1. No category

Summative Assesment-II
TOPPER SAMPLE PAPER IV
MATHEMATICS
CLASS X
Time : 3- 3
1
2
Hrs
M.M: 80
GENERAL INSTRUCTIONS :
1. All questions are compulsory.
2. The question paper is divided into four sections
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)
3. 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.
4. Use of calculators is not allowed.
SECTION A
Q1.
The values of k for which the quadratic equation 2x 2-kx+k=0 has equal roots
is
(a) 0 only
(b) 0,4
(c) 8 only
(d) 0,8
Q2.
In the given figure, the respective values of y and x are
Q3.
(a) 60o and 30o
(b) 45o and 60o
(c) 60o and 45
(d) 30o and 45o
The next term of the given series
(a)
50
(b)
70
2  8  18  32  ............ would be
(c)
80
(d)
90
Q4.
To draw a pair of tangents to a circle which are inclined to each other at an
angle of 35, it is required to draw tangents at the end points of those two
radii of the circle, the angle between which is
(a) 105
(b) 70
(c) 140
(d) 145
Q5.
If a die is thrown once the probability of getting a prime number is
1
1
1
1
(a)
(b)
(c)
(d)
2
4
3
5
Q6.
The ordinate of a point is twice its abscissa. If its distance from the point
(4,3) is 10 , then the coordinates of the point are
(a) (1,2) or (3,5)
(b) (1,2) or (3,6)
(c) (2,1) or (6,3)
(d) (2,1) or (3,6)
Q7.
The ratio in which the line segment joining A(3,4) and B(-2,1) is divided by
the y-axis is
(a) 1:2
(b) 2:3
(c) 3:2
(d) 2:5
Q8.
The given figure shows a sector of a circle of radius 10.5 cm. The perimeter of
the sector is
(a) 28 cm
(b) 30 cm
(c) 32 cm
(d) 34 cm
Q9.
The radius of a cylindrical tank is 28m. If its capacity is equal to that of a
rectangular tank of size 28mx16mx11m then, its depth is
(a) 2m
(b) 3m
(c) 4m
(d) 5m
Q10.
In given figure, a circle of radius 7.5cm is inscribed in a square, the remaining
area of the square is
(a) 48.91 cm sq.
(b) 48.375 cm sq.
(c) 46 cm sq.
(d) 52.32 cm sq.
SECTION B
Q11.
Q12.
If the 3rd and 6th term of an AP are 7 and 13 respectively, find the 10th term.
In the given figure, XP and XQ are tangents from X to the circle with centre
O. R is a point on the circle. Prove that XA+AR=XB+BR.
Q13.
A letter is chosen at random from the letters of the word ‘ASSASSINATION’.
Find the probability that the letter chosen is a (i) vowel (ii) consonant
OR
Rekha tosses a one rupee coin three times and wins the game if she gets the
same outcome every time. What are her chances to win?
Q14.
In given figure, ABCD is a square of side 14 cm and four congruent circles are
inscribed in it. Find the remaining area of the square.
Q15.
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter
of the wheel.
Determine the set of values of p for which the quadratic equation
px2+6x+1=0 has real roots.
Find the condition that the point (x,y) may lie on the line joining (3,4) and (5,-6).
Two concentric circles are of radii 5cm and 3cm. Find the length of the chord
of the larger circle which touches the smaller circle.
Q16.
Q17.
Q18.
SECTION C
Q19.
Solve :
1 1 1
1
 

