1. No category

Mathematics
Class X
TOPPER SAMPLE PAPER-1
Maximum Marks: 80
Time: 3 Hrs
1. All questions are compulsory.
2. The question paper consist of 30questions divided into four sections
A, B,C and D. Section A comprises of 10 questions of one mark
each, section B comprises of 5 questions of two marks each ,section
C comprises of 10 questions of three marks each and section D
comprises of 5 questions of six marks each.
3. All questions in Section A are to be answered in one word, one
sentence or as per the exact requirement of the question.
4. In question on construction, the drawing should be neat and exactly
as per the given measurements.
5. Use of calculators is not permitted. You may ask for mathematical
tables, if required.
6. There is no overall choice. However, internal choice has been
provided in one question of 02 marks each, three questions of 03
marks each and two questions of 06 marks each. You have to
attempt only one of the alternatives in all such questions.
Section A
Q1
If HCF (252, 378) = 126, find their LCM.
Q2
Find the polynomial shown in the graph.
Q3
For what value of ‘k’ will the equations 13x+23y-1=0 and kx–46y-2=0
represents intersecting lines?
Q4
QM ⊥ RP and PR 2 − PQ 2 = QR 2 . If ∠ QPM = 300, find ∠ MQR.
Q
P
M
R
Q5
Find the length of PN if OM = 9 cm.
M
N
Q
P
O
Q6
If the median of a data which represents the weight of 150 students in
a school is 45.5 kg, find the point of intersection of the less than and
more than ogive curves.
Q7
If two coins are tossed simultaneously, find the probability of getting
exactly two heads.
Q8
If three times the third term of an AP is four times the fourth term ,
find the seventh term.
Q9
If sin α + cos α = 2 cos(90 − α ) , find cot α .
Q10 Find the perimeter of the figure , where AC is the diameter of the semi
circle and AB ⊥ BC
A
6cm
B
8cm
C
SECTION B
Q11 A and B are points (1, 2) and (4, 5) . Find the coordinates of a point P
on AB if AP =
2
AB.
5
Q12
∆ ABC is right angled at C. Let BC = a , AB = c and AC = b. p is the
length of the perpendicular from C to AB. Prove that
1
1
1
= 2+ 2
2
p
a
b
Q13 Solve for x and y : 3(2x+y) = 7xy, 3(x +3y) = 11xy
Q14 Find the probability of getting 5 Wednesdays in the month of August.
Q15
If sin ( A + B ) = 1 and cos ( A - B)=
3
, 0 ≤ A + B ≤ 90° , A >B , find
2
A and B.
OR
If tan A =
7
, evaluate
24
1 − cos A
1 + cos A
SECTION C
Q16 Prove that
5 is an irrational number.
Q17 Find the coordinates of a point(s) whose distance from (0,5) is 5 units
and from (0,1 ) is 3 units.
Q18 Prove
(sin A + cos ecA)2 + (cos A + sec A)2 = tan 2 A + cot 2 A + 7
OR
Prove (1 + cot θ − cos ecθ )(1 + tan θ + secθ ) = 2
Q19
Solve for x and y :
10
4
15
7
+
= −2 ,
−
= 10,
x+ y y−x
x+ y y−x
x + y ≠ 0, x ≠ y
OR
For what values of ‘m’ will 2mx 2 − 2(1 + 2m) x + (3 + 2m ) = 0 have real and
distinct roots?
Q20 Find the area of a triangle whose sides have (10, 5), (8, 5) and (6, 6)
as the midpoints.
Q21 If α , β are the zeroes of the polynomial 3 x 2 − 11x + 14 , find the value of
α2 + β2 .
Q22 Prove that a parallelogram circumscribing a circle is a rhombus.
OR
Prove that the opposite sides of a quadrilateral circumscribing a circle
subtend supplementary angles at the centre of the circle.
Q23 Draw a circle of radius 3.5 cm. Construct two tangents to the circle
which are inclined to each other at 120° .
