1. No category

TOPPER SAMPLE PAPER 5
Summative Assessment-II
MATHEMATICS
CLASS X
M.M: 80
TIME : 3-3
1
Hrs.
2
GENERAL INSTRUCTIONS :
1.
2.
3.
4.
All questions are compulsory.
The question paper consists of 34 questions divided into four
sections, namely
Section A : 10 questions (1 mark each)
Section B : 8 questions (2 marks each)
Section C : 10 questions (3 marks each)
Section D : 6 questions (4 marks each)
There is no overall choice. However, internal choice has been
provided in 1 question of two marks, 3 questions of three marks
and 2 questions of four marks each.
Use of calculators is not allowed.
SECTION A
Q1.
If x=b is a solution of x2- (a + b) x + p=0, then the value of p is
(a) ab
Q2.
(b)
a + b (c)
a-b
(d)
a
b
The ratio of the volume of a cube to that of a sphere which will exactly fit
inside the cube is
(a) :8
(b) 
(c) 8: 
(d) 6: 
Q3.
If A (1,2) , B (4,y), c (x,6) and D (3,5) are the vertices of a parallelogram
taken in order then the values of x and y are
(a)
(b)
(c)
(d)
Q4.
6 and 5
6 and 3
2 and 3
5 and 2
In a lottery there are 5 prizes and 20 blanks. The probability of getting a prize
is
(a)
Q5.
1
2
(b)
1
3
(c)
1
4
(d)
1
5
If x silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted
to form a cuboid 11cmx10cmx7cm, then the value of x is
(a) 1200
Q6.
1600
(d) 1800
(b)
450
(c)
300
(d) 900
In given AP 210 is ............. term: 2,6,10........
(a) 50th
Q8.
(c)
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a
diameter of the circle such that  POR=1200, then  OPQ is
(a) 600
Q7.
(b) 1400
(b)
52nd
(c)
53rd
(d) 54th
A kite is flying, attached to a thread which is 165m long. The thread makes
an angle of 300 with the ground. The height of the kite from the ground,
assuming that there is no slack in the thread is
(a) 80 m
(b)
81.5 m
(c)
82.5 m
(d)
84 m
Q9.
The points (a, b + c), (b, c + a) and (c, a + b) are
(a) Collinear
(b) Non-collinear
(c) Concurrent
(d) All of the above
Q10.
If the area of a circle is 154 cm 2, then its perimeter is
(a) 11cm
(b)
22cm
(c)
44cm
(d)
55cm
SECTION B
Q11.
Find the area of a square ABCD, whose vertices are A(5,6), B(1,5), C(2,1)
and D(6,2).
Q12.
Q13.
Q14.
The height of a right triangle is 7 cm less than its base. If the hypotenuse is
13 cm, find the other two sides.
Prove that the angle between two tangents drawn from an external point to a
circle is supplementary to the angle subtended by the line segments joining
the points of contact at the centre.
The minute hand of a clock is 10 cm long. Find the area of the face of the
clock described by the minute hand between 9AM and 9:35 AM.
Q15.
Four equal circles are described about the four corners of a square so that
each touches two of the others as shown in figure. Find the area of the square
not included in the circles if each side of the square measures 14cm.
Q16.
A number x is selected from the numbers 1, 2, 3 and then a second number y
is randomly selected from the numbers 1, 4, 9. What is the probability that
the product xy of the two numbers will be less than 9?
OR
Three unbiased coins are tossed together. Find the probability of getting
atmost two tails.
Q17.
Q18.
Determine the 10th term from the end of the AP 4,9,14,........,254.
Show that the tangents at the end points of a diameter of a circle are parallel.
SECTION C
Q19.
The area of a triangle is 5. Two of its vertices are (2, 1) and (3,-2) and the
third vertex lies on y=x+3. Find the third vertex.
Q20.
If the roots of the equation (a-b)x2+(b-c)x+(c-a)=0 are equal then prove that
2a=b+c.
OR
Solve :
Q21.
1 1 1
1
  
a b x abx
Draw a pair of tangents to a circle of radius 5cm which are inclined to each
other at angle of 600. Also, write the steps of construction.
OR
Construct a  ABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Then,
CA  BA  BC 2
construct a triangle ABC such that