a b x
abx
Q20.
Find the area of the quadrilateral ABCD whose vertices are A(1,1), B(7,3),C(12,2) and D(7,21) respectively.
OR
One end of a line of length 10 units is at the point (-3,2). If the ordinate of the other
end is 10, then find the abscissa of other end.
Q21. A boy standing on a horizontal plane finds a bird flying at a distance of 100m
from him at an elevation of 30o. A girl standing on the roof of a 20m high
building, finds the angle of elevation of the same bird to be 45o. the boy and
the girl are on the opposite sides of the bird. Find the distance of the bird
from the girl.
OR
A tower is 50m high. Its shadow is x metres shorter when the sun’s altitude is
45o than when it was 300. Find the value of x.
Q22. If the point (x,y) is equidistant from the points (a+b,b-a) and (a-b,a+b), then
prove that bx = ay.
Q23. Find the sum of the first 25 terms of an AP whose nth term is given by 3n+2.
Q24. Draw a circle of any convinient radius. Draw a pair of tangents to the circle
such that they are inclined to each other at an angle of 90°.
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.
The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280.
The field is to be ploughed at the rate of Rs. 0.50 per m 2. Find the cost of
22
ploughing the field. (Take =
)
7
OR
The area of an equilateral triangle ABC is 17320.5 cm 2. With each vertex of
triangle as the centre, a circle is drawn with radius equal to half the length of
the side of the triangle. Find the area of the triangle not included in the
circles. (use =3.14 and 3 = 1.73205).
Q27.
The ratio of the radii of two spheres is 1:2. The two spheres are melted
together to form a cylinder of height which is 12 times its radius . So what is
the ratio of the radius of the smaller sphere and the cylinder ?
Q28.
If h,c and V respectively are the height, the curved surface area and the
volume of cone, then prove that 3Vh3-c2h2+9V2=0
SECTION D
Q29.
The difference of two numbers is 5 and the difference of their reciprocals is
1
. Find the two numbers.
10
OR
The difference of square of two natural numbers is 45. The square of the
smaller number is four times the larger number. Find the two numbers.
Q30.
The sum of four consecutive numbers in an AP is 32 and the ratio of the
product of the first and the last terms to the product of the two middle terms
is 7:15. Find the numbers.
OR
If S1, S2 and S3 respectively be the sum of n, 2n and 3n terms of an AP, show
that S3=3(S2-S1).
Q31.
From a window, h metres high above the ground, of a house in a street, the
angles of elevation and depression of the top and the foot of another house
on the opposite side of the street are α and β respectively. Show that the
height of the opposite house is h(1+tanα.cotβ) metres.
The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F
1
respectively. Show that AF+BD+CE = AE+BF+CD=
(Perimeter of ∆ABC)
2
Prove that the tangent of a circle is perpendicular to the radius at the point of
contact.
A building is in the form of a cylinder surmounted by a hemispherical dome.
2
The base diameter of the dome is equal to
of the total height of the
3
1
building. Find the height of the building, if it contains 67
m3 of air.
21
Q32.
Q33.
Q34.
SOLUTIONS
Section A
Ans1. Option (d)
2x2-kx+k=0
For equal roots, D=0
b2-4ac=0
(-k)2-4(2)(k)=0
k2-8k=0
k(k-8)=0
k=0 or k-8=0
k=0, 8
1mark
Ans2. Option (a)
In given figure, ABǁCO
 ABD +  DBO =90
 ABD + 30 =90
 ABD = 60
(alternate interior angles)
 ABD =  BDO = y = 60
Also,  ABC =  BCO = x = 30
(alternate interior angles)
Thus, the value of y and x are 60 and 30 respectively.
1mark
Ans3. Option (a)
The given series can be written as 2  2 2  3 2  4 2  ............
So, the next term would be 5 2 , i.e.,
50
1mark
Ans4. Option (d)
The angle between a pair of tangents to a circle which are inclined to each
other at an angle is supplementary to the angle between the two radii of the
circle.
Thus, the angle between the radii of the circle = 180-35=145
1mark
Ans5. Option (a)
When a die is thrown once the outcomes are 1,2,3,4,5,6 out which 2,3 and 5
are prime numbers.
3
1
Therefore, P(prime number) =
1mark

6
2
Ans6. Option (b)
Let the coordinates of the point be P(x,2x). Let Q be the point (4,3).
PQ2=(4-x)2+(3-2x)2 = 10
16+x2-8x+9+4x2-12x=10
 5x2-20x+15=0
 x2-4x+3=0
 (x-3)(x-1)=0
 x=1 or x=3
So, 2x = 2 or 6
Hence, the coordinates of the required point are (1,2) or (3,6).
1mark
Ans7. Option (c)
Let the line segment AB cut the y-axis at the point P(0,y).
Let the ratio in which AB is divided by the point P be k:1.
2k  3
3
0=
 k
k 1
2
3
Hence the required ratio is
: 1, i.e., 3:2.
2
Ans8. Option (c)
Required perimeter=
1mark
2 r 
2  22  10.5  60
+2(10.5)
 2r 
360
7  360
= 11+21= 32cm
1mark
Ans9. Option (a)
Let d be the depth of the cylindrical tank.
According to the given information,
(28)2xd=28x16x11
 d=2
Thus, the depth of the cylindrical tank is 2 m.
1mark
Ans10.Option (b)
Area of the circle = r2 = 3.14  (7.5)2= 176.625 cm2
Side of square = diameter of circle = 15cm
Area of square = side2 = 152=225 cm2
Remaining area = area of square – area of circle = 225 - 176.625=48.375cm2
1mark
SECTION B
Ans11.Here, t3=a + 2d = 7 and t6 = a + 5d = 13
Solving the two equations, we get,
a = 3; d = 2
Therefore, 10th term = a + 9d = 3 + 9 x 2 = 21
1
mark
2
1
mark
2
1mark
Ans12. Since the lengths of tangents from an exterior point to a circle are equal.
Therefore, XP=XQ (from X) ……(i)
AP=AR (from A) ………..(ii)
BQ=BR (from B) ………..(iii)
Now, XP=XQ
 XA+AP=XB+BQ
1mark
 XA+AR=XB+BR
1mark
(Using (ii) and (iii))
Ans13.There are 13 letters in the word ‘ASSASSINATION’.
Total number of outcomes = 13
(i) There are 6 vowels in the word. (A,A,A,I,I,O)
Number of favourable outcomes = 6
6
P(vowel) =
13
1mark
(ii) Number of consonants in the word = 13-6 =7
7
P(consonant) =
13
1mark
OR
Tossing of a coin results in either a Head or a Tail.
After 3 tosses, the possible outcomes are:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, which are 8 in number.
The outcomes which are in her favour to win are HHH or TTT, which are 2 in
number.
1 mark
Let E be the event that Rekha wins.
Favourable outcomes 2 1
P(E) =