Q24 A grassy plot is in the form of a triangle with sides 45m, 32m and 35m.
One horse is tied at each vertex of the plot with a rope of length 14m.
Find the area grazed by the three horses.
Q25 The 46th term of an AP is 25. Find the sum of first 91 terms.
SECTION D
Q26 Prove that ratio of the areas of two similar triangles is equal to the
ratio of the squares of their corresponding sides.
Using this theorem find the ratio of the area of the triangle drawn on
the diagonal of a square and the triangle drawn on one side of the
square.
OR
State and prove the Basic Proportionality Theorem. If PQ II BC, find
PQ.
B
6cm
11.4 cm
P
3cm
C
A
Q
Q27 The area of a rectangle remains the same if the length is increased by
7m and the breadth is decreased by 3 m. The area of the rectangle
remains the same if the length is decreased by 7m and the breadth is
increased by 5 m. Find the dimensions of the rectangle and the area of
the rectangle.
Q28 A boy is standing on the ground and flying a kite with a string of 150m
at an angle of 30° . Another boy is standing on the roof of a 25m high
building and flying a kite at an angle of 45° . Both boys are on the
opposite sides of the kites. Find the length of the string the second boy
must have so that the two kites meet.
OR
At a point on level ground, the angle of elevation of a vertical tower is
such that its tangent is
5
. On walking 192 m towards the tower, the
12
tangent of the angle of elevation is
3
. Find the height of the tower.
4
Q29 A solid consists of a cylinder with a cone on one end and a hemisphere
on the other end. If the length of the entire solid is 12.8cm and the
diameter and height of the cylinder are 7cm and 6.5 cm respectively,
find the total surface area of the solid.
Q30 Draw a less than ogive of the following data and find the median from
the graph. Verify the result by using the formula.
Marks
No.
girls
Less
Less
Less
Less
Less
Less
than
than
than
than
than
than
140
145
150
155
160
165
11
29
40
46
51
of 4
Mathematics
Class X
TOPPER SAMPLE PAPER-1
SOLUTIONS
Ans1
HCF x LCM = Product of the 2 numbers
126 x LCM = 252 x 378
LCM = 756
Ans2
(1 Mark)
The zeroes are –1, 4
∴ p ( x) = ( x + 1)( x − 4 ) = x 2 − 3 x − 4
Ans3
(1 Mark)
For intersecting lines:
a1 b1
13 23
≠
⇒
≠
a2 b2
k −46
(1 Mark)
⇒ k ≠ −26
Ans4
2
2
Since PR − PQ = QR
2
⇒ PR 2 = QR 2 + PQ 2
⇒∠RQP = 90o (Converse of Pythagoras Theorem)
Therefore, In ∆PQM
Since ∠QPM = 30o and ∠QMP = 90o
So ∠MQP = 60o
Hence, ∠ MQR= 30o
(1 Mark)
Ans5
M
N
Q
P
O
OM = MQ + QO
= QP + QN [Since Tangents from external point are equal]
= PN = 9cm
(1 Mark)
Ans6
The two curves namely less than and more than ogives
intersect at the median so the point of intersection is (45.5, 75)
(1 Mark)
Ans7
Total outcomes = HH, TT, HT, TH
Favourable outcomes = HH
P ( E : Both Heads ) =
Ans8
1
4
(1 Mark)
Let a3 and a4 be the third and fourth term of the AP
According to given Condition
3.a3 = 4.a4
⇒ 3 ( a + 2d ) = 4 ( a + 3d )
⇒ a = −6d
⇒ a + 6d = 0
⇒ a7 = 0
(1 Mark)
Ans9
sin α + cos α = 2 sin α
cos α = sin α
(
)
2 −1
cos α
= 2 −1
sin α
(1 Mark)
cot α = 2 − 1
Ans10
A
6cm
B
8cm
C
AC =
=
AB2 + BC2
(U sin g Pythagoras Theorem)
82 + 62
= 64 + 36
= 10 cm
Circumference of semi circle = π r
= 3.14 × 5
= 15.70cm
∴ Perimeter = 6 + 8 + 15 .7
= 29.7 cm
(1 Mark)
SECTION B
Ans11
Since, AP =
2
AB
5
So AP: PB =2: 3
P divides AB in 2:3 ratios
(1 mark)
 2 × 4 + 3 ×1 2 × 5 + 3 × 2 
P
,