 .
CA
BA
BC 3
Q22.
The m th term of an AP is n and the n th term is m. find the sum of (m + n)
terms.
Q23.
Two pillars of equal height are on either side of a road, which is 100m wide.
The angles of elevation of the top of the pillars are 60 0 and 300 at a point on
the road between the pillars. Find the height of the pillars.
OR
As observed from the top of a lighthouse, 100 metres high above sea level,
the angle of depression of a ship moving directly towards it, changes from 30 o
to 60o. Determine the distance travelled by the ship during the period of
observation.
Q24.
Find the coordinates of the vertices of a ∆ABC with A (1,-4) and the mid point
of sides through A being (2,-1) and (0,-1).
Q25.
One card is drawn from a well shuffled pack of 52 cards. Calculate the
probability of getting
(i) A king or a queen
(ii) Neither a heart nor a red king
Q26.
Find the area of the shaded region if PQ=24cm, PR=7cm and O is the centre
of the circle.
Q27.
An iron pole consisting of a cylindrical portion 110cm high and of base
diameter 12cm is surmounted by a cone 9cm high. Find the mass of the pole,
given that 1cu cm of iron has 8 gm mass (approx.)
Q28.
Water flows at the rate of 10m per minute through a cylindrical pipe having
its diameter as 5mm. how much time will it take to fill a conical vessel whose
diameter of base is 40cm and depth is 24cm?
SECTION D
Q29.
Prove that the coordinates of the centroid of a triangle whose vertices are
  x  x2  x3     y1  y2  y3  
(x1,y1),(x2,y2)and (x3,y3) is given by  1
,
.
3
3

 

Q30.
A vertical tower stands on a horizontal plane and is surmounted by a vertical
flagstaff of height h. At a point on the plane, the angles of elevation of the
bottom and the top of the flagstaff are  and β respectively. Prove that the
tan 
height of the tower is h
.
tan   tan 
Q31.
By increasing the list price of a book by Rs10, a person can buy 10 less books
for Rs.1200. Find the original list price of the book.
OR
Two years ago a man’s age was three times the square of his son’s age.
Three years hence his age will be four times his son’s age. Find their present
ages.
Q32.
Prove that the lengths of tangents drawn from an external point to a circle are
equal.
OR
Prove that the radius of a circle is perpendicular to the tangent at the point of
contact.
Q33.
A circle is touching the side BC of ∆ABC at P and touching AB and AC
produced at Q and R respectively. Prove that
AQ =
Q34.
1
(perimeter of ∆ABC).
2
A bucket is in the form of a frustum of a cone of height 30 cm with radii of its
lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity
and surface area of the bucket. Also, find the cost of the milk which can
completely fill the container, at the rate of Rs 25 per litre (use  = 3.14).
SOLUTIONS
SECTION A
Ans1. Option (a) ab
Given x=b therefore it satisfies the given equation
b2-(a + b) b+p=0
 p=ab
Ans2. Option (d) 6:
Let x be the edge of cube. Then, x is also the diameter of the sphere.
Ratio of the volume of the cube to the volume of sphere
= x3 :
4
x3
4
 
 1:
6: 
3
8
24
Ans3. Option (b) 6 and 3
We know that the diagonals of a parallelogram bisect each other.
So, coordinates of mid point of AC = coordinates of mid point of BD.
 x 1 6  2 3  4 5  y 
 2 , 2  2 , 2 

 

 x 1  7 5  y
 2 ,4   2 , 2 

 

x 1 7
5y
 and 4 
2
2
2
X=6 and y=3.
Ans4. Option (d)
1
5
5 prizes and 20 blanks
Total number of outcomes = 25
Number of favourable outcomes =5
P(getting a prize) =
5
1