Total outcomes
8 4
Ans14.Area of square = 14cmx14cm=196 cm2
Diameter of each circle =
7
2
Area of four circles = 4x r2
14
=7
2
1 mark
1
mark
2
Radius of each circle =
2
22  7 
  
7
2
2
= 154 cm
=4x
Remaining area = 196-154 = 42 cm
2
Ans15.Let r be the radius of the wheel.
Distance covered in 1 revolution = 2r
Distance covered in 5000 revolution = 5000  2r = 11km
5000  22
(2r) =11  1000 metres
7
7
2r =
metres
10
1mark
1
mark
2
1mark
7
m or 70 cm.
10
Ans16.The given equation will have real roots, if b2-4ac≥0.
 (6)2-4(p)(1)≥0
 36-4p≥0
 36≥4p
 p≤9
Hence diameter of the wheel =
1mark
1mark
1mark
Ans17. Since the point P(x,y) lies on the line joining A(3,4) and B(-5,-6), therefore,
P, A and B are collinear points.
1
So, area of triangle PAB = 0
mark
2
0=
1
x1(y2  y3 )  x2 (y3  y1 )  x3 (y1  y2 )
2
[x(4+6)+3(-6-y)-5(y-4)]=0
10x-18-3y-5y+20=0
10x-8y+2=0
5x-4y+1=0, which is the required condition.
1mark
1
mark
2
Ans18.
Let PQ be the chord of the larger circle which touches the smaller circle at the
point L.
Since PQ is tangent at the point L to the smaller circle with centre O.
Therefore, OL is perpendicular to PQ.
Since PQ is a chord of the bigger circle and OL is perpendicular to PQ, OL
bisects PQ.
So, PQ=2PL
1mark
In right ∆OPL,
PL2=OP2-OL2
PL2= 25-9=16
PL=4cm
PQ=2PL= 2(4cm) =8cm
Hence, the length of the chord PQ is 8cm.
1mark
Ans19.
SECTION C
1 1 1
1
 

a b x
abx
1 1
1
1



a b
abx x
ab
x abx


ab
a  b  x  x







1mark
 a  b
ab

ab
a  b  x  x
ax+bx+x2=-ab
x2+ax+bx+ab=0
x(x+a)+b(x+a)=0
(x+a)(x+b)=0
x+a=0 or x+b=0
1mark
x=-a,-b
1mark
Ans20. Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
Area of triangle =
1
x1(y2  y3 )  x2 (y3  y1 )  x3 (y1  y2 )
2
1
[1(-3-2) + 7(2-1) +12(1+3)]
2
1
=
[-5+7+48] = 25 sq. units
1mark
2
1
Area of ∆ACD =
[1(2-21) + 12(21-1) +7(1-2)]
2
1
=
[-19+240-7] = 107 sq. units
1mark
2
Therefore, area of quadrilateral ABCD = 25 +107 = 132 sq. units.
1mark
Area of ∆ABC =
OR
Let the abscissa of other end be x.
Given, distance of A(-3,2) from B(x,10) is 10 units, i.e., AB =10
AB =
10 =
 3  x 
2
 2  10
2
9  x 2  6x  64
Squaring both the sides, we get,
100 = x2 + 6x + 73
x2 + 6x -27 = 0
x2 + 9x – 3x -27 = 0
x(x+9)-3(x+9) =0
(x+9)(x-3) =0
x = -9 or 3
Thus, the abscissa of other end is –9 or 3.
Ans21.
1mark
1mark
1mark
1mark
Let O be the position of the bird, B be the position of the boy, G be the
position of the girl and FG be the building at which the girl is standing.
BO= 100m , FG= 20m
In ∆OLB,
OL
OL
1
= sin30 