5
5


1

 2 mark 


 11 16 
P , 
5 5
1

 2 mark 


1
AC . CB
2
1
= a.b
2
1
Also, area ( ∆ACB ) = . AB .CD
2
1

 2 mark 


Ans12
Area ( ∆ACB ) =
=
1
cp
2
⇒
⇒
1
1
ab = cp
2
2
ab = cp
1

 2 mark 


1
1 b2 + a2
Now 2 + 2 =
a b
a 2b 2
c2
a2b2
c2
= 2 2
ab
=
c2
= 2 2
c p
1
= 2
p
( By Pythagoras theorem)
( Since, ab = cp )
Hence Proved
Ans13
(1 Mark)
3 ( 2 x + y ) = 7 xy
⇒ 6 x + 3 y = 7 xy
(1)
3 ( x + 3 y ) = 11xy
⇒ 3 x + 9 y = 11xy
(2)
Eq (2) × 2 gives :
6 x + 18 y = 22 xy
(3)
When x ≠ 0 and y ≠ 0 eq(1) − eq (3) gives
−15y = − 15 xy
⇒
x =1
⇒
y=
3
2
Also x = 0, y = 0 is a solution.
1

 2 mark 


1

 2 mark 


1

 2 mark 


1

 2 mark 


Ans14
August has 31 days
⇒ 4 weeks and 3 days.
So 4 weeks means 4 Wednesdays
Now remaining 3 days can be
SMT
T W Th
Th F Sa
MTW
W Th F
F Sa S
Sa. S M
(1 Mark)
Favorable outcomes are = M T W
T W Th
1

 2 mark 


W Th F
∴ P (3 Wednesdays) =
Ans15
3
7
1

 2 mark 


sin (A + B) = 1
o
Since sin 90 = 1
A + B = 90o
cos ( A − B ) =
1

 2 mark 


(1)
3
2
since cos30o =
3
2
A – B = 30o
1

 2 mark 


(2)
Solving (1) and (2)
A = 60o
1

 2 mark 


B = 30o
1

 2 mark 


OR
tan A =
7
24
So the ratio of adjacent and opposite side of the triangle is in
the ratio 7:24
Let the common ratio term be k
Using Pythagoras Theorem
1

 2 mark 


AC = 25 k.
Consider
1 − cos A
1 + cos A
24
25
=
24
1+
25
1−
(1 Mark)
1
1
=
49 7
=
1

 2 mark 


SECTION C
Ans16
Let us assume
⇒ 5=
5 is rational.
p
Where p and q are co prime integers and q ≠ 0
q
1

 2 mark 


⇒
5q = p
⇒
5q 2 = p 2
⇒
⇒
5divides p 2
5divides p (1)
1

 2 mark 


1

 2 mark 


So p =5a for some integer a
2
Substituting p = 5a in 5q = p
2
5q 2 = 25a 2
⇒
q 2 = 5a 2
⇒
⇒
5divides q 2
5divides q
1

 2 mark 


(2)
From (1) & (2) 5 is a common factor to p and q which
contradicts the fact that P and q are co prime
∴ Our assumption is wrong and hence
Ans17
1
5 is irrational.  mark 
2

Let A ( x, y ) be the required point which is at a distance of 5
units from the point P(0,5) and 3 units from Q(0,1)
So AP =5 and AQ = 3
⇒
2
( x − 0 ) + ( y − 5)
2
2
⇒
( x − 0 ) + ( y − 5)
⇒
x 2 + y 2 − 10 y = 0
2
( x − 0 ) + ( y − 1)
2
2
=5
1

 2 mark 


= 25
(1)
1

 2 mark 


1

 2 mark 


=3
2
x 2 + ( y − 1) = 9
x2 + y 2 − 2 y − 8 = 0
(2)
Equation (1) – Equation (2) gives:
−8 y + 8 = 0 ⇒ y = 1
1