25
5
Ans5. Option (c) 1600
Required number of coins =
=
volume of cuboid
volume of eachcoin
770  7
22  0.875  0. 875  .2
= 40x40 =1600
Ans6. Option (c) 300
Ans7. Option (c) 53rd
an=a+(n-1)d
 210=2+(p-1)4
 4(p-1)=208
 P-1=52
 P=53
Hence 53rd term is 210.
Ans8. Option (c) 82.5 m
Let A be the position of the kite. Let O be the observer and OA be the thread.
Given, OA =165m,  BOA = 300 and let AB = h m. In right ∆OBA,
AB
OA
= sin30
h
1

165 2
h=
165
= 82.5
2
Thus, the height of the kite from the ground is 82.5 m.
Ans9. Option (a) collinear
Area of a triangle =
1
[{a(c+a)+b(a+b)+c(b+C)}-{b(b+c)+c(c+a)+a(a+b)}]
2
=
1
[ac+a2+ab+b2+bc+c2-b2-bc-c2-ca-a2-ab]
2
=
1
(0) = 0
2
Since, the area of triangle is 0, therefore, the given points are collinear.
Ans10. Option (c) 44cm
Area of a circle = πr2
154 =
22 2
r
7
r2=49 or r=7
perimeter = 2πr = 2x
22
x 7 = 44cm.
7
SECTION B
Ans11. In a square, all the sides are equal. So, AB=BC=CD=DA
Distance AB=
1  52
 (5  6)2

17 units
1 mark
Area of Square ABCD = AB 2
=

17

2
= 17 sq. Units.
1 mark
Ans12. let base = x cm and height = (x – 7) cm
Hypotenuse = 13 cm
By Pythagoras theorem,
X2 + (x – 7)2 = 132
X2 + 49 + x2 – 14x = 169
2x2 – 14x – 120 = 0
X2 – 12x + 5x – 60 = 0
1
mark
2
X( x – 12) + 5 (x – 12) = 0
1
mark
2
(x – 12) (x + 5) = 0
x – 12 = 0 or x + 5 = 0
x=12 or x=-5 (rejecting)
The other two sides are 12 cm and 5 cm.
1
mark
2
1
mark
2
Ans13.
Let PA and PB be two tangents drawn from an external point P to a circle with
centre O. We have to prove that angles  AOB and  APB are supplementary
i.e.  AOB+  APB=1800.
In right triangles OAP and OBP, we have
PA=PB (tangents drawn from an external point to a circle are equal in length)
OA=OB (radii of same circle)
OP=OP (common)
∆OAP  ∆OBP
(by SSS)
 OPA=  OPB and  AOP=  BOP
(by cpct)
1 mark
  APB = 2  OPA and  AOB =2  AOP
But,  AOP=90-  OPA
 2  AOP =180-2  OPA
  AOB =180-  APB
  AOB+  APB=180
1mark
Ans14. Angle described by the minute hand in 1 minute=60
Angle described by the minute hand in 35 minutes =6x35=2100
Area swept by the minute hand in 35 minutes=
210  22
x (10)2
360  7
= 183.3 cm2.
Ans15. Area of the square = (14)2 = 196 cm2
1
mark
2
1 mark
1
mark
2
1
mark
2
Area of four quadrants = area of one circle = πr 2
=
22
x (7)2 =154 cm2
7
Required area = 196-154 = 42 cm2.
1 mark
1
mark
2
Ans16. Number x can be selected in three ways and corresponding to each such way
there are three ways of selecting number y. therefore,the two numbers can
be selected in 9 ways as listed below:
(1,1),(1,4),(1,9),(2,1),(2,4),(2,9),(3,1),(3,4),(3,9)
Total number of outcomes =9
1 mark
The product xy will be less than 9, if x and y are chosen in one of the
following ways:
1
mark
2
(1,1),(1,4),(2,1),(2,4), (3,1)
Number of favourable outcomes =5
P(product less than 9) =
1
mark
2
5
.
9
OR
S=[HHH,HHT,HTH,THH,HTT,THT,TTH,TTT]
Total number of outcomes = 8
1
mark
2
At most two tails = HHH, HHT, HTH, THH, HTT, THT, TTH
Number of favourable outcomes =7
P(at most two tails) =
1 mark
1
mark
2
7
8
Ans17. L= last term =254 and d=5
Nth term from the end = l- (n-1)d
1
mark
2
1
mark
2
tenth term from the end = 254- (10-1)5
=254-45 = 209.
1 mark
Ans18.
1
mark
2
Let AB be a diameter of a given circle, and let PQ and RS be the tangent lines
drawn to the circle at points A and B respectively. Since tangent at a point to
a circle is perpendicular to the radius through the point.
Therefore, AB is perpendicular to both PQ and RS.
  PAB=90 and  ABS =90
  PAB=  ABS
But, these are a pair of alternate interior angles.
Therefore, PQ is parallel to RS.
1 mark
1
mark
2
SECTION C
Ans19. Let the third vertex be A(x, y). Other two vertices of the triangle are B(2,1)
and C(3,-2).
Area of ∆BC = 5 sq. Units
1
{x (1+2) + 2 (-2-y) + 3 (y-1)} = 5
2
1 mark
1
{3x-4-2y + 3y-3} = 5
2
{3x+ y-7} = ± 10
3x + y -17 = 0 (i)
or 3x +y +3 = 0 (ii)
1 mark
Given that, A(x, y) lies on y = x + 3 (iii)
On solving (i) and (iii) we get, x =
On solving (ii) and (iii) we get, x =
7
13
and y =
and,
2
2
3
3
and y = .
2
2
1 mark
Ans20. Here, (a-b)x2+(b-c)x+(c-a)=0
The given equation will have equal roots, if
(b-c)2-4(a-b)(c-a)=0
1 mark
b2+c2-2bc-4(ac-bc-a2+ab) =0
b2+c2+4a2+2bc-4ab-4ac=0
(b+c-2a) 2 =0
b+c-2a=0
1 mark
b+c=2a
1 mark
OR
1 1 1
1
 