 OL =50m
BO
100
2
OM = OL-ML = OL-FG = 50-20 = 30m
In ∆OMG,
OM
1
 sin 45o 
 OG  30 2 m  42.3 m
OG
2
Thus, the distance of the bird from the girl is 42.3 m.
1mark
1mark
OR
1mark
Let PQ be the tower and let PA and PB be its shadows when the altitudes of
the sun are 45o and 30o respectively. Then,  PAQ = 45o,  PBQ = 30o, PQ =
50m, AB =x m.
In ∆APQ,
cot 45 =
AP
PQ
1=
AP
50
 AP  50 m
1mark
In ∆BPQ,
BP
PQ
x  50
3 =
50
cot 30 =
x= 50( 3  1 )
1mark
Ans22.Let P(x,y), Q(a + b, b – a) and R(a – b, a + b) be the given points.
It is given that PQ = PR
1
PQ2 = PR2
mark
2
{x – (a + b)}2 +{y – (b – a)}2 = {x – (a - b)}2 +{y – (a + b)}2
1mark
 x2 – 2 x (a + b) + (a + b)2 + y2 – 2y (b – a) + (b-a)2
= x2 + (a – b)2 – 2x (a - b) + y2 – 2y (a + b) + (a + b)2
 -2 x (a + b) – 2 y (b – a) = -2x (a – b) – 2 y (a + b)
 -ax - bx - by + ay = -ax + bx - ay - by
 2 bx = 2 ay
 bx = ay
Ans23.a1=3(1)+2=5
a2=3(2)+2=8
a3=3(3)+2=11
The AP with nth term as 3n+2 is 5,8,11,14,………
Here, a=5 and d=3
S25=
25
25
[2x5+(25-1)3] =
[10+72] =1025
2
2
1
1
marks
2
1
marks
2
1
1 marks
2
1
Ans24. The steps of construction are as follows:
1. Draw a circle of any convinient radius and mark its centre as O.
2. Draw one of its radius OA and draw a line perpendicular to OA at A.
3. Draw an angle of 90º at O such that it intersects the circle at B.
4. Draw a line perpendicular to OB at B and let both perpendiculars intersect
at point P.
Thus, PA and PB are the required tangents.
1mark
2 marks
Ans25. Total number of outcomes =52
(i) Number of kings=4
Number of queens = 4
P(king or queen) =
 4  4
52
(ii) Number of hearts =13

8
2

52
13
1
Number of red kings =2 (out of these 1 ia a heart)
13  2  1 38
19
P(neither a heart nor a red king) = 1–


52
52
26
Ans26.For Rs. 24 , the length of fencing =1m
For Rs. 5280, the length of fencing =
OR
1
1
 5280 = 220 metres.
24
Circumference of the field = 220 m
2r =220
22
220  7
2x
 r
 35m
 r =220
7
44
Area of the field = r2 =  (35)2 =1225 m2
Cost of ploughing = Rs. 0.50 per m2
Total cost of ploughing the field = Rs. 1225  x 0.50
1225  22  1
= Rs.
 175  11  Rs.1925
72
1
marks
2
1
marks
2
1mark
1mark
1mark
Let ABC be the equilateral triangle whose each side is of length ‘a’.
1.73205
3

 a2 = 4  10000 or
 a2 = 17320.5
 a2 = 17320.5
4
4
a=200
1mark
1
Radius of each circle = 100cm
mark
2
Area of the required region = area of equilateral – area of 3 sectors
1
60
= 17320.5 – 3 
 3.14  10000
  (100)2 = 17320.5 –
2
360
1
= 17320.5 – 1.57 x 10000 = 17320.5 -15700 = 1620.5 cm2
1 marks
2
Ans27. Let the radius of the smaller sphere be r , so the radius of the larger sphere
is 2r
Let the base radius of the cylinder be R
So, height of the cylinder=12R...given
1mark
ATQ,
4 3 4
πr + π(2r)3 = πR 2 (12R)
1mark
3
3
 12r 3 = 12R 3
 r=R
1mark
 The ratio of the radii is 1:1
Ans28.Curved surface area of a cone , c= rl
1 2
Volume of a cone , V =
r h
3
L.H.S.= 3Vh3-c2h2+9V2
1mark
2
1

