75 ...(1)
x + 15 y = 110 ...(2) ½
Subtracting equation (2) from equation (1), we get
– 5 y = – 35
y = 7
Putting y = 7 in equation (1), we get x = 5 ½
Total charges for travelling a distance of 25 km.
= x + 25 y= Rs. (5 + 25 × 7)
= Rs. (5 + 175)
= Rs. 180 1(ii) Pair of Linear equations in two variables. ½
(iii) We should always be justified in our dealings. ½
21. Mode = l + f f
f f f
1 0
1 0 22
× h 1
Here, modal class : 30 – 40, l = 30, f 1 = 25, f 0 = 20, f 2 = 12, h = 10 ½
Mode = 30 +25 20
50 20 12
–
– –×10 1
= 30 +5 10
18
= 30 + 277 = 3277 (Modal age). ½
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30 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
22. Given : ABC, right angled at A.
B A
C
M
L
BL and CM are medians.
To prove : 4(BL2 + LM2) = 5BC2.
Proof : In ABL,
BL2 = AB2 + AL2 1
= AB2 + AC
2
F H G
I K J
2
[BL is median]
In ACM, CM2 = AC2 + AM2
= AC2 + AB
2
2F H G
I K J , [CM is median] 1
BL2 + CM2 = AB2 + AC2 + AC
4
AB
4
2 2
4(BL2 + CM2) = 5AB2 + 5AC2
= 5(AB2 + AC2)
= 5BC2. Proved. 1
Or
Given : In ABC, ADBC and AD2 = BD × CD.
To prove : ABC is a right triangle.
Proof : In right triangle ABD and ACD, Applying Pythagoras theorem
AB2 = AD2 + BD2
and AC2 = AD2 + CD2 1
AB2 + AC2 = 2AD2 + BD2 + CD2
= 2(BD × CD) + BD2 +CD2, given AD2 = BD × CD 1
= (BD + CD)2 = BC2
i.e. AB2 + AC2 = BC2
ABC is a right triangle. 1
23. tan + cot = 2
tan +1
tan = 2
2tan 1
tan
= 2 1
tan2 + 1 = 2tan tan2 – 2tan + 1 = 0 (tan – 1)2 = 0 tan – 1 = 0 tan = 1 = tan 45°
= 45° 1
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Solutions | 31
Now, tan7 + cot7 = tan7 45° + cot7 45°
= (tan 45°)7 + (cot 45°)7
= (1)7 + (1)7
= 1 + 1
= 2 1
24.
Class f Class x u = x a
h
fu
0 – 10 7 5 – 2 – 14
10 – 20 12 15 – 1 – 12
20 – 30 13 25 0 0
30 – 40 10 35 1 10
40 – 50 8 45 2 16
Total f = 50 fu = 0 2
Let a = 25 and h = 10
Mean, x–
= a + fu
f × h ½
= 25 +0
50× 10 = 25 (Mean) ½
25. Let us assume that there is a positive integer n for which n n 1 1 is rational and equal
toP
Q, where P and Q are positive integers (Q 0) 1
n n 1 1 = PQ ...(i)
Q
P=
1
– 1 +1n n
=n n
n n n n
– 1 +1
– 1 +1 – 1 +1
( )( )
=n n
n n
n n– 1 +1 – 1 +1
( – ) – ( )1 1 21
But n n+1 – 1 =2Q
P...(ii)
Apply (i) + (ii), we get 2 n+1 =
P
Q
2Q
P
P 2Q
PQ
2 2
n +1 =P 2Q
2PQ
2 2...(iii)
Apply [(i) – (ii)], we get
n – 1 =P 2Q
2PQ
2 2...(iv)1
From (iii) and (iv) we can say n +1 and n – 1 both are rational because P and Q both are
rational. But it is possible only when n + 1 and n – 1 both are perfect squares. But they differ by 2and two perfect squares never differ by 2. So both n + 1 and n – 1 not be perfect squares, hence
there is no positive integer n for which n n– 1 +1 is rational. 1
26. Let the speed of the boat be x km/hr and speed of the stream be y km/hr.
According to the question,
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32 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
32 36
x y x y
= 7
and40 48
x y x y
= 9
Let 1 x y = A, 1
x y = B, we get
32A + 36B = 7
and 40A + 48B = 9
Solving these equations, we get A =1
8, B =
1
21
Hence A =1
8=
1
x y
x – y = 8 ...(1)½
and B =1
12
1
x y
x + y = 12 ...(2)½
Adding equations (1) and (2), we get 2 x = 20
x = 10 1
Putting this value of x in eqn.(1), we get
y = x – 8 = 10 – 8 = 2
Hence, the speed of the boat = 10 km/hr and speed of the stream = 2 km/hr. 1
27. Given 5 cos = 4 tan =45
1
Now5sin 3cos
5sin 2cos
5sincos 3cos
5sincos 2cos
1
5tan 35tan 2
=
45 3545 25
=4 34 2
=
16
1
28.
Class c.f.