a b x
abx
1 1
1
1



a b
abx
x
x  (a  b  x)
ab

ab
a  b  x  x
1 mark
 a  b
ab

ab
a  b  x  x
ax+bx+x2=-ab
x2+ax+bx+ab=0
x(x+a)+b(x+b)=0
(x+a)(x+b)=0
x+a=0 or x+b=0
1 mark
x=-a,-b
Ans21. The tangents on given circle as follows:
1 mark
1. Draw a circle of 5 cm radius and with centre O.
2. Take a point P on circumference of this circle. Extend OP to Q such that OP
= PQ.
3. Midpoint of OQ is P. Draw a circle with radius OP with centre as P.
Let it intersect our circle at R and S. Join QR and QS. QR and QS are required
tangents.
1 mark
2 marks
OR
CA  BA  BC 2


 . This
CA
BA
BC 3
means that the triangle ABC is similar to the triangle ABC with scale factor as
2
.
3
ABC is the required triangle.
A triangle ABC is to be constructed such that
3 marks
Ans22. Here,
And,
a+(m-1)d =n …………(i)
a+(n-1)d =m ………..(ii)
1 mark
On solving (i) and (ii), we get,
d = -1 and a= m+n-1
Sm+n=
Ans23.
mn
mn
[2(m+n-1) + (m+n-1)(-1)] =
(m+n-1)
2
2
1 mark
1 mark
1
mark
2
Let AB and CD be two pillars of equal heights, say, h metres. Let P be a point
on road such that AP= x m and PC = 100 –x.  APB= 600 and  CPD =300.
In right ∆PAB,
AB
AP
Tan60 =
3 
h
x
 h 
3x ……….. (1)
1 mark
In right ∆PCD,
tan30 =
1
3

CD
PC
h
100  x 
…………….(2)
1
mark
2
From (1) and (2) we get,
3x=100-x or x=25
From (1) ,
h=25 3
1 mark
Thus, the height of the pillars is 25 3 metres.
OR
1
mark
2
Let A and B be the two positions of the ship. Let d be the distance travelled
by the ship during the period of observation, i.e., AB=d metres.
Suppose that the observer is at the point P. given that PC = 100m.
Let h be the distance (in metres) from B to C.
From right triangle PCA, we have
dh
 cot 30  3
100
d+h = 100 3 ..(i)
Again in triangle PCB, we have,
h
1
 cot 60 
100
3
1 mark
h=
100
3
metres.
1
mark
2
Putting the value of h in (i) we get,