1

= 3h3  r2h  -(rl)2h2+9  r2h 
3

3

1

= 2r2h4- 2r2h2l2 +9  2r 4h2 
9


= 2r2h4 - 2r2h2(r2+h2) + 2r4h2
(Since, l2=r2+h2)
2 2 4
2 4 2
2 2 4
2 4 2
=  r h -  r h -  r h +  r h =0 = R.H.S.
1mark
1mark
SECTION D
Ans29.Let the numbers be x
1
1
1
Given,
 
x 5 x
10
and x-5.
1
 1
x 5  x


1
1
marks
2
x x 5
1

 x  5 x 10
 (x-5)x=50
 x2-5x-50=0
 (x-10)(x+5)=0
1mark
 x=10 or x=-5
1
mark
2
When x=10, x-5=5
When x=-5, x-5=-5-5 = -10
Thus the required numbers are 10 and 5 or -5 and -10.
1mark
OR
Let x and y be the two natural numbers and x>y.
Given, x2-y2=45 ………(i)
y2=4x …………….(ii)
Using (ii) in (i), we get,
1mark
1
mark
2
1
mark
2
x2-4x-45=0
 (x-9)(x+5)=0
 x=9 or x=-5
1mark
Rejecting x=-5 because x is a natural number.
Therefore, x=9
From (ii),
y2=4x = 4x9=36
y = ±6
But y is a natural number. Therefore, y=-6 is rejected.
Thus, y=6.
Hence the required natural numbers are 9 and 6.
1mark
Ans30.Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.
So, a-3d+a-d+a+d+a+3d=32
 4a=32
 a=8
Also,
or,
or,
or,
or,
So,
 a  3d  a  3d
 a  d  a  d
1

7
15
1mark
15(a2-9d2)=7(a2-d2)
15a2-135d2=7a2-7d2
8a2-128d2=0
d2=4 or d=±2
when a=8 and d=2, the numbers are 2,6,10,14.
1mark
1
mark
2
When a=8,d=-2 the numbers are 14,10,6,2.
n
Here, S1 =
[2a+(n-1)d]
2
2n
S2 =
[2a+(2n-1)d]
2
3n
S3 =
[2a+(3n-1)d]
2
1
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2
OR
1
1
marks
2
S2-S1= 2an+n(2n-1)d –an –
n n  1 d
2
n 1
=an +[(2n-1) –
]nd
2
n 3n  1 d
= an +
2
3n 3n  1 d
3(S2-S1)=3an +
= S3
2
1
1
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2
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Ans31.
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Let B be the window of a house AB and let CD be the other house. Then, AB=
= EC =h metres.
Let CD = H metres. Then, ED= (H-h) m
In ∆BED,
BE
cotα =
ED
BE = (H-h) cotα
In ∆ACB,
AC
= cotβ
AB
AC=h.cotβ
…………….. (a)
…………………………. (b)
But BE=AC
(H-h) cotα = hcotβ
 cot   cot  
H= h
cot 
H = h(1+tanα cotβ) m
Hence proved.
Ans32.
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1
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2
Since lengths of the tangents drawn from an external point to a circle are
equal.
Therefore, AF=AE (From A)…………(i)
BD=BF (From B) …………(ii)
CE=CD (From C) ……..(iii)
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Adding equations (i), (ii) and (iii), we get
1
AF+BD+CE = AE+BF+CD
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2
Now, Perimeter of ∆ABC = AB+BC+CA
Perimeter of ∆ABC = (AF+FB)+(BD+CD)+(EC+AE)
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= (AF+AE)+(BD+BF)+(EC+CD)
= 2(AF+BD+CE)
AF+BD+CE =
1
(Perimeter of ∆ABC)
2
1
(Perimeter of ∆ABC)
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2
Ans33. 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.
Hence, AF+BD+CE = AE+BF+CD =
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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.
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Ans34. Let the radius of the hemispherical dome be r metres and the total height of
the building be h metres.
Since the base diameter of the dome is equal to
2
of the total height, 2r =
3
2h
.
3
h
. Let H metres be the height of the cylindrical portion.
3
This implies r =
Therefore, H= h –
h
2h
=
metres.
3
3
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Volume of the air inside the building
= volume of air inside the dome+ volume of the air inside the cylinder
3
=
2
2 h
8
 h   2h 
       
h3 cu.meters
3 3
81
3  3 
Volume of the air inside the building is 67
This gives h=6m
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1
1408
8
m3.therefore,
h3 =
21
21
81
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