 2 mark 


Substituting y = 1 in (1)
x2 − 9 = 0 ⇒ x = ± 3
1

 2 mark 


∴ The required points are (3, 1) and (-3, 1)
Ans18
( sin A + cos ecA)
2
+ ( cos A + sec A )
2
= sin 2 A + cosec 2 A + 2 sin A cos ec A + cos 2 A
1

 2 mark 


2
+ sec A + 2 cos A s ec A
= ( sin 2 A + cos 2 A) + 2 + 2 + cosec 2 A + sec 2
(Since sin A.cosec A=1 and cosA.secA =1)
(1Mark)
= 1 + 2 + 2 + 1 + cot 2 A + 1 + tan 2 A
2
2
(1 Mark)
2
2
(Since, cosec A = 1 + cot A and sec A = 1 + tan A )
= 7 + cot 2 A + tan 2 A
1

 2 mark 


=RHS
OR
(1+ cotθ - cosec θ )(1+ tanθ + secθ )
1 
 cosθ 1  sinθ
=  1+
+
 1+

 sinθ sinθ  cosθ cosθ 
1

 2 mark 


 sinθ + cosθ -1  cosθ + sinθ +1 
=


sinθ
cosθ



2
2
( sinθ + cos θ ) - (1)
=
sinθ cosθ
1

 2 mark 


sin 2 θ + cos 2 θ + 2 sinθ cosθ -1
sinθ cosθ
1 + 2sin θcosθ -1
=
siin cosθ
1

 2 mark 


=
1

 2 mark 


1

 2 mark 


=
2sin θ cosθ
=2
sin θ cosθ
Ans19 Let
1

 2 mark 


1
1
= a,
=b
x+ y
y−x
10a + 4b = −2
→
(1)
15a − 7b = 10
→
(2)
1

 2 mark 


(1) × 3 and (2) × 2 gives
30 a + 12 b = − 6
30 a − 14 b = 20
26b = −26
(1 mark)
⇒ b = −1
Substituting b = -1 in
(1):
10 A − 4 = −2
1

 2 mark 


⇒ 10a = 2
1
⇒a=
5
∴
x+ y =5
− x + y = −1
2y = 4 ⇒ y = 2
(1 mark)
∴ x=3
OR
For real and distinct roots:
D>0
1

 2 mark 


Discriminant D = b2-4ac
2
−
 2 (1 + 2m )  − 4 ( 2m )( 3 + 2m ) > 0
2
4 (1 + 2m ) − 4 ( 2m )( 3 + 2m ) > 0
1

 2 mark 


1 + 4m 2 + 4m − 6m − 4m 2 > 0
1

 2 mark 


1 − 2m > 0
⇒
⇒
⇒
1

 2 mark 


1 > 2m
1
>m
2
1
m<
2
(1mark )
Ans20
(1 mark)
We know that area of triangle formed by joining the midpoint of
sides of a triangle is
1
th the area of the triangle.
4
1
(ar ∆ABC)
4
1
ar(∆PQR) = 10 ( 6 − 5 ) + 6 ( 5 − 5 ) + 8 ( 5 − 6 ) 
2
1
= [10 − 8]
2
= 1 sq unit
ar(∆PQR) =
So ar (∆ABC) = 4 sq unit
1

 2 mark 


(1 mark)
1

 2 mark 


Ans21
3x 2 − 11x + 14
11
14
α + β = , αβ =
3
3
2
2
α + β = ( α + β ) − 2αβ
2
 11 
 14 
=  −2  
3
 3
121 28
=
−
9
3
37
=
9
Ans22
(1 mark)
(1 mark)
1

 2 mark 


1

 2 mark 


We know that tangents drawn from an external point are equal.
∴
Let AP = AQ = a
BP= BS = b
CS = CR = C
DQ = DR = d
1

 2 mark 


Since ABCD is a parallelogram, opposite sides are equal.
a +b = c +d
a +d = c +b
on subtracting , we get
1