200
1 
d= 100  3 
= 115.47 (approx.)
 =
3
3


Thus, the distance travelled by the ship from A to B is 115.47 m(approx.)
1 mark
Ans24. Let the coordinates of A, B and C be (1, -4), (a, b) and (x, y) resp.
 1  a 4  b 
Then, 
= (2,-1)
,
2 
 2
1 mark
Therefore, 1+a=4, -4+b= -2
a=3, b=2
1 mark
 1  x 4  y 
Also, 
= (0,-1)
,
2 
 2
Therefore, 1+x=0, -4+y= -2
x= -1 , y=2
Therefore, the coordinates of the vertices of ∆ABC are A (1,-4), B (3, 2)
and C (-1, 2)
Ans25. Total number of outcomes =52
(i) Number of kings = 4
Number of queens = 4
1 mark
 4  4
8
2

52
52
13
(ii) Number of cards of hearts =13
P(king or queen) =

1
marks
2
1
Number of red kings =2 ( out of these 1 is already in hearts)
Neither a heart nor a red king = 52 – (13+2-1) =52-14 =38
P(neither a heart nor a red king) =
38
19

52
26
1
1
marks
2
Ans26
.
In ∆QPR, by Pythagoras theorem, we have,
QR2=PR2+PQ2 = 72+242=49+576=625
QR=25cm
Diameter of the circle = 25cm
Radius of the circle =
Area of semicircle =
25
2
r2
22  25  25
11  625
cm2


2
2722
28
1 mark
Area of ∆PQR =
1
1
PR  PQ   7  24  84 cm2
2
2
Area of shaded region =
1 mark
11  625
6875  2352
4523
 84 

28
28
28
= 161.54 cm2
1 mark
Ans27.
Volume of the pole = volume of the cylinder + volume of the cone
1


=    62  110   62  9  cu cm
3


1 mark
= [3960+108] cu cm
= 4068  cu cm
1 mark
Mass of the pole = 4068 x3.14x8
= 102188.16 grams = 102.19 kg
1 mark
Ans28. Amount of water required to fill the conical vessel = volume of the conical
vessel
=
2
1  40 
(24) cu cm =3200  cu cm ……….(i)

3  2 
1 mark
Amount of water that flows out of the cylindrical pipe in 1 minute
2
 5 
= 
 (10x100) = 62.5  cu cm …………..(ii)
 20 
1 mark
From (i) and (ii) , we get,
Time taken to fill the vessel =
3200 
minutes = 51.2 minutes
62.5 
1 mark
SECTION D
Ans29.
1
mark
2
Let A(x1,y1), B(x2,y2) and C(x3,y3) be the vertices of ∆ABC whose medians are
AD,BE and CF respectively. So, D,E and F are respectively the mid points of
BC,CA and AB.
1 mark
 x  x2   y3  y2  
Coordinates of D are  3
,

2
2
 


1
mark
2
Coordinates of a point G(x, y) dividing AD in the ratio 2 : 1 are
1
mark
2
2(
(
(
x 2  x3
y  y3
)  x1 2( 2
)  y1
2
2
,
)
2 1
2 1
1 mark
x1 + x 2  x3 y1 + y 2  y3
,
)
3
3
1
mark
2
Ans30.
1 mark
Let AB be the tower and BC be the flagstaff. Let OA=x metres, AB= y metres
and BC= h metres.
In right ∆OAB,
Tan  =
AB
OA
In right ∆OAC,
 y  x tan  or x 
y
………………………..(i)
tan 
1mark
Tanβ =
yh
x
 x
 y  h
tan 
………………………………….(ii)
1mark
From (i) and (ii),
 y  h
y

tan 
tan 
y( tanβ
y=
- tan  ) = h tan 
h tan 
tan
   tan  
Thus, the height of the tower is
h tan 
.
 tan   tan  
Ans31. Let list price of the book = Rs x
And increased price of the book = Rs (x+10)
According to question,
1200 1200
=10