Less than 30 3Less than 40 7
Less than 50 17
Less than 60 25
Less than 70 34
Less than 80 42
Less than 90 54
Less than 100 60
1½ + 1½
10 20 30 40 50 60 70
10
20
30
40
50
60
70
80
80 90
Median - 65 (approx)
X
Y
O 100
Scale
x-axis 1 cm = 10 unitsy-axis 1 cm = 10 units
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Solutions | 33
29. Since 3 x2 – 5 divides f ( x) completely
(3 x2 – 5) is a factor of f ( x)
3( x2 –5
3) is a factor of f ( x)
x x
F
H G
I
K J
F
H G
I
K J
5
3
5
3 is a factor of f ( x)
5
3and –
5
3are zeroes of f ( x) 1
3 x2 – 5 ) 3 x4 + 3 x3 – 11 x2 – 5 x + 10 ( x2 + x – 2
3 x4 – 5 x2
– +
3 x3 – 6 x2 – 5 x
3 x3 – 5 x 1
– +
– 6 x2 + 10
– 6 x2 + 10
+ –×
( x2 + x – 2) is a factor of p( x)
( x2 + 2 x – x – 2) is a factor of p( x)
( x + 2) ( x – 1) is a factor of p( x)
– 2 and 1 are zeroes of p( x) 1
all the zeroes of p( x) are5
3, –
5
3, – 2 and 1. 1
Or
x – y = 1 y = x – 1
x 2 3 – 1
y 1 2 – 2
2 x + y = 8 y = 8 – 2 x
x 2 4 0
y 4 0 8
Putting the above points and drawing a line joining them, we get the graph of the above equation
x – y =1 and x + y = 8.
–4 –3 –2 –1 1 2 3 4 5 6 7
y'
–1
–2
–3
1
2
3
5
4
x' x
y
(0, – 1)
6
7
8
A
(3, 2)
x
y –
=1
2 + = 8 x y
0
(0, 8)
(2, 4)
(2, 1)(4, 0)
(–1, –2)
2
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34 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
Clearly, the two lines interesect at point A (3, 2). 1
Solution of given equations is x = 3, y = 2.
Again x – y = 1 intersects y-axis at (0, –1) 1
and 2 x + y = 8 intersects y-axis at (0, 8).
30. AB || CQ, BC is transversl 1 = 2 Also, A C || PQ, BC is transversal 3 = 4
ABC ~ QCP
ar (ABC)ar (QCP) =
2
2
BC
CP1
= 2
2
(BC)
2 BC3
1 2as BP = BC CP = BC3 3
1
=2
2
9BC
4BC The required ratio is 9 : 4. 1
Or
Let BC = 2 x and QR = x
AB =14
y and PQ = y 1
ar (ABC)ar (PQR) =
1 BC × AB21 QR × PQ2
1
=
124
x y
x y
=24 =
12 1
The required ratio is 1 : 2. 1
31. acot + bcosec = x2
bcot + acosec = y2
L.H.S. : x4 – y4 = ( x2 – y2)( x2 + y 2) ½
= (a cot + b cosec – b cot – a cosec )(a cot + b cosec + b cot + a cosec ) 1
= a2 (cot2 – cosec2– b2 (cot2 – cosec2 ) 1
= a2 (– 1) – b2 (– 1) ½
= – a2 + b2 ½
= b2 – a2 ½
32. In ABC, DE || AC [Given]
BD
DA =
BE
EC....(1) [BPT]1
In ABL, DC || AL [Given]
BD
DA =
BC
CL...(2) [BPT]1
From (1) and (2), we getBE
EC=
BC
CL
4
2=
6
CL
CL = 3 cm. ½
A
B LC
D
E4 2
B P C
A
O
4 2
31
A
B C2 x
y1
4
P
Q R x
y
1
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Solutions | 35
33.cos A – sin A +1cos A +sin A – 1
=cot A – 1 + cosec A cot A + 1 – cosecA ½
[Dividing numerator and denominator by sin ]
=
(cot A + cosec A) – 1
cot A + 1 – cosecA ½
=2 2(cot A + cosec A) – (cosec A – cot A)
cot A + 1 – cosecA 1
=(cot A + cosec A) – [1 – cosecA + cotA]
cot A + 1 – cosec A 1
=cosA sinA +
1sinA =
1 cosA sinA
= R.H.S
Proved. 1
34.
Weight (in g) Cumulative Frequency
More than or equal to 0 120
More than or equal to 10 106
More than or equal to 20 89More than or equal to 30 67
More than or equal to 40 41
More than or equal to 50 18 2
Plotting the points
120
110
100
90
80
70
60
50
40
30
20
10
(0, 120)
(10, 106)
(20, 89)
(30, 67)
(40, 41)
(50, 18)
0 10 20 30 40 50 x
y
C u m u l a t i v e f r e q u e n c y
Scale
-axis 2 cm = 10 units
-axis 1 cm = 10 units
x
y
More than ogive
Lower Lts
60
2
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36 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
1. (D) mn = 32 = 25
Hence n = 5k, and m = 2
Now nm + n = (5)5+2
= 57 1
2. (A), From the graph it is clear that the curve cuts the x-axis at four places, so number of zeroes
is 4. 1
3. (A),ar ( ABC)
ar ( PQR
AC
PR
2
2
)
= AC
PR
2F H G
I K J
81
169=
7.2
PR
F H G
I K J
2
7.2
PR
F H G
I K J
2
= 29
13
7.2PR
=9
13
PR = 13 7 29 .
= 10.4 cm. 1
4. (D), 2 tan22 1 sin 47ºcot68 2 cos43
= 2
tan22 1 sin47ºcot (90 22 ) 2 cos (90 47 )
= 2tan22 1 sin 47ºtan22 2 sin 47
= 2 +12
=52
1
5. (A), a = 5, b = 13 16. (C), 2 x + y = 7
6 x – py = 21For the condition of infinite number of solutions
a
a
1
2
=b
b
c
c
1
2
1
2
2
6=
1 7
21
p
1
3=
1
p
p = – 3. 1
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Solutions | 37
7. (A),2 30
1 302
tan º
tan º=
21
3
11
3
2
3
11
3
2
34
3
2
F H G
I K J
F H G
I K J
=
2
3
3
4
=
3
2
= sin 60º. 1
8. (C), The upper limit of modal class is 175. 1
9. Any positive integer ending with the digit zero is divisible by 5. So prime factorization mustcontain 5.
(12)n = (22 × 3)n 1
By uniqueness of fundamental theoren 5 does not occur in the prime factorization of (12) n. ½
(12)n does not end with the digit zero. ½
10.
3 – 4 + 2 x x2
3 + 5 – 7 + 5 + 3 x x x x4 3 2 x x2 + 3 + 1
3 + 9 + 3 x x x4 3 2
– – –
– 4 – 10 + 5 + 3 x x x3 2
– 4 – 12 – 4 x x x3 2+ + +
2 + 9 + 3 x x2
2 + 6 + 2 x x2
– – –
3 + 1 = ( ) x r x
– (3 x + 1) is to be added so that the polynomial is exactly divisible by x2 + 3 x + 1. ½
11.
FED ~ STU
F = S ½E = T ½
D = U ½
EDTU
=FD EFSU TS
are proportional
andDE EFST TU
is wrong ½
These are not their corresponding sides. ½
12. 2 sin 45° + 3 tan 30° – tan 45° = 21
2
+ 31
3
– 1 1
= 1 + 1 – 1 = 1 ½
1½
E D T U
SF
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38 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
13. p( x) = ax2 + bx + c ½
constant = 0 = c
p( x) = ax2 + bx ½
Product of zeroes =0
a= 0 ½
One of the zero must be zero, then only product is zero. ½
14. 2 mean = 3 median – mode
Mean =3 median – mode
21
=3 28 – 35
2
=84–35
2=
492
1
Mean = 245
Or
Class Fequency Less then cf
65 - 85 4 4
85 - 105 5 9
105 - 125 13 22
125 - 145 20 42 ½
145 - 165 14 56
165 - 185 7 63
185 - 205 5 68
½
Upper limit of median class = 145 ½
Lower limit of modalclass = 125 ½Difference = 20
15. Since XY || QR
PX
XQ=
PY
YR1
1
2=
PY
PR PY =
4
PR 4
RR – 4 = 8 PR = 12 cm
In right PQR, QR2 = PR2 – PQ2 1
= 122 – 62 = 144 –36 = 108
QR = 6 3 cm 1
16. Let and are zeroes
=
23
½
=23
+ = 15 ½
RQ
P
X Y
4 cm
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Solutions | 39
23
= 15
53
= 15 ½
= 15 35 = 9
so =23
× 9 = 6 ½
Now, sum of zeroes = + = 6 + 9 = 15
product of zeroes = = 6 × 9 = 54 ½
Hence polynomial = x 2 – (sum of zeroes) x + product of zeroes
= x2 – 15 x + 54 ½
17. Euclid's algorithm
a = bq + r
0 r b
if b = 6
a = 6q + r
then r = 0, 1, 2, 3, 4, 5.
The odd positive integers n can be expressed as 6q + 1, 6q + 3, 6q + 5 ½
n = 6q + 1
n 2 = (6q + 1)2
= 36q2 + 12q + 1 ½
= 6 (6q 2 + 2q) + 1
= 6m + 1 where m = 6q2 + 2q
n = 6q + 3 ½n2 = (6q + 3)2
= 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
= 6m + 3where m = 6q2 + 6q + 1 ½
n = 6q + 5
n2 = (6q + 5)2
= 36q2 + 60q + 25 ½
= 6(6q2 + 10q + 4) + 1
= 6m + 1 where m = 6q2 + 10q + 4
Thus square of an odd positive integer is either of the form 6m + 1 (or) 6m + 3. ½
Or
(i) In order to arrage the books as required, we have to find the largest number that devides96, 240 and 336 exactly, clearly, such a number is their HCF.We have,
96 = 25 × 3240 = 24 × 3 × 5
and 336 = 24 × 3× 7 HCF of 96, 240, and 336 is 24 × 3 = 48So, there must be 48 books in each stack. 1
Number of stacks of English books =96
48= 2
Number of stacks of Hindi books = 24048
= 5
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40 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
Number of stacks of sociology books =336
48= 7 1
(ii) HCF of numbers. ½(iii) Cleanlines has been discussed in this question, it is a good habit that leads to good health. ½
18. In ABC, A + B + C = 180°
2
= 90° – C2
1
tan A + B2 = cot
C2
L.H.S. = tan A + B2 . tan
C2
1
= cotC2
.tanC2
= 1 1
19. and are the zeroes of p ( x) p ( x) = 3 x2 – x – 4 ½
+ = – ba = –
13
= 13
=ca
= –43
½
43 + 34 = 33 ( + )= ()3 ( + ) ½
= 34 1
3 3
½
= –64 1
27 3
4
3
+ 3
4
= ––6481 1
20. Since BC || DE and BE || CD with BC CD, BCDE is a rectangle.
Opposite sides are equal.
i.e., BE = CD x + y = 5 ...(i)
and DE = BC = x – y 1
Since perimeter of ABCDE is 21
AB + BC + CD + DE + EA = 21
3 + x – y + x + y + x – y + 3 = 21
6 + 3 x – y = 21
3 x – y = 15 1
Adding (i) and (ii), we get 4 x = 20 ...(ii)
x = 5
On putting the value of x in (i), we get
y = 0
x = 5 and y = 0. 1
Or
Let the number of red balls be x and white balls be y.
According to the question,1
2 y =
1
3 x or 2 x – 3 y = 0 ...(1)
and 3( x + y) – 7 y = 6 1
or, 3 x – 4 y = 6 ...(2)
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Solutions | 41
Mutliplying eqn. (1) by 3 and eqn. (2) by 2 and then subtracting, we get
6 x – 9 y = 0 1
6 x – 8 y = 12
– y = – 12
Subtracting from (1), y = 12
2 x – 36 = 0 x = 18 1 x = 18, y = 12
Hence, No. of red balls = 18 and No. of white balls = 12. 1
21. The chief motivational factor for .....ent.
Class Frequency
0 - 100 8
100 - 200 12
200 - 300 x
300 - 400 20
400 - 500 14
500 - 600 7
Since 340 is mode
300 - 400 is modal class ½
mode l +1 0
1 0 22
f f
f f f
× h ½
340 = 300 +20
40 14 x
x
× 100 1
40 = 2026
x x
× 100
52 – 2 x = 100 – 5 x ½
3 x = 48 x = 16 ½
22.
AM BCDN BC ½
In AMO and DNO
M = N = 90°
AOM = DON [Vertically opposite s] ½
AMO ~ DNO
AMDN
= AODO
...(1) ½
ar ( ABC)ar ( DBC)
=
1 × BC × AM21 × BC × DN2
½
= AMDN
Substituting from (1)
ar ( ABC)
ar ( DBC)
= AO
DO
Proved. 1
Or
M
O NCC
A
D
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42 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
In PQR, CA || PR
PC
CQ=
RA
AQ[By BPT]
PC
15 =
20
12
PC =15 20
12
= 25
PC = 25 cm 1
In PQR, CB || QR
PC
CQ=
PB
BR
25
15=
15
BR1
BR =15 15
25
= 9 cm. 1
23. LHS = 2sec2 – sec4 – 2cosec2 + cosec4
= 2 sec2 – (sec2 )2 – 2 cosec2 + (cosec2 )2
= 2(1 + tan2 ) – (1 + tan2 )2 – 2 (1 +cot2 ) + (1 + cot2 )2 1
= 2 + 2tan2 – 1 – 2 tan2 – tan4 – 2 – 2cot2 + 1 + 2cot2 + cot4 1
= cot4 – tan4 RHS Proved 1
24.
C.I. f x v =i x a
h
fv
10 – 20 5 15 – 2 – 10
20 – 30 11 25 – 1 – 11
30 – 40 19 35 0 0
40 – 50 30 45 1 30
50 – 60 15 55 2 30
Total 80 fv = 39 2
x = a + hi i
i
f x
f
= 35 +3980
× 10
= 35 + 4875
= 39875 1
25. Let a = 4q + r, 0 r < 4
a = 4q, 4q + 1, 4q + 2 or 4q + 3 1
Case I : a2 = (4q)2 = 16q2 = 4(4q2) = 4m, m = 4q2 ½Case II : a2 = (4q + 1)2 = 16q2 + 8q + 1 = 4(4q2 + 2q) + 1
= 4m + 1, where m = 4q2 + 2q ½
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Solutions | 43
Cases III : a2 = (4q + 2)2 = 16q2 + 16q + 4
= 4(4q2 + 4q + 1) = 4m, where m = 4q2 + 4q + 1 ½Cases IV : a2 = (4q + 3)2 = 16q2 + 24q + 9
= 4(4q2 + 6q + 2) + 1
= 4m +1, where m = 4q2 + 6q + 2 ½
From cases I, II, III and IV, we conclude that the square of any +ve integer is of the form 4 m or4m +1. 1
26. x + 3 y = 6 ...(1)
y =6
3
x
x 3 6 0
y 1 0 2 ½
2 x – 3 y = 12 ...(2)
y =2 12
3
x
x 0 6 3
y – 4 0 – 2 ½
Putting the above points and drawing a line joing them, we get the graphs of the equations x + 3 y
= 6 and 2 x – 3 y = 12.
–3 –2 –1 1 2 3 4 5 6
(6, 0)
(3, 1)(0, 2)
–1–2
–3
–4
y'
x x'
y
1
2
2–3
=1 2
x
y
x
y + 3
= 6
(0, –4)
A
(3, –2)
0B
Clearly, the two lines intersect at point B (6, 0).
Hence, x = 6 and y = 0 is the solution of the system. 1
Again OAB is the region bounded by the linee 2 x – 3 y = 12 and both the co-ordinate axes. ½
27. sin 62° = sin (90 – 28) = cos 28° ½
tan 73° = tan (90 – 17) = cot 17° ½sin 28° sec 62° = sin (90 – 62)° sec 62°
= cos 62° sec 62° = 1
= 7 [sec2 (90 – 58) – cot2 58] 1
= 7 [cosec2 58 – cot2 58] = 7 1
=5(1)cos 28 3 cot17 –
cos28 cot17 7(1) ½
1 + 3 –57
=(28– 5)
7
=
23
7 ½
1½
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44 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
28.
Class Fequency Class mark x fx
0 - 20 5 10 50
20 - 40 8 30 240
40 - 60 f 50 50 f 60 - 80 12 70 840
80 - 100 7 90 630
100 - 120 8 110 880
Total 40 + f 2640 + 50 f 1
x = fx
f
½
628 =2640 50
40 f
2512 + 628 f = 2640 + 50 f ½
628 f – 50 f = 2640 – 2512128 f = 128
f = 10
Mode = l f f
f f f h+
–
2 –0
0
1
1 2– ½
= 60 +12 10
24 10 7
× 20 ½
= 60 +27
× 20 = 60 + 571 1
Mode = 6571
29. p( x) = x3 + 10 x2 + px + q
1 and – 2 are zeroes p (1) = 1 + 10 + p + q = 0
p + q = – 11 ...(1) ½
p (– 2) = (– 2)3 + 10 (–2)2 + p (– 2) + q = 0
– 8 + 40 – 2 p + q = 0
– 2 p + q = – 32 ...(2)
p + q = – 11
– 2 p + q = – 32 7 + q = – 11 1
+ – +
+ 3 p = + 21
p = 7 q = – 18 p ( x) = x3 + 10 x2 + 7 x – 18 ½
x + 9
x x x3 + 10 + 7 – 182 x x2 + – 2
– – +
9x x2 + 9 – 18
9 + – 18 x x2 2– – +
0
x x x3 + – 22
Third zero is – 9 2
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Solutions | 45
Or
Let the speed of the car I from A = x km/hr.
Speed of the car II from B = y km/hr.
Same dirrection :
Distance covered by car I = 150 + (distance covered by car II) 15 x = 150 + 15 y
15 x – 15 y = 150 x – y = 10 ...(1) 1
Opposite direction :
Distance covered by car I + distance covered by car II = 150 km
x + y = 150 ...(2)1
Adding eqns. (1) and (2), we get 2 x = 160 x = 80
Putting x = 80 in eqn. (1), we get y = 70 1
speed of the car I from A = 80 km/hr
speed of the car II from B = 70 km/hr. 1
30. ABC is a equilateral triangle so AB = AC = BC
Since BD =13
BC
DC =23
BC
BE = EC =23
BC
AD2 = AE2 + DE2
= AC2 – EC2 + (BE – BD)2 ½
AB2 – EC2 + BE2 + BD2 – 2BEBD ½
= AB2 + 2
1 BC3 – 2 12 BC 13 BC 1
= AB2 + BC2 1 1–9 3 1
= AB2 –29
BC2
= AD2 = AB2 –29
AB2
=27AB
9½
9AD2 = 7AB2 ½
Or
In ABC and APQ, BC || PQ [Given]
CBA = AQP, [Alternate angles]
BAC = PAQ, [Vertically opposite angles]
ABC ~ AQP [AA ] 1
AB
AQ=
BC
QP=
AC
AP1
65 AQ
=8
4=
AC28
1
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46 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
AQ =652
= 3.25 cm, AC = 2 × 28 = 56 cm. 1
P
Q A
B
C
6.5 cm
4 cm
8 cm
2 . 8 c m
31. L.H.S. =3 3
2 2
tan cot
1 tan 1 cot
=3 3
2 2
tan cot
sec cosec
1
=
3 3
3 3
2 2
sin coscos sin
1 1cos sin
½
=3 3sin cos
cos sin
=
4 4sin coscos sin
½
=2 2 2 2 2(sin cos ) 2sin cos
cos sin
½
=2 21 2sin cos
cos sin
½
=sec .cosec 2sin .cos
= R.H.S 1
32. Let PQR be a right triangle right-angled at Q such that QR = b and A = Area of PQR.
Draw QN perpendicular to PR.
We have A = Area of PQR
=12
(QR × PQ)
=12
(b × PQ)
PQ =2A b
...(i)
Now, in 's PNQ and PQR, we have
PNQ = PQRQPN = QPR
So by AA – Criterion of similarity, we have
PNQ ~ PQQ
PQPR
=NQ
QR ...(ii)
Applyiong pythagoras theorem in PQR, we have
PQ2 + QR2 = PR2
2
22
4A b
b = PR2
PR =2 4
24A bb 2 4
4A bb 1
N
P
Q R
1
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Solutions | 47
from (i) and (ii) we have
2A × PRb
=NQ
b
NQ =2A PR
= 2 4
2A
4A
b
b1
33. p = sec + tan
=1 sin 1 sin
cos cos cos
½
p2 =2
2
1 sin 2sin
cos
½
p2 + 1 =2 2
2
1 sin 2sin cos
cos
= 2
2 2sin
cos
1
p2 + 1 = 2 2
(1 sin )
cos
1
L.H.S.2 1
2
p
p
=
2
2(1 sin )
cos 12(1 sin ) cos
cos
1
= sec = R.H.S.
34.
Less than cf More than cf
40 3 38 35
42 5s 40 32
44 9 40 30
46 14 40 26
48 28 40 21
50 32 40 7
52 35 40 3 ½+½
2
40 42 44 46 48 50 52
5
10
15
20
25
30
35
40
Classes
X
Y
O
Median = 47
c.f.
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48 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
1. (A),6243
2 53 4=
6243 2
2 54 4
=12486
104
Hence, decimal expansion of the rational number will terminate after 4 places of decimal. 1
2. (B), f ( x) = x2 – 7 x – 8
Let other zero be k, then – 1 + k = ––7
1
F H G
I K J
= 7
k = 8 1
3. (C), Given 2AB = DE and BC = 8 cm
ABC DEF
So, AB
BC=
DE
EF
AB
8=
2AB
EF
EF = 2 × 8 = 16 cm. 1
4. (D),11 11
2 2cot cos = 11
12
2 2
sin
cos cos
F
H G
I
K J
= 111
112
2
2
2
sin
cos
sin
cos
F
H G
I
K J
F
H G
I
K J
= – 11cos
cos
2
2
F
H G
I
K J = – 11. 1
5. (C), The smallest prime number = 2 and smallest composite number is 22.
Hence, required HCF (4, 2) = 2 1
6. (C), ad bc
ac
bd
Hence, the pair of linear equation has unique solution. 1
7. (A), cosec – cot =1
4
(cos cot )(cos cot )
(cos cot )
ec ec
ec
=
1
4
A
B C
D
E F
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Solutions | 49
2 2cosec cotcosec cot
=1
4
2 21 cot cot
cosec cot
=
1
4
cosec + cot = 4. 1
8. (B), Median 1
9. If the number 4n, were to end with the digit zero, then it should be divisible by 5. ½
But 4n = 22n ½
Only prime in the factorization of 4n is 2.
So by fundamental theorem of Arithmetic, there are no other primes in the factorisation of 4n.½
4n can never end with the digit zero. ½
10. Let, f ( x) = 2 x2 – 5 x – 3
Let the zeroes of polynomial are and then
Sum of zeroes + =5
2, product of zeroes = –
3
2
According to question, zeros of x2 + px + q are 2 and 2
Sum of zeroes =coeff. of
coeff. of 2
x
x=
p
1
= 2 + 2 = 2 ( + ) = 2 ×5
2= 5 p = – 5 1
Product of zeros =constant
coeff. of 2
x=
q
1
= 2 × 2 = 4 = 4 F H G
I K J
3
2= – 6
Hence,
p = – 5 and q = – 6. 1
11. Since G is mid point of PQ. PG = GQPG
GQ= 1 1
According to question, GH || QR
PG
GQ =
PH
HR [BPT]
1 =PH
HR
PH = HR
Hence, H is mid points of PR.
12. 2 cosec2 30° + 3sin2 60° –34
tan2 30°. 1
= 2 × (2)2 + 3
2 23 3 1
2 4 3
1
= 8 +
9
4 –
1
4 = 10 ½+½
P
Q R
G H
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50 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
13. Let p ( x) = 3 x2 + 11 x – 4
= 3 x2 + 12 x – x – 4
= 3 x ( x + 4) – 1 ( x + 4) ½
= (3 x – 1) ( x + 4)
So, zeroes are m = 12
and n = – 4. ½
Now,m
n
n
m =
1
3
4
4
1
3
F H G
I K J
F H G
I K J
=1
12– 12 ½
=145
12. ½
14.
Class Fequency Cumulative Frequency
( f ) (c.f.)
100 - 200 11 11
200 - 300 12 p
300 - 400 10 33
400 - 500 q 46
500 - 600 20 66
600 - 700 14 80
From above p + 10 = 33
p = 33 – 10 = 23 ½
33 + q = 46
q = 46 – 33 = 13 ½
Median Class = 400-500 ½
Model class = 500-600 ½
Or
C.I. f 1
100 - 120 12
120 - 140 14
140 - 160 8
160 - 180 6
180 - 200 1050
Modal Class : 120 – 140 ½
Mode = L +0 1
0 1 22
f f
f f f
× h ½
= 120 +14 12
28 12 8
× 20
= 120 + 5 = 125 1
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Solutions | 51
15. AP
AB=
35105
=1
3
AQ AC
= 39
= 13
1
In ABC, AP
AB=
AQ
ACand A is common 1
APQ ~ ABC [SAS]
AP
AB=
PQ
BC1
3=
45BC
BC = 135 cm. 1
A
B
P
C
Q
3.5
7 6
3
4.5
16. Let and are the zeroes of the polynomial, then as per question
= 7
+ 7 = 8 = – F H G
I K J
8
3½
=
1
3 ½
and × 7 =2 1
3
k
72 =2 1
3
k
71
3
2F H G
I K J =
2 1
3
k 1
7 ×1
9=
2 1
3
k
7
3– 1 = 2k
2
3= k. 1
17. We have 117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0 1
Hence HCF = 13
65m – 117 = 13
65m = 117 + 13 = 130
m =130
65= 2 1
Now, 65 = 13 × 5
117 = 32 × 13 1
LCM = 13 × 5 × 32
= 585Or
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52 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
(i) The number of rooms will be minimum if each room accomodates maximum number of participants. Since in each room the same number of participants are to be seated and all of themmust be of the same subject. Therefore, the number of paricipants in each room must be the HCFof 60, 84 and 108. The prime factorisations of 60, 84 and 108 are as under
60 = 22 × 3 × 5
84 = 2
2
× 3 × 7108 = 22 × 33
Hence, HCF = 22 × 3 = 12Therefore, in each room 12 participants can be seated. 1
Number of rooms required =Total number of participants
12
=60 84 108
12
=
252
12= 21. 1
(ii) HCF of numbers. ½(iii) Liberty and equality are the pay marks of democracy. ½
18.
cos ( º ) cos ( º– )
tan( º ) tan( º– )
2 245 45
60 30
+ cosec (75º + ) – sec (15º – )
=cos ( º ) sin ( º–45º )
tan( º ) cot( º–30º )
2 245 90
60 90
+ cosec (75º + ) – cosec (90º – 15º + ) 1
=cos ( º ) sin ( º )
tan( º ).cot( º )
2 245 45
60 60
+ cosec (75º + ) – cosec (75º + ) 1
=1
1= 1. 1
19.
4 3 2 8 14 2
8 6 4
8 2
8 6 4
4 4
4 3 2
7 2
2 2 12 4 3 2
4 3 2
3 2
3 2
2
2
2
x x x x x ax b
x x x
x x ax
x x x
x a x b
x x
a x b
x x
( )
( )
For exact division, remainder is zero, then(a + 7) x + b – 2 = 0 2
a + 7 = 0, b – 2 = 0
a = – 7, b = 2. 1
20. 141 x + 93 y = 189 ...(1)
93 x + 141 y = 45 ...(2)
Adding equations (1) and (2), we get
234 x + 234 y = 234
x + y = 1 ...(3)
Subtracting eqn. (2) from eqn. (1), we get
48 x – 48 y = 144
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Solutions | 53
x – y = 3 ...(4) 1
Adding eqns. (3) and (4), we get 2 x = 4
x = 2 1
Put the value of x in eqn. (3), we get 2 + y = 1
y = 1 – 2 = –1
Hence, x = 2 and y = –1. 1
Or
Let the speed of bus be x km/hr and the speed of the train be y km/hr.
According to question,240 120
x y = 8
and120 240
x y = 7
Let1
x= a,
1
y= b, then 1
240a + 120b = 8 ...(1)
120a + 240b = 7 ...(2)
Apply [(1) × 2 – (2)], we get
480a + 240b = 16
120a + 240b = 7 ...(2)
– – –
On subtracting 360a = 9
a =9
360
1
40 1
Putting this value of a in eqn.(1), we get
b =1
60
b =1
60
1
y
y = 60
a =1
40
1
x x = 40 1
Hence, Speed of bus = 60km/hr and Speed of train = 40 km/hr.
21.
xi f i xi f i
3 10 30
9 p 9 p
15 4 60
21 7 147
27 q 27q
33 4 13239 1 39
Total f i = 26 + p + q xi f i = 408 + 9 p + 27q
Given, f i = 40,
26 + p + q = 40
p + q = 14 ...(i)½
Mean, x–
=
x f
f
i i
i½
147 =408 9 27
40
p q
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54 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
588 = 408 + 9 p + 27q 1
180 = 9 p + 27q
p + 3q = 20 ...(2)
Subtracting eq. (1) from eq. (2), we get
2q = 6
q = 3 ½
Putting this value of q in eq. (1), we get
p = 14 – q = 14 – 3 = 11 ½
Or
(i) Let the assumed mean, A = 1400 and h = 400.
Calculation of Mean
Height (in m) x1 No. of Villlages f 1 D = xi –1400 ui = xi 1400
400 f i ui
200 142 –1200 – 3 –426
600 265 –800 – 2 –5301000 560 –400 – 1 –560
1400 271 0 0 0
1800 89 400 1 89
2200 16 800 2 32
Total N= f i= 1343 f iui = –1395 1
We have A = 1400, h = 400, f iui = –1395 and N = 1343.
Mean = A + h f ui i1
N RST
UVW
= 1400 + 400 ×
–1395
1343
F
H G
I
K J
= 1400 – 41549 = 98451 1
(ii) Mean by assumed mean method. ½
(iii) Villages are much necessary to keep a balance between the ecological problems. ½
22. Given, DBBC, DE AB and ACBC.
To prove :BE
DE=
AC
BC½
Proof : In ABC, 1 + 2 = 90º [C = 90º] 1
But 23 = 90º [Given]
1 = 3 ½
In ABC and BDE, 1 = 3 [Proved] ACB = DEB = 90º [Given]
ABC ~ BDE [AA] 1
AC
BC=
BE
DE
23. LHS =cos sin
cos sin
cos – sin
cos – sin
3 3 3 3
=(cos sin )(cos sin sin cos )
(cos sin )
(cos – sin )(cos sin sin cos )
(cos – sin )
2 2 2 2
1
= (1 – sin cos ) + (1 + sin cos ) 1
CB
A
D
E
1
23
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Solutions | 55
= 2 – sin cos + sin cos 1
= 2 = RHS Proved
24.
CI f c.f.
50 - 60 8 8
60 - 70 10 18
70 - 80 12 30
80 - 90 16 46
90 - 100 18 64
100 - 110 14 78
110 - 120 12 90
120 - 130 10 100 1
Here N = 100 N2
= 50. So, median class is 90-100.
Median = l +
N . .2
c f h
f
½
= 90 + 50 4610
18 ½
= 90 +4018
= 922
Median weight = 922 gm. 1
25. n3 – n = n(n2 – 1) = n(n + 1)(n – 1) = (n – 1)n(n + 1)
= product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
Let a, a + 1, a + 2 be any three consecutive integers. ½
Case I. If a = 3q.
a(a + 1) (a + 2) = 3q(3q + 1) (3q + 2)
= 3q (even number, say 2r) = 6qr,( Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer)
which is divisible by 6. 1
Case II. If a = 3q + 1.
a(a + 1) (a + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)
= 6r (q + 1), 1
which is divisible by 6.
Case III. If a = 3q + 2.
a(a + 1) (a + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q
= 6r (say), 1
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56 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6. ½
26. 2 x + 3 y = 12 y =12 2
3
x
x 0 6 3
y 4 0 2
x – y = 1 y = x – 1
x 0 1 3
y – 1 0 2
Putting the above points and drawing a line joining them, we get the graph of the equations 2 x +
3 y = 12 and x – y = 1.
–5 –4 –3 –2 –1 1 2 3 4 5 6 7 8
y'
–1
–2
–3
–4
1
2
3
5
4
x' x
y
B
P(3, 2)
x – y = 1
(0, – 1)
2 3 = 12 x + y
A
(0, 4)
M
(1, 0) (6, 0)
Clearly, the two lines interesect at point P (3,2).
2
Hence x = 3 and y = 2 is the solution of the system 1
Area of shaded region = area of PAB
=1
2× base × height =
1
2× AB × PM
=1
2× 5 × 3
= 75 square unit. 1
27. sin A =1
5, cos A = 1 sin A 2
cos A = 11
5
2
F H G
I K J
= 11
5 =
2
51
sin B =1
10, cos B = 1 2 sin B
cos B = 11
10
2
F H G
I K J
= 11
10 =
3
101
Now cos ( A + B) = cos A cos B – sin A sin B
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Solutions | 57
=2
5
3
10
1
5
1
10 –
=6
50
1
50 =
5
50=
5 1
5 2 2 1
cos (A + B) =1
2= cos 45º
A + B = 45º. 1
Or
2 52
38
70
20
sin º
cos º
tan º
cot º + cos2 63º + sin 63º cos 27º +
sin ºsec º
tan º.tan º.tan º
55 35
7 45 83
=2cos (90 – 52)º cot (90 – 70)º
cos38º cot20º + cos2 63º + sin 63º × sin (90º – 27º) +
sin55º cosec (90º –35º)
tan7ºtan45º cot(90º–83º)
1
=2cos38º cot20ºcos 38º cot 20º
+ cos2 63º + sin2 63º +sin55º cosec 55º
tan7º.tan45º.cot7º
1
= 2 –1 + 1 +
1sin55ºsin55º
1tan 7º 1tan7º
1
= 2 –1 + 1 +1
1= 3. 1
28. The mean of the following distribution is 628.
C.I. f
5 - 15 2
15 - 25 3
25 - 35 f
35 - 45 7
45 - 55 4
55 - 65 2
65 - 75 2
Modal class : 35-45 ½
Mode = l +1 0
1 0 22
f f h
f f f
½
39 = 35 +7
14 4
f
f
10 1
4 =70 10
10
f
f
½
40 – 4 f = 70 – 10 f ½
6 f = 30 f = 5 1
29. Since and are the zeroes of polynomial 3 x2 + 2 x + 1.
Hence + = 2
3
and =1
3
1
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58 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
Now for the new polynomial
Sum of the zeroes =1
1
1
1
=
( ) ( )
( )( )
1 1
1 1
=2 2
1
22
3
1
2
3
1
3
Sum of zeroes =
4
32
3
= 2 1
Product of zeroes =1
1 1
F H G
I K J
F H G
I K J
1 –
=( )( )
( )( )
1 1
1 1
=1
1
1
1
( )
( )
Product of zeroes =1
2
3
1
3
12
3
1
3
6
32
3
3
1
Hence, required polynomial = [ x2 – (Sum of zeroes) x + Product of zeroes]
= x2 – 2 x + 3. 1
Or
Let two digit number is 10 x + y
According to question, 8( x + y) –5 = 10 x + y
2 x –7 y + 5 = 0 ...(1)
and 16( x – y) + 3 = 10 x + y
6 x –17 y + 3 = 0 ...(2) 1
Solving eqns. (1) and (2) by cross-multiplication method, we get x
(–7)( ) – (–17)( )3 5=
y
( )( ) – ( )( )5 6 2 3=
1
2 6( )(–7) – ( )(–7)1
x
21 85=
y
30 6
1
34 42
x
64=
y
24
1
8
x
8=
y
3= 1 1
Hence x = 8, y = 3
So required number = 10 × 8 + 3 = 83. 1
30. Proof : In POQ, AB || PQ, [given]
OA
AP=
OB
BQ...(i) [BPT] 1
In OPR, AC || PR, (given)
OA
AP=
OC
CR...(ii) [BPT ] 1
From (i) and (ii), we getOB
BQ=
OC
CR1
BC || QR, [By converse of BPT] 1
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Solutions | 59
Or
31. LHS. =1
1
1
1
sin
sin
sin
sin
=( sin )
( sin )
( sin )
( sin )
( sin )
( sin )
( sin )
( sin )
1
1
1
1
1
1
1
1
1
=( sin )
sin
( – sin )
sin
1
1
1
1
2
2
2
2
1
=( sin )
cos
( sin )
cos
1 12
2
2
2
1
=1 1
sin
cos
sin
cos
=2
cos= 2 sec = RHS Proved 1
32. In ABC, DE || AC, [Given]
BD
DA =
BE
EC[BPT] ...(1)1
In ABE, DF || AC, (Given)
BD
DA =
BF
FE[BPT] ...(2)1
From (1) and (2), we haveBF
FE=
BE
EC Proved 1
33. sin =3
4 cos =
7
4and tan =
3
7½+½
2 2
2
cosec cot 72cot cos
sec 1 x
1
27
3
7 7
42tan
x1
1 2 7
3
7 7
4tan
x
7
3
2 7
3
7
4
7
x1
4 7 7
4
7
x
3 74
7 x x =4
3 1
1
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60 | CBSE (CCE) Sample Question Papers Term-I, Mathematics Class X
34. Draw the ‘less then’ ogive and ‘more than’ ogive sumultaneously.
Less than c.f. More than c.f.
30 10 20 10040 18 30 90
50 30 40 8260 54 50 7070 60 60 4680 85 70 40
90 100 80 15 2
100
90
80
70
60
50
40
30
20
10
0 10 20 30 40 50 x
y
C u m u l a t i v e f r e q u e n c y
From the graph median = 58.3
60 70 80 90
(20, 100)
(30, 90)
(40, 82)
(50, 70)
(60, 46)
(70, 40)
(80, 15)
(90, 10)
(80, 85)
(70, 60)
(60, 54)
(50, 30)
(40, 18)
(30, 10)
More than ogive
Less than ogive
2