 2 mark 


b−d =
d −b
⇒
⇒
2b = 2d
b=d
∴
AB = a + b
1

 2 mark 


= a+d
= AD
Since adjacent sides are equal ABCD is a rhombus.
1

 2 mark 


OR
We know that the two tangents drawn from an external point are
1

equally inclined to the line joining the point and centre
 2 mark 


1

 2 mark 


∴
Let ∠OAP = ∠OAS = a ∠OCQ = ∠OCR = c
∠OBP = ∠OBQ = b ∠ODR = ∠ODS = d
In ∆AOB : a + b + x = 1800
In ∆ COD : c + d + y = 180
0
On adding a + b + c + d + x + y = 360
1

 2 mark 


1

 2 mark 


1

 2 mark 


⇒ 180 + x + y = 360
(Using angle sum property of quadrilateral 2a+2b+2c+2d=360)
So x + y = 180o
1

 2 mark 


Hence proved.
Ans23
Ans24
Construction of circle and 2 radii OA,OB at an angle of 60o
Construction of tangents through the points on the circle
(1 mark)
(2 marks)
Let the angles of triangle be x, y, z.
Area grazed by the three horses
=
=
x
y
z
π r2 +
π r2 +
π r2
360
360
360
π r2
360
( x + y + z)
(1 mark)
1

 2 mark 


=
π r2
1

 2 mark 


× 180
360
22
1
=
× 14 × 14 ×
7
2
1

 2 mark 


1

 2 mark 


= 308 m 2
Ans25
a46 = 25
1

 2 mark 


⇒ a + 45d = 25
S91 =
91
[ 2a + 90 d ]
2
(1mark)
1

 2 mark 


1

 2 mark 


= 91( a + 45d )
= 91 × 25
1

 2 mark 


= 2275
Section D
Ans26
Given: ∆ ABC ∼ ∆ DEF
ar ( ∆ ABC ) AB2 BC2 AC2
To Prove:
=
=
=
ar ( ∆ DEF ) DE 2 EF2 DF2
Construction: Draw AM ⊥ BC and DN ⊥ EF
Proof: In ∆ ABC and ∆ DEF
(1mark)
1
ar ( ∆ ABC ) 2 × BC × AM BC AM
=
=
.
1
ar ( ∆ DEF )
EF
DN
× EF× DN
2
∴
∴
∴
∴
...(i)
1


Area
of
∆
=
× base ×corresponding altitude 

2

∆ ABC ∼ ∆ DEF
...(Given)
AB BC
=
DE EF
∠B = ∠E
...(Sides are proportional)...(ii)
... (∵ ∆ ABC ∼ ∆ DEF )
...(each 90o )
...(AASimilarity)
∠M = ∠N
∆ ABM ∼ ∆ DEN
AB AM
=
DE DN
...(iii )[Sides are proportional]
From (ii) and (iii), we have
BC AM
=
DE DN
(1 mark)
From (i) and (iv), we have
ar ( ∆ ABC ) BC BC BC 2
=
.
=
ar ( ∆ DEF ) EF EF EF2
Similarly, we can prove that
ar ( ∆ ABC ) AB2 AC 2
=
=
ar ( ∆ DEF ) DE 2 DF2
∴
ar ( ∆ ABC ) AB 2 BC2 AC2
=
=
=
ar ( ∆ DEF ) DE 2 EF 2 DF 2
(2 marks)
∆BCQ and ∆ACP are equilateral triangles and therefore similar.
(1 mark)
AC 2 = AB 2 + BC 2 = 2 BC 2
(By Pythagoras theorem)
1

 2 mark 


Using the above theorem
area ACP AC 2 2 BC 2
=
=
=2
area BCQ BC 2
BC 2
1

 2 mark 


OR
Statement: If a line is drawn parallel to one side of a triangle to
intersect the other two sides in distinct points, the other two sides
are divided in the same ratio.
(1 Mark)
Given: In ∆ABC, DE BC
To prove:
AD AE
=
DB EC
Construction: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D
(1 mark)
Proof: In ∆ADE and ∆BDE
1
ar ( ∆ADE ) 2 × AD × EM AD
=
=
ar ( ∆BDE ) 1 × DB× EM DB
2
[Area of ∆ =
...(i )
1
× base× corresponding altitude]
2
In ∆ADE and ∆CDE
1
ar ( ∆ADE ) 2 × AE × DN AE
=
=
ar ( ∆CDE ) 1 × EC × DN EC
2
...(ii )
∵
DE BC
...(Given)
∴
ar ( ∆BDE ) = ar ( ∆CDE )
...(iii )
(∵ ∆s on the same base and between the same parallel sides are
equal in area)
From (i), (ii) and (iii)
AD AE
=
DB EC
Since PQ || BC
∆APQ ∼ ∆ABC
∴
⇒
⇒
AP PQ
=
AB BC
3 PQ
=
9 11.4
34.2
PQ =
= 3.8 cm
9
(2 marks)
(By AA condition)
(1 mark)
(1 mark)
Ans27
Let length of rectangle = x m
Breadth = y m
Area
1

 2 mark 


= xy m 2 .
( x + 7 )( y − 3) = xy
(1 mark)
⇒ − 3 x + 7 y − 21 = 0 → (1)
1

 2 mark 


( x − 7 )( y + 5) = xy
(1 mark)
5 x − 7 y − 35 = 0 →
(2)
1

 2 mark 


(1) + (2) gives:
2 x − 56 = 0
⇒
x = 28m
(1 mark)
On substituting x = 28 m in equation (2), we get y = 15 m
The length is 28 m and the breadth is 15m.
1

 2 mark 


Therefore, area is 420m2
(1 mark)
Ans28
A and R are the positions of the two boys. P is the point where the two
1

kites meet.
 2 mark 


(1 mark)
In ∆ ABP
sin 30o =
PB
AP
1 PB
=
2 150
⇒
PB = 75 m
 1

1 2 mark 


and QB = 25m
⇒ PQ = 50m
(1mark )
In ∆ PQR
sin 45o =
PQ
PR
1
50
=
2 PR
⇒ 50 2 = PR
 1

1 2 mark 


∴The boy should have a string of length 70.7m
1

 2 mark 


OR
(1 mark)
D is the initial point of observation and C is the next point of
observation.
AB is the tower of height h. Let BC = x
AB
BD
5
h
=
12 x + 192
1

 2 mark 


⇒ 12h − 5 x − 960 = 0 → (1)
(1 mark)
tan α =
tan p =
AB
BC
1

 2 mark 


3 h
=
4 x
⇒ 3x = 4h → (2)
From
(2):
(1 mark)
12h = 9 x and substituting in (1):
9 x − 5 x = 960
⇒
4 x = 960
x = 240
(1mark )
3 × 240
from (2)
4
= 180
∴h =
1

 2 mark 


∴ The height of the tower is 180 m.
1

 2 mark 


Ans29
Height of cone = 12.8 − ( 6.5 + 3.5 )
= 2.8 c
Slant height l =
(1 mark)
2
( 3.5) + ( 2.8)
2
= 12.25 + 7.84
= 20.09
= 4.48
TSA = 2π r 2 + 2π rh + π rl
= π r ( 2r + 2h + l )
 1

1 2 mark 


=
22 7
× ( 7 + 13 + 4.48 )
7 2
= 11× 24.48
= 269.28
∴ The surface area of the solid is 269.28 cm2
(1 mark)
 1

1 2 mark 


1

 2 mark 


Ans30
CI
f
C. f
Less than 140
4
4
140 - 145
7
11
145 -150
150 − 155
18
11
29
40
155 − 160
160 − 165
6
5
51
46
51
(2 marks)
(2 marks)
n
= 25.5
2
Median class = 145 − 150
n

 2 − cf 
Median = l + 
h
 f 


n = 51 ⇒
 25.5 − 11 
= 145 + 
5
 18 
14.5 × 5
= 145 +
18
72.5
= 145 +
18
= 149.02
(2 marks)