x
x  10
1 mark
1 mark
1 
1
(1200)  
 =10
 x x  10 
  x  10  x  
 =10
(1200) 
  x  x  10  
1
mark
2
1200=x(x+10)
x2+10x-1200=0
(x+40)(x-30)=0
X=-40
X=30
But x is the cost of the book and hence can’t be negative.
Therefore x=30
1 mark
1 mark
1
mark
2
List price of book=Rs 30
OR
Two years ago, let son’s age = x years
Therefore man’s age = 3x2
Three years hence, son’s age = x+2+3=(x+5) years and man’s age = 3x2+5
By the given condition
3x2+5=4(x+5)
1 mark
3x2-4x+5-20=0
3x2-4x-15=0
1 mark
2
3x -9x+5x-15=0
3x(x-3)+5(x-3)=0
(x-3)(3x+5)=0
x=3
[Because 3x+5=0 means x =
5
, not possible)
3
1 mark
Therefore son’s present age = x+2=3+2=5 years
Man’s present age = 3x2+2=3(3)2+2
= 27+2=29 years
1 mark
Ans32. Given: A circle with centre O; PA and PB are two tangents to the circle drawn
from an external point P.
To prove: PA = PB
Construction: Join OA, OB, and OP.
2 marks
It is known that a tangent at any point of a circle is perpendicular to the
radius through the point of contact.
 OA  PA and OB  PB
... (1)
In OPA and OPB:
OAP = OBP
(Using (1))
OA = OB
(Radii of the same circle)
OP = PO
(Common side)
Therefore, OPA  OPB
(RHS congruency criterion)
PA = PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external
point to a circle are equal.
2 marks
OR
Given: A circle C (O, r) and a tangent AB at a point P.
To Prove: OP is perpendicular to AB.
Construction: Take any point Q, other than P, on the tangent AB. Join OQ.
2 marks
Since, Q is a point on the tangent AB, other than the point of contact P, so Q
will be outside the circle.
Let OQ intersect the circle at R.
Then, OQ=OR+RQ
 OQ>OR
(OR=OP=radius)
 OQ>OP
Thus, OP<OQ, i.e., OP is shorter than any other segment joining O to any
point of AB.
But, among all the line segments, joining the point O to a point on AB, the
shortest one is the perpendicular from O on AB.
Hence, OP is perpendicular to AB.
2 marks
Ans33.
1 mark
Since tangents from an exterior point to a circle are equal in length.
BP=BQ (From B)
..............(i)
CP=CR (From C) ...............(ii)
And, AQ=AR (From A) ..........(iii)
From (iii) we have,
1 mark
AQ=AR
 AB+BQ=AC+CR
 AB+BP=AC+CP ...........(iv)
 Perimeter of ∆ABC = AB+BC+CA
= AB+(BP+PC)+AC
= (AB+BP) + (AC+PC)
= 2(AB+BP)
= 2(AB+BQ)
= 2AQ
1
(perimeter of ∆ABC).
 AQ =
2
Ans34. Capacity (or volume) of the bucket =
1 mark
h
[r21 + r22 + r1r2]
3
Here, h = 30 cm, r1 = 20 cm and r2 = 10 cm
So, the capacity of bucket = 3.14 x
1 mark
1
mark
2
30
[202 + 102 + 20 x 10] cm3
3
= 21980 cu cm = 21.980 litres.
1 mark
Cost of 1 litre of milk = Rs 25
Cost of 21980 litres of milk = Rs 21.980 x 25 = Rs 549.50
Surface area of the bucket
1
mark
2
= curved surface area of the bucket + surface area of the bottom
= l(r1 + r2) + r22
Now, l =
where l=
h2  (r1  r2 )2
900  100 cm = 31.62 cm
1
mark
2
1
mark
2
Therefore, surface area of the bucket = 3.14 x 31.62 (20+10) + 3.14x 102
= 3.14 x 1048.6 cm2
= 3292.6 cm2 (approx.)
1